Chapter 7 Integrals

We need to integrate the given expression with respect to $x$. The expression is:

$\int \left( \frac{2a}{\sqrt{x}} − \frac{b}{x^2} + 3c \sqrt[3]{x^2} \right) dx$

First, let's rewrite the terms using exponents:

$\frac{2a}{\sqrt{x}} = 2a x^{-1/2}$

$\frac{b}{x^2} = b x^{-2}$

$3c \sqrt[3]{x^2} = 3c x^{2/3}$

Now, we can integrate each term separately using the power rule for integration, which states that $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$. Also, the integral of a sum is the sum of the integrals:

$\int \left( 2a x^{-1/2} − b x^{-2} + 3c x^{2/3} \right) dx = \int 2a x^{-1/2} dx - \int b x^{-2} dx + \int 3c x^{2/3} dx$

Integrate the first term:

$\int 2a x^{-1/2} dx = 2a \int x^{-1/2} dx = 2a \frac{x^{-1/2 + 1}}{-1/2 + 1} + C_1 = 2a \frac{x^{1/2}}{1/2} + C_1 = 2a \times 2 x^{1/2} + C_1 = 4a x^{1/2} + C_1$

This can be written as $4a\sqrt{x} + C_1$.

Integrate the second term:

$\int -b x^{-2} dx = -b \int x^{-2} dx = -b \frac{x^{-2 + 1}}{-2 + 1} + C_2 = -b \frac{x^{-1}}{-1} + C_2 = b x^{-1} + C_2$

This can be written as $\frac{b}{x} + C_2$.

Integrate the third term:

$\int 3c x^{2/3} dx = 3c \int x^{2/3} dx = 3c \frac{x^{2/3 + 1}}{2/3 + 1} + C_3 = 3c \frac{x^{5/3}}{5/3} + C_3 = 3c \times \frac{3}{5} x^{5/3} + C_3 = \frac{9c}{5} x^{5/3} + C_3$

Combining the results and the constants of integration ($C = C_1 + C_2 + C_3$):

$\int \left( \frac{2a}{\sqrt{x}} − \frac{b}{x^2} + 3c \sqrt[3]{x^2} \right) dx = 4a x^{1/2} + \frac{b}{x} + \frac{9c}{5} x^{5/3} + C$

The final answer is:

$\int \left( \frac{2a}{\sqrt{x}} − \frac{b}{x^2} + 3c \sqrt[3]{x^2} \right) dx = 4a\sqrt{x} + \frac{b}{x} + \frac{9c}{5}\sqrt[3]{x^5} + C$

We want to evaluate the integral:

$\int \frac{3ax}{b^2 + c^2x^2} dx$

This integral can be solved using the method of substitution.

Let us substitute the denominator:

Let $u = b^2 + c^2x^2$

Now, we find the differential $du$ by differentiating $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(b^2 + c^2x^2)$

$\frac{du}{dx} = 0 + c^2(2x)$

From equation (i), we can express $x \, dx$ in terms of $du$:

$du = 2c^2x \, dx$

$x \, dx = \frac{du}{2c^2}$

Now, substitute $u$ and $x \, dx$ into the original integral:

$\int \frac{3ax}{b^2 + c^2x^2} dx = \int \frac{3a}{b^2 + c^2x^2} (x \, dx)$

$= \int \frac{3a}{u} \left(\frac{du}{2c^2}\right)$

$= \int \frac{3a}{2c^2} \frac{1}{u} du$

Since $\frac{3a}{2c^2}$ is a constant (assuming $a$ and $c$ are constants and $c \neq 0$), we can take it out of the integral:

$= \frac{3a}{2c^2} \int \frac{1}{u} du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log|u| + C'$, where $C'$ is the constant of integration.

$= \frac{3a}{2c^2} (\log|u| + C')$

$= \frac{3a}{2c^2} \log|u| + \frac{3a}{2c^2}C'$

Now, substitute back $u = b^2 + c^2x^2$. Let $C = \frac{3a}{2c^2}C'$ be the new constant of integration:

$= \frac{3a}{2c^2} \log|b^2 + c^2x^2| + C$

Since $b^2 \ge 0$ and $c^2x^2 \ge 0$ for real values of $b, c, x$, the term $b^2 + c^2x^2$ is always non-negative. Assuming $b^2 + c^2x^2 > 0$ in the domain of the function, we can remove the absolute value:

$= \frac{3a}{2c^2} \log(b^2 + c^2x^2) + C$

The final answer is:

$\int \frac{3ax}{b^2 + c^2x^2} dx = \frac{3a}{2c^2} \log(b^2 + c^2x^2) + C$

To verify the given integration using the concept of integration as an antiderivative, we need to differentiate the right-hand side of the equation with respect to $x$ and check if the result is equal to the integrand.

The given equation is:

$\int \frac{x^3}{x + 1} dx = x − \frac{x^2}{2} + \frac{x^3}{3} - \log | x + 1 | + C$

Let the right-hand side (excluding the constant of integration initially) be $g(x)$.

$g(x) = x − \frac{x^2}{2} + \frac{x^3}{3} - \log | x + 1 |$

We need to find the derivative of $g(x)$ with respect to $x$, i.e., $g'(x) = \frac{d}{dx} \left( x − \frac{x^2}{2} + \frac{x^3}{3} - \log | x + 1 | \right)$.

Differentiating each term:

$\frac{d}{dx}(x) = 1$

$\frac{d}{dx}\left(\frac{x^2}{2}\right) = \frac{1}{2} \frac{d}{dx}(x^2) = \frac{1}{2}(2x) = x$

$\frac{d}{dx}\left(\frac{x^3}{3}\right) = \frac{1}{3} \frac{d}{dx}(x^3) = \frac{1}{3}(3x^2) = x^2$

$\frac{d}{dx}(\log | x + 1 |) = \frac{1}{x + 1}$

The derivative of the constant of integration $C$ is $0$.

So, the derivative of the right-hand side is:

$g'(x) = 1 - x + x^2 - \frac{1}{x + 1}$

Now, let's combine the terms in $g'(x)$ by finding a common denominator, which is $(x+1)$.

$g'(x) = (1 - x + x^2) - \frac{1}{x + 1}$

$g'(x) = \frac{(1 - x + x^2)(x + 1)}{x + 1} - \frac{1}{x + 1}$

$g'(x) = \frac{(x^2 - x + 1)(x + 1) - 1}{x + 1}$

Let's expand the product $(x^2 - x + 1)(x + 1)$ in the numerator. This is a standard factorization for the sum of cubes, $x^3 + 1^3 = (x+1)(x^2 - x + 1)$.

Alternatively, we can multiply it out:

$(x^2 - x + 1)(x + 1) = x^2(x+1) - x(x+1) + 1(x+1)$

$= (x^3 + x^2) - (x^2 + x) + (x + 1)$

$= x^3 + x^2 - x^2 - x + x + 1$

$= x^3 + 1$

Substitute this back into the numerator of $g'(x)$:

$g'(x) = \frac{(x^3 + 1) - 1}{x + 1}$

$g'(x) = \frac{x^3}{x + 1}$

The result of the differentiation, $g'(x)$, is equal to the integrand $\frac{x^3}{x+1}$.

Therefore, the given integration is verified using the concept of integration as an antiderivative.

We want to evaluate the integral:

$\int \sqrt{\frac{1 + x}{1 − x}} dx$

The integrand requires $\frac{1+x}{1-x} \ge 0$, and the question states $x \neq 1$. This implies we are working in the domain $-1 \le x < 1$.

Method 1: Trigonometric Substitution

Let us use the substitution $x = \cos \theta$. Since $-1 \le x < 1$, we can choose $\theta \in (0, \pi]$. Specifically, for $x = \cos \theta$ to cover the interval $[-1, 1)$, $\theta$ can be restricted to $(0, \pi]$. For $x=-1$, $\theta = \pi$. For $x=1$, $\theta = 0$ (not included). The interval $-1 < x < 1$ corresponds to $\theta \in (0, \pi)$.

Differentiating both sides with respect to $\theta$:

So, $dx = -\sin \theta \, d\theta$.

Now, let's transform the integrand using trigonometric identities:

$1 + x = 1 + \cos \theta = 2 \cos^2(\theta/2)$

$1 - x = 1 - \cos \theta = 2 \sin^2(\theta/2)$

$\sqrt{\frac{1 + x}{1 − x}} = \sqrt{\frac{2 \cos^2(\theta/2)}{2 \sin^2(\theta/2)}} = \sqrt{\cot^2(\theta/2)}$

Since $\theta \in (0, \pi)$, we have $\theta/2 \in (0, \pi/2)$. In this interval, $\cot(\theta/2) > 0$.

Therefore, $\sqrt{\cot^2(\theta/2)} = |\cot(\theta/2)| = \cot(\theta/2)$.

Substitute $x$ and $dx$ into the integral:

$\int \sqrt{\frac{1 + x}{1 − x}} dx = \int \cot(\theta/2) (-\sin \theta) d\theta$

We can use the double angle identity for $\sin \theta$: $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$.

$= \int \frac{\cos(\theta/2)}{\sin(\theta/2)} (-2 \sin(\theta/2) \cos(\theta/2)) d\theta$

$= \int -2 \cos^2(\theta/2) d\theta$

Using the identity $2 \cos^2 A = 1 + \cos(2A)$, with $A = \theta/2$:

$= \int -(1 + \cos \theta) d\theta$

$= -\int 1 \, d\theta - \int \cos \theta \, d\theta$

Now, perform the integration with respect to $\theta$:

$= -(\theta) - (\sin \theta) + C'$

$= -\theta - \sin \theta + C'$

Substitute back from $\theta$ to $x$. From (1), $\theta = \arccos x$.

To find $\sin \theta$ in terms of $x$, we use the identity $\sin^2 \theta + \cos^2 \theta = 1$.

$\sin^2 \theta = 1 - \cos^2 \theta = 1 - x^2$.

Since $\theta \in (0, \pi)$, $\sin \theta \ge 0$. Thus, $\sin \theta = \sqrt{1 - x^2}$.

Substituting back:

$= -\arccos x - \sqrt{1 - x^2} + C'$

Method 2: Rationalization and Substitution

Multiply the numerator and denominator inside the square root by $(1+x)$:

$\sqrt{\frac{1 + x}{1 − x}} = \sqrt{\frac{(1 + x)(1 + x)}{(1 − x)(1 + x)}} = \sqrt{\frac{(1+x)^2}{1-x^2}}$

For $-1 < x < 1$, we have $1+x > 0$ and $1-x^2 > 0$. So, $\sqrt{(1+x)^2} = |1+x| = 1+x$, and $\sqrt{1-x^2}$ is real and positive.

The integrand becomes $\frac{1+x}{\sqrt{1-x^2}}$.

Now, split the integral into two parts:

$\int \frac{1+x}{\sqrt{1-x^2}} dx = \int \frac{1}{\sqrt{1-x^2}} dx + \int \frac{x}{\sqrt{1-x^2}} dx$

The first integral is a standard form:

$\int \frac{1}{\sqrt{1-x^2}} dx = \arcsin x + C_1$

For the second integral, $\int \frac{x}{\sqrt{1-x^2}} dx$, let us use a substitution.

Differentiating with respect to $x$:

So, $du = -2x \, dx$, which means $x \, dx = -\frac{1}{2} du$.

The second integral becomes:

$\int \frac{1}{\sqrt{1-x^2}} x \, dx = \int \frac{1}{\sqrt{u}} \left(-\frac{1}{2} du\right) = -\frac{1}{2} \int u^{-1/2} du$

Integrating with respect to $u$:

$-\frac{1}{2} \frac{u^{-1/2 + 1}}{-1/2 + 1} + C_2 = -\frac{1}{2} \frac{u^{1/2}}{1/2} + C_2 = -u^{1/2} + C_2 = -\sqrt{u} + C_2$

Substitute back $u = 1-x^2$:

$-\sqrt{1-x^2} + C_2$

Combining the results for the two parts of the integral from Method 2:

$\int \frac{1+x}{\sqrt{1-x^2}} dx = (\arcsin x + C_1) + (-\sqrt{1-x^2} + C_2)$

$= \arcsin x - \sqrt{1-x^2} + C''$, where $C'' = C_1 + C_2$.

Comparison of Results

Method 1 gave $-\arccos x - \sqrt{1 - x^2} + C'$.

Method 2 gave $\arcsin x - \sqrt{1 - x^2} + C''$.

Using the identity $\arcsin x + \arccos x = \frac{\pi}{2}$, we have $\arcsin x = \frac{\pi}{2} - \arccos x$ or $-\arccos x = \arcsin x - \frac{\pi}{2}$.

Substituting this into the result from Method 1:

$(\arcsin x - \frac{\pi}{2}) - \sqrt{1 - x^2} + C' = \arcsin x - \sqrt{1 - x^2} + (C' - \frac{\pi}{2})$.

Since $C'$ is an arbitrary constant, $C' - \frac{\pi}{2}$ is also an arbitrary constant. Let $C = C' - \frac{\pi}{2} = C''$. The two results are equivalent.

The final answer can be written in either form. We choose the form from Method 2 as it is often considered simpler.

$\int \sqrt{\frac{1 + x}{1 − x}} dx = \arcsin x - \sqrt{1-x^2} + C$

We want to evaluate the integral:

$\int \frac{dx}{\sqrt{(x − α) (β − x)}}$

Given $\beta > \alpha$, the expression $(x - \alpha)(\beta - x)$ is positive when $\alpha < x < \beta$. This is the domain where the square root is real and non-zero.

We can rewrite the product inside the square root by expanding it:

$(x - \alpha)(\beta - x) = x\beta - x^2 - \alpha\beta + \alpha x = -x^2 + (\alpha + \beta)x - \alpha\beta$

The integral is of the form $\int \frac{dx}{\sqrt{Ax^2 + Bx + C}}$. We complete the square in the denominator for the expression $-x^2 + (\alpha + \beta)x - \alpha\beta$.

$-x^2 + (\alpha + \beta)x - \alpha\beta = -(x^2 - (\alpha + \beta)x + \alpha\beta)$

Complete the square for $x^2 - (\alpha + \beta)x$: add and subtract $\left(\frac{\alpha + \beta}{2}\right)^2$ inside the parenthesis.

$= -\left[x^2 - (\alpha + \beta)x + \left(\frac{\alpha + β}{2}\right)^2 - \left(\frac{\alpha + β}{2}\right)^2 + αβ\right]$

$= -\left[\left(x - \frac{\alpha + β}{2}\right)^2 - \frac{(\alpha + β)^2}{4} + \frac{4αβ}{4}\right]$

$= -\left[\left(x - \frac{\alpha + β}{2}\right)^2 - \frac{\alpha^2 + 2\alphaβ + β^2 - 4αβ}{4}\right]$

$= -\left[\left(x - \frac{\alpha + β}{2}\right)^2 - \frac{\alpha^2 - 2\alphaβ + β^2}{4}\right]$

$= -\left[\left(x - \frac{\alpha + β}{2}\right)^2 - \frac{(\alpha - β)^2}{4}\right]$

Since $(\alpha - \beta)^2 = (\beta - \alpha)^2$, we have:

$= -\left[\left(x - \frac{\alpha + β}{2}\right)^2 - \left(\frac{β - α}{2}\right)^2\right]$

$= \left(\frac{β - α}{2}\right)^2 - \left(x - \frac{\alpha + β}{2}\right)^2$

So, the expression inside the square root is $\left(\frac{β - α}{2}\right)^2 - \left(x - \frac{\alpha + β}{2}\right)^2$. Let $a = \frac{β - α}{2}$ and $y = x - \frac{\alpha + β}{2}$. Since $\beta > \alpha$, $a > 0$. The integral becomes:

$\int \frac{dx}{\sqrt{\left(\frac{β - α}{2}\right)^2 - \left(x - \frac{\alpha + β}{2}\right)^2}}$

Let $y = x - \frac{\alpha + β}{2}$. Then $dy = dx$. The integral is:

$\int \frac{dy}{\sqrt{a^2 - y^2}}$

This is a standard integral form: $\int \frac{dz}{\sqrt{a^2 - z^2}} = \sin^{-1}\left(\frac{z}{a}\right) + C$.

Applying this formula with $z = y$ and the value of $a$:

$\int \frac{dy}{\sqrt{a^2 - y^2}} = \sin^{-1}\left(\frac{y}{a}\right) + C$

Substitute back $y = x - \frac{\alpha + β}{2}$ and $a = \frac{β - α}{2}$:

$\frac{y}{a} = \frac{x - \frac{\alpha + β}{2}}{\frac{β - α}{2}} = \frac{\frac{2x - (\alpha + β)}{2}}{\frac{β - α}{2}} = \frac{2x - \alpha - β}{β - α}$

So, the integral is:

$\sin^{-1}\left(\frac{2x - \alpha - β}{β - α}\right) + C$

The final answer is:

$\int \frac{dx}{\sqrt{(x − α) (β − x)}} = \sin^{-1}\left(\frac{2x - \alpha - β}{β - α}\right) + C$

We want to evaluate the integral:

$\int \tan^8 x \; \sec^4 x \;dx$

This integral is of the form $\int \tan^m x \sec^n x \, dx$. Since the power of the secant ($n=4$) is even and positive, we can save a factor of $\sec^2 x$ for the differential $du$ when we use the substitution $u = \tan x$. The remaining factors of $\sec x$ are converted to powers of $\tan x$ using the identity $\sec^2 x = 1 + \tan^2 x$.

Rewrite the integrand by separating $\sec^2 x$:

$\int \tan^8 x \; \sec^2 x \; \sec^2 x \;dx$

Use the identity $\sec^2 x = 1 + \tan^2 x$ to express one $\sec^2 x$ term in terms of $\tan x$:

$\int \tan^8 x \; (1 + \tan^2 x) \; \sec^2 x \;dx$

Let us use the substitution:

Let $u = \tan x$

Differentiating both sides with respect to $x$ gives the differential $du$:

$\frac{du}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x$

So, $du = \sec^2 x \;dx$.

Substitute $u$ and $du$ into the integral:

$\int u^8 (1 + u^2) du$

Expand the integrand by multiplying $u^8$ through the parenthesis:

$\int (u^8 \cdot 1 + u^8 \cdot u^2) du$

$\int (u^8 + u^{10}) du$

Now, integrate term by term using the power rule for integration, which states $\int z^k dz = \frac{z^{k+1}}{k+1}$ for $k \neq -1$. Remember to add the constant of integration $C$ at the end.

$\int u^8 du = \frac{u^{8+1}}{8+1} = \frac{u^9}{9}$

$\int u^{10} du = \frac{u^{10+1}}{10+1} = \frac{u^{11}}{11}$

So, the integral is:

$\frac{u^9}{9} + \frac{u^{11}}{11} + C$

Substitute back $u = \tan x$ to express the result in terms of $x$:

$\frac{(\tan x)^9}{9} + \frac{(\tan x)^{11}}{11} + C$

This can be written as:

$\frac{\tan^9 x}{9} + \frac{\tan^{11} x}{11} + C$

The final answer is:

$\int \tan^8 x \; \sec^4 x \;dx = \frac{\tan^9 x}{9} + \frac{\tan^{11} x}{11} + C$

We want to evaluate the integral:

$\int \frac{x^3}{x^4 + 3x^2 + 2} \;dx$

We can use a substitution method. Notice that the denominator $x^4 + 3x^2 + 2$ is a quadratic expression in terms of $x^2$. The numerator contains $x^3$. Let's consider a substitution involving $x^2$.

Let $t = x^2$.

Differentiating both sides with respect to $x$ gives $\frac{dt}{dx} = 2x$. So, $dt = 2x \, dx$.

The numerator $x^3 \, dx$ can be rewritten as $x^2 \cdot x \, dx$. Using our substitution, $x^2 = t$ and $x \, dx = \frac{1}{2} dt$.

Thus, $x^3 \, dx = t \cdot \frac{1}{2} dt = \frac{1}{2} t \, dt$.

Now, rewrite the integral in terms of $t$. The denominator becomes $t^2 + 3t + 2$. The numerator becomes $\frac{1}{2} t \, dt$.

The integral is:

$\int \frac{\frac{1}{2} t \, dt}{t^2 + 3t + 2} = \frac{1}{2} \int \frac{t}{t^2 + 3t + 2} dt$

The denominator $t^2 + 3t + 2$ can be factored:

$t^2 + 3t + 2 = (t+1)(t+2)$

So, we need to evaluate $\frac{1}{2} \int \frac{t}{(t+1)(t+2)} dt$. This integral can be solved using partial fraction decomposition.

Let $\frac{t}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2}$.

Multiplying both sides by $(t+1)(t+2)$:

$t = A(t+2) + B(t+1)$

To find $A$, set $t = -1$:

$-1 = A(-1+2) + B(-1+1) \implies -1 = A(1) + B(0) \implies A = -1$

To find $B$, set $t = -2$:

$-2 = A(-2+2) + B(-2+1) \implies -2 = A(0) + B(-1) \implies -2 = -B \implies B = 2$

So, the integrand can be written as:

$\frac{t}{(t+1)(t+2)} = \frac{-1}{t+1} + \frac{2}{t+2}$

The integral becomes:

$\frac{1}{2} \int \left(\frac{-1}{t+1} + \frac{2}{t+2}\right) dt$

Split the integral into two parts:

$= \frac{1}{2} \left( \int \frac{-1}{t+1} dt + \int \frac{2}{t+2} dt \right)$

$= \frac{1}{2} \left( -\int \frac{1}{t+1} dt + 2\int \frac{1}{t+2} dt \right)$

Integrate each term. The integral of $\frac{1}{z} dz$ is $\log|z|$.

$= \frac{1}{2} \left( -\log|t+1| + 2\log|t+2| \right) + C'$

Substitute back $t = x^2$:

$= \frac{1}{2} \left( -\log|x^2+1| + 2\log|x^2+2| \right) + C'$

Since $x^2+1 \ge 1 > 0$ and $x^2+2 \ge 2 > 0$ for real values of $x$, the absolute values can be removed:

$= \frac{1}{2} \left( -\log(x^2+1) + 2\log(x^2+2) \right) + C'$

Using logarithm properties ($k \log a = \log a^k$ and $\log a - \log b = \log(a/b)$):

$= -\frac{1}{2}\log(x^2+1) + \log((x^2+2)^2) + C'$

$= \log(x^2+2) - \frac{1}{2}\log(x^2+1) + C'$

or

$= \frac{1}{2} \left( \log((x^2+2)^2) - \log(x^2+1) \right) + C'$

$= \frac{1}{2} \log\left(\frac{(x^2+2)^2}{x^2+1}\right) + C'$

A common form is:

$\log(x^2+2) - \frac{1}{2}\log(x^2+1) + C$

The final answer is:

$\int \frac{x^3}{x^4 + 3x^2 + 2} \;dx = \log(x^2+2) - \frac{1}{2}\log(x^2+1) + C$

We want to evaluate the integral:

$\int \frac{dx}{2 \sin^2 x + 5 \cos^2 x}$

This integral is of the form $\int \frac{dx}{a \sin^2 x + b \cos^2 x}$. To solve this type of integral, we typically divide both the numerator and the denominator by $\cos^2 x$.

Divide the numerator by $\cos^2 x$:

$\frac{1}{\cos^2 x} = \sec^2 x$

Divide the denominator by $\cos^2 x$:

$\frac{2 \sin^2 x + 5 \cos^2 x}{\cos^2 x} = \frac{2 \sin^2 x}{\cos^2 x} + \frac{5 \cos^2 x}{\cos^2 x} = 2 \tan^2 x + 5$

The integral becomes:

$\int \frac{\sec^2 x}{2 \tan^2 x + 5} dx$

Now, we use the substitution method. Let $u = \tan x$.

Let $u = \tan x$

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x$

So, $du = \sec^2 x \, dx$.

Substitute $u$ and $du$ into the integral:

$\int \frac{du}{2 u^2 + 5}$

Factor out 2 from the denominator:

$\frac{1}{2} \int \frac{du}{u^2 + \frac{5}{2}}$

This integral is in the standard form $\int \frac{dz}{z^2 + a^2}$, where $z = u$ and $a^2 = \frac{5}{2}$, so $a = \sqrt{\frac{5}{2}} = \frac{\sqrt{5}}{\sqrt{2}}$.

The integral of $\frac{1}{z^2 + a^2}$ with respect to $z$ is $\frac{1}{a} \tan^{-1}\left(\frac{z}{a}\right) + C'$.

Applying this formula:

$\frac{1}{2} \left( \frac{1}{\sqrt{5/2}} \tan^{-1}\left(\frac{u}{\sqrt{5/2}}\right) \right) + C$

$= \frac{1}{2} \left( \frac{\sqrt{2}}{\sqrt{5}} \tan^{-1}\left(\frac{u\sqrt{2}}{\sqrt{5}}\right) \right) + C$

$= \frac{\sqrt{2}}{2\sqrt{5}} \tan^{-1}\left(\frac{\sqrt{2}u}{\sqrt{5}}\right) + C$

We can rationalize the constant $\frac{\sqrt{2}}{2\sqrt{5}}$:

$\frac{\sqrt{2}}{2\sqrt{5}} = \frac{\sqrt{2}}{2\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{10}}{2 \times 5} = \frac{\sqrt{10}}{10}$

So the constant is $\frac{1}{\sqrt{10}}$.

The integral is $\frac{1}{\sqrt{10}} \tan^{-1}\left(\frac{u \sqrt{2}}{\sqrt{5}}\right) + C$.

Substitute back $u = \tan x$:

$= \frac{1}{\sqrt{10}} \tan^{-1}\left(\frac{\sqrt{2} \tan x}{\sqrt{5}}\right) + C$

The final answer is:

$\int \frac{dx}{2 \sin^2 x + 5 \cos^2 x} = \frac{1}{\sqrt{10}} \tan^{-1}\left(\frac{\sqrt{2}}{\sqrt{5}} \tan x\right) + C$

We want to evaluate the definite integral $\int_{-1}^2(7x−5) \;dx$ as a limit of sums.

The definition of the definite integral as a limit of Riemann sums is given by:

$\int_{a}^b f(x) dx = \lim\limits_{n \to \infty} \sum\limits_{i=1}^n f(x_i) \Delta x$

where $\Delta x = \frac{b - a}{n}$ and $x_i = a + i \Delta x$ for a right Riemann sum.

In this problem, we have:

$a = -1$

$b = 2$

$f(x) = 7x - 5$

First, calculate $\Delta x$:

$\Delta x = \frac{b - a}{n} = \frac{2 - (-1)}{n} = \frac{3}{n}$

Next, calculate $x_i = a + i \Delta x$:

$x_i = -1 + i \left(\frac{3}{n}\right) = -1 + \frac{3i}{n}$

Now, evaluate $f(x_i) = f\left(-1 + \frac{3i}{n}\right)$:

$f\left(-1 + \frac{3i}{n}\right) = 7\left(-1 + \frac{3i}{n}\right) - 5$

$= -7 + \frac{21i}{n} - 5$

$= -12 + \frac{21i}{n}$

Now, set up the sum $\sum\limits_{i=1}^n f(x_i) \Delta x$:

$\sum\limits_{i=1}^n \left(-12 + \frac{21i}{n}\right) \left(\frac{3}{n}\right)$

$= \sum\limits_{i=1}^n \left( -12 \cdot \frac{3}{n} + \frac{21i}{n} \cdot \frac{3}{n} \right)$

$= \sum\limits_{i=1}^n \left( \frac{-36}{n} + \frac{63i}{n^2} \right)$

Split the sum using the property $\sum (c_i + d_i) = \sum c_i + \sum d_i$:

$= \sum\limits_{i=1}^n \frac{-36}{n} + \sum\limits_{i=1}^n \frac{63i}{n^2}$

Factor out constants and use summation formulas $\sum\limits_{i=1}^n c = nc$ and $\sum\limits_{i=1}^n i = \frac{n(n+1)}{2}$:

$= \frac{-36}{n} \sum\limits_{i=1}^n 1 + \frac{63}{n^2} \sum\limits_{i=1}^n i$

$= \frac{-36}{n} (n) + \frac{63}{n^2} \frac{n(n+1)}{2}$

$= -36 + \frac{63n(n+1)}{2n^2}$

$= -36 + \frac{63(n+1)}{2n}$

$= -36 + \frac{63}{2} \left(\frac{n+1}{n}\right)$

$= -36 + \frac{63}{2} \left(1 + \frac{1}{n}\right)$

Finally, evaluate the limit as $n \to \infty$:

$\int_{-1}^2 (7x - 5) dx = \lim\limits_{n \to \infty} \left( -36 + \frac{63}{2} \left(1 + \frac{1}{n}\right) \right)$

$= \lim\limits_{n \to \infty} (-36) + \lim\limits_{n \to \infty} \left( \frac{63}{2} \left(1 + \frac{1}{n}\right) \right)$

$= -36 + \frac{63}{2} \lim\limits_{n \to \infty} \left(1 + \frac{1}{n}\right)$

As $n \to \infty$, $\frac{1}{n} \to 0$.

$= -36 + \frac{63}{2} (1 + 0)$

$= -36 + \frac{63}{2}$

$= \frac{-72}{2} + \frac{63}{2}$

$= \frac{-72 + 63}{2}$

$= \frac{-9}{2}$

The value of the integral is $-\frac{9}{2}$.

$\int\limits_{−1}^2(7x−5) \;dx = -\frac{9}{2}$

Let the given integral be $I$.

We use the property of definite integrals: $\int\limits_0^a f(x) dx = \int\limits_0^a f(a-x) dx$.

Here, $a = \frac{π}{2}$ and $f(x) = \frac{\tan^7 x}{\cot^7 x + \tan^7 x}$.

Applying the property to equation (1), we replace $x$ with $\frac{π}{2} - x$:

$I = \int\limits_0^{\frac{π}{2}} \frac{\tan^7 (\frac{π}{2}-x)}{\cot^7 (\frac{π}{2}-x) + \tan^7 (\frac{π}{2}-x)} \; dx$

Using the identities $\tan(\frac{π}{2}-x) = \cot x$ and $\cot(\frac{π}{2}-x) = \tan x$, we get:

Adding equation (1) and equation (2):

$I + I = \int\limits_0^{\frac{π}{2}} \frac{\tan^7 x}{\cot^7 x + \tan^7 x} \; dx + \int\limits_0^{\frac{π}{2}} \frac{\cot^7 x}{\tan^7 x + \cot^7 x} \; dx$

$2I = \int\limits_0^{\frac{π}{2}} \left( \frac{\tan^7 x}{\cot^7 x + \tan^7 x} + \frac{\cot^7 x}{\tan^7 x + \cot^7 x} \right) \; dx$

$2I = \int\limits_0^{\frac{π}{2}} \frac{\tan^7 x + \cot^7 x}{\cot^7 x + \tan^7 x} \; dx$

The numerator and denominator are the same, so the integrand simplifies to 1:

$2I = \int\limits_0^{\frac{π}{2}} 1 \; dx$

Now, we evaluate the simple integral:

$2I = [x]_0^{\frac{π}{2}}$

$2I = \frac{π}{2} - 0$

$2I = \frac{π}{2}$

Solving for $I$:

$I = \frac{π}{4}$

Thus, the value of the integral is $\frac{π}{4}$.

Let the given integral be $I$.

We use the property of definite integrals: $\int\limits_a^b f(x) dx = \int\limits_a^b f(a+b-x) dx$.

Here, $a=2$ and $b=8$, so $a+b = 2+8=10$.

Applying the property to equation (1), we replace $x$ with $a+b-x$, which is $10-x$:

$I = \int\limits_2^8 \frac{\sqrt{10 − (10-x)}}{\sqrt{10-x} + \sqrt{10 − (10-x)}} \;dx$

Simplify the terms inside the square roots:

$I = \int\limits_2^8 \frac{\sqrt{10 - 10 + x}}{\sqrt{10-x} + \sqrt{10 - 10 + x}} \;dx$

Now, add equation (1) and equation (2):

$I + I = \int\limits_2^8 \frac{\sqrt{10 − x}}{\sqrt{x} + \sqrt{10 − x}} \;dx + \int\limits_2^8 \frac{\sqrt{x}}{\sqrt{10-x} + \sqrt{x}} \;dx$

$2I = \int\limits_2^8 \left( \frac{\sqrt{10 − x}}{\sqrt{x} + \sqrt{10 − x}} + \frac{\sqrt{x}}{\sqrt{10-x} + \sqrt{x}} \right) \;dx$

Since the denominators are the same, we can combine the numerators:

$2I = \int\limits_2^8 \frac{\sqrt{10 − x} + \sqrt{x}}{\sqrt{x} + \sqrt{10 − x}} \;dx$

The numerator and the denominator are identical, so the fraction simplifies to 1:

$2I = \int\limits_2^8 1 \;dx$

Evaluate the simple integral:

$2I = [x]_2^8$

$2I = 8 - 2$

$2I = 6$

Solve for $I$:

$I = \frac{6}{2}$

$I = 3$

Thus, the value of the integral is 3.

Let the given integral be $I$.

$I = \int\limits_0^{\frac{π}{4}} \sqrt{1+ \sin 2x} \;dx$

We use the trigonometric identity $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$.

So, the expression inside the square root can be rewritten as:

$1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x$

$1 + \sin 2x = (\sin x + \cos x)^2$

Substitute this back into the integral:

$I = \int\limits_0^{\frac{π}{4}} \sqrt{(\sin x + \cos x)^2} \;dx$

$I = \int\limits_0^{\frac{π}{4}} |\sin x + \cos x| \;dx$

For $x \in [0, \frac{π}{4}]$, both $\sin x$ and $\cos x$ are non-negative, and their sum $\sin x + \cos x$ is positive.

Therefore, $|\sin x + \cos x| = \sin x + \cos x$ for $x \in [0, \frac{π}{4}]$.

The integral becomes:

$I = \int\limits_0^{\frac{π}{4}} (\sin x + \cos x) \;dx$

Now, integrate term by term:

$I = [-\cos x + \sin x]_0^{\frac{π}{4}}$

Apply the limits of integration:

$I = (-\cos \frac{π}{4} + \sin \frac{π}{4}) - (-\cos 0 + \sin 0)$

Substitute the values of the trigonometric functions:

$\cos \frac{π}{4} = \frac{1}{\sqrt{2}}$

$\sin \frac{π}{4} = \frac{1}{\sqrt{2}}$

$\cos 0 = 1$

$\sin 0 = 0$

$I = (-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (-1 + 0)$

$I = 0 - (-1)$

$I = 1$

Thus, the value of the integral is 1.

Let the integral be $I$.

$I = \int x^2 \tan^{-1} x \;dx$

We will use the method of integration by parts, which states $\int u \; dv = uv - \int v \; du$.

According to the ILATE rule, we choose:

$u = \tan^{-1} x$

$dv = x^2 \;dx$

Now, we find $du$ by differentiating $u$ and $v$ by integrating $dv$.

Differentiating $u = \tan^{-1} x$ with respect to $x$:

$du = \frac{1}{1+x^2} \;dx$

Integrating $dv = x^2 \;dx$ with respect to $x$:

$v = \int x^2 \;dx = \frac{x^3}{3}$

Now, substitute these into the integration by parts formula:

$I = (\tan^{-1} x) \left(\frac{x^3}{3}\right) - \int \left(\frac{x^3}{3}\right) \left(\frac{1}{1+x^2}\right) \;dx$

$I = \frac{x^3}{3} \tan^{-1} x - \frac{1}{3} \int \frac{x^3}{1+x^2} \;dx$

Now, we need to evaluate the integral $\int \frac{x^3}{1+x^2} \;dx$. We can perform algebraic manipulation on the integrand:

$\frac{x^3}{1+x^2} = \frac{x^3 + x - x}{1+x^2} = \frac{x(x^2+1) - x}{1+x^2} = \frac{x(x^2+1)}{1+x^2} - \frac{x}{1+x^2} = x - \frac{x}{1+x^2}$

So, the integral becomes:

$\int \frac{x^3}{1+x^2} \;dx = \int \left(x - \frac{x}{1+x^2}\right) \;dx = \int x \;dx - \int \frac{x}{1+x^2} \;dx$

The first part is $\int x \;dx = \frac{x^2}{2}$.

For the second part, $\int \frac{x}{1+x^2} \;dx$, we use a substitution. Let $t = 1+x^2$. Then $dt = 2x \;dx$, which means $x \;dx = \frac{1}{2} dt$.

$\int \frac{x}{1+x^2} \;dx = \int \frac{1}{t} \left(\frac{1}{2} dt\right) = \frac{1}{2} \int \frac{1}{t} \;dt = \frac{1}{2} \log|t| + C_1$

Substitute back $t = 1+x^2$. Since $1+x^2 > 0$, we have $\log|1+x^2| = \log(1+x^2)$.

$\int \frac{x}{1+x^2} \;dx = \frac{1}{2} \log(1+x^2) + C_1$

Combining the parts of $\int \frac{x^3}{1+x^2} \;dx$:

$\int \frac{x^3}{1+x^2} \;dx = \frac{x^2}{2} - \frac{1}{2} \log(1+x^2) + C_1$

Now substitute this result back into the main equation for $I$:

$I = \frac{x^3}{3} \tan^{-1} x - \frac{1}{3} \left(\frac{x^2}{2} - \frac{1}{2} \log(1+x^2)\right) + C$

Distribute the $-\frac{1}{3}$:

$I = \frac{x^3}{3} \tan^{-1} x - \frac{x^2}{6} + \frac{1}{6} \log(1+x^2) + C$

The integral is $\frac{x^3}{3} \tan^{-1} x - \frac{x^2}{6} + \frac{1}{6} \log(1+x^2) + C$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \sqrt{10−4x+4x^2} \;dx$

First, we rewrite the quadratic expression inside the square root by completing the square. We rearrange the terms as $4x^2 - 4x + 10$.

Factor out the coefficient of $x^2$ (which is 4):

$4x^2 - 4x + 10 = 4(x^2 - x) + 10$

Complete the square for the expression inside the parenthesis, $x^2 - x$. To do this, we add and subtract $(\frac{1}{2} \times \text{coefficient of } x)^2 = (\frac{1}{2} \times -1)^2 = (-\frac{1}{2})^2 = \frac{1}{4}$.

$x^2 - x = x^2 - x + \frac{1}{4} - \frac{1}{4} = (x - \frac{1}{2})^2 - \frac{1}{4}$

Substitute this back into the expression:

$4(x^2 - x) + 10 = 4\left((x - \frac{1}{2})^2 - \frac{1}{4}\right) + 10$

$= 4(x - \frac{1}{2})^2 - 4 \times \frac{1}{4} + 10$

$= 4(x - \frac{1}{2})^2 - 1 + 10$

$= 4(x - \frac{1}{2})^2 + 9$

So the integral becomes:

$I = \int \sqrt{4(x - \frac{1}{2})^2 + 9} \;dx$

Factor out 4 from the expression inside the square root:

$\sqrt{4(x - \frac{1}{2})^2 + 9} = \sqrt{4\left((x - \frac{1}{2})^2 + \frac{9}{4}\right)}$

$= \sqrt{4} \sqrt{(x - \frac{1}{2})^2 + \left(\frac{3}{2}\right)^2}$

$= 2 \sqrt{(x - \frac{1}{2})^2 + \left(\frac{3}{2}\right)^2}$

The integral is now:

$I = \int 2 \sqrt{(x - \frac{1}{2})^2 + \left(\frac{3}{2}\right)^2} \;dx$

$I = 2 \int \sqrt{(x - \frac{1}{2})^2 + \left(\frac{3}{2}\right)^2} \;dx$

Let $u = x - \frac{1}{2}$. Then $du = dx$. The integral becomes:

$I = 2 \int \sqrt{u^2 + \left(\frac{3}{2}\right)^2} \;du$

This integral is in the form $\int \sqrt{x^2 + a^2} \;dx$. We use the standard formula:

$\int \sqrt{x^2 + a^2} \;dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| + C$

Applying this formula with $x=u$ and $a=\frac{3}{2}$:

$I = 2 \left[ \frac{u}{2}\sqrt{u^2+\left(\frac{3}{2}\right)^2} + \frac{\left(\frac{3}{2}\right)^2}{2}\log\left|u+\sqrt{u^2+\left(\frac{3}{2}\right)^2}\right| \right] + C$

$I = 2 \left[ \frac{u}{2}\sqrt{u^2+\frac{9}{4}} + \frac{\frac{9}{4}}{2}\log\left|u+\sqrt{u^2+\frac{9}{4}}\right| \right] + C$

$I = 2 \left[ \frac{u}{2}\sqrt{u^2+\frac{9}{4}} + \frac{9}{8}\log\left|u+\sqrt{u^2+\frac{9}{4}}\right| \right] + C$

Now substitute back $u = x - \frac{1}{2}$. Also note that $u^2+\frac{9}{4} = (x - \frac{1}{2})^2 + \frac{9}{4} = \frac{4x^2 - 4x + 10}{4}$.

$I = 2 \left[ \frac{x - \frac{1}{2}}{2}\sqrt{\frac{4x^2 - 4x + 10}{4}} + \frac{9}{8}\log\left|x - \frac{1}{2}+\sqrt{\frac{4x^2 - 4x + 10}{4}}\right| \right] + C$

$I = 2 \left[ \frac{\frac{2x-1}{2}}{2}\frac{\sqrt{4x^2 - 4x + 10}}{2} + \frac{9}{8}\log\left|\frac{2x-1}{2}+\frac{\sqrt{4x^2 - 4x + 10}}{2}\right| \right] + C$

$I = 2 \left[ \frac{(2x-1)\sqrt{4x^2 - 4x + 10}}{8} + \frac{9}{8}\log\left|\frac{2x-1 + \sqrt{4x^2 - 4x + 10}}{2}\right| \right] + C$

$I = \frac{(2x-1)\sqrt{4x^2 - 4x + 10}}{4} + \frac{9}{4}\log\left|\frac{2x-1 + \sqrt{4x^2 - 4x + 10}}{2}\right| + C$

We can simplify the logarithm term using $\log|\frac{A}{B}| = \log|A| - \log|B|$:

$\log\left|\frac{2x-1 + \sqrt{4x^2 - 4x + 10}}{2}\right| = \log\left|2x-1 + \sqrt{4x^2 - 4x + 10}\right| - \log|2|$

So, $I = \frac{(2x-1)\sqrt{4x^2 - 4x + 10}}{4} + \frac{9}{4}\left(\log\left|2x-1 + \sqrt{4x^2 - 4x + 10}\right| - \log 2\right) + C$

$I = \frac{(2x-1)\sqrt{4x^2 - 4x + 10}}{4} + \frac{9}{4}\log\left|2x-1 + \sqrt{4x^2 - 4x + 10}\right| - \frac{9}{4}\log 2 + C$

We can absorb the constant term $-\frac{9}{4}\log 2$ into the constant of integration. Let $C_{new} = C - \frac{9}{4}\log 2$.

$I = \frac{(2x-1)\sqrt{4x^2 - 4x + 10}}{4} + \frac{9}{4}\log\left|2x-1 + \sqrt{4x^2 - 4x + 10}\right| + C_{new}$

We can drop the subscript on the constant.

The integral is $\frac{(2x-1)\sqrt{4x^2 - 4x + 10}}{4} + \frac{9}{4}\log\left|2x-1 + \sqrt{4x^2 - 4x + 10}\right| + C$.

Let the given integral be $I$.

$I = \int \frac{x^2}{x^4+x^2 −2} \;dx$

First, we factor the denominator $x^4+x^2-2$. Let $y=x^2$. The denominator becomes $y^2+y-2$.

$y^2+y-2 = (y+2)(y-1)$

Substituting back $y=x^2$, the denominator is $(x^2+2)(x^2-1)$.

So, the integrand is $\frac{x^2}{(x^2+2)(x^2-1)}$.

We use the method of partial fraction decomposition for the integrand.

Consider the rational expression $\frac{x^2}{(x^2+2)(x^2-1)}$. We can set up the partial fraction decomposition by treating $x^2$ as a variable temporarily. Let $y=x^2$.

$\frac{y}{(y+2)(y-1)} = \frac{A}{y+2} + \frac{B}{y-1}$

Multiply by $(y+2)(y-1)$: $y = A(y-1) + B(y+2)$.

To find $A$, set $y=-2$:

$-2 = A(-2-1) + B(-2+2)$

$-2 = -3A + 0$

$A = \frac{-2}{-3} = \frac{2}{3}$

To find $B$, set $y=1$:

$1 = A(1-1) + B(1+2)$

$1 = 0 + 3B$

$B = \frac{1}{3}$

So, $\frac{y}{(y+2)(y-1)} = \frac{2/3}{y+2} + \frac{1/3}{y-1}$.

Substitute back $y=x^2$:

$\frac{x^2}{(x^2+2)(x^2-1)} = \frac{2/3}{x^2+2} + \frac{1/3}{x^2-1}$

The integral becomes:

$I = \int \left(\frac{2}{3(x^2+2)} + \frac{1}{3(x^2-1)}\right) \;dx$

$I = \frac{2}{3} \int \frac{1}{x^2+2} \;dx + \frac{1}{3} \int \frac{1}{x^2-1} \;dx$

We evaluate each integral separately.

For the first integral, $\int \frac{1}{x^2+2} \;dx$, we use the formula $\int \frac{1}{x^2+a^2} \;dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$. Here $a^2=2$, so $a=\sqrt{2}$.

$\int \frac{1}{x^2+2} \;dx = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right)$

For the second integral, $\int \frac{1}{x^2-1} \;dx$, we use the formula $\int \frac{1}{x^2-a^2} \;dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C$. Here $a^2=1$, so $a=1$.

$\int \frac{1}{x^2-1} \;dx = \frac{1}{2(1)}\log\left|\frac{x-1}{x+1}\right| = \frac{1}{2}\log\left|\frac{x-1}{x+1}\right|$

Now substitute these results back into the expression for $I$ and add the constant of integration $C$:

$I = \frac{2}{3} \left(\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right)\right) + \frac{1}{3} \left(\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|\right) + C$

$I = \frac{2}{3\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{1}{6}\log\left|\frac{x-1}{x+1}\right| + C$

Simplify the coefficient $\frac{2}{3\sqrt{2}}$:

$\frac{2}{3\sqrt{2}} = \frac{2\sqrt{2}}{3\sqrt{2}\sqrt{2}} = \frac{2\sqrt{2}}{3 \times 2} = \frac{\sqrt{2}}{3}$

So, the final result is:

$I = \frac{\sqrt{2}}{3}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{1}{6}\log\left|\frac{x-1}{x+1}\right| + C$

The value of the integral is $\frac{\sqrt{2}}{3}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{1}{6}\log\left|\frac{x-1}{x+1}\right| + C$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{x^3+ x}{x^4 - 9} \;dx$

We can factor the denominator: $x^4 - 9 = (x^2)^2 - 3^2 = (x^2-3)(x^2+3)$.

The numerator is $x^3+x = x(x^2+1)$.

So the integrand is $\frac{x(x^2+1)}{(x^2-3)(x^2+3)}$.

We can use a substitution to simplify the integral.

Let $u = x^2$.

Then, the differential $du$ is $du = \frac{d}{dx}(x^2) \;dx = 2x \;dx$.

This means $x \;dx = \frac{1}{2} du$.

The numerator can be written as $(x^2+1)x\;dx$. Substitute $u$ and $du$ into the integral:

$I = \int \frac{(x^2+1)}{(x^2-3)(x^2+3)} (x \;dx)$

$I = \int \frac{u+1}{(u-3)(u+3)} \frac{1}{2} du$

$I = \frac{1}{2} \int \frac{u+1}{(u-3)(u+3)} du$

Now we use the method of partial fraction decomposition for the rational function in terms of $u$.

$\frac{u+1}{(u-3)(u+3)} = \frac{A}{u-3} + \frac{B}{u+3}$

Multiply both sides by $(u-3)(u+3)$: $u+1 = A(u+3) + B(u-3)$.

To find $A$, set $u = 3$:

$3+1 = A(3+3) + B(3-3)$

$4 = 6A + 0$

$A = \frac{4}{6} = \frac{2}{3}$

To find $B$, set $u = -3$:

$-3+1 = A(-3+3) + B(-3-3)$

$-2 = 0 + B(-6)$

$B = \frac{-2}{-6} = \frac{1}{3}$

So the partial fraction decomposition is:

$\frac{u+1}{(u-3)(u+3)} = \frac{2/3}{u-3} + \frac{1/3}{u+3}$

Substitute this back into the integral for $I$ in terms of $u$:

$I = \frac{1}{2} \int \left(\frac{2/3}{u-3} + \frac{1/3}{u+3}\right) du$

$I = \frac{1}{2} \left( \frac{2}{3} \int \frac{1}{u-3} du + \frac{1}{3} \int \frac{1}{u+3} du \right)$

Integrate each term:

$\int \frac{1}{u-3} du = \log|u-3|$

$\int \frac{1}{u+3} du = \log|u+3|$

Substitute these back into the expression for $I$:

$I = \frac{1}{2} \left( \frac{2}{3} \log|u-3| + \frac{1}{3} \log|u+3| \right) + C'$

$I = \frac{1}{3} \log|u-3| + \frac{1}{6} \log|u+3| + C'$

Finally, substitute back $u = x^2$:

$I = \frac{1}{3} \log|x^2-3| + \frac{1}{6} \log|x^2+3| + C$

(We use $C$ as the constant of integration).

The value of the integral is $\frac{1}{3} \log|x^2-3| + \frac{1}{6} \log|x^2+3| + C$, where $C$ is the constant of integration.

Let the given integral be $I$.

We use the property of definite integrals: $\int\limits_0^a f(x) dx = \int\limits_0^a f(a-x) dx$.

Here, $a = \frac{π}{2}$. Applying the property to equation (1), we replace $x$ with $\frac{π}{2} - x$:

$I = \int\limits_0^{\frac{π}{2}} \frac{\sin^2 (\frac{π}{2}-x)}{\sin (\frac{π}{2}-x) + \cos (\frac{π}{2}-x)} \; dx$

Using the identities $\sin(\frac{π}{2}-x) = \cos x$ and $\cos(\frac{π}{2}-x) = \sin x$, we get:

Adding equation (1) and equation (2):

$I + I = \int\limits_0^{\frac{π}{2}} \frac{\sin^2 x}{\sin x + \cos x} \; dx + \int\limits_0^{\frac{π}{2}} \frac{\cos^2 x}{\cos x + \sin x} \; dx$

$2I = \int\limits_0^{\frac{π}{2}} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} \; dx$

Using the identity $\sin^2 x + \cos^2 x = 1$:

$2I = \int\limits_0^{\frac{π}{2}} \frac{1}{\sin x + \cos x} \; dx$

Now we evaluate the integral $\int\limits_0^{\frac{π}{2}} \frac{1}{\sin x + \cos x} \; dx$.

We rewrite the denominator $\sin x + \cos x$ in the form $R \sin(x+\alpha)$.

$\sin x + \cos x = \sqrt{1^2+1^2} \sin(x + \tan^{-1}(\frac{1}{1})) = \sqrt{2} \sin(x + \frac{π}{4})$

So, the integral becomes:

$2I = \int\limits_0^{\frac{π}{2}} \frac{1}{\sqrt{2} \sin(x + \frac{π}{4})} \; dx$

$2I = \frac{1}{\sqrt{2}} \int\limits_0^{\frac{π}{2}} \text{cosec}(x + \frac{π}{4}) \; dx$

Let $u = x + \frac{π}{4}$. Then $du = dx$.

When $x=0$, $u = 0 + \frac{π}{4} = \frac{π}{4}$.

When $x=\frac{π}{2}$, $u = \frac{π}{2} + \frac{π}{4} = \frac{3π}{4}$.

The integral limits change from $[0, \frac{π}{2}]$ to $[\frac{π}{4}, \frac{3π}{4}]$.

$2I = \frac{1}{\sqrt{2}} \int\limits_{\frac{π}{4}}^{\frac{3π}{4}} \text{cosec}(u) \; du$

Using the integral formula $\int \text{cosec } u \; du = \log|\text{cosec } u - \cot u|$, we get:

$2I = \frac{1}{\sqrt{2}} [\log|\text{cosec } u - \cot u|]_{\frac{π}{4}}^{\frac{3π}{4}}$

Evaluate at the limits:

$2I = \frac{1}{\sqrt{2}} \left[ \log\left|\text{cosec } \frac{3π}{4} - \cot \frac{3π}{4}\right| - \log\left|\text{cosec } \frac{π}{4} - \cot \frac{π}{4}\right| \right]$

We know that $\text{cosec } \frac{3π}{4} = \sqrt{2}$, $\cot \frac{3π}{4} = -1$, $\text{cosec } \frac{π}{4} = \sqrt{2}$, and $\cot \frac{π}{4} = 1$.

$2I = \frac{1}{\sqrt{2}} \left[ \log|\sqrt{2} - (-1)| - \log|\sqrt{2} - 1| \right]$

$2I = \frac{1}{\sqrt{2}} \left[ \log(\sqrt{2} + 1) - \log(\sqrt{2} - 1) \right]$

Using the property $\log A - \log B = \log(\frac{A}{B})$:

$2I = \frac{1}{\sqrt{2}} \log\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)$

Rationalize the denominator inside the logarithm:

$\frac{\sqrt{2} + 1}{\sqrt{2} - 1} = \frac{(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2})^2 - 1^2} = \frac{2 + 1 + 2\sqrt{2}}{2 - 1} = 3 + 2\sqrt{2}$

However, we can also notice that $(\sqrt{2}+1)(\sqrt{2}-1) = 2-1=1$, so $\sqrt{2}-1 = \frac{1}{\sqrt{2}+1}$.

Thus, $\frac{\sqrt{2} + 1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{1/(\sqrt{2} + 1)} = (\sqrt{2} + 1)^2$. Using this simplifies the next step.

$2I = \frac{1}{\sqrt{2}} \log\left((\sqrt{2} + 1)^2\right)$

Using the property $\log A^p = p \log A$:

$2I = \frac{1}{\sqrt{2}} \times 2 \log(\sqrt{2} + 1)$

$2I = \sqrt{2} \log(\sqrt{2} + 1)$

Solving for $I$:

$I = \frac{\sqrt{2}}{2} \log(\sqrt{2} + 1)$

Since $\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{1}{\sqrt{2}}$:

$I = \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1)$

Thus, we have shown that $\int\limits_0^{\frac{π}{2}} \frac{\sin^2 x}{\sin x + \cos x} \; dx = \frac{1}{\sqrt{2}} \log (\sqrt{2}+1)$.

Let the given integral be $I$.

We will evaluate this integral using integration by parts, which is given by $\int u \; dv = uv - \int v \; du$ for indefinite integrals, and $[uv]_a^b - \int_a^b v \; du$ for definite integrals.

According to the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential), we choose $u = (\tan^{-1} x)^2$ and $dv = x \;dx$.

Find $du$ by differentiating $u$ with respect to $x$:

$u = (\tan^{-1} x)^2$

$du = \frac{d}{dx}((\tan^{-1} x)^2) \;dx = 2(\tan^{-1} x) \cdot \frac{d}{dx}(\tan^{-1} x) \;dx$

$du = 2(\tan^{-1} x) \cdot \frac{1}{1+x^2} \;dx$

Find $v$ by integrating $dv$ with respect to $x$:

$dv = x \;dx$

$v = \int x \;dx = \frac{x^2}{2}$

Now apply the integration by parts formula to equation (1) with limits $a=0$ and $b=1$:

$I = \left[ uv \right]_0^1 - \int\limits_0^1 v \; du$

$I = \left[ \frac{x^2}{2}(\tan^{-1} x)^2 \right]_0^1 - \int\limits_0^1 \frac{x^2}{2} \left( 2(\tan^{-1} x) \frac{1}{1+x^2} \right) \;dx$

$I = \left[ \frac{x^2}{2}(\tan^{-1} x)^2 \right]_0^1 - \int\limits_0^1 \frac{x^2 \tan^{-1} x}{1+x^2} \;dx$

First, evaluate the boundary term $\left[ \frac{x^2}{2}(\tan^{-1} x)^2 \right]_0^1$:

At $x=1$: $\frac{1^2}{2}(\tan^{-1} 1)^2 = \frac{1}{2}\left(\frac{π}{4}\right)^2 = \frac{1}{2} \cdot \frac{π^2}{16} = \frac{π^2}{32}$

At $x=0$: $\frac{0^2}{2}(\tan^{-1} 0)^2 = 0 \cdot (0)^2 = 0$

Boundary term value $= \frac{π^2}{32} - 0 = \frac{π^2}{32}$.

The integral becomes $I = \frac{π^2}{32} - \int\limits_0^1 \frac{x^2 \tan^{-1} x}{1+x^2} \;dx$.

Let's evaluate the remaining integral $J = \int\limits_0^1 \frac{x^2 \tan^{-1} x}{1+x^2} \;dx$.

We can rewrite the integrand by performing algebraic division or manipulation:

$\frac{x^2}{1+x^2} = \frac{x^2+1-1}{1+x^2} = 1 - \frac{1}{1+x^2}$

So, the integral $J$ becomes:

$J = \int\limits_0^1 \left(1 - \frac{1}{1+x^2}\right) \tan^{-1} x \;dx$

$J = \int\limits_0^1 \tan^{-1} x \;dx - \int\limits_0^1 \frac{\tan^{-1} x}{1+x^2} \;dx$

Let's evaluate the first part $\int\limits_0^1 \tan^{-1} x \;dx$. We use integration by parts again with $u = \tan^{-1} x$ and $dv = dx$.

$u = \tan^{-1} x \implies du = \frac{1}{1+x^2} \;dx$

$dv = dx \implies v = x$

$\int\limits_0^1 \tan^{-1} x \;dx = [x \tan^{-1} x]_0^1 - \int\limits_0^1 x \cdot \frac{1}{1+x^2} \;dx$

Evaluate the boundary term $[x \tan^{-1} x]_0^1$:

At $x=1$: $1 \cdot \tan^{-1} 1 = 1 \cdot \frac{π}{4} = \frac{π}{4}$

At $x=0$: $0 \cdot \tan^{-1} 0 = 0 \cdot 0 = 0$

Boundary term value $= \frac{π}{4} - 0 = \frac{π}{4}$.

Now evaluate the integral $\int\limits_0^1 \frac{x}{1+x^2} \;dx$. Use substitution: Let $w = 1+x^2$. Then $dw = 2x \;dx$, so $x \;dx = \frac{1}{2} dw$.

Limits of integration for $w$: When $x=0$, $w = 1+0^2 = 1$. When $x=1$, $w = 1+1^2 = 2$.

$\int\limits_0^1 \frac{x}{1+x^2} \;dx = \int\limits_1^2 \frac{1}{w} \frac{1}{2} dw = \frac{1}{2} \int\limits_1^2 \frac{1}{w} dw = \frac{1}{2} [\log|w|]_1^2$

$= \frac{1}{2} (\log 2 - \log 1) = \frac{1}{2} (\log 2 - 0) = \frac{1}{2} \log 2$

So, $\int\limits_0^1 \tan^{-1} x \;dx = \frac{π}{4} - \frac{1}{2} \log 2$.

Now, evaluate the second part of $J$, $\int\limits_0^1 \frac{\tan^{-1} x}{1+x^2} \;dx$. Use substitution: Let $w = \tan^{-1} x$. Then $dw = \frac{1}{1+x^2} \;dx$.

Limits of integration for $w$: When $x=0$, $w = \tan^{-1} 0 = 0$. When $x=1$, $w = \tan^{-1} 1 = \frac{π}{4}$.

$\int\limits_0^1 \frac{\tan^{-1} x}{1+x^2} \;dx = \int\limits_0^{\frac{π}{4}} w \;dw$

$= \left[ \frac{w^2}{2} \right]_0^{\frac{π}{4}} = \frac{(π/4)^2}{2} - \frac{0^2}{2} = \frac{π^2/16}{2} - 0 = \frac{π^2}{32}$

Combine the parts of $J$:

$J = \int\limits_0^1 \tan^{-1} x \;dx - \int\limits_0^1 \frac{\tan^{-1} x}{1+x^2} \;dx$

$J = \left(\frac{π}{4} - \frac{1}{2} \log 2\right) - \frac{π^2}{32}$

Finally, substitute the value of $J$ back into the expression for $I$:

$I = \frac{π^2}{32} - J$

$I = \frac{π^2}{32} - \left(\frac{π}{4} - \frac{1}{2} \log 2 - \frac{π^2}{32}\right)$

$I = \frac{π^2}{32} - \frac{π}{4} + \frac{1}{2} \log 2 + \frac{π^2}{32}$

$I = \frac{2π^2}{32} - \frac{π}{4} + \frac{1}{2} \log 2$

$I = \frac{π^2}{16} - \frac{π}{4} + \frac{1}{2} \log 2$

The value of the integral is $\frac{π^2}{16} - \frac{π}{4} + \frac{1}{2} \log 2$.

The function is given by $f(x) = |x + 1| + |x| + |x – 1|$.

The absolute value functions change their definition at the points where the expressions inside them become zero. These points are $x+1=0 \implies x=-1$, $x=0$, and $x-1=0 \implies x=1$.

These points divide the interval of integration $[-1, 2]$ into subintervals: $[-1, 0]$, $[0, 1]$, and $[1, 2]$. We define $f(x)$ on each subinterval.

Case 1: For $-1 \leq x \leq 0$:

$x+1 \geq 0 \implies |x+1| = x+1$

$x \leq 0 \implies |x| = -x$

$x-1 \leq 0 \implies |x-1| = -(x-1) = 1-x$

So, $f(x) = (x+1) + (-x) + (1-x) = x+1-x+1-x = 2-x$ for $-1 \leq x \leq 0$.

Case 2: For $0 \leq x \leq 1$:

$x+1 \geq 0 \implies |x+1| = x+1$

$x \geq 0 \implies |x| = x$

$x-1 \leq 0 \implies |x-1| = -(x-1) = 1-x$

So, $f(x) = (x+1) + x + (1-x) = x+1+x+1-x = x+2$ for $0 \leq x \leq 1$.

Case 3: For $1 \leq x \leq 2$:

$x+1 \geq 0 \implies |x+1| = x+1$

$x \geq 0 \implies |x| = x$

$x-1 \geq 0 \implies |x-1| = x-1$

So, $f(x) = (x+1) + x + (x-1) = x+1+x+x-1 = 3x$ for $1 \leq x \leq 2$.

We can write the integral as a sum of integrals over the subintervals:

$\int\limits_{−1}^2 f(x) \;dx = \int\limits_{-1}^0 f(x) \;dx + \int\limits_0^1 f(x) \;dx + \int\limits_1^2 f(x) \;dx$

$\int\limits_{−1}^2 f(x) \;dx = \int\limits_{-1}^0 (2-x) \;dx + \int\limits_0^1 (x+2) \;dx + \int\limits_1^2 (3x) \;dx$

Evaluate the first integral:

$\int\limits_{-1}^0 (2-x) \;dx = \left[2x - \frac{x^2}{2}\right]_{-1}^0$

$= \left(2(0) - \frac{0^2}{2}\right) - \left(2(-1) - \frac{(-1)^2}{2}\right)$

$= (0 - 0) - \left(-2 - \frac{1}{2}\right) = 0 - \left(-\frac{4}{2} - \frac{1}{2}\right) = - \left(-\frac{5}{2}\right) = \frac{5}{2}$

Evaluate the second integral:

$\int\limits_0^1 (x+2) \;dx = \left[\frac{x^2}{2} + 2x\right]_0^1$

$= \left(\frac{1^2}{2} + 2(1)\right) - \left(\frac{0^2}{2} + 2(0)\right)$

$= \left(\frac{1}{2} + 2\right) - (0 + 0) = \frac{1}{2} + \frac{4}{2} = \frac{5}{2}$

Evaluate the third integral:

$\int\limits_1^2 (3x) \;dx = \left[\frac{3x^2}{2}\right]_1^2$

$= \left(\frac{3(2)^2}{2}\right) - \left(\frac{3(1)^2}{2}\right)$

$= \left(\frac{3 \times 4}{2}\right) - \left(\frac{3 \times 1}{2}\right) = \frac{12}{2} - \frac{3}{2} = 6 - \frac{3}{2} = \frac{12-3}{2} = \frac{9}{2}$

Sum the results of the three integrals:

$\int\limits_{−1}^2 f(x) \;dx = \frac{5}{2} + \frac{5}{2} + \frac{9}{2}$

$= \frac{5+5+9}{2} = \frac{19}{2}$

Thus, the value of the integral is $\frac{19}{2}$.

Let the given integral be $I$.

$I = \int e^x (\cos x − \sin x) \;dx$

We can recognize that the integrand is in the form $e^x (f(x) + f'(x))$.

Consider $f(x) = \cos x$.

Then, the derivative of $f(x)$ is $f'(x) = \frac{d}{dx}(\cos x) = -\sin x$.

So, the integrand is $e^x (\cos x + (-\sin x)) = e^x (f(x) + f'(x))$.

We use the standard integration formula:

$\int e^x (f(x) + f'(x)) \;dx = e^x f(x) + C$

Applying this formula with $f(x) = \cos x$:

$I = e^x \cos x + C$

Alternatively, we can evaluate this integral using integration by parts on one term.

$I = \int e^x \cos x \;dx - \int e^x \sin x \;dx$

Let's evaluate $\int e^x \cos x \;dx$ using integration by parts $\int u \; dv = uv - \int v \; du$.

Let $u = \cos x$ and $dv = e^x \;dx$.

Then $du = -\sin x \;dx$ and $v = e^x$.

$\int e^x \cos x \;dx = e^x \cos x - \int e^x (-\sin x) \;dx$

$\int e^x \cos x \;dx = e^x \cos x + \int e^x \sin x \;dx$

Substitute this back into the original integral $I$:

$I = \left(e^x \cos x + \int e^x \sin x \;dx\right) - \int e^x \sin x \;dx$

$I = e^x \cos x + \int e^x \sin x \;dx - \int e^x \sin x \;dx$

The two integral terms cancel out:

$I = e^x \cos x + C$

where $C$ is the constant of integration.

Comparing this result with the given options, we see that it matches option (A).

The correct answer is (A).

Let the given integral be $I$.

$I = \int \frac{1}{\sin^2 x \; \cos^2 x} \; dx$

We use the trigonometric identity $1 = \sin^2 x + \cos^2 x$. Substitute this into the numerator:

$I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \; \cos^2 x} \; dx$

Split the fraction into two terms:

$I = \int \left(\frac{\sin^2 x}{\sin^2 x \; \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \; \cos^2 x}\right) \; dx$

Cancel the common terms in the numerator and denominator of each fraction:

$I = \int \left(\frac{1}{\cos^2 x} + \frac{1}{\sin^2 x}\right) \; dx$

Using the reciprocal identities $\frac{1}{\cos x} = \sec x$ and $\frac{1}{\sin x} = \text{cosec } x$, we get:

$I = \int (\sec^2 x + \text{cosec}^2 x) \; dx$

Now, we can integrate term by term:

$I = \int \sec^2 x \; dx + \int \text{cosec}^2 x \; dx$

Using the standard integral formulas $\int \sec^2 x \; dx = \tan x$ and $\int \text{cosec}^2 x \; dx = -\cot x$, we have:

$I = \tan x - \cot x + C$

where $C$ is the constant of integration.

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C).

The given problem states that $\int\limits \frac{3e^x − 5e^{−x}}{4e^x + 4e^{−x}} \;dx = ax + b \log | 4e^x + 5e^{-x}| + C$.

Let the integral be $I = \int \frac{3e^x − 5e^{−x}}{4e^x + 4e^{−x}} \;dx$.

The target form of the result involves $\log | 4e^x + 5e^{-x}|$. The derivative of $\log | 4e^x + 5e^{-x}|$ is $\frac{4e^x - 5e^{-x}}{4e^x + 5e^{-x}}$.

The integrand given is $\frac{3e^x − 5e^{−x}}{4e^x + 4e^{−x}}$. There is a discrepancy between the denominator of the integrand ($4e^x + 4e^{-x}$) and the argument of the logarithm in the target form ($4e^x + 5e^{-x}$).

Based on the structure of the target form and the multiple-choice options, it is highly probable that there is a typographical error in the denominator of the integrand in the question, and it should have been $4e^x + 5e^{-x}$. We will proceed assuming the intended integral is $\int\limits \frac{3e^x − 5e^{−x}}{4e^x + 5e^{−x}} \;dx$.

Let the assumed integral be $I_{assumed} = \int \frac{3e^x − 5e^{−x}}{4e^x + 5e^{−x}} \;dx$.

We are given that $I_{assumed} = ax + b \log | 4e^x + 5e^{-x}| + C$.

Taking the derivative of the right side with respect to $x$, we should obtain the integrand:

$\frac{d}{dx} (ax + b \log | 4e^x + 5e^{-x}| + C) = \frac{d}{dx}(ax) + b \frac{d}{dx}(\log | 4e^x + 5e^{-x}|) + \frac{d}{dx}(C)$

$= a + b \cdot \frac{1}{4e^x + 5e^{-x}} \cdot \frac{d}{dx}(4e^x + 5e^{-x}) + 0$

$= a + b \cdot \frac{4e^x - 5e^{-x}}{4e^x + 5e^{-x}}$

This derivative must be equal to the integrand $\frac{3e^x − 5e^{−x}}{4e^x + 5e^{−x}}$.

$a + \frac{b(4e^x - 5e^{-x})}{4e^x + 5e^{-x}} = \frac{3e^x − 5e^{−x}}{4e^x + 5e^{−x}}$

Combine the terms on the left side by finding a common denominator:

$\frac{a(4e^x + 5e^{-x})}{4e^x + 5e^{-x}} + \frac{b(4e^x - 5e^{-x})}{4e^x + 5e^{-x}} = \frac{3e^x − 5e^{−x}}{4e^x + 5e^{−x}}$

$\frac{a(4e^x + 5e^{-x}) + b(4e^x - 5e^{-x})}{4e^x + 5e^{-x}} = \frac{3e^x − 5e^{−x}}{4e^x + 5e^{−x}}$

Since the denominators are equal, the numerators must be equal:

$a(4e^x + 5e^{-x}) + b(4e^x - 5e^{-x}) = 3e^x − 5e^{−x}$

$4ae^x + 5ae^{-x} + 4be^x - 5be^{-x} = 3e^x − 5e^{−x}$

Group the terms with $e^x$ and $e^{-x}$:

$(4a+4b)e^x + (5a-5b)e^{-x} = 3e^x − 5e^{−x}$

By comparing the coefficients of $e^x$ and $e^{-x}$ on both sides of the equation, we obtain a system of linear equations:

Simplify equation (ii) by dividing by 5:

Now we solve the system of equations (i) and (iii).

Multiply equation (iii) by 4:

$4(a - b) = 4(-1)$

$4a - 4b = -4$

Add this modified equation (iii) to equation (i):

$8a = -1$

$a = -\frac{1}{8}$

Substitute the value of $a$ into equation (iii):

$-\frac{1}{8} - b = -1$

$-b = -1 + \frac{1}{8}$

$-b = -\frac{8}{8} + \frac{1}{8}$

$-b = -\frac{7}{8}$

$b = \frac{7}{8}$

So, the values of $a$ and $b$ that satisfy the condition (assuming the corrected integrand) are $a = -\frac{1}{8}$ and $b = \frac{7}{8}$.

Comparing these values with the given options:

(A) a = $\frac{-1}{8}$, b = $\frac{7}{8}$

(B) a = $\frac{1}{8}$, b = $\frac{7}{8}$

(C) a = $\frac{−1}{8}$, b = $\frac{−7}{8}$

(D) a = $\frac{1}{8}$, b = $\frac{−7}{8}$

The values $a = -\frac{1}{8}$ and $b = \frac{7}{8}$ match option (A).

The correct answer is (A).

Let the given integral be $I$.

$I = \int\limits_{a+c}^{b+c} f(x) \;dx$

We want to evaluate this integral and find an equivalent expression among the given options. We can use a substitution to change the limits of integration.

Let's choose a substitution $u = x - c$.

Then, we can express $x$ in terms of $u$: $x = u + c$.

Now, find the differential $dx$ in terms of $du$:

$du = \frac{d}{dx}(x - c) \;dx = 1 \;dx$

So, $dx = du$.

Next, we need to change the limits of integration according to the substitution:

When $x = a+c$ (the lower limit), the new lower limit for $u$ is:

$u = (a+c) - c = a$

When $x = b+c$ (the upper limit), the new upper limit for $u$ is:

$u = (b+c) - c = b$

Now, substitute $x = u+c$, $dx = du$, and the new limits into the integral $I$:

$I = \int\limits_{u=a}^{u=b} f(u+c) \;du$

Since the variable of integration is a dummy variable, we can replace $u$ with $x$:

$I = \int\limits_{a}^{b} f(x+c) \;dx$

Comparing our result $\int\limits_{a}^{b} f(x+c) \;dx$ with the given options:

(A) $\int\limits_a^b f(x−c) \;dx$

(B) $\int\limits_a^b f(x+c) \;dx$

(C) $\int\limits_a^b f(x) \;dx$

(D) $\int\limits_{a−c}^{b−c} f(x) \;dx$

Our result matches option (B).

The correct answer is (B).

Let the given integral be $I$.

We are given the properties:

From the second property, we can write $g(a-x) = a - g(x)$.

We use the property of definite integrals: $\int\limits_0^a h(x) dx = \int\limits_0^a h(a-x) dx$.

Let $h(x) = f(x)g(x)$. Then $h(a-x) = f(a-x)g(a-x)$.

Using the given properties, we substitute $f(a-x) = f(x)$ and $g(a-x) = a - g(x)$ into the expression for $h(a-x)$:

$h(a-x) = f(x) (a - g(x))$

$h(a-x) = a f(x) - f(x)g(x)$

Now, apply the property $\int\limits_0^a h(x) dx = \int\limits_0^a h(a-x) dx$ to the integral $I$:

Substitute the derived expression for $f(a-x)g(a-x)$:

$I = \int\limits_0^a (a f(x) - f(x)g(x)) \;dx$

Using the linearity of integration, split the integral into two parts:

$I = \int\limits_0^a a f(x) \;dx - \int\limits_0^a f(x)g(x) \;dx$

Factor out the constant $a$ from the first integral:

$I = a \int\limits_0^a f(x) \;dx - \int\limits_0^a f(x)g(x) \;dx$

Notice that the second integral on the right side is the original integral $I$ (from equation 1).

So, we have the equation:

$I = a \int\limits_0^a f(x) \;dx - I$

Now, solve this equation for $I$:

Add $I$ to both sides of the equation:

$I + I = a \int\limits_0^a f(x) \;dx$

$2I = a \int\limits_0^a f(x) \;dx$

Divide both sides by 2:

$I = \frac{a}{2} \int\limits_0^a f(x) \;dx$

Comparing this result with the given options, we find that it matches option (B).

The correct answer is (B).

We are given the relationship $x = \int\limits_0^y \frac{dt}{\sqrt{1 + 9t^2}}$.

To find $\frac{dy}{dx}$, we first differentiate the given equation with respect to $y$.

Using the Fundamental Theorem of Calculus, if $x = \int\limits_0^y g(t) dt$, then $\frac{dx}{dy} = g(y)$.

In this case, $g(t) = \frac{1}{\sqrt{1 + 9t^2}}$. So, evaluating at $t=y$ gives:

$\frac{dx}{dy} = \frac{1}{\sqrt{1 + 9y^2}}$

Now, we find $\frac{dy}{dx}$ by taking the reciprocal of $\frac{dx}{dy}$:

$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{\frac{1}{\sqrt{1 + 9y^2}}} = \sqrt{1 + 9y^2}$

Next, we need to find the second derivative, $\frac{d^2y}{dx^2}$. This is the derivative of $\frac{dy}{dx}$ with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\sqrt{1 + 9y^2}\right)$

Since $\sqrt{1 + 9y^2}$ is a function of $y$, and $y$ is a function of $x$, we use the Chain Rule:

$\frac{d}{dx}\left(\sqrt{1 + 9y^2}\right) = \frac{d}{dy}\left(\sqrt{1 + 9y^2}\right) \cdot \frac{dy}{dx}$

First, calculate $\frac{d}{dy}\left(\sqrt{1 + 9y^2}\right)$. Let $u = 1+9y^2$. Then $\frac{d}{dy}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dy}$.

$\frac{d}{dy}(1 + 9y^2) = 18y$.

So, $\frac{d}{dy}\left(\sqrt{1 + 9y^2}\right) = \frac{1}{2\sqrt{1 + 9y^2}} \cdot 18y = \frac{9y}{\sqrt{1 + 9y^2}}$

Now, substitute this and the expression for $\frac{dy}{dx}$ (from equation 1) into the formula for $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = \left(\frac{9y}{\sqrt{1 + 9y^2}}\right) \cdot \left(\sqrt{1 + 9y^2}\right)$

The term $\sqrt{1 + 9y^2}$ in the numerator and denominator cancels out:

$\frac{d^2y}{dx^2} = 9y$

We are given that $\frac{d^2y}{dx^2} = ay$.

Comparing our result with the given equation:

Assuming $y$ is not identically zero, we can divide both sides by $y$:

$a = 9$

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C).

Let the given integral be $I$.

$I = \int\limits_{−1}^1 \frac{x^3 + |x| + 1}{x^2 + 2 |x| + 1} \;dx$

The interval of integration is symmetric about 0, i.e., $[-1, 1]$. We can use the properties of definite integrals for even and odd functions over a symmetric interval $\int\limits_{-a}^a f(x) dx = \int\limits_{-a}^a f_{odd}(x) dx + \int\limits_{-a}^a f_{even}(x) dx = 0 + 2 \int\limits_0^a f_{even}(x) dx$.

Let the integrand be $f(x) = \frac{x^3 + |x| + 1}{x^2 + 2 |x| + 1}$.

We can split the numerator and rewrite the integrand as:

$f(x) = \frac{x^3}{x^2 + 2 |x| + 1} + \frac{|x| + 1}{x^2 + 2 |x| + 1}$

Note that $x^2 = |x|^2$. So the denominator is $x^2 + 2|x| + 1 = |x|^2 + 2|x| + 1 = (|x|+1)^2$.

$f(x) = \frac{x^3}{(|x|+1)^2} + \frac{|x|+1}{(|x|+1)^2} = \frac{x^3}{(|x|+1)^2} + \frac{1}{|x|+1}$

Let's examine the two parts of the integrand for being even or odd.

Consider $g(x) = \frac{x^3}{(|x|+1)^2}$.

$g(-x) = \frac{(-x)^3}{(|-x|+1)^2} = \frac{-x^3}{(|x|+1)^2} = - \frac{x^3}{(|x|+1)^2} = -g(x)$.

So, $g(x)$ is an odd function.

Consider $h(x) = \frac{1}{|x|+1}$.

$h(-x) = \frac{1}{|-x|+1} = \frac{1}{|x|+1} = h(x)$.

So, $h(x)$ is an even function.

The integral can be written as:

$I = \int\limits_{−1}^1 \left(\frac{x^3}{(|x|+1)^2} + \frac{1}{|x|+1}\right) \;dx$

$I = \int\limits_{−1}^1 \frac{x^3}{(|x|+1)^2} \;dx + \int\limits_{−1}^1 \frac{1}{|x|+1} \;dx$

Since $\frac{x^3}{(|x|+1)^2}$ is odd and the integration interval is $[-1, 1]$, the first integral is 0:

$\int\limits_{−1}^1 \frac{x^3}{(|x|+1)^2} \;dx = 0$

Since $\frac{1}{|x|+1}$ is even and the integration interval is $[-1, 1]$, the second integral is $2 \int\limits_0^1 \frac{1}{|x|+1} \;dx$.

For $x \in [0, 1]$, $|x| = x$. So $\frac{1}{|x|+1} = \frac{1}{x+1}$.

$I = 0 + 2 \int\limits_0^1 \frac{1}{x+1} \;dx$

$I = 2 \int\limits_0^1 \frac{1}{x+1} \;dx$

Now, evaluate the definite integral $\int\limits_0^1 \frac{1}{x+1} \;dx$.

The antiderivative of $\frac{1}{x+1}$ is $\log|x+1|$.

$I = 2 [\log|x+1|]_0^1$

Evaluate the limits:

$I = 2 (\log|1+1| - \log|0+1|)$

$I = 2 (\log 2 - \log 1)$

Since $\log 1 = 0$:

$I = 2 (\log 2 - 0) = 2 \log 2$

Comparing this result with the given options, we find that it matches option (B).

The correct answer is (B).

Let the integral we need to evaluate be $I$.

$I = \int\limits_0^1 \frac{e^t}{(1 + t)^2} \;dt$

We are given $a = \int\limits_0^1 \frac{e^t}{1 + t} \;dt$.

We can evaluate $I$ using integration by parts for definite integrals: $\int_a^b u \; dv = [uv]_a^b - \int_a^b v \; du$.

Let's choose $u = e^t$ and $dv = \frac{1}{(1 + t)^2} \;dt$.

To find $du$, differentiate $u$ with respect to $t$:

$u = e^t \implies du = e^t \;dt$

To find $v$, integrate $dv$ with respect to $t$:

$dv = (1 + t)^{-2} \;dt$

$v = \int (1 + t)^{-2} \;dt = \frac{(1 + t)^{-1}}{-1} = -\frac{1}{1 + t}$

Now apply the integration by parts formula to $I$ with limits from 0 to 1:

$I = \left[ u v \right]_0^1 - \int\limits_0^1 v \; du$

$I = \left[ e^t \left(-\frac{1}{1 + t}\right) \right]_0^1 - \int\limits_0^1 \left(-\frac{1}{1 + t}\right) e^t \;dt$

$I = \left[ -\frac{e^t}{1 + t} \right]_0^1 + \int\limits_0^1 \frac{e^t}{1 + t} \;dt$

Evaluate the boundary term $\left[ -\frac{e^t}{1 + t} \right]_0^1$:

At the upper limit $t=1$: $-\frac{e^1}{1 + 1} = -\frac{e}{2}$

At the lower limit $t=0$: $-\frac{e^0}{1 + 0} = -\frac{1}{1} = -1$

Boundary term value $= \left(-\frac{e}{2}\right) - (-1) = 1 - \frac{e}{2}$.

The integral term in the expression for $I$ is $\int\limits_0^1 \frac{e^t}{1 + t} \;dt$, which is given as $a$.

So, substitute the values back into the expression for $I$:

$I = \left(1 - \frac{e}{2}\right) + a$

$I = a + 1 - \frac{e}{2}$

Comparing this result with the given options:

(A) $a - 1 + \frac{e}{2}$

(B) $a + 1 - \frac{e}{2}$

(C) $a - 1 - \frac{e}{2}$

(D) $a + 1 + \frac{e}{2}$

Our result matches option (B).

The correct answer is (B).

Let the given integral be $I$.

$I = \int\limits_{−2}^2 |x \cos πx|\; dx$

The interval of integration is symmetric about 0, $[-2, 2]$. Let the integrand be $f(x) = |x \cos πx|$.

We check if $f(x)$ is an even or odd function:

$f(-x) = |(-x) \cos(π(-x))| = |-x \cos(-πx)|$

Since $\cos(-\theta) = \cos \theta$, we have $\cos(-πx) = \cos(πx)$.

$f(-x) = |-x \cos πx| = |-(x \cos πx)| = |x \cos πx| = f(x)$.

Since $f(-x) = f(x)$, the integrand $f(x) = |x \cos πx|$ is an even function.

For an even function over a symmetric interval $[-a, a]$, we use the property $\int\limits_{-a}^a f(x) dx = 2 \int\limits_0^a f(x) dx$.

Here $a=2$, so $I = 2 \int\limits_0^2 |x \cos πx|\; dx$.

Now we need to evaluate $\int\limits_0^2 |x \cos πx|\; dx$. On the interval $[0, 2]$, $x \geq 0$, so $|x \cos πx| = x |\cos πx|$.

We need to consider the sign of $\cos πx$ on the interval $[0, 2]$. The values of $πx$ range from $0$ to $2π$.

$\cos \theta$ is non-negative for $\theta \in [0, \frac{π}{2}] \cup [\frac{3π}{2}, 2π]$.

So, $\cos πx \geq 0$ when $πx \in [0, \frac{π}{2}] \implies x \in [0, \frac{1}{2}]$.

And $\cos πx \geq 0$ when $πx \in [\frac{3π}{2}, 2π] \implies x \in [\frac{3}{2}, 2]$.

$\cos \theta$ is non-positive for $\theta \in [\frac{π}{2}, \frac{3π}{2}]$.

So, $\cos πx \leq 0$ when $πx \in [\frac{π}{2}, \frac{3π}{2}] \implies x \in [\frac{1}{2}, \frac{3}{2}]$.

We split the integral $\int\limits_0^2 |x \cos πx|\; dx$ based on the sign of $\cos πx$:

$\int\limits_0^2 |x \cos πx|\; dx = \int\limits_0^{1/2} x \cos πx \;dx + \int\limits_{1/2}^{3/2} x (-\cos πx) \;dx + \int\limits_{3/2}^2 x \cos πx \;dx$

$= \int\limits_0^{1/2} x \cos πx \;dx - \int\limits_{1/2}^{3/2} x \cos πx \;dx + \int\limits_{3/2}^2 x \cos πx \;dx$

We evaluate the indefinite integral $\int x \cos πx \;dx$ using integration by parts: $\int u \; dv = uv - \int v \; du$.

Let $u = x$ and $dv = \cos πx \;dx$.

Then $du = dx$ and $v = \int \cos πx \;dx = \frac{1}{π} \sin πx$.

$\int x \cos πx \;dx = x \left(\frac{1}{π} \sin πx\right) - \int \left(\frac{1}{π} \sin πx\right) dx$

$= \frac{x}{π} \sin πx - \frac{1}{π} \int \sin πx \;dx$

$\int \sin πx \;dx = -\frac{1}{π} \cos πx$.

So, $\int x \cos πx \;dx = \frac{x}{π} \sin πx - \frac{1}{π} \left(-\frac{1}{π} \cos πx\right) = \frac{x}{π} \sin πx + \frac{1}{π^2} \cos πx$.

Let $F(x) = \frac{x}{π} \sin πx + \frac{1}{π^2} \cos πx$.

Evaluate the definite integrals using the Fundamental Theorem of Calculus:

$\int\limits_0^{1/2} x \cos πx \;dx = [F(x)]_0^{1/2} = F(\frac{1}{2}) - F(0)$

$F(\frac{1}{2}) = \frac{1/2}{π} \sin(\frac{π}{2}) + \frac{1}{π^2} \cos(\frac{π}{2}) = \frac{1}{2π}(1) + \frac{1}{π^2}(0) = \frac{1}{2π}$

$F(0) = \frac{0}{π} \sin(0) + \frac{1}{π^2} \cos(0) = 0 + \frac{1}{π^2}(1) = \frac{1}{π^2}$

$\int\limits_0^{1/2} x \cos πx \;dx = \frac{1}{2π} - \frac{1}{π^2}$.

$\int\limits_{1/2}^{3/2} x \cos πx \;dx = [F(x)]_{1/2}^{3/2} = F(\frac{3}{2}) - F(\frac{1}{2})$

$F(\frac{3}{2}) = \frac{3/2}{π} \sin(\frac{3π}{2}) + \frac{1}{π^2} \cos(\frac{3π}{2}) = \frac{3}{2π}(-1) + \frac{1}{π^2}(0) = -\frac{3}{2π}$

$F(\frac{1}{2}) = \frac{1}{2π}$

$\int\limits_{1/2}^{3/2} x \cos πx \;dx = -\frac{3}{2π} - \frac{1}{2π} = -\frac{4}{2π} = -\frac{2}{π}$.

$\int\limits_{3/2}^{2} x \cos πx \;dx = [F(x)]_{3/2}^{2} = F(2) - F(\frac{3}{2})$

$F(2) = \frac{2}{π} \sin(2π) + \frac{1}{π^2} \cos(2π) = \frac{2}{π}(0) + \frac{1}{π^2}(1) = \frac{1}{π^2}$

$F(\frac{3}{2}) = -\frac{3}{2π}$

$\int\limits_{3/2}^{2} x \cos πx \;dx = \frac{1}{π^2} - (-\frac{3}{2π}) = \frac{1}{π^2} + \frac{3}{2π}$.

Now substitute these values back into the expression for $\int\limits_0^2 |x \cos πx|\; dx$:

$\int\limits_0^2 |x \cos πx|\; dx = \left(\frac{1}{2π} - \frac{1}{π^2}\right) - \left(-\frac{2}{π}\right) + \left(\frac{1}{π^2} + \frac{3}{2π}\right)$

$= \frac{1}{2π} - \frac{1}{π^2} + \frac{2}{π} + \frac{1}{π^2} + \frac{3}{2π}$

$= \left(\frac{1}{2π} + \frac{4}{2π} + \frac{3}{2π}\right) + \left(-\frac{1}{π^2} + \frac{1}{π^2}\right)$

$= \frac{1+4+3}{2π} + 0 = \frac{8}{2π} = \frac{4}{π}$

Finally, calculate the original integral $I$:

$I = 2 \int\limits_0^2 |x \cos πx|\; dx = 2 \times \frac{4}{π} = \frac{8}{π}$.

Comparing this result with the given options, we find that it matches option (A).

The correct answer is (A).

Let the given integral be $I$.

$I = \int \frac{\sin^6 x}{\cos^8 x} \;dx$

We can rewrite the integrand by separating the terms involving sine and cosine:

$\frac{\sin^6 x}{\cos^8 x} = \frac{\sin^6 x}{\cos^6 x \cdot \cos^2 x}$

Using the trigonometric identities $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$, we can write:

$\frac{\sin^6 x}{\cos^6 x \cdot \cos^2 x} = \left(\frac{\sin x}{\cos x}\right)^6 \cdot \frac{1}{\cos^2 x} = (\tan x)^6 \cdot \sec^2 x$

So the integral becomes:

$I = \int \tan^6 x \sec^2 x \;dx$

We can evaluate this integral using a substitution.

Let $u = \tan x$.

Then the differential $du$ is the derivative of $\tan x$ with respect to $x$, multiplied by $dx$:

$du = \frac{d}{dx}(\tan x) \;dx = \sec^2 x \;dx$

Substitute $u$ and $du$ into the integral:

$I = \int u^6 \;du$

Now, integrate the power of $u$ with respect to $u$:

$I = \frac{u^{6+1}}{6+1} + C = \frac{u^7}{7} + C$

Substitute back $u = \tan x$ to express the result in terms of $x$:

$I = \frac{(\tan x)^7}{7} + C = \frac{\tan^7 x}{7} + C$

Thus, $\int \frac{\sin^6 x}{\cos^8 x} \;dx = \frac{\tan^7 x}{7} + C$, where $C$ is the constant of integration.

The expression to fill the blank is $\frac{\tan^7 x}{7} + C$.

This question asks to complete a statement about the property of definite integrals over a symmetric interval $[-a, a]$.

The property states that if $f$ is an odd function, then its integral over the interval $[-a, a]$ is zero.

Recall that a function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain.

The property is:

$\int\limits_{-a}^a f(x) \;dx = 0$ if $f(x)$ is an odd function.

Also, for comparison, the property for an even function is:

$\int\limits_{-a}^a f(x) \;dx = 2 \int\limits_0^a f(x) \;dx$ if $f(x)$ is an even function (i.e., $f(-x) = f(x)$).

Since the given statement says the integral is 0, the function $f$ must be an odd function.

The blank should be filled with the word "odd".

This question asks to complete a statement about a property of definite integrals.

The property relates the integral over the interval $[0, 2a]$ to the integral over the interval $[0, a]$.

The property states that $\int\limits_0^{2a} f(x) \;dx = 2\int\limits_0^a f(x) \;dx$ if the function $f(x)$ satisfies the condition $f(2a - x) = f(x)$ for all $x$ in the interval $[0, 2a]$.

If, instead, $f(2a - x) = -f(x)$ for all $x$ in the interval $[0, 2a]$, then $\int\limits_0^{2a} f(x) \;dx = 0$.

Since the statement specifies that the integral is $2\int\limits_0^a f(x) \;dx$, the condition on $f(2a-x)$ must be $f(2a-x) = f(x)$.

The blank should be filled with $f(x)$.

Let the given integral be $I$.

We use the property of definite integrals: $\int\limits_0^a f(x) dx = \int\limits_0^a f(a-x) dx$.

Here, $a = \frac{π}{2}$. Applying the property to equation (1), we replace $x$ with $\frac{π}{2} - x$:

$I = \int\limits_0^{\frac{π}{2}} \frac{\sin^n (\frac{π}{2}-x)}{\sin^n (\frac{π}{2}-x) + \cos^n (\frac{π}{2}-x)} \;dx$

Using the identities $\sin(\frac{π}{2}-x) = \cos x$ and $\cos(\frac{π}{2}-x) = \sin x$, we get:

Add equation (1) and equation (2):

$I + I = \int\limits_0^{\frac{π}{2}} \frac{\sin^n x}{\sin^n x + \cos^n x} \;dx + \int\limits_0^{\frac{π}{2}} \frac{\cos^n x}{\cos^n x + \sin^n x} \;dx$

$2I = \int\limits_0^{\frac{π}{2}} \left(\frac{\sin^n x}{\sin^n x + \cos^n x} + \frac{\cos^n x}{\sin^n x + \cos^n x}\right) \;dx$

Since the denominators are the same, we combine the numerators:

$2I = \int\limits_0^{\frac{π}{2}} \frac{\sin^n x + \cos^n x}{\sin^n x + \cos^n x} \;dx$

The numerator and denominator are identical, simplifying the integrand to 1:

$2I = \int\limits_0^{\frac{π}{2}} 1 \;dx$

Now, evaluate the definite integral:

$2I = [x]_0^{\frac{π}{2}}$

$2I = \frac{π}{2} - 0$

$2I = \frac{π}{2}$

Solve for $I$:

$I = \frac{π}{4}$

Thus, the value of the integral is $\frac{π}{4}$.

The expression to fill the blank is $\frac{π}{4}$.

Solution:

We are asked to verify the given integration formula:

$\int\limits \frac{2x − 1}{2x + 3} \;dx = x - \log |(2x+3)^2| + C$

Let's evaluate the left-hand side (LHS) integral:

$\int\limits \frac{2x − 1}{2x + 3} \;dx$

We can rewrite the integrand by manipulating the numerator to include the denominator:

$\frac{2x - 1}{2x + 3} = \frac{(2x + 3) - 4}{2x + 3} = \frac{2x + 3}{2x + 3} - \frac{4}{2x + 3} = 1 - \frac{4}{2x + 3}$

Now substitute this back into the integral:

$\int\limits \left(1 - \frac{4}{2x + 3}\right) \;dx$

We can split this into two separate integrals:

$\int\limits 1 \;dx - \int\limits \frac{4}{2x + 3} \;dx$

The first integral is straightforward:

$\int\limits 1 \;dx = x$

For the second integral, $\int\limits \frac{4}{2x + 3} \;dx$, we can use a substitution. Let $u = 2x + 3$. Then, the differential $du = 2 \;dx$, which means $dx = \frac{1}{2} du$.

Substitute these into the second integral:

$\int\limits \frac{4}{u} \cdot \frac{1}{2} \;du = \int\limits \frac{2}{u} \;du$

Now integrate with respect to $u$:

$2 \int\limits \frac{1}{u} \;du = 2 \log |u| + C'$

Substitute back $u = 2x + 3$:

$2 \log |2x + 3| + C'$

Combine the results from both integrals:

$x - (2 \log |2x + 3|) + C$

Using the logarithm property $n \log a = \log a^n$, we can rewrite $2 \log |2x + 3|$ as $\log |(2x + 3)^2|$.

$x - \log |(2x + 3)^2| + C$

Comparing this result with the given right-hand side (RHS) of the equation, we see that they match.

LHS = $x - \log |(2x + 3)^2| + C$

RHS = $x - \log |(2x + 3)^2| + C$

Since LHS = RHS, the given integration formula is verified.

Solution:

We are asked to verify the given integration formula:

$\int\limits \frac{2x+3}{x^2 +3x} \;dx = \log |x^2 + 3x| + C$

Let's evaluate the left-hand side (LHS) integral:

$\int\limits \frac{2x+3}{x^2 +3x} \;dx$

We observe that the numerator, $2x+3$, is the derivative of the denominator, $x^2+3x$.

Let $f(x) = x^2 + 3x$.

Then, $f'(x) = \frac{d}{dx}(x^2 + 3x) = 2x + 3$.

The integral is in the form $\int\limits \frac{f'(x)}{f(x)} \;dx$.

The standard formula for such integrals is $\int\limits \frac{f'(x)}{f(x)} \;dx = \log |f(x)| + C$.

Applying this formula to our integral:

$\int\limits \frac{2x+3}{x^2 +3x} \;dx = \log |x^2 + 3x| + C$

This result matches the given right-hand side (RHS) of the equation.

LHS = $\log |x^2 + 3x| + C$

RHS = $\log |x^2 + 3x| + C$

Since LHS = RHS, the given integration formula is verified.

Solution:

To evaluate the integral $\int\limits \frac{(x^2 + 2)}{x + 1} \;dx$, since the degree of the numerator ($x^2$) is greater than the degree of the denominator ($x$), we first perform polynomial long division.

Divide $x^2 + 2$ by $x + 1$:

$\begin{array}{r} x-1 \\ x+1{\overline{\smash{\big)}\,x^2+0x+2\phantom{)}}} \\ \underline{-~\phantom{(}(x^2+x)\phantom{-b)}} \\ 0-x+2\phantom{)} \\ \underline{-~\phantom{()}(-x-1)} \\ 0+3\phantom{)} \end{array}$

So, we can write the integrand as:

$\frac{x^2 + 2}{x + 1} = x - 1 + \frac{3}{x + 1}$

Now, we can integrate the simplified expression:

$\int\limits \frac{x^2 + 2}{x + 1} \;dx = \int\limits \left(x - 1 + \frac{3}{x + 1}\right) \;dx$

We can integrate each term separately:

$\int\limits x \;dx = \frac{x^2}{2} + C_1$

$\int\limits -1 \;dx = -x + C_2$

$\int\limits \frac{3}{x + 1} \;dx = 3 \int\limits \frac{1}{x + 1} \;dx$

For the last integral, let $u = x + 1$. Then $du = dx$. The integral becomes:

$3 \int\limits \frac{1}{u} \;du = 3 \log |u| + C_3 = 3 \log |x + 1| + C_3$

Combining the results of the individual integrals, and letting $C = C_1 + C_2 + C_3$ be the constant of integration:

$\int\limits \frac{x^2 + 2}{x + 1} \;dx = \frac{x^2}{2} - x + 3 \log |x + 1| + C$

Solution:

To evaluate the integral $\int\limits \frac{e^{6 \log x} − e^{5 \log x}}{e^{4 \log x}− e^{3 \log x}} \;dx$, we first simplify the integrand using the properties of logarithms and exponents. Recall that $a \log x = \log x^a$ and $e^{\log y} = y$.

Numerator:

$e^{6 \log x} = e^{\log x^6} = x^6$

$e^{5 \log x} = e^{\log x^5} = x^5$

So the numerator is $x^6 - x^5 = x^5(x-1)$.

Denominator:

$e^{4 \log x} = e^{\log x^4} = x^4$

$e^{3 \log x} = e^{\log x^3} = x^3$

So the denominator is $x^4 - x^3 = x^3(x-1)$.

The integrand becomes:

$\frac{x^6 - x^5}{x^4 - x^3} = \frac{x^5(x-1)}{x^3(x-1)}$

Assuming $x \neq 1$ (which is required for the denominator to be non-zero after factoring), we can cancel the $(x-1)$ term:

$\frac{x^5}{x^3} = x^{5-3} = x^2$

Thus, the integral simplifies to:

$\int\limits x^2 \;dx$

Now, we integrate $x^2$ using the power rule for integration $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$:

$\int\limits x^2 \;dx = \frac{x^{2+1}}{2+1} + C = \frac{x^3}{3} + C$

The value of the integral is $\frac{x^3}{3} + C$.

Solution:

To evaluate the integral $\int\limits \frac{(1 + \cos x)}{x + \sin x} \;dx$, we examine the relationship between the numerator and the denominator.

Let $u = x + \sin x$ be the denominator.

Now, let's find the differential $du$ by differentiating $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x + \sin x)$

$\frac{du}{dx} = 1 + \cos x$

So, $du = (1 + \cos x) \;dx$.

We can see that the numerator $(1 + \cos x) \;dx$ is exactly $du$, and the denominator $x + \sin x$ is $u$.

Substitute $u$ and $du$ into the integral:

$\int\limits \frac{du}{u}$

This is a standard integral:

$\int\limits \frac{1}{u} \;du = \log |u| + C$

Now, substitute back $u = x + \sin x$:

$\log |x + \sin x| + C$

The value of the integral is $\log |x + \sin x| + C$.

Solution 1:

To evaluate the integral $\int\limits \frac{dx}{1 + \cos x}$, we can multiply the numerator and the denominator by the conjugate of the denominator, which is $1 - \cos x$.

$\int\limits \frac{1}{1 + \cos x} \cdot \frac{1 - \cos x}{1 - \cos x} \;dx$

The denominator becomes $(1 + \cos x)(1 - \cos x) = 1^2 - \cos^2 x$.

Using the identity $\sin^2 x + \cos^2 x = 1$, we have $1 - \cos^2 x = \sin^2 x$.

So the integral becomes:

$\int\limits \frac{1 - \cos x}{\sin^2 x} \;dx$

Now, we can split the fraction into two terms:

$\int\limits \left(\frac{1}{\sin^2 x} - \frac{\cos x}{\sin^2 x}\right) \;dx$

Using trigonometric identities, $\frac{1}{\sin^2 x} = \text{cosec}^2 x$ and $\frac{\cos x}{\sin^2 x} = \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = \cot x \text{cosec} x$.

The integral is now:

$\int\limits (\text{cosec}^2 x - \cot x \text{cosec} x) \;dx$

We can integrate each term separately:

$\int\limits \text{cosec}^2 x \;dx = -\cot x + C_1$

$\int\limits \cot x \text{cosec} x \;dx = -\text{cosec} x + C_2$

Combining these results:

$\int\limits (\text{cosec}^2 x - \cot x \text{cosec} x) \;dx = (-\cot x) - (-\text{cosec} x) + C$

$= -\cot x + \text{cosec} x + C$

$= \text{cosec} x - \cot x + C$

Solution 2 (Alternate):

We can use the half-angle identity for the denominator: $1 + \cos x = 2 \cos^2 \frac{x}{2}$.

The integral becomes:

$\int\limits \frac{dx}{2 \cos^2 \frac{x}{2}}$

We can rewrite $\frac{1}{\cos^2 \frac{x}{2}}$ as $\sec^2 \frac{x}{2}$.

$\int\limits \frac{1}{2} \sec^2 \frac{x}{2} \;dx$

Let $u = \frac{x}{2}$. Then, the differential $du = \frac{1}{2} dx$, which means $dx = 2 du$.

Substitute $u$ and $dx$ into the integral:

$\int\limits \frac{1}{2} \sec^2 u \cdot (2 \;du)$

$\int\limits \sec^2 u \;du$

Now integrate with respect to $u$:

$\int\limits \sec^2 u \;du = \tan u + C$

Substitute back $u = \frac{x}{2}$:

$\tan \frac{x}{2} + C$

Both results $\text{cosec} x - \cot x + C$ and $\tan \frac{x}{2} + C$ are equivalent.

Solution:

To evaluate the integral $\int\limits \tan^2 x \; \sec^4 x \;dx$, we can rewrite the integrand using trigonometric identities.

We have $\sec^4 x = \sec^2 x \cdot \sec^2 x$.

Using the identity $\sec^2 x = 1 + \tan^2 x$, we can replace one of the $\sec^2 x$ terms:

$\int\limits \tan^2 x (1 + \tan^2 x) \sec^2 x \;dx$

Now, let's use substitution. Let $u = \tan x$.

The differential $du$ is the derivative of $u$ with respect to $x$ multiplied by $dx$:

$\frac{du}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x$

So, $du = \sec^2 x \;dx$.

Substitute $u = \tan x$ and $du = \sec^2 x \;dx$ into the integral:

$\int\limits u^2 (1 + u^2) \;du$

Expand the expression inside the integral:

$\int\limits (u^2 + u^4) \;du$

Now, integrate term by term using the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C$:

$\int\limits u^2 \;du = \frac{u^{2+1}}{2+1} + C_1 = \frac{u^3}{3} + C_1$

$\int\limits u^4 \;du = \frac{u^{4+1}}{4+1} + C_2 = \frac{u^5}{5} + C_2$

Combining the results and letting $C = C_1 + C_2$ be the constant of integration:

$\int\limits (u^2 + u^4) \;du = \frac{u^3}{3} + \frac{u^5}{5} + C$

Finally, substitute back $u = \tan x$:

$\frac{(\tan x)^3}{3} + \frac{(\tan x)^5}{5} + C$

The evaluated integral is $\frac{\tan^3 x}{3} + \frac{\tan^5 x}{5} + C$.

Solution:

To evaluate the integral $\int\limits \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \;dx$, we first simplify the expression under the square root in the denominator.

We use the trigonometric identity $1 = \sin^2 x + \cos^2 x$ and the double angle identity $\sin 2x = 2 \sin x \cos x$.

$1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x$

This is the expansion of a perfect square:

$1 + \sin 2x = (\sin x + \cos x)^2$

Now, the denominator becomes:

$\sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2}$

The square root of a square is the absolute value:

$\sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|$

So the integral is $\int\limits \frac{\sin x + \cos x}{|\sin x + \cos x|} \;dx$.

The integrand $\frac{\sin x + \cos x}{|\sin x + \cos x|}$ is equal to $1$ when $\sin x + \cos x > 0$ and $-1$ when $\sin x + \cos x < 0$. This means the integrand is $\text{sgn}(\sin x + \cos x)$.

The integral $\int \text{sgn}(\sin x + \cos x) \;dx$ is a piecewise defined function.

In intervals where $\sin x + \cos x > 0$ (i.e., $x \in (-\frac{\pi}{4} + 2k\pi, \frac{3\pi}{4} + 2k\pi)$ for integer $k$), the integral is $\int 1 \;dx = x + C_k$.

In intervals where $\sin x + \cos x < 0$ (i.e., $x \in (\frac{3\pi}{4} + 2k\pi, \frac{7\pi}{4} + 2k\pi)$ for integer $k$), the integral is $\int -1 \;dx = -x + D_k$.

Assuming a context where $\sin x + \cos x > 0$, or if the problem intends a simpler form ignoring the absolute value in the denominator (a common simplification in some contexts, though mathematically $|\sqrt{A^2}| = |A|$), we can write:

$\int\limits \frac{\sin x + \cos x}{\sin x + \cos x} \;dx = \int\limits 1 \;dx$

$\int\limits 1 \;dx = x + C$

Thus, the integral evaluates to $x + C$, keeping in mind the domain restrictions or piecewise nature implied by the absolute value.

Solution:

To evaluate the integral $\int\limits \sqrt{1 + \sin x} \;dx$, we can simplify the expression under the square root using trigonometric identities.

We know the identity $1 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$ and the double angle identity $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$.

Substitute these into the expression $1 + \sin x$:

$1 + \sin x = (\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}) + (2 \sin \frac{x}{2} \cos \frac{x}{2})$

This is the expansion of a perfect square $(\sin \frac{x}{2} + \cos \frac{x}{2})^2$:

$1 + \sin x = (\sin \frac{x}{2} + \cos \frac{x}{2})^2$

Now, the integral becomes:

$\int\limits \sqrt{(\sin \frac{x}{2} + \cos \frac{x}{2})^2} \;dx$

The square root of a square is the absolute value: $\sqrt{a^2} = |a|$.

So, $\sqrt{(\sin \frac{x}{2} + \cos \frac{x}{2})^2} = |\sin \frac{x}{2} + \cos \frac{x}{2}|$.

The integral we need to evaluate is $\int\limits |\sin \frac{x}{2} + \cos \frac{x}{2}| \;dx$.

The absolute value function requires us to consider the sign of the expression inside it. The integral will be piecewise depending on whether $\sin \frac{x}{2} + \cos \frac{x}{2}$ is positive or negative.

Let's find the integral of $(\sin \frac{x}{2} + \cos \frac{x}{2})$:

$\int\limits (\sin \frac{x}{2} + \cos \frac{x}{2}) \;dx = \int\limits \sin \frac{x}{2} \;dx + \int\limits \cos \frac{x}{2} \;dx$

Using the standard integrals $\int \sin(ax) dx = -\frac{1}{a} \cos(ax) + C$ and $\int \cos(ax) dx = \frac{1}{a} \sin(ax) + C$ with $a = \frac{1}{2}$:

$\int\limits \sin \frac{x}{2} \;dx = -\frac{1}{1/2} \cos \frac{x}{2} + C_1 = -2 \cos \frac{x}{2} + C_1$

$\int\limits \cos \frac{x}{2} \;dx = \frac{1}{1/2} \sin \frac{x}{2} + C_2 = 2 \sin \frac{x}{2} + C_2$

So, $\int\limits (\sin \frac{x}{2} + \cos \frac{x}{2}) \;dx = 2 \sin \frac{x}{2} - 2 \cos \frac{x}{2} + C$.

Now, we define the integral based on the sign of $\sin \frac{x}{2} + \cos \frac{x}{2}$.

Case 1: $\sin \frac{x}{2} + \cos \frac{x}{2} \geq 0$

In this case, $|\sin \frac{x}{2} + \cos \frac{x}{2}| = \sin \frac{x}{2} + \cos \frac{x}{2}$.

The integral is $\int\limits (\sin \frac{x}{2} + \cos \frac{x}{2}) \;dx = 2 \sin \frac{x}{2} - 2 \cos \frac{x}{2} + C_1$.

Case 2: $\sin \frac{x}{2} + \cos \frac{x}{2} < 0$

In this case, $|\sin \frac{x}{2} + \cos \frac{x}{2}| = -(\sin \frac{x}{2} + \cos \frac{x}{2})$.

The integral is $\int\limits -(\sin \frac{x}{2} + \cos \frac{x}{2}) \;dx = -(2 \sin \frac{x}{2} - 2 \cos \frac{x}{2}) + C_2 = 2 \cos \frac{x}{2} - 2 \sin \frac{x}{2} + C_2$.

The condition $\sin \frac{x}{2} + \cos \frac{x}{2} \geq 0$ is equivalent to $\sqrt{2} \sin(\frac{x}{2} + \frac{\pi}{4}) \geq 0$, which occurs when $\frac{x}{2} + \frac{\pi}{4}$ is in the intervals $[2n\pi, (2n+1)\pi]$ for integer $n$. This simplifies to $x \in [4n\pi - \frac{\pi}{2}, (4n+2)\pi - \frac{\pi}{2}]$.

The condition $\sin \frac{x}{2} + \cos \frac{x}{2} < 0$ occurs when $\frac{x}{2} + \frac{\pi}{4}$ is in the intervals $((2n+1)\pi, (2n+2)\pi)$ for integer $n$. This simplifies to $x \in ((4n+2)\pi - \frac{\pi}{2}, (4n+4)\pi - \frac{\pi}{2})$.

Thus, the integral is given by the piecewise function:

Alternatively, this can be written as:

$\int\limits \sqrt{1 + \sin x} \;dx = \begin{cases} 2(\sin\frac{x}{2} - \cos\frac{x}{2}) + C & , & \text{if } x \in [4n\pi - \frac{\pi}{2}, (4n+2)\pi - \frac{\pi}{2}] \\ 2(\cos\frac{x}{2} - \sin\frac{x}{2}) + C & , & \text{if } x \in ((4n+2)\pi - \frac{\pi}{2}, (4n+4)\pi - \frac{\pi}{2}) \end{cases}$ for integer $n$.

Solution:

To evaluate the integral $\int\limits \frac{x}{\sqrt{x} + 1} \;dx$, we use the substitution suggested in the hint.

Let $z = \sqrt{x}$.

Squaring both sides gives $z^2 = x$.

Now, we find the differential $dx$ in terms of $z$ and $dz$. Differentiate $x = z^2$ with respect to $z$:

$\frac{dx}{dz} = \frac{d}{dz}(z^2) = 2z$

So, $dx = 2z \;dz$.

Substitute $x = z^2$, $\sqrt{x} = z$, and $dx = 2z \;dz$ into the integral:

$\int\limits \frac{z^2}{z + 1} (2z) \;dz$

Simplify the integrand:

$\int\limits \frac{2z^3}{z + 1} \;dz$

Since the degree of the numerator ($z^3$) is greater than the degree of the denominator ($z$), we perform polynomial long division of $2z^3$ by $z + 1$.

$\begin{array}{r} 2z^2 - 2z + 2 \\ z+1{\overline{\smash{\big)}\,2z^3 + 0z^2 + 0z + 0\phantom{)}}} \\ \underline{-~\phantom{(}(2z^3 + 2z^2)\phantom{-b)}} \\ 0 - 2z^2 + 0z\phantom{)} \\ \underline{-~\phantom{()}(-2z^2 - 2z)} \\ 0 + 2z + 0\phantom{)} \\ \underline{-~\phantom{()}(2z + 2)} \\ 0 - 2\phantom{)} \end{array}$

So, $\frac{2z^3}{z + 1} = 2z^2 - 2z + 2 - \frac{2}{z + 1}$.

Now, integrate the resulting expression with respect to $z$:

$\int\limits \left(2z^2 - 2z + 2 - \frac{2}{z + 1}\right) \;dz$

Integrate each term separately:

$\int\limits 2z^2 \;dz = 2 \int\limits z^2 \;dz = 2 \left(\frac{z^{2+1}}{2+1}\right) + C_1 = \frac{2z^3}{3} + C_1$

$\int\limits -2z \;dz = -2 \int\limits z \;dz = -2 \left(\frac{z^{1+1}}{1+1}\right) + C_2 = -2 \left(\frac{z^2}{2}\right) + C_2 = -z^2 + C_2$

$\int\limits 2 \;dz = 2z + C_3$

$\int\limits -\frac{2}{z + 1} \;dz = -2 \int\limits \frac{1}{z + 1} \;dz = -2 \log |z + 1| + C_4$

Combine the results, letting $C = C_1 + C_2 + C_3 + C_4$:

$\int\limits \frac{2z^3}{z + 1} \;dz = \frac{2z^3}{3} - z^2 + 2z - 2 \log |z + 1| + C$

Finally, substitute back $z = \sqrt{x}$:

$\frac{2(\sqrt{x})^3}{3} - (\sqrt{x})^2 + 2\sqrt{x} - 2 \log |\sqrt{x} + 1| + C$

Simplify the powers of $\sqrt{x}$ (recall $\sqrt{x} = x^{1/2}$, so $(\sqrt{x})^3 = (x^{1/2})^3 = x^{3/2}$ and $(\sqrt{x})^2 = x$):

$\frac{2x^{3/2}}{3} - x + 2\sqrt{x} - 2 \log (\sqrt{x} + 1) + C$

Note that since $x \ge 0$ for $\sqrt{x}$ to be real, $\sqrt{x} + 1$ is always positive. Thus, the absolute value around $\sqrt{x} + 1$ can be removed.

Solution:

To evaluate the integral $\int\limits \sqrt{\frac{a + x}{a − x}} \;dx$, we can manipulate the integrand. Assuming $a > 0$ and $-a \le x < a$ for the expression to be real and defined, we can multiply the numerator and denominator under the square root by $(a+x)$.

$\sqrt{\frac{a + x}{a − x}} = \sqrt{\frac{a + x}{a − x} \cdot \frac{a + x}{a + x}} = \sqrt{\frac{(a + x)^2}{(a − x)(a + x)}} = \sqrt{\frac{(a + x)^2}{a^2 - x^2}}$

Using the property $\sqrt{y^2} = |y|$, we have $\sqrt{(a+x)^2} = |a+x|$. For the domain $-a \le x < a$ (with $a>0$), $a+x \ge 0$, so $|a+x| = a+x$.

The integrand becomes:

$\frac{|a + x|}{\sqrt{a^2 - x^2}} = \frac{a + x}{\sqrt{a^2 - x^2}}$

The integral is now:

$\int\limits \frac{a + x}{\sqrt{a^2 - x^2}} \;dx$

We can split this integral into two parts:

$\int\limits \left(\frac{a}{\sqrt{a^2 - x^2}} + \frac{x}{\sqrt{a^2 - x^2}}\right) \;dx = \int\limits \frac{a}{\sqrt{a^2 - x^2}} \;dx + \int\limits \frac{x}{\sqrt{a^2 - x^2}} \;dx$

Evaluate the first integral: $\int\limits \frac{a}{\sqrt{a^2 - x^2}} \;dx$

$a \int\limits \frac{1}{\sqrt{a^2 - x^2}} \;dx$

This is a standard integral form $\int \frac{1}{\sqrt{a^2 - u^2}} du = \arcsin(u/a) + C$.

$a \int\limits \frac{1}{\sqrt{a^2 - x^2}} \;dx = a \arcsin\left(\frac{x}{a}\right) + C_1$

Evaluate the second integral: $\int\limits \frac{x}{\sqrt{a^2 - x^2}} \;dx$

Let $u = a^2 - x^2$. Differentiating with respect to $x$ gives $\frac{du}{dx} = -2x$. So, $x \;dx = -\frac{1}{2} du$.

Substitute $u$ and $x \;dx$ into the integral:

$\int\limits \frac{1}{\sqrt{u}} \left(-\frac{1}{2} \;du\right) = -\frac{1}{2} \int\limits u^{-1/2} \;du$

Using the power rule for integration $\int u^n \;du = \frac{u^{n+1}}{n+1} + C$:

$-\frac{1}{2} \cdot \frac{u^{-1/2 + 1}}{-1/2 + 1} + C_2 = -\frac{1}{2} \cdot \frac{u^{1/2}}{1/2} + C_2 = -\sqrt{u} + C_2$

Substitute back $u = a^2 - x^2$:

$-\sqrt{a^2 - x^2} + C_2$

Combine the results of the two integrals, letting $C = C_1 + C_2$:

$\int\limits \sqrt{\frac{a + x}{a − x}} \;dx = \left(a \arcsin\left(\frac{x}{a}\right) + C_1\right) + \left(-\sqrt{a^2 - x^2} + C_2\right)$

$= a \arcsin\left(\frac{x}{a}\right) - \sqrt{a^2 - x^2} + C$

This result is valid for $-a < x < a$ (where $\sqrt{a^2 - x^2} \neq 0$).

Solution:

To evaluate the integral $\int\limits \frac{x^{\frac{1}{2}}}{1 + x^{\frac{3}{4}}} \;dx$, we use the substitution suggested by the hint.

Let $x = z^4$.

From this substitution, we can express the terms involving $x$ in terms of $z$:

$x^{\frac{1}{2}} = (z^4)^{\frac{1}{2}} = z^{4 \times \frac{1}{2}} = z^2$

$x^{\frac{3}{4}} = (z^4)^{\frac{3}{4}} = z^{4 \times \frac{3}{4}} = z^3$

Next, we find the differential $dx$ in terms of $z$ and $dz$. Differentiate $x = z^4$ with respect to $z$:

$\frac{dx}{dz} = \frac{d}{dz}(z^4) = 4z^3$

So, $dx = 4z^3 \;dz$.

Now, substitute $x^{\frac{1}{2}} = z^2$, $x^{\frac{3}{4}} = z^3$, and $dx = 4z^3 \;dz$ into the integral:

$\int\limits \frac{z^2}{1 + z^3} (4z^3) \;dz$

Simplify the integrand:

$\int\limits \frac{4z^5}{1 + z^3} \;dz$

Since the degree of the numerator ($z^5$) is greater than the degree of the denominator ($z^3$), we perform polynomial long division of $4z^5$ by $z^3 + 1$.

$\begin{array}{r} 4z^2 \phantom{-2z+2)} \\ z^3+1{\overline{\smash{\big)}\,4z^5+0z^4+0z^3+0z^2+0z+0\phantom{)}}} \\ \underline{-~\phantom{(}(4z^5+4z^2)\phantom{-b)}}\\ -4z^2\phantom{2)} \end{array}$

So, $4z^5 = 4z^2(z^3 + 1) - 4z^2$. Dividing by $(z^3 + 1)$, we get:

$\frac{4z^5}{z^3 + 1} = \frac{4z^2(z^3 + 1) - 4z^2}{z^3 + 1} = 4z^2 - \frac{4z^2}{z^3 + 1}$

Now, rewrite the integral using this result:

$\int\limits \left(4z^2 - \frac{4z^2}{z^3 + 1}\right) \;dz = \int\limits 4z^2 \;dz - \int\limits \frac{4z^2}{z^3 + 1} \;dz$

Evaluate the first integral:

$\int\limits 4z^2 \;dz = 4 \int\limits z^2 \;dz = 4 \cdot \frac{z^{2+1}}{2+1} + C_1 = \frac{4z^3}{3} + C_1$

Evaluate the second integral, $\int\limits \frac{4z^2}{z^3 + 1} \;dz$. Let $u = z^3 + 1$. Then, $du = 3z^2 \;dz$, which means $z^2 \;dz = \frac{1}{3} du$.

Substitute $u$ and $z^2 \;dz$ into the second integral:

$\int\limits 4 \cdot \frac{1}{u} \cdot \frac{1}{3} \;du = \frac{4}{3} \int\limits \frac{1}{u} \;du$

This is a standard integral:

$\frac{4}{3} \int\limits \frac{1}{u} \;du = \frac{4}{3} \log |u| + C_2$

Substitute back $u = z^3 + 1$:

$\frac{4}{3} \log |z^3 + 1| + C_2$

Combine the results of the two integrals, letting $C = C_1 - C_2$:

$\int\limits \frac{4z^5}{z^3 + 1} \;dz = \frac{4z^3}{3} - \left(\frac{4}{3} \log |z^3 + 1|\right) + C$

$= \frac{4}{3}z^3 - \frac{4}{3} \log |z^3 + 1| + C$

Finally, substitute back $z = x^{1/4}$ (since $x = z^4$). Thus $z^3 = (x^{1/4})^3 = x^{3/4}$.

$\frac{4}{3}x^{3/4} - \frac{4}{3} \log |x^{3/4} + 1| + C$

Since $x = z^4$ and $x^{\frac{1}{2}}$ is in the integrand, we require $x \ge 0$. If $x \ge 0$, then $x^{3/4} \ge 0$, so $x^{3/4} + 1 > 0$. Thus, the absolute value is not strictly necessary.

The evaluated integral is $\frac{4}{3}x^{3/4} - \frac{4}{3} \log (x^{3/4} + 1) + C$.

Solution:

To evaluate the integral $\int\limits \frac{\sqrt{1 + x^2}}{x^4} \;dx$, we use a trigonometric substitution suitable for expressions involving $\sqrt{a^2 + x^2}$. Let $a=1$.

Let $x = \tan \theta$.

Then, the differential $dx$ is the derivative of $\tan \theta$ with respect to $\theta$ multiplied by $d\theta$:

$dx = \sec^2 \theta \;d\theta$

Now, express the terms in the integrand in terms of $\theta$:

$\sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta}$

Using the identity $1 + \tan^2 \theta = \sec^2 \theta$:

$\sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta} = |\sec \theta|$

Assuming the principal value for the substitution, $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, where $\sec \theta > 0$, we have $\sqrt{1 + x^2} = \sec \theta$.

The term $x^4$ becomes:

$x^4 = (\tan \theta)^4 = \tan^4 \theta$

Substitute these into the integral:

$\int\limits \frac{\sec \theta}{\tan^4 \theta} (\sec^2 \theta \;d\theta)$

Simplify the integrand:

$\int\limits \frac{\sec^3 \theta}{\tan^4 \theta} \;d\theta$

Rewrite the integrand in terms of $\sin \theta$ and $\cos \theta$:

$\sec \theta = \frac{1}{\cos \theta}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

$\frac{\sec^3 \theta}{\tan^4 \theta} = \frac{(1/\cos \theta)^3}{(\sin \theta / \cos \theta)^4} = \frac{1/\cos^3 \theta}{\sin^4 \theta / \cos^4 \theta} = \frac{1}{\cos^3 \theta} \cdot \frac{\cos^4 \theta}{\sin^4 \theta} = \frac{\cos \theta}{\sin^4 \theta}$

The integral becomes:

$\int\limits \frac{\cos \theta}{\sin^4 \theta} \;d\theta$

Use substitution. Let $u = \sin \theta$.

Then, the differential $du = \cos \theta \;d\theta$.

Substitute $u = \sin \theta$ and $du = \cos \theta \;d\theta$ into the integral:

$\int\limits \frac{1}{u^4} \;du = \int\limits u^{-4} \;du$

Integrate with respect to $u$ using the power rule $\int u^n \;du = \frac{u^{n+1}}{n+1} + C$:

$\int\limits u^{-4} \;du = \frac{u^{-4 + 1}}{-4 + 1} + C = \frac{u^{-3}}{-3} + C = -\frac{1}{3u^3} + C$

Substitute back $u = \sin \theta$:

$-\frac{1}{3\sin^3 \theta} + C$

Now, substitute back $\theta$ in terms of $x$. Since $x = \tan \theta$, we can construct a right triangle with opposite side $x$ and adjacent side 1. The hypotenuse is $\sqrt{x^2 + 1}$.

So, $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2 + 1}}$.

Substitute this expression for $\sin \theta$ back into the result:

$-\frac{1}{3\left(\frac{x}{\sqrt{x^2 + 1}}\right)^3} + C$

$= -\frac{1}{3\left(\frac{x^3}{(x^2 + 1)^{3/2}}\right)} + C$

$= -\frac{1}{3} \cdot \frac{(x^2 + 1)^{3/2}}{x^3} + C$

The evaluated integral is $-\frac{(x^2 + 1)^{3/2}}{3x^3} + C$.

Solution:

To evaluate the integral $\int\limits \frac{dx}{\sqrt{16 − 9x^2}}$, we recognize that the integrand is in the form of a standard integral.

The integral $\int\limits \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C$.

In our integral, the denominator is $\sqrt{16 - 9x^2}$. We can rewrite this to match the standard form $\sqrt{a^2 - u^2}$.

Identify $a^2$ and $u^2$ from the term $16 - 9x^2$:

$a^2 = 16 \implies a = \sqrt{16} = 4$ (assuming $a > 0$)

$u^2 = 9x^2 \implies u = \sqrt{9x^2} = |3x|$. For simplicity in integration, we usually take $u=3x$.

Now, we need to find the differential $du$. If $u = 3x$, then $du = \frac{d}{dx}(3x) \;dx = 3 \;dx$.

The numerator of the given integral is $dx$. To match the standard form which requires $du$, we need $3 \;dx$. We can achieve this by multiplying the numerator by 3 and dividing the entire integral by 3.

$\int\limits \frac{dx}{\sqrt{16 − 9x^2}} = \int\limits \frac{\frac{1}{3} (3 \;dx)}{\sqrt{4^2 − (3x)^2}}$

$= \frac{1}{3} \int\limits \frac{3 \;dx}{\sqrt{4^2 − (3x)^2}}$

Now, substitute $u = 3x$, $du = 3 \;dx$, and $a = 4$ into the integral:

$\frac{1}{3} \int\limits \frac{du}{\sqrt{a^2 − u^2}}$

Apply the standard integral formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin(\frac{u}{a}) + C$:

$\frac{1}{3} \left(\arcsin\left(\frac{u}{a}\right)\right) + C'$

Substitute back $u = 3x$ and $a = 4$:

$\frac{1}{3} \arcsin\left(\frac{3x}{4}\right) + C$

Where $C$ is the constant of integration.

The evaluated integral is $\frac{1}{3} \arcsin\left(\frac{3x}{4}\right) + C$.

Solution:

To evaluate the integral $\int\limits \frac{dx}{\sqrt{16 − 9x^2}}$, we recognize that the integrand is in the form of a standard integral.

The integral $\int\limits \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$.

In our integral, the denominator is $\sqrt{16 - 9x^2}$. We can rewrite this to match the standard form $\sqrt{a^2 - u^2}$.

Identify $a^2$ and $u^2$ from the term $16 - 9x^2$:

$a^2 = 16 \implies a = \sqrt{16} = 4$ (assuming $a > 0$)

$u^2 = 9x^2 \implies u = \sqrt{9x^2} = |3x|$. For simplicity in integration, we usually take $u=3x$.

Now, we need to find the differential $du$. If $u = 3x$, then $du = \frac{d}{dx}(3x) \;dx = 3 \;dx$.

The numerator of the given integral is $dx$. To match the standard form which requires $du$, we need $3 \;dx$. We can achieve this by multiplying the numerator by 3 and dividing the entire integral by 3.

$\int\limits \frac{dx}{\sqrt{16 − 9x^2}} = \int\limits \frac{\frac{1}{3} (3 \;dx)}{\sqrt{4^2 − (3x)^2}}$

$= \frac{1}{3} \int\limits \frac{3 \;dx}{\sqrt{4^2 − (3x)^2}}$

Now, substitute $u = 3x$, $du = 3 \;dx$, and $a = 4$ into the integral:

$\frac{1}{3} \int\limits \frac{du}{\sqrt{a^2 − u^2}}$

Apply the standard integral formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + C$:

$\frac{1}{3} \left(\sin^{-1}\left(\frac{u}{a}\right)\right) + C'$

Substitute back $u = 3x$ and $a = 4$:

$\frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C$

Where $C$ is the constant of integration.

The evaluated integral is $\frac{1}{3} \sin^{-1}\left(\frac{3x}{4}\right) + C$.

Solution:

To evaluate the integral $\int\limits \frac{3x + 1}{\sqrt{x^2 + 9}} \;dx$, we can split the integrand into two parts by separating the terms in the numerator:

$\frac{3x + 1}{\sqrt{x^2 + 9}} = \frac{3x}{\sqrt{x^2 + 9}} + \frac{1}{\sqrt{x^2 + 9}}$

So, the integral becomes the sum of two integrals:

$\int\limits \frac{3x}{\sqrt{x^2 + 9}} \;dx + \int\limits \frac{1}{\sqrt{x^2 + 9}} \;dx$

Let's evaluate the first integral: $\int\limits \frac{3x}{\sqrt{x^2 + 9}} \;dx$.

We can use a substitution method here. Let $u = x^2 + 9$.

Differentiate $u$ with respect to $x$ to find the differential $du$:

$du = \frac{d}{dx}(x^2 + 9) \;dx = 2x \;dx$

We have $x \;dx$ in the numerator of the integral, so we rearrange the differential expression to solve for $x \;dx$:

$x \;dx = \frac{1}{2} du$

Substitute $u = x^2 + 9$ and $x \;dx = \frac{1}{2} du$ into the first integral:

$\int\limits \frac{3}{\sqrt{u}} \cdot \frac{1}{2} \;du = \frac{3}{2} \int\limits u^{-1/2} \;du$

Now, integrate with respect to $u$ using the power rule for integration, $\int u^n \;du = \frac{u^{n+1}}{n+1} + C$:

$\frac{3}{2} \cdot \frac{u^{-1/2 + 1}}{-1/2 + 1} + C_1 = \frac{3}{2} \cdot \frac{u^{1/2}}{1/2} + C_1 = \frac{3}{2} \cdot 2 \sqrt{u} + C_1 = 3 \sqrt{u} + C_1$

Substitute back $u = x^2 + 9$ to express the result in terms of $x$:

$3\sqrt{x^2 + 9} + C_1$

Now, let's evaluate the second integral: $\int\limits \frac{1}{\sqrt{x^2 + 9}} \;dx$.

This integral is in a standard form. The standard integral formula for $\int \frac{dx}{\sqrt{x^2 + a^2}}$ is $\log |x + \sqrt{x^2 + a^2}| + C$.

Comparing our integral $\int\limits \frac{dx}{\sqrt{x^2 + 9}}$ with the standard form, we identify $a^2 = 9$, which means $a = \sqrt{9} = 3$ (assuming $a > 0$).

Apply the standard integral formula directly:

$\int\limits \frac{dx}{\sqrt{x^2 + 3^2}} = \log |x + \sqrt{x^2 + 9}| + C_2$

Finally, combine the results from both integrals. The total integral is the sum of the results of the two parts. Let $C = C_1 + C_2$ be the overall constant of integration.

$\int\limits \frac{3x + 1}{\sqrt{x^2 + 9}} \;dx = \left(3\sqrt{x^2 + 9} + C_1\right) + \left(\log |x + \sqrt{x^2 + 9}| + C_2\right)$

$\int\limits \frac{3x + 1}{\sqrt{x^2 + 9}} \;dx = 3\sqrt{x^2 + 9} + \log |x + \sqrt{x^2 + 9}| + C$

The evaluated integral is $3\sqrt{x^2 + 9} + \log |x + \sqrt{x^2 + 9}| + C$.

Solution:

To evaluate the integral $\int\limits \sqrt{5−2x+x^2} \;dx$, we first complete the square for the quadratic expression under the square root.

The quadratic expression is $x^2 - 2x + 5$.

To complete the square for $x^2 - 2x$, we take half of the coefficient of $x$ (which is -2), square it ($(-1)^2 = 1$), and add and subtract it:

$x^2 - 2x + 5 = (x^2 - 2x + 1) + 5 - 1$

Group the terms to form a perfect square trinomial:

$(x - 1)^2 + 4$

We can write $4$ as $2^2$. So the expression is $(x - 1)^2 + 2^2$.

The integral becomes:

$\int\limits \sqrt{(x - 1)^2 + 2^2} \;dx$

This integral is in the standard form $\int\limits \sqrt{u^2 + a^2} \;du$, where $u = x - 1$ and $a = 2$.

Let $u = x - 1$. Then, the differential $du = \frac{d}{dx}(x - 1) \;dx = 1 \;dx = dx$. The differential $du$ matches $dx$.

Now, apply the standard integral formula for $\int\limits \sqrt{u^2 + a^2} \;du$, which is:

$\int\limits \sqrt{u^2 + a^2} \;du = \frac{u}{2}\sqrt{u^2+a^2} + \frac{a^2}{2}\log|u+\sqrt{u^2+a^2}| + C$

Substitute $u = x - 1$ and $a = 2$ into the formula:

$\frac{x - 1}{2}\sqrt{(x - 1)^2 + 2^2} + \frac{2^2}{2}\log|(x - 1)+\sqrt{(x - 1)^2 + 2^2}| + C$

Simplify the expression under the square root and the coefficient of the logarithmic term:

$(x - 1)^2 + 2^2 = x^2 - 2x + 1 + 4 = x^2 - 2x + 5$

$\frac{2^2}{2} = \frac{4}{2} = 2$

So the evaluated integral is:

$\frac{x - 1}{2}\sqrt{x^2 - 2x + 5} + 2\log|(x - 1)+\sqrt{x^2 - 2x + 5}| + C$

Where $C$ is the constant of integration.

Solution:

To evaluate the integral $\int\limits \frac{x}{x^4 −1} \;dx$, we can use a substitution.

Let $u = x^2$.

Now, find the differential $du$ by differentiating $u$ with respect to $x$:

$du = \frac{d}{dx}(x^2) \;dx = 2x \;dx$

We have $x \;dx$ in the numerator of the integral. From the differential, we have $x \;dx = \frac{1}{2} du$.

Also, the denominator $x^4 - 1$ can be written in terms of $u$ as $(x^2)^2 - 1 = u^2 - 1$.

Substitute $u = x^2$ and $x \;dx = \frac{1}{2} du$ into the integral:

$\int\limits \frac{\frac{1}{2} \;du}{u^2 - 1}$

Take the constant $\frac{1}{2}$ outside the integral:

$\frac{1}{2} \int\limits \frac{du}{u^2 - 1}$

This integral is in the standard form $\int\limits \frac{du}{u^2 - a^2}$.

Comparing $\int\limits \frac{du}{u^2 - 1}$ with the standard form, we have $a^2 = 1$, which means $a = 1$ (assuming $a > 0$).

The standard integral formula is:

$\int\limits \frac{du}{u^2 - a^2} = \frac{1}{2a} \log \left|\frac{u-a}{u+a}\right| + C'$

Apply the formula with $a=1$ and $u$:

$\frac{1}{2} \left( \frac{1}{2(1)} \log \left|\frac{u-1}{u+1}\right| \right) + C$

$\frac{1}{2} \left( \frac{1}{2} \log \left|\frac{u-1}{u+1}\right| \right) + C$

$\frac{1}{4} \log \left|\frac{u-1}{u+1}\right| + C$

Finally, substitute back $u = x^2$ to express the result in terms of $x$:

$\frac{1}{4} \log \left|\frac{x^2-1}{x^2+1}\right| + C$

Where $C$ is the constant of integration.

The evaluated integral is $\frac{1}{4} \log \left|\frac{x^2-1}{x^2+1}\right| + C$.

Solution:

To evaluate the integral $\int\limits \frac{x^2}{1 − x^4} \;dx$, we can first analyze the structure of the integrand. The integrand is a rational function of $x$. We can observe that the numerator contains $x^2$ and the denominator $1 - x^4$ is a function of $x^2$, specifically $1 - (x^2)^2$.

The hint suggests using the substitution $x^2 = t$. While this substitution doesn't directly simplify the entire integral using $dt$, it guides us to consider the integrand as a rational function of $x^2$.

Let $y = x^2$. Then the integrand can be written as:

$\frac{y}{1 - y^2}$

We can decompose this rational expression into partial fractions. The denominator $1 - y^2$ factors as $(1-y)(1+y)$.

So, we set up the partial fraction decomposition:

$\frac{y}{1 - y^2} = \frac{y}{(1 - y)(1 + y)} = \frac{A}{1 - y} + \frac{B}{1 + y}$

To find the constants $A$ and $B$, we multiply both sides by $(1-y)(1+y)$:

$y = A(1 + y) + B(1 - y)$

Now, we can substitute values of $y$ to solve for $A$ and $B$.

Set $y = 1$: $1 = A(1 + 1) + B(1 - 1) \implies 1 = 2A + 0 \implies A = \frac{1}{2}$.

Set $y = -1$: $-1 = A(1 - 1) + B(1 - (-1)) \implies -1 = 0 + B(2) \implies B = -\frac{1}{2}$.

So the partial fraction decomposition is:

$\frac{y}{1 - y^2} = \frac{1/2}{1 - y} - \frac{1/2}{1 + y}$

Now, substitute back $y = x^2$:

$\frac{x^2}{1 - x^4} = \frac{1/2}{1 - x^2} - \frac{1/2}{1 + x^2} = \frac{1}{2} \left(\frac{1}{1 - x^2} - \frac{1}{1 + x^2}\right)$

Now, we can integrate the decomposed expression:

$\int\limits \frac{x^2}{1 - x^4} \;dx = \int\limits \frac{1}{2} \left(\frac{1}{1 - x^2} - \frac{1}{1 + x^2}\right) \;dx$

Take the constant $\frac{1}{2}$ outside the integral and split the integral into two parts:

$\frac{1}{2} \left(\int\limits \frac{1}{1 - x^2} \;dx - \int\limits \frac{1}{1 + x^2} \;dx\right)$

Evaluate the first integral $\int\limits \frac{1}{1 - x^2} \;dx$. This is a standard integral of the form $\int\limits \frac{1}{a^2 - x^2} \;dx = \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right| + C$. With $a=1$, we have:

$\int\limits \frac{1}{1 - x^2} \;dx = \frac{1}{2(1)}\log\left|\frac{1+x}{1-x}\right| + C_1 = \frac{1}{2}\log\left|\frac{1+x}{1-x}\right| + C_1$

Evaluate the second integral $\int\limits \frac{1}{1 + x^2} \;dx$. This is a standard integral of the form $\int\limits \frac{1}{a^2 + x^2} \;dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$. With $a=1$, we have:

$\int\limits \frac{1}{1 + x^2} \;dx = \frac{1}{1}\tan^{-1}\left(\frac{x}{1}\right) + C_2 = \tan^{-1}(x) + C_2$

Substitute these results back into the expression for the original integral:

$\frac{1}{2} \left(\left(\frac{1}{2}\log\left|\frac{1+x}{1-x}\right| + C_1\right) - \left(\tan^{-1}(x) + C_2\right)\right)$

$\frac{1}{2} \left(\frac{1}{2}\log\left|\frac{1+x}{1-x}\right| - \tan^{-1}(x)\right) + \frac{1}{2}(C_1 - C_2)$

Let $C = \frac{1}{2}(C_1 - C_2)$ be the constant of integration.

The evaluated integral is:

$\frac{1}{4}\log\left|\frac{1+x}{1-x}\right| - \frac{1}{2}\tan^{-1}(x) + C$

Solution:

To evaluate the integral $\int\limits \sqrt{2ax − x^2} \;dx$, we first complete the square for the quadratic expression under the square root.

The expression is $2ax - x^2$. We can rewrite this as $-x^2 + 2ax$.

Factor out -1 from the terms involving $x$:

$-(x^2 - 2ax)$

To complete the square for $x^2 - 2ax$, take half of the coefficient of $x$ (which is $-2a$), square it ($(-a)^2 = a^2$), and add and subtract it inside the parenthesis:

$-(x^2 - 2ax + a^2 - a^2)$

Group the first three terms to form a perfect square trinomial $(x - a)^2$:

$-((x - a)^2 - a^2)$

Distribute the -1:

$-(x - a)^2 + a^2 = a^2 - (x - a)^2$

So, the expression under the square root is $a^2 - (x - a)^2$. The integral becomes:

$\int\limits \sqrt{a^2 - (x - a)^2} \;dx$

This integral is in the standard form $\int\limits \sqrt{a^2 - u^2} \;du$, where $u = x - a$ and the constant is $a$.

Let $u = x - a$.

Find the differential $du$ by differentiating $u$ with respect to $x$:

$du = \frac{d}{dx}(x - a) \;dx = 1 \;dx = dx$

The differential $du$ matches $dx$, so no adjustment is needed.

Now, apply the standard integral formula for $\int\limits \sqrt{a^2 - u^2} \;du$, which is:

$\int\limits \sqrt{a^2 - u^2} \;du = \frac{u}{2}\sqrt{a^2-u^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{u}{a}\right) + C'$

Substitute $u = x - a$ and $a$ back into the formula:

$\frac{x - a}{2}\sqrt{a^2 - (x - a)^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x - a}{a}\right) + C$

Simplify the expression under the square root. As we found during completing the square, $a^2 - (x - a)^2 = 2ax - x^2$.

So the evaluated integral is:

$\frac{x - a}{2}\sqrt{2ax - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x - a}{a}\right) + C$

Where $C$ is the constant of integration.

Solution:

To evaluate the integral $\int\limits \frac{\sin^{−1} x}{(1 − x^2)^{\frac{3}{2}}} \;dx$, we can rewrite the integrand.

Note that $(1 − x^2)^{\frac{3}{2}} = (1 − x^2) (1 − x^2)^{\frac{1}{2}} = (1 − x^2) \sqrt{1 − x^2}$.

The integrand can be written as:

$\frac{\sin^{-1} x}{(1 − x^2) \sqrt{1 − x^2}}$

We can rearrange this expression as:

$\sin^{-1} x \cdot \frac{1}{\sqrt{1 - x^2}} \cdot \frac{1}{1 - x^2}$

Now, we use the substitution method. Let $u = \sin^{-1} x$.

Differentiating $u$ with respect to $x$, we get the differential $du$:

$du = \frac{d}{dx}(\sin^{-1} x) \;dx = \frac{1}{\sqrt{1 - x^2}} \;dx$

From the substitution $u = \sin^{-1} x$, we also have $x = \sin u$.

Then, $1 - x^2 = 1 - \sin^2 u = \cos^2 u$.

Substitute $u = \sin^{-1} x$, $\frac{1}{\sqrt{1 - x^2}} \;dx = du$, and $1 - x^2 = \cos^2 u$ into the integral:

$\int\limits u \cdot \frac{1}{1 - x^2} \cdot \left(\frac{1}{\sqrt{1 - x^2}} \;dx\right) = \int\limits u \cdot \frac{1}{\cos^2 u} \;du$

Using the identity $\frac{1}{\cos^2 u} = \sec^2 u$, the integral becomes:

$\int\limits u \sec^2 u \;du$

We can evaluate this integral using integration by parts, which has the formula $\int p \;dq = pq - \int q \;dp$.

Let $p = u$ and $dq = \sec^2 u \;du$.

Then, $dp = \frac{d}{du}(u) \;du = 1 \;du = du$.

And, $q = \int \sec^2 u \;du = \tan u$.

Apply the integration by parts formula:

$\int\limits u \sec^2 u \;du = u \tan u - \int\limits \tan u \;du$

The integral of $\tan u$ is $-\log |\cos u| + C'$.

$\int\limits u \sec^2 u \;du = u \tan u - (-\log |\cos u|) + C'$

$= u \tan u + \log |\cos u| + C'$

Now, substitute back $u = \sin^{-1} x$. We also need to express $\tan u$ and $\cos u$ in terms of $x$.

Since $u = \sin^{-1} x$, $\sin u = x$. Consider a right triangle where $\sin u = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{1}$. The adjacent side is $\sqrt{1^2 - x^2} = \sqrt{1 - x^2}$.

Then, $\tan u = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{\sqrt{1 - x^2}}$.

And, $\cos u = \frac{\text{adjacent}}{\text{hypotenuse}} = \sqrt{1 - x^2}$.

Substitute these back into the result:

$(\sin^{-1} x) \left(\frac{x}{\sqrt{1 - x^2}}\right) + \log |\sqrt{1 - x^2}| + C$

Since $\sqrt{1 - x^2} \ge 0$ for real $x$ in the domain of $\sin^{-1} x$, $|\sqrt{1 - x^2}| = \sqrt{1 - x^2}$.

Also, $\log (\sqrt{1 - x^2}) = \log ((1 - x^2)^{1/2}) = \frac{1}{2} \log (1 - x^2)$.

The evaluated integral is:

$\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} + \frac{1}{2} \log (1 - x^2) + C$

Where $C$ is the constant of integration.

Solution:

To evaluate the integral $\int\limits \frac{(\cos 5x + \cos 4x)}{1 − 2 \cos 3x} \;dx$, we first simplify the integrand using trigonometric identities.

Consider the numerator $(\cos 5x + \cos 4x)$. We can express the angles $5x$ and $4x$ in terms of $3x$ and other simple angles like $x$ and $2x$.

We use the identity $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$.

Let $A = 3x$. We want to find $B_1$ and $B_2$ such that $3x + B_1 = 5x$ and $3x - B_1 = ?$ for $\cos 5x$, and $3x + B_2 = 4x$ and $3x - B_2 = ?$ for $\cos 4x$. This doesn't directly give a simple form.

Instead, let's use the identity relating $\cos(nx)$ terms:

$\cos(n+k)x + \cos(n-k)x = 2 \cos nx \cos kx$

Let $n=3$ and $k=2$: $\cos(3+2)x + \cos(3-2)x = 2 \cos 3x \cos 2x$

$\cos 5x + \cos x = 2 \cos 3x \cos 2x$

Let $n=3$ and $k=1$: $\cos(3+1)x + \cos(3-1)x = 2 \cos 3x \cos x$

$\cos 4x + \cos 2x = 2 \cos 3x \cos x$}

Add these two results:

$(\cos 5x + \cos x) + (\cos 4x + \cos 2x) = 2 \cos 3x \cos 2x + 2 \cos 3x \cos x$

$\cos 5x + \cos 4x + \cos x + \cos 2x = 2 \cos 3x (\cos 2x + \cos x)$

Rearrange the terms to isolate the numerator of the integral:

$\cos 5x + \cos 4x = 2 \cos 3x (\cos 2x + \cos x) - (\cos x + \cos 2x)$

Factor out the common term $(\cos x + \cos 2x)$ on the right side:

$\cos 5x + \cos 4x = (2 \cos 3x - 1)(\cos x + \cos 2x)$

Now substitute this back into the integrand:

$\frac{(2 \cos 3x - 1)(\cos x + \cos 2x)}{1 − 2 \cos 3x}$

Recognize that the term $(1 − 2 \cos 3x)$ in the denominator is the negative of the term $(2 \cos 3x - 1)$ in the numerator.

$\frac{(2 \cos 3x - 1)(\cos x + \cos 2x)}{-(2 \cos 3x - 1)}$

Assuming $(2 \cos 3x - 1) \neq 0$, we can cancel this term from the numerator and the denominator:

$\frac{\cos x + \cos 2x}{-1} = -(\cos x + \cos 2x)$

The integral simplifies to:

$\int\limits -(\cos x + \cos 2x) \;dx$

Integrate term by term:

$\int\limits -(\cos x + \cos 2x) \;dx = \int\limits -\cos x \;dx + \int\limits -\cos 2x \;dx$

$\int\limits -\cos x \;dx = -\sin x + C_1$

For the second integral, let $u = 2x$, so $du = 2 \;dx$, which means $dx = \frac{1}{2} du$:

$\int\limits -\cos 2x \;dx = -\int\limits \cos u \left(\frac{1}{2} \;du\right) = -\frac{1}{2} \int\limits \cos u \;du = -\frac{1}{2} \sin u + C_2$

Substitute back $u = 2x$:

$-\frac{1}{2} \sin 2x + C_2$

Combining the results, and letting $C = C_1 + C_2$ be the constant of integration:

$\int\limits \frac{(\cos 5x + \cos 4x)}{1 − 2 \cos 3x} \;dx = -\sin x - \frac{1}{2} \sin 2x + C$

This solution is valid for $x$ such that $1 - 2 \cos 3x \neq 0$, i.e., $\cos 3x \neq \frac{1}{2}$.

Solution:

To evaluate the integral $\int\limits \frac{\sin^6 x + \cos^6 x}{\sin^2 x \; \cos^2 x} \;dx$, we first simplify the numerator using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Here, we let $a = \sin^2 x$ and $b = \cos^2 x$.

$\sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3$

$= (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2)$

Using the identity $\sin^2 x + \cos^2 x = 1$, the first factor is 1:

$= (1)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$

$= \sin^4 x + \cos^4 x - \sin^2 x \cos^2 x$

Now, we can rewrite $\sin^4 x + \cos^4 x$ using the identity $(\sin^2 x + \cos^2 x)^2 = \sin^4 x + 2 \sin^2 x \cos^2 x + \cos^4 x$.

So, $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1^2 - 2 \sin^2 x \cos^2 x = 1 - 2 \sin^2 x \cos^2 x$.

Substitute this back into the simplified numerator:

Numerator = $(1 - 2 \sin^2 x \cos^2 x) - \sin^2 x \cos^2 x = 1 - 3 \sin^2 x \cos^2 x$

Now, substitute this simplified numerator back into the integrand:

$\frac{1 - 3 \sin^2 x \cos^2 x}{\sin^2 x \; \cos^2 x}$

Split the fraction into two terms:

$\frac{1}{\sin^2 x \; \cos^2 x} - \frac{3 \sin^2 x \cos^2 x}{\sin^2 x \; \cos^2 x}$

The second term simplifies to -3.

For the first term, we can use the identity $\sin^2 x + \cos^2 x = 1$ in the numerator:

$\frac{\sin^2 x + \cos^2 x}{\sin^2 x \; \cos^2 x} - 3$

Split this fraction further:

$\frac{\sin^2 x}{\sin^2 x \; \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \; \cos^2 x} - 3$

Cancel the common terms in each fraction:

$\frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} - 3$

Using the reciprocal identities $\frac{1}{\cos^2 x} = \sec^2 x$ and $\frac{1}{\sin^2 x} = \text{cosec}^2 x$, we get:

$\sec^2 x + \text{cosec}^2 x - 3$

Now, integrate the simplified expression:

$\int\limits (\sec^2 x + \text{cosec}^2 x - 3) \;dx$

We can integrate each term separately:

$\int\limits \sec^2 x \;dx = \tan x + C_1$

$\int\limits \text{cosec}^2 x \;dx = -\cot x + C_2$

$\int\limits -3 \;dx = -3x + C_3$

Combine these results, letting $C = C_1 + C_2 + C_3$ be the constant of integration:

$\int\limits \frac{\sin^6 x + \cos^6 x}{\sin^2 x \; \cos^2 x} \;dx = (\tan x) + (-\cot x) + (-3x) + C$

$= \tan x - \cot x - 3x + C$

The evaluated integral is $\tan x - \cot x - 3x + C$. This is valid for values of $x$ where $\sin x \neq 0$ and $\cos x \neq 0$.

Solution:

To evaluate the integral $\int\limits \frac{\sqrt{x}}{\sqrt{a^3−x^3}} \;dx$, we observe the terms involving $x$ and $a$ under the square roots. The term $x^3$ in the denominator suggests a substitution involving $x^{3/2}$.

Let $u = x^{\frac{3}{2}}$.

Now, find the differential $du$ by differentiating $u$ with respect to $x$:

$du = \frac{d}{dx}\left(x^{\frac{3}{2}}\right) \;dx = \frac{3}{2} x^{\frac{3}{2}-1} \;dx = \frac{3}{2} x^{\frac{1}{2}} \;dx = \frac{3}{2} \sqrt{x} \;dx$

The numerator of the integrand is $\sqrt{x} \;dx$. From the differential $du$, we can express $\sqrt{x} \;dx$ in terms of $du$:

$\sqrt{x} \;dx = \frac{2}{3} du$

Next, express the term $x^3$ in the denominator in terms of $u$. Since $u = x^{3/2}$, squaring both sides gives:

$u^2 = \left(x^{\frac{3}{2}}\right)^2 = x^{3/2 \times 2} = x^3$

The denominator $\sqrt{a^3 - x^3}$ becomes $\sqrt{a^3 - u^2}$. Note that $a^3$ can be written as $(a^{3/2})^2$. Assuming $a > 0$, we let $A = a^{3/2}$. The denominator is $\sqrt{A^2 - u^2}$.

Substitute $\sqrt{x} \;dx = \frac{2}{3} du$ and $\sqrt{a^3 - x^3} = \sqrt{a^3 - u^2}$ into the integral:

$\int\limits \frac{\frac{2}{3} du}{\sqrt{a^3 - u^2}}$

Take the constant $\frac{2}{3}$ outside the integral:

$\frac{2}{3} \int\limits \frac{du}{\sqrt{a^3 - u^2}}$

This integral is in the standard form $\int\limits \frac{du}{\sqrt{A^2 - u^2}} = \sin^{-1}\left(\frac{u}{A}\right) + C'$. Here, $A^2 = a^3$, so $A = a^{3/2}$.

Apply the standard integral formula:

$\frac{2}{3} \left(\sin^{-1}\left(\frac{u}{a^{3/2}}\right)\right) + C'$

Substitute back $u = x^{3/2}$ to express the result in terms of $x$:

$\frac{2}{3} \sin^{-1}\left(\frac{x^{3/2}}{a^{3/2}}\right) + C$

The term $\frac{x^{3/2}}{a^{3/2}}$ can be written as $\left(\frac{x}{a}\right)^{3/2}$.

The evaluated integral is:

$\frac{2}{3} \sin^{-1}\left(\left(\frac{x}{a}\right)^{\frac{3}{2}}\right) + C$

Where $C$ is the constant of integration. This is valid for $0 \le x < a$ (assuming $a>0$) so that $a^3 - x^3 > 0$ and $x \ge 0$.

Solution:

To evaluate the integral $\int\limits \frac{\cos x − \cos 2x}{1 − \cos x} \;dx$, we first simplify the integrand using trigonometric identities.

We use the double angle identity for $\cos 2x$: $\cos 2x = 2 \cos^2 x - 1$.

Substitute this identity into the numerator of the integrand:

Numerator $= \cos x - \cos 2x = \cos x - (2 \cos^2 x - 1)$

$= \cos x - 2 \cos^2 x + 1$

We can rewrite the numerator as $1 + \cos x - 2 \cos^2 x$.

Let $y = \cos x$. The numerator is $1 + y - 2y^2$. We can factor this quadratic expression:

$1 + y - 2y^2 = -(2y^2 - y - 1)$

Factor the quadratic $2y^2 - y - 1$: find two numbers that multiply to $2 \times -1 = -2$ and add to -1. These numbers are -2 and 1.

$2y^2 - 2y + y - 1 = 2y(y - 1) + 1(y - 1) = (2y + 1)(y - 1)$

So the numerator is $-(2y + 1)(y - 1) = (2y + 1)(1 - y)$.

Substitute back $y = \cos x$:

Numerator $= (2 \cos x + 1)(1 - \cos x)$

Now, substitute this back into the integrand:

$\frac{(2 \cos x + 1)(1 - \cos x)}{1 − \cos x}$

Assuming the denominator $1 - \cos x \neq 0$, which means $x \neq 2k\pi$ for any integer $k$, we can cancel the term $(1 - \cos x)$ from the numerator and the denominator.

The integrand simplifies to $2 \cos x + 1$.

The integral becomes:

$\int\limits (2 \cos x + 1) \;dx$

Now, integrate term by term:

$\int\limits 2 \cos x \;dx = 2 \int\limits \cos x \;dx = 2 \sin x + C_1$

$\int\limits 1 \;dx = x + C_2$

Combine these results, letting $C = C_1 + C_2$ be the constant of integration:

$\int\limits \frac{\cos x − \cos 2x}{1 − \cos x} \;dx = (2 \sin x + C_1) + (x + C_2)$

$= 2 \sin x + x + C$

The evaluated integral is $2 \sin x + x + C$, valid for $x \neq 2k\pi$ for any integer $k$.

Solution:

To evaluate the integral $\int\limits \frac{dx}{x\sqrt{x^4 − 1}}$, we use the substitution suggested by the hint.

Let $x^2 = \sec \theta$.

From this substitution, we have $x^4 = (\sec \theta)^2 = \sec^2 \theta$.

The term under the square root becomes $x^4 - 1 = \sec^2 \theta - 1$. Using the identity $\sec^2 \theta - 1 = \tan^2 \theta$, we have:

$\sqrt{x^4 - 1} = \sqrt{\sec^2 \theta - 1} = \sqrt{\tan^2 \theta}$

Assuming $x > 1$ (so that $x^4 - 1 > 0$ and $\sec \theta = x^2 > 1$, implying $\theta$ is in a range where $\tan \theta \ge 0$, such as $0 < \theta < \frac{\pi}{2}$), we have $\sqrt{\tan^2 \theta} = \tan \theta$.

Now, we need to find the differential $dx$ in terms of $\theta$ and $d\theta$. Differentiate $x^2 = \sec \theta$ with respect to $x$ implicitly:

$\frac{d}{dx}(x^2) = \frac{d}{dx}(\sec \theta)$

$2x = \frac{d}{d\theta}(\sec \theta) \cdot \frac{d\theta}{dx}$

$2x = \sec \theta \tan \theta \frac{d\theta}{dx}$

Rearranging to solve for $dx$:

$dx = \frac{\sec \theta \tan \theta}{2x} \;d\theta$

Substitute $x^2 = \sec \theta$ and $\sqrt{x^4 - 1} = \tan \theta$ into the original integral, and also substitute $dx$:

$\int\limits \frac{1}{x \cdot \sqrt{x^4 - 1}} \;dx = \int\limits \frac{1}{x \cdot \tan \theta} \left(\frac{\sec \theta \tan \theta}{2x} \;d\theta\right)$

Simplify the integrand:

$= \int\limits \frac{\sec \theta \tan \theta}{2x^2 \tan \theta} \;d\theta$

Assuming $\tan \theta \neq 0$ (which is true for $x > 1$), we can cancel $\tan \theta$:

$= \int\limits \frac{\sec \theta}{2x^2} \;d\theta$

Substitute $x^2 = \sec \theta$ into the denominator:

$= \int\limits \frac{\sec \theta}{2 \sec \theta} \;d\theta$

Assuming $\sec \theta \neq 0$ (which is true for $x^2 > 1$), we can cancel $\sec \theta$:

$= \int\limits \frac{1}{2} \;d\theta$

Now, integrate with respect to $\theta$:

$\int\limits \frac{1}{2} \;d\theta = \frac{1}{2} \theta + C$

Finally, substitute back $\theta$ in terms of $x$. From the substitution $x^2 = \sec \theta$, we get $\theta = \sec^{-1}(x^2)$.

The evaluated integral is:

$\frac{1}{2} \sec^{-1}(x^2) + C$

Where $C$ is the constant of integration. This result is valid for $|x| > 1$.

To evaluate $\int\limits_0^2 (x^2+3) \;dx$ as the limit of sums, we use the definition:

$\int\limits_a^b f(x) \;dx = \lim\limits_{n \to \infty} \Delta x \sum\limits_{i=1}^{n} f(a+i\Delta x)$

Here, we have $f(x) = x^2+3$, $a = 0$, and $b = 2$.

The width of each subinterval is given by:

The points in the partition are given by $x_i = a + i\Delta x$:

Now we evaluate $f(x_i) = f\left(\frac{2i}{n}\right)$:

Now we set up the sum $\Delta x \sum\limits_{i=1}^{n} f(a+i\Delta x)$:

$\Delta x \sum\limits_{i=1}^{n} f\left(\frac{2i}{n}\right) = \frac{2}{n} \sum\limits_{i=1}^{n} \left(\frac{4i^2}{n^2} + 3\right)$

$= \frac{2}{n} \sum\limits_{i=1}^{n} \frac{4i^2}{n^2} + \frac{2}{n} \sum\limits_{i=1}^{n} 3$

$= \frac{8}{n^3} \sum\limits_{i=1}^{n} i^2 + \frac{6}{n} \sum\limits_{i=1}^{n} 1$

Using the standard summation formulas $\sum\limits_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum\limits_{i=1}^{n} 1 = n$:

$= \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{6}{n} \cdot n$

$= \frac{4}{3n^2} (n+1)(2n+1) + 6$

$= \frac{4}{3n^2} (2n^2 + n + 2n + 1) + 6$

$= \frac{4}{3n^2} (2n^2 + 3n + 1) + 6$

$= \frac{8n^2 + 12n + 4}{3n^2} + 6$

$= \frac{8n^2}{3n^2} + \frac{12n}{3n^2} + \frac{4}{3n^2} + 6$

$= \frac{8}{3} + \frac{4}{n} + \frac{4}{3n^2} + 6$

$= \left(\frac{8}{3} + 6\right) + \frac{4}{n} + \frac{4}{3n^2}$

$= \left(\frac{8+18}{3}\right) + \frac{4}{n} + \frac{4}{3n^2}$

$= \frac{26}{3} + \frac{4}{n} + \frac{4}{3n^2}$

Finally, we take the limit as $n \to \infty$:

$\int\limits_0^2 (x^2+3) \;dx = \lim\limits_{n \to \infty} \left(\frac{26}{3} + \frac{4}{n} + \frac{4}{3n^2}\right)$

$= \lim\limits_{n \to \infty} \frac{26}{3} + \lim\limits_{n \to \infty} \frac{4}{n} + \lim\limits_{n \to \infty} \frac{4}{3n^2}$

As $n \to \infty$, $\frac{4}{n} \to 0$ and $\frac{4}{3n^2} \to 0$.

$= \frac{26}{3} + 0 + 0$

$= \frac{26}{3}$

Thus, the value of the integral evaluated as the limit of sums is $\frac{26}{3}$.

Final Answer:

$\int\limits_0^2 (x^2+3) \;dx = \frac{26}{3}$

To evaluate $\int\limits_0^2 e^x \;dx$ as the limit of sums, we use the definition:

$\int\limits_a^b f(x) \;dx = \lim\limits_{n \to \infty} \Delta x \sum\limits_{i=1}^{n} f(a+i\Delta x)$

Here, we have $f(x) = e^x$, $a = 0$, and $b = 2$.

The width of each subinterval is given by:

The points in the partition are given by $x_i = a + i\Delta x$:

Now we evaluate $f(x_i) = f\left(\frac{2i}{n}\right)$:

Now we set up the sum $\Delta x \sum\limits_{i=1}^{n} f(a+i\Delta x)$:

$\Delta x \sum\limits_{i=1}^{n} f\left(\frac{2i}{n}\right) = \frac{2}{n} \sum\limits_{i=1}^{n} e^{\frac{2i}{n}}$

This is the sum of a geometric series with first term $a_1 = e^{\frac{2}{n}}$ (for $i=1$) and common ratio $r = e^{\frac{2}{n}}$. The sum of the first $n$ terms of a geometric series is given by $S_n = a_1 \frac{r^n - 1}{r-1}$.

So, $\sum\limits_{i=1}^{n} e^{\frac{2i}{n}} = e^{\frac{2}{n}} \frac{\left(e^{\frac{2}{n}}\right)^n - 1}{e^{\frac{2}{n}}-1} = e^{\frac{2}{n}} \frac{e^2 - 1}{e^{\frac{2}{n}}-1}$.

Therefore, the sum is:

$\frac{2}{n} \sum\limits_{i=1}^{n} e^{\frac{2i}{n}} = \frac{2}{n} \cdot e^{\frac{2}{n}} \frac{e^2 - 1}{e^{\frac{2}{n}}-1}$

Finally, we take the limit as $n \to \infty$:

$\int\limits_0^2 e^x \;dx = \lim\limits_{n \to \infty} \left(\frac{2}{n} \cdot e^{\frac{2}{n}} \frac{e^2 - 1}{e^{\frac{2}{n}}-1}\right)$

$= (e^2 - 1) \lim\limits_{n \to \infty} \left(\frac{e^{\frac{2}{n}}}{e^{\frac{2}{n}}-1} \cdot \frac{2}{n}\right)$

Let $h = \frac{2}{n}$. As $n \to \infty$, $h \to 0$. The expression becomes:

$= (e^2 - 1) \lim\limits_{h \to 0} \left(\frac{e^{h}}{e^{h}-1} \cdot h\right)$

$= (e^2 - 1) \lim\limits_{h \to 0} e^{h} \cdot \lim\limits_{h \to 0} \frac{h}{e^{h}-1}$

We know that $\lim\limits_{h \to 0} e^h = e^0 = 1$ and $\lim\limits_{h \to 0} \frac{e^{h}-1}{h} = 1$, which implies $\lim\limits_{h \to 0} \frac{h}{e^{h}-1} = 1$.

$= (e^2 - 1) \cdot 1 \cdot 1$

$= e^2 - 1$

Thus, the value of the integral evaluated as the limit of sums is $e^2 - 1$.

Final Answer:

$\int\limits_0^2 e^x \;dx = e^2 - 1$

We need to evaluate the definite integral $\int\limits_0^1 \frac{dx}{e^x+e^{-x}}$.

First, we can rewrite the integrand by multiplying the numerator and denominator by $e^x$:

$\frac{1}{e^x+e^{-x}} = \frac{1}{e^x+\frac{1}{e^x}} = \frac{1}{\frac{e^{2x}+1}{e^x}} = \frac{e^x}{e^{2x}+1}$

So the integral becomes:

$I = \int\limits_0^1 \frac{e^x}{e^{2x}+1} \;dx$

Now, we use a substitution method.

Let $u = e^x$.

Differentiating with respect to $x$, we get $du = e^x \;dx$.

We also need to change the limits of integration:

When $x = 0$, $u = e^0 = 1$.

When $x = 1$, $u = e^1 = e$.

Substituting $u$ and $du$ into the integral, and changing the limits, we get:

$I = \int\limits_1^e \frac{du}{u^2+1}$

The integral of $\frac{1}{u^2+1}$ with respect to $u$ is $\tan^{-1}(u)$.

$I = [\tan^{-1}(u)]_1^e$

Now, we apply the limits of integration:

$I = \tan^{-1}(e) - \tan^{-1}(1)$

We know that $\tan^{-1}(1) = \frac{\pi}{4}$.

$I = \tan^{-1}(e) - \frac{\pi}{4}$

Thus, the value of the definite integral is $\tan^{-1}(e) - \frac{\pi}{4}$.

Final Answer:

$\int\limits_0^1 \frac{dx}{e^x+e^{-x}} = \tan^{-1}(e) - \frac{\pi}{4}$

We need to evaluate the definite integral $I = \int\limits_0^{\frac{\pi}{2}} \frac{\tan x}{1+m^2 \tan^2 x} \;dx$.

First, we rewrite the integrand in terms of $\sin x$ and $\cos x$:

$\frac{\tan x}{1+m^2 \tan^2 x} = \frac{\frac{\sin x}{\cos x}}{1+m^2 \frac{\sin^2 x}{\cos^2 x}} = \frac{\frac{\sin x}{\cos x}}{\frac{\cos^2 x + m^2 \sin^2 x}{\cos^2 x}} = \frac{\sin x}{\cos x} \cdot \frac{\cos^2 x}{\cos^2 x + m^2 \sin^2 x} = \frac{\sin x \cos x}{\cos^2 x + m^2 \sin^2 x}$.

So, the integral becomes $I = \int\limits_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\cos^2 x + m^2 \sin^2 x} \;dx$.

We use the substitution method.

Let $u = \cos^2 x + m^2 \sin^2 x$.

Differentiating $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\cos^2 x) + m^2 \frac{d}{dx}(\sin^2 x)$

$\frac{du}{dx} = 2\cos x (-\sin x) + m^2 (2\sin x \cos x)$

$\frac{du}{dx} = -2\sin x \cos x + 2m^2 \sin x \cos x$

$\frac{du}{dx} = 2 \sin x \cos x (m^2 - 1)$

So, $du = 2 \sin x \cos x (m^2 - 1) \;dx$. This substitution requires $m^2 \neq 1$.

$\sin x \cos x \;dx = \frac{1}{2(m^2-1)} du$ for $m^2 \neq 1$.

We need to change the limits of integration according to the substitution $u = \cos^2 x + m^2 \sin^2 x$:

When $x = 0$, $u = \cos^2(0) + m^2 \sin^2(0) = 1^2 + m^2(0^2) = 1 + 0 = 1$.

When $x = \frac{\pi}{2}$, $u = \cos^2\left(\frac{\pi}{2}\right) + m^2 \sin^2\left(\frac{\pi}{2}\right) = 0^2 + m^2(1^2) = 0 + m^2 = m^2$.

Substituting $u$ and $du$ into the integral, and changing the limits (assuming $m^2 \neq 1$ and $m \neq 0$):

$I = \int\limits_1^{m^2} \frac{1}{u} \cdot \frac{1}{2(m^2-1)} \;du$

$I = \frac{1}{2(m^2-1)} \int\limits_1^{m^2} \frac{1}{u} \;du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log|u|$.

$I = \frac{1}{2(m^2-1)} [\log|u|]_1^{m^2}$

$I = \frac{1}{2(m^2-1)} (\log|m^2| - \log|1|)$

Since $\log|1| = 0$ and $\log|m^2| = \log(m^2)$ for $m \neq 0$, we have:

$I = \frac{1}{2(m^2-1)} (\log(m^2) - 0)$

$I = \frac{\log(m^2)}{2(m^2-1)} = \frac{2\log|m|}{2(m^2-1)} = \frac{\log|m|}{m^2-1}$, for $m \neq 0$ and $m^2 \neq 1$.

Now let's consider the special case when $m^2 = 1$, i.e., $m=1$ or $m=-1$.

If $m^2=1$, the original integral becomes:

$I = \int\limits_0^{\frac{\pi}{2}} \frac{\tan x}{1+ \tan^2 x} \;dx$

Using the identity $1+\tan^2 x = \sec^2 x$, we get:

$I = \int\limits_0^{\frac{\pi}{2}} \frac{\tan x}{\sec^2 x} \;dx = \int\limits_0^{\frac{\pi}{2}} \tan x \cos^2 x \;dx$

$I = \int\limits_0^{\frac{\pi}{2}} \frac{\sin x}{\cos x} \cos^2 x \;dx = \int\limits_0^{\frac{\pi}{2}} \sin x \cos x \;dx$

Let $v = \sin x$. Then $dv = \cos x \;dx$. The limits change from $x=0$ to $v=\sin(0)=0$, and from $x=\frac{\pi}{2}$ to $v=\sin(\frac{\pi}{2})=1$.

$I = \int\limits_0^1 v \;dv$

$I = \left[\frac{v^2}{2}\right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} - 0 = \frac{1}{2}$.

So, for $m^2=1$, the integral is $\frac{1}{2}$. Note that the general formula $\frac{\log|m|}{m^2-1}$ approaches $\frac{1}{2}$ as $m^2 \to 1$ (by L'Hopital's rule, as shown in thinking process).

Let's consider the case when $m=0$.

The integral becomes $I = \int\limits_0^{\frac{\pi}{2}} \tan x \;dx$.

$I = [-\log|\cos x|]_0^{\frac{\pi}{2}} = \lim\limits_{x \to \frac{\pi}{2}^-} (-\log|\cos x|) - (-\log|\cos 0|)$

$I = \lim\limits_{x \to \frac{\pi}{2}^-} (-\log|\cos x|) - (-\log|1|)$

$I = \lim\limits_{x \to \frac{\pi}{2}^-} (-\log|\cos x|) - 0$.

As $x \to \frac{\pi}{2}^-$, $\cos x \to 0^+$, so $\log|\cos x| \to -\infty$. Thus, $-\log|\cos x| \to \infty$.

The integral diverges when $m=0$.

Combining the results:

If $m^2 \neq 1$ and $m \neq 0$, $I = \frac{\log|m|}{m^2-1}$.

If $m^2 = 1$ (i.e., $m=1$ or $m=-1$), $I = \frac{1}{2}$.

If $m=0$, the integral diverges.

Assuming the question implies a convergent integral, the result depends on $m^2$.

Final Answer:

$\int\limits_0^{\frac{π}{2}} \frac{\tan x \;dx}{1+m^2 \tan^2 x} = \begin{cases} \frac{\log|m|}{m^2-1} & , & m \neq 0, m^2 \neq 1 \\ \frac{1}{2} & , & m^2 = 1 \\ \text{Diverges} & , & m = 0 \end{cases}$

We need to evaluate the definite integral $I = \int\limits_1^2 \frac{dx}{\sqrt{(x−1) (2−x)}}$.

First, simplify the expression inside the square root:

$(x-1)(2-x) = 2x - x^2 - 2 + x = -x^2 + 3x - 2$

Now, complete the square for the quadratic expression $-x^2 + 3x - 2$:

$-x^2 + 3x - 2 = -(x^2 - 3x + 2)$

$= -\left(x^2 - 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + 2\right)$

$= -\left(\left(x - \frac{3}{2}\right)^2 - \frac{9}{4} + \frac{8}{4}\right)$

$= -\left(\left(x - \frac{3}{2}\right)^2 - \frac{1}{4}\right)$

$= \frac{1}{4} - \left(x - \frac{3}{2}\right)^2$

$= \left(\frac{1}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2$

The integral becomes:

$I = \int\limits_1^2 \frac{dx}{\sqrt{\left(\frac{1}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2}}$

This integral is of the form $\int \frac{du}{\sqrt{a^2 - u^2}}$. We use the substitution $u = a \sin \theta$.

Let $x - \frac{3}{2} = \frac{1}{2} \sin \theta$.

Differentiating with respect to $\theta$:

$dx = \frac{1}{2} \cos \theta \;d\theta$

Change the limits of integration:

When $x = 1$:

When $x = 2$:

Substitute $x - \frac{3}{2} = \frac{1}{2} \sin \theta$ and $dx = \frac{1}{2} \cos \theta \;d\theta$ into the integral, and change the limits:

The denominator $\sqrt{\left(\frac{1}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2} = \sqrt{\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2} \sin \theta\right)^2} = \sqrt{\frac{1}{4} - \frac{1}{4} \sin^2 \theta} = \sqrt{\frac{1}{4}(1-\sin^2 \theta)} = \sqrt{\frac{1}{4}\cos^2 \theta} = \frac{1}{2}|\cos \theta|$.

For $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\cos \theta \ge 0$, so $|\cos \theta| = \cos \theta$.

The integral becomes:

$I = \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\frac{1}{2} \cos \theta}{\frac{1}{2} \cos \theta} \;d\theta$

$I = \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \;d\theta$

Evaluate the integral:

$I = [\theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$I = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right)$

$I = \frac{\pi}{2} + \frac{\pi}{2}$

$I = \pi$

Thus, the value of the definite integral is $\pi$.

Final Answer:

$\int\limits_1^2 \frac{dx}{\sqrt{(x−1) (2−x)}} = \pi$

We need to evaluate the definite integral $I = \int\limits_0^1 \frac{x}{\sqrt{1 + x^2}} \;dx$.

We use the substitution method.

Let $u = 1 + x^2$.

Differentiating $u$ with respect to $x$:

So, $du = 2x \;dx$, which implies $x \;dx = \frac{1}{2} du$.

We need to change the limits of integration according to the substitution $u = 1 + x^2$:

When $x = 0$, $u = 1 + (0)^2 = 1$.

When $x = 1$, $u = 1 + (1)^2 = 1 + 1 = 2$.

Substituting $u$ and $x \;dx$ into the integral, and changing the limits, we get:

$I = \int\limits_1^2 \frac{\frac{1}{2} du}{\sqrt{u}}$

$I = \frac{1}{2} \int\limits_1^2 u^{-1/2} \;du$

Now, evaluate the integral using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$:

$I = \frac{1}{2} \left[\frac{u^{-1/2+1}}{-1/2+1}\right]_1^2$

$I = \frac{1}{2} \left[\frac{u^{1/2}}{1/2}\right]_1^2$

$I = \frac{1}{2} [2\sqrt{u}]_1^2$

$I = [\sqrt{u}]_1^2$

Apply the limits of integration:

$I = \sqrt{2} - \sqrt{1}$

$I = \sqrt{2} - 1$

Thus, the value of the definite integral is $\sqrt{2} - 1$.

Final Answer:

$\int\limits_0^1 \frac{x \;dx}{\sqrt{1 + x^2}} = \sqrt{2} - 1$

We need to evaluate the definite integral $I = \int\limits_0^\pi x \sin x \cos^2 x \;dx$.

We use the property of definite integrals: $\int\limits_0^a f(x) \;dx = \int\limits_0^a f(a-x) \;dx$.

Here, $a = \pi$ and $f(x) = x \sin x \cos^2 x$.

Let's find $f(\pi - x)$:

$f(\pi - x) = (\pi - x) \sin(\pi - x) \cos^2(\pi - x)$

Using the trigonometric identities $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$, we have:

$f(\pi - x) = (\pi - x) (\sin x) (-\cos x)^2$

$f(\pi - x) = (\pi - x) \sin x \cos^2 x$

$f(\pi - x) = \pi \sin x \cos^2 x - x \sin x \cos^2 x$

Applying the property:

$I = \int\limits_0^\pi (\pi \sin x \cos^2 x - x \sin x \cos^2 x) \;dx$

$I = \int\limits_0^\pi \pi \sin x \cos^2 x \;dx - \int\limits_0^\pi x \sin x \cos^2 x \;dx$

$I = \pi \int\limits_0^\pi \sin x \cos^2 x \;dx - I$

Adding $I$ to both sides, we get:

Now, let's evaluate the integral $J = \int\limits_0^\pi \sin x \cos^2 x \;dx$.

We use the substitution method.

Let $u = \cos x$.

Differentiating with respect to $x$, we get $du = -\sin x \;dx$. So, $\sin x \;dx = -du$.

We need to change the limits of integration:

When $x = 0$, $u = \cos(0) = 1$.

When $x = \pi$, $u = \cos(\pi) = -1$.

Substituting $u$ and $\sin x \;dx$ into the integral $J$, and changing the limits, we get:

$J = \int\limits_1^{-1} u^2 (-du)$

$J = -\int\limits_1^{-1} u^2 \;du$

Using the property $\int\limits_a^b f(x) \;dx = -\int\limits_b^a f(x) \;dx$:

$J = \int\limits_{-1}^1 u^2 \;du$

Evaluate the integral $J$ using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$:

$J = \left[\frac{u^{2+1}}{2+1}\right]_{-1}^1 = \left[\frac{u^3}{3}\right]_{-1}^1$

$J = \frac{(1)^3}{3} - \frac{(-1)^3}{3}$

$J = \frac{1}{3} - \left(-\frac{1}{3}\right)$

$J = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

Substitute the value of $J$ back into equation (i):

Divide by 2 to find $I$:

Thus, the value of the definite integral is $\frac{\pi}{3}$.

Final Answer:

$\int\limits_0^π x \sin x \; \cos^2 x \;dx = \frac{\pi}{3}$

We need to evaluate the definite integral $I = \int\limits_0^{\frac{1}{2}} \frac{dx}{(1+x^2) \sqrt{1−x^2}}$.

Following the hint, we use the substitution $x = \sin \theta$.

Differentiating with respect to $\theta$, we get $dx = \cos \theta \;d\theta$.

We need to change the limits of integration according to the substitution $x = \sin \theta$:

When $x = 0$:

When $x = \frac{1}{2}$:

The new limits are from $0$ to $\frac{\pi}{6}$.

Now, substitute $x = \sin \theta$ and $dx = \cos \theta \;d\theta$ into the integrand:

$1+x^2 = 1+\sin^2 \theta$

$\sqrt{1-x^2} = \sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = |\cos \theta|$

For the interval $0 \le \theta \le \frac{\pi}{6}$, $\cos \theta \ge 0$, so $|\cos \theta| = \cos \theta$.

The integrand becomes $\frac{\cos \theta \;d\theta}{(1+\sin^2 \theta) \cos \theta}$.

Assuming $\cos \theta \neq 0$ (which is true for $0 \le \theta < \frac{\pi}{2}$), we can cancel $\cos \theta$:

$\frac{1}{1+\sin^2 \theta} \;d\theta$

The integral in terms of $\theta$ is:

$I = \int\limits_0^{\frac{\pi}{6}} \frac{d\theta}{1+\sin^2 \theta}$

To evaluate this integral, divide the numerator and denominator by $\cos^2 \theta$:

$I = \int\limits_0^{\frac{\pi}{6}} \frac{\frac{1}{\cos^2 \theta}}{\frac{1}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta}} \;d\theta = \int\limits_0^{\frac{\pi}{6}} \frac{\sec^2 \theta}{\sec^2 \theta + \tan^2 \theta} \;d\theta$

Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$:

$I = \int\limits_0^{\frac{\pi}{6}} \frac{\sec^2 \theta}{(1+\tan^2 \theta) + \tan^2 \theta} \;d\theta = \int\limits_0^{\frac{\pi}{6}} \frac{\sec^2 \theta}{1+2\tan^2 \theta} \;d\theta$

Now, use another substitution. Let $v = \tan \theta$.

Differentiating with respect to $\theta$, we get $dv = \sec^2 \theta \;d\theta$.

Change the limits for $v$:

When $\theta = 0$, $v = \tan(0) = 0$.

When $\theta = \frac{\pi}{6}$, $v = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$.

The integral in terms of $v$ is:

$I = \int\limits_0^{\frac{1}{\sqrt{3}}} \frac{dv}{1+2v^2}$

To evaluate this integral, we can use the standard formula $\int \frac{du}{a^2+b^2 u^2} = \frac{1}{ab} \tan^{-1}\left(\frac{bu}{a}\right)$. Here $a=1, b=\sqrt{2}$.

$I = \left[\frac{1}{1 \cdot \sqrt{2}} \tan^{-1}\left(\frac{\sqrt{2}v}{1}\right)\right]_0^{\frac{1}{\sqrt{3}}}$

$I = \left[\frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}v)\right]_0^{\frac{1}{\sqrt{3}}}$

Apply the limits of integration:

$I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\sqrt{2} \cdot \frac{1}{\sqrt{3}}\right) - \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2} \cdot 0)$

$I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\sqrt{2}}{\sqrt{3}}\right) - \frac{1}{\sqrt{2}} \tan^{-1}(0)$

$I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\sqrt{6}}{3}\right) - 0$

$I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\sqrt{6}}{3}\right)$

We can rationalize the denominator:

$I = \frac{\sqrt{2}}{2} \tan^{-1}\left(\frac{\sqrt{6}}{3}\right)$

Thus, the value of the definite integral is $\frac{\sqrt{2}}{2} \tan^{-1}\left(\frac{\sqrt{6}}{3}\right)$.

Final Answer:

$\int\limits_0^{\frac{1}{2}} \frac{dx}{(1+x^2) \sqrt{1−x^2}} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\sqrt{6}}{3}\right)$ or $\frac{\sqrt{2}}{2} \tan^{-1}\left(\frac{\sqrt{6}}{3}\right)$

We need to evaluate the indefinite integral $I = \int \frac{x^2 \;dx}{x^4−x^2−12}$.

First, factor the denominator:

$x^4 - x^2 - 12 = (x^2)^2 - (x^2) - 12$

Let $y = x^2$. The expression becomes $y^2 - y - 12$. Factoring this quadratic in $y$, we get $(y-4)(y+3)$.

Substituting $x^2$ back for $y$, the denominator is $(x^2-4)(x^2+3)$.

Further factoring $x^2-4$ as a difference of squares, we get $(x-2)(x+2)$.

So, the denominator is $(x-2)(x+2)(x^2+3)$.

The integrand is $\frac{x^2}{(x^2-4)(x^2+3)}$.

We use partial fraction decomposition. We set up the decomposition as:

$\frac{x^2}{(x^2-4)(x^2+3)} = \frac{A}{x^2-4} + \frac{B}{x^2+3}$

Multiply both sides by $(x^2-4)(x^2+3)$ to clear the denominators:

$x^2 = A(x^2+3) + B(x^2-4)$

$x^2 = Ax^2 + 3A + Bx^2 - 4B$

$x^2 = (A+B)x^2 + (3A-4B)$

Comparing the coefficients of $x^2$ on both sides:

Comparing the constant terms:

From equation (2), $3A = 4B$, so $A = \frac{4}{3}B$.

Substitute this into equation (1):

$\frac{4}{3}B + B = 1$

$\frac{4B+3B}{3} = 1$

$\frac{7B}{3} = 1$

$7B = 3 \implies B = \frac{3}{7}$

Now find $A$ using $A = \frac{4}{3}B$:

$A = \frac{4}{3} \cdot \frac{3}{7} = \frac{4}{7}$

So the partial fraction decomposition is:

$\frac{x^2}{(x^2-4)(x^2+3)} = \frac{4/7}{x^2-4} + \frac{3/7}{x^2+3}$

Now we integrate the partial fractions:

$I = \int \left(\frac{4/7}{x^2-4} + \frac{3/7}{x^2+3}\right) \;dx$

$I = \frac{4}{7} \int \frac{dx}{x^2-4} + \frac{3}{7} \int \frac{dx}{x^2+3}$

We use the standard integral formulas:

$\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log\left|\frac{x-a}{x+a}\right| + C$

$\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$

For the first integral, $a^2=4$, so $a=2$:

$\int \frac{dx}{x^2-4} = \frac{1}{2(2)} \log\left|\frac{x-2}{x+2}\right| = \frac{1}{4} \log\left|\frac{x-2}{x+2}\right|$

For the second integral, $a^2=3$, so $a=\sqrt{3}$:

$\int \frac{dx}{x^2+3} = \int \frac{dx}{x^2+(\sqrt{3})^2} = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right)$

Substitute these back into the expression for $I$:

$I = \frac{4}{7} \left(\frac{1}{4} \log\left|\frac{x-2}{x+2}\right|\right) + \frac{3}{7} \left(\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right)\right) + C$

$I = \frac{1}{7} \log\left|\frac{x-2}{x+2}\right| + \frac{3}{7\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C$

Rationalize the denominator of the coefficient of the $\tan^{-1}$ term:

$\frac{3}{7\sqrt{3}} = \frac{3\sqrt{3}}{7\sqrt{3}\sqrt{3}} = \frac{3\sqrt{3}}{7 \cdot 3} = \frac{\sqrt{3}}{7}$

So the integral is:

$I = \frac{1}{7} \log\left|\frac{x-2}{x+2}\right| + \frac{\sqrt{3}}{7} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C$

Thus, the value of the indefinite integral is $\frac{1}{7} \log\left|\frac{x-2}{x+2}\right| + \frac{\sqrt{3}}{7} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C$.

Final Answer:

$\int \frac{x^2 \;dx}{x^4−x^2−12} = \frac{1}{7} \log\left|\frac{x-2}{x+2}\right| + \frac{\sqrt{3}}{7} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C$

We need to evaluate the indefinite integral $I = \int \frac{x^2 \;dx}{(x^2+a^2) (x^2+b^2)}$.

We use the method of partial fraction decomposition for the integrand $\frac{x^2}{(x^2+a^2) (x^2+b^2)}$.

Assume $a^2 \neq b^2$. We can write:

$\frac{x^2}{(x^2+a^2) (x^2+b^2)} = \frac{A}{x^2+a^2} + \frac{B}{x^2+b^2}$

Multiply both sides by $(x^2+a^2)(x^2+b^2)$ to clear the denominators:

$x^2 = A(x^2+b^2) + B(x^2+a^2)$

$x^2 = Ax^2 + Ab^2 + Bx^2 + Ba^2$

$x^2 = (A+B)x^2 + (Ab^2 + Ba^2)$

Comparing the coefficients of $x^2$ and the constant terms on both sides, we get a system of linear equations:

Comparing coefficients of $x^2$:

Comparing constant terms:

From equation (1), $B = 1-A$. Substitute this into equation (2):

$Ab^2 + (1-A)a^2 = 0$

$Ab^2 + a^2 - Aa^2 = 0$

$A(b^2 - a^2) = -a^2$

Assuming $b^2 - a^2 \neq 0$, we can solve for A:

$A = \frac{-a^2}{b^2 - a^2} = \frac{a^2}{a^2 - b^2}$

Now, substitute the value of A back into $B = 1-A$:

$B = 1 - \frac{a^2}{a^2 - b^2} = \frac{(a^2 - b^2) - a^2}{a^2 - b^2} = \frac{a^2 - b^2 - a^2}{a^2 - b^2} = \frac{-b^2}{a^2 - b^2} = \frac{b^2}{b^2 - a^2}$

So, the partial fraction decomposition is:

$\frac{x^2}{(x^2+a^2) (x^2+b^2)} = \frac{\frac{a^2}{a^2 - b^2}}{x^2+a^2} + \frac{\frac{b^2}{b^2 - a^2}}{x^2+b^2}$

$= \frac{1}{a^2 - b^2} \left( \frac{a^2}{x^2+a^2} \right) + \frac{1}{b^2 - a^2} \left( \frac{b^2}{x^2+b^2} \right)$

Note that $\frac{1}{b^2 - a^2} = -\frac{1}{a^2 - b^2}$.

$= \frac{1}{a^2 - b^2} \left( \frac{a^2}{x^2+a^2} - \frac{b^2}{x^2+b^2} \right)$

Now, we integrate this expression:

$I = \int \frac{1}{a^2 - b^2} \left( \frac{a^2}{x^2+a^2} - \frac{b^2}{x^2+b^2} \right) \;dx$

$I = \frac{1}{a^2 - b^2} \left( a^2 \int \frac{dx}{x^2+a^2} - b^2 \int \frac{dx}{x^2+b^2} \right)$

Using the standard integral formula $\int \frac{dx}{x^2+c^2} = \frac{1}{c} \tan^{-1}\left(\frac{x}{c}\right) + C$, assuming $a>0$ and $b>0$:

$\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$

$\int \frac{dx}{x^2+b^2} = \frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right)$

Substitute these results back into the integral for $I$:

$I = \frac{1}{a^2 - b^2} \left( a^2 \cdot \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) - b^2 \cdot \frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right) \right) + C$

$I = \frac{1}{a^2 - b^2} \left( a \tan^{-1}\left(\frac{x}{a}\right) - b \tan^{-1}\left(\frac{x}{b}\right) \right) + C$

This result is valid provided that $a^2 \neq b^2$. If $a=0$ or $b=0$, the original integral is different. If $a^2=b^2 > 0$, the integral $\int \frac{x^2}{(x^2+a^2)^2} dx$ requires a different approach (e.g., trigonometric substitution).

Thus, the value of the indefinite integral is $\frac{1}{a^2 - b^2} \left( a \tan^{-1}\left(\frac{x}{a}\right) - b \tan^{-1}\left(\frac{x}{b}\right) \right) + C$, assuming $a^2 \neq b^2$, $a \neq 0$, $b \neq 0$.

Final Answer:

$\int \frac{x^2 \;dx}{(x^2+a^2) (x^2+b^2)} = \frac{1}{a^2 - b^2} \left( a \tan^{-1}\left(\frac{x}{a}\right) - b \tan^{-1}\left(\frac{x}{b}\right) \right) + C$, for $a^2 \neq b^2$

Let the given integral be $I = \int\limits_0^\pi \frac{x}{1 + \sin x} \;dx$.

We use the property of definite integrals: $\int\limits_0^a f(x) \;dx = \int\limits_0^a f(a-x) \;dx$.

Here, $a = \pi$ and $f(x) = \frac{x}{1 + \sin x}$.

So, $f(\pi - x) = \frac{\pi - x}{1 + \sin(\pi - x)}$.

Using the trigonometric identity $\sin(\pi - x) = \sin x$, we have:

Applying the property, we can write $I$ as:

Split the integrand:

$I = \int\limits_0^\pi \left(\frac{\pi}{1 + \sin x} - \frac{x}{1 + \sin x}\right) \;dx$

$I = \int\limits_0^\pi \frac{\pi}{1 + \sin x} \;dx - \int\limits_0^\pi \frac{x}{1 + \sin x} \;dx$

Notice that the second integral on the right side is the original integral $I$.

$I = \pi \int\limits_0^\pi \frac{1}{1 + \sin x} \;dx - I$

Adding $I$ to both sides, we get:

Let $J = \int\limits_0^\pi \frac{1}{1 + \sin x} \;dx$. We need to evaluate $J$.

We use the property $\int\limits_0^{2a} f(x) dx = 2\int\limits_0^a f(x) dx$ if $f(2a-x) = f(x)$.

Here, $2a = \pi$, so $a = \pi/2$. Let $g(x) = \frac{1}{1+\sin x}$.

$g(\pi - x) = \frac{1}{1+\sin(\pi - x)} = \frac{1}{1+\sin x} = g(x)$.

Since $g(\pi - x) = g(x)$, we can use the property:

Let's evaluate the integral $\int\limits_0^{\pi/2} \frac{1}{1 + \sin x} \;dx$. Multiply the numerator and denominator by $(1-\sin x)$:

$\int\limits_0^{\pi/2} \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} \;dx = \int\limits_0^{\pi/2} \frac{1 - \sin x}{1 - \sin^2 x} \;dx$

$= \int\limits_0^{\pi/2} \frac{1 - \sin x}{\cos^2 x} \;dx$

$= \int\limits_0^{\pi/2} \left(\frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x}\right) \;dx$

$= \int\limits_0^{\pi/2} (\sec^2 x - \tan x \sec x) \;dx$

$= [\tan x - \sec x]_0^{\pi/2}$

This is an improper integral at the upper limit $x = \frac{\pi}{2}$. We evaluate it using a limit:

$\lim\limits_{b \to \frac{\pi}{2}^-} [\tan x - \sec x]_0^b = \lim\limits_{b \to \frac{\pi}{2}^-} (\tan b - \sec b) - (\tan 0 - \sec 0)$

$= \lim\limits_{b \to \frac{\pi}{2}^-} \left(\frac{\sin b}{\cos b} - \frac{1}{\cos b}\right) - (0 - 1)$

$= \lim\limits_{b \to \frac{\pi}{2}^-} \left(\frac{\sin b - 1}{\cos b}\right) + 1$

The limit $\lim\limits_{b \to \frac{\pi}{2}^-} \left(\frac{\sin b - 1}{\cos b}\right)$ is of the form $\frac{0}{0}$. Apply L'Hopital's rule:

$\lim\limits_{b \to \frac{\pi}{2}^-} \frac{\frac{d}{db}(\sin b - 1)}{\frac{d}{db}(\cos b)} = \lim\limits_{b \to \frac{\pi}{2}^-} \frac{\cos b}{-\sin b} = \frac{\cos(\pi/2)}{-\sin(\pi/2)} = \frac{0}{-1} = 0$

So, the value of the integral from $0$ to $\frac{\pi}{2}$ is $0 + 1 = 1$.

Substitute the value from (v) into (iv):

Substitute the value of $J$ from (vi) into (iii):

Divide by 2:

Thus, the value of the definite integral is $\pi$.

Final Answer:

$\int\limits_0^π \frac{x}{1 + \sin x} \;dx = \pi$

We need to evaluate the indefinite integral $I = \int \frac{2x − 1}{(x−1)(x+2)(x−3)} \;dx$.

We use the method of partial fraction decomposition. We write the integrand as:

$\frac{2x − 1}{(x−1)(x+2)(x−3)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-3}$

Multiply both sides by $(x-1)(x+2)(x-3)$:

$2x - 1 = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2)$

To find the constants A, B, and C, substitute the roots of the factors:

Set $x = 1$:

Set $x = -2$:

Set $x = 3$:

Substitute the values of A, B, and C back into the partial fraction decomposition:

$\frac{2x − 1}{(x−1)(x+2)(x−3)} = \frac{-1/6}{x-1} + \frac{-1/3}{x+2} + \frac{1/2}{x-3}$

Now, integrate term by term:

$I = \int \left(\frac{-1/6}{x-1} + \frac{-1/3}{x+2} + \frac{1/2}{x-3}\right) \;dx$

$I = -\frac{1}{6} \int \frac{dx}{x-1} - \frac{1}{3} \int \frac{dx}{x+2} + \frac{1}{2} \int \frac{dx}{x-3}$

Using the standard integral formula $\int \frac{dx}{ax+b} = \frac{1}{a} \log|ax+b| + C$:

$\int \frac{dx}{x-1} = \log|x-1|$

$\int \frac{dx}{x+2} = \log|x+2|$

$\int \frac{dx}{x-3} = \log|x-3|$

Substitute these results back into the expression for $I$:

$I = -\frac{1}{6} \log|x-1| - \frac{1}{3} \log|x+2| + \frac{1}{2} \log|x-3| + C$

Thus, the value of the indefinite integral is $-\frac{1}{6} \log|x-1| - \frac{1}{3} \log|x+2| + \frac{1}{2} \log|x-3| + C$.

Final Answer:

$\int \frac{2x − 1}{(x−1)(x+2)(x−3)} \;dx = -\frac{1}{6} \log|x-1| - \frac{1}{3} \log|x+2| + \frac{1}{2} \log|x-3| + C$

We need to evaluate the indefinite integral $I = \int e^{\tan^{−1} x} \left( \frac{1+x+x^2}{1+x^2} \right) \;dx$.

First, let's rewrite the fraction within the integrand by splitting the numerator:

$\frac{1+x+x^2}{1+x^2} = \frac{(1+x^2) + x}{1+x^2} = \frac{1+x^2}{1+x^2} + \frac{x}{1+x^2} = 1 + \frac{x}{1+x^2}$

So, the integral becomes:

$I = \int e^{\tan^{−1} x} \left( 1 + \frac{x}{1+x^2} \right) \;dx$

Consider the derivative of the product of two functions, $u(x)v(x)$. The product rule states that $\frac{d}{dx}(u v) = u'v + uv'$.

We observe that the integrand contains the term $e^{\tan^{-1} x}$ and a sum of two terms, one of which is $\frac{x}{1+x^2}$. The derivative of $\tan^{-1} x$ is $\frac{1}{1+x^2}$. This suggests looking for an integrand of the form $e^{g(x)}(g'(x)f(x) + f'(x))$.

Let's consider the function $y = x e^{\tan^{-1} x}$. We compute its derivative using the product rule with $u = x$ and $v = e^{\tan^{-1} x}$.

$u' = \frac{d}{dx}(x) = 1$

$v' = \frac{d}{dx}(e^{\tan^{-1} x}) = e^{\tan^{-1} x} \cdot \frac{d}{dx}(\tan^{-1} x) = e^{\tan^{-1} x} \cdot \frac{1}{1+x^2}$

Now, apply the product rule to find the derivative of $y = x e^{\tan^{-1} x}$:

$\frac{dy}{dx} = u'v + uv'$

$\frac{dy}{dx} = 1 \cdot e^{\tan^{-1} x} + x \cdot \left(e^{\tan^{-1} x} \cdot \frac{1}{1+x^2}\right)$

$\frac{dy}{dx} = e^{\tan^{-1} x} + \frac{x e^{\tan^{-1} x}}{1+x^2}$

Factor out $e^{\tan^{-1} x}$:

$\frac{dy}{dx} = e^{\tan^{-1} x} \left( 1 + \frac{x}{1+x^2} \right)$

Comparing this derivative with the integrand of our integral $I$, we see that they are identical:

Integrand $= e^{\tan^{−1} x} \left( 1 + \frac{x}{1+x^2} \right)$

Derivative of $x e^{\tan^{-1} x}$ $= e^{\tan^{−1} x} \left( 1 + \frac{x}{1+x^2} \right)$

Since the integrand is the derivative of $x e^{\tan^{-1} x}$, the integral is simply $x e^{\tan^{-1} x}$ plus the constant of integration.

Thus, the value of the indefinite integral is $x e^{\tan^{-1} x} + C$.

Final Answer:

$\int e^{tan^{−1} x} \left( \frac{1+x+x^2}{1+x^2} \right) \;dx = x e^{\tan^{-1} x} + C$

We need to evaluate the indefinite integral $I = \int \sin^{−1} \sqrt{\frac{x}{a+x}} \;dx$.

Following the hint, we use the substitution $x = a \tan^2 \theta$. Assume $a > 0$ and $x \ge 0$, which implies $\theta \in [0, \pi/2)$.

Differentiating $x = a \tan^2 \theta$ with respect to $\theta$:

$dx = \frac{d}{d\theta}(a \tan^2 \theta) \;d\theta = a \cdot 2 \tan \theta \cdot \sec^2 \theta \;d\theta$

$dx = 2a \tan \theta \sec^2 \theta \;d\theta$

Now, simplify the term inside the square root: $\sqrt{\frac{x}{a+x}}$

$\frac{x}{a+x} = \frac{a \tan^2 \theta}{a + a \tan^2 \theta} = \frac{a \tan^2 \theta}{a(1 + \tan^2 \theta)}$

Using the identity $1 + \tan^2 \theta = \sec^2 \theta$:

$\frac{x}{a+x} = \frac{\tan^2 \theta}{\sec^2 \theta} = \frac{\sin^2 \theta / \cos^2 \theta}{1 / \cos^2 \theta} = \sin^2 \theta$

So, $\sqrt{\frac{x}{a+x}} = \sqrt{\sin^2 \theta}$. Since $\theta \in [0, \pi/2)$, $\sin \theta \ge 0$, so $\sqrt{\sin^2 \theta} = \sin \theta$.

The term $\sin^{-1} \sqrt{\frac{x}{a+x}}$ becomes $\sin^{-1}(\sin \theta)$. For $\theta \in [0, \pi/2)$, $\sin^{-1}(\sin \theta) = \theta$.

So, $\sin^{-1} \sqrt{\frac{x}{a+x}} = \theta$.

Substitute $\sin^{-1} \sqrt{\frac{x}{a+x}} = \theta$ and $dx = 2a \tan \theta \sec^2 \theta \;d\theta$ into the integral:

$I = \int \theta \cdot (2a \tan \theta \sec^2 \theta) \;d\theta$

$I = 2a \int \theta \tan \theta \sec^2 \theta \;d\theta$

We evaluate this integral using integration by parts, $\int u \;dv = uv - \int v \;du$.

Let $u = \theta$ and $dv = \tan \theta \sec^2 \theta \;d\theta$.

Then $du = d\theta$.

To find $v$, we integrate $dv$: $v = \int \tan \theta \sec^2 \theta \;d\theta$. Let $w = \tan \theta$, then $dw = \sec^2 \theta \;d\theta$. $\int w \;dw = \frac{w^2}{2}$.

So, $v = \frac{\tan^2 \theta}{2}$.

Apply the integration by parts formula:

$I = 2a \left[ \theta \cdot \frac{\tan^2 \theta}{2} - \int \frac{\tan^2 \theta}{2} \;d\theta \right]$

$I = a \theta \tan^2 \theta - a \int \tan^2 \theta \;d\theta$

Evaluate the integral of $\tan^2 \theta$ using the identity $\tan^2 \theta = \sec^2 \theta - 1$:

$\int \tan^2 \theta \;d\theta = \int (\sec^2 \theta - 1) \;d\theta = \int \sec^2 \theta \;d\theta - \int 1 \;d\theta = \tan \theta - \theta + C'$

Substitute this result back into the expression for $I$:

$I = a \theta \tan^2 \theta - a (\tan \theta - \theta) + C$

$I = a \theta \tan^2 \theta - a \tan \theta + a \theta + C$

Now, substitute back to express the result in terms of $x$.

From $x = a \tan^2 \theta$, we get $\tan^2 \theta = \frac{x}{a}$.

From $\tan^2 \theta = \frac{x}{a}$ and $\theta \in [0, \pi/2)$, $\tan \theta = \sqrt{\frac{x}{a}}$.

From $\sin^{-1} \sqrt{\frac{x}{a+x}} = \theta$, we have $\theta = \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right)$.

Substitute these back into the expression for $I$:

$I = a \left(\sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right)\right) \left(\frac{x}{a}\right) - a \sqrt{\frac{x}{a}} + a \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) + C$

$I = x \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) - a \frac{\sqrt{x}}{\sqrt{a}} + a \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) + C$

$I = x \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) - \sqrt{ax} + a \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) + C$

$I = (x+a) \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) - \sqrt{ax} + C$

Thus, the value of the indefinite integral is $(x+a) \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) - \sqrt{ax} + C$.

Final Answer:

$\int \sin^{−1} \sqrt{\frac{x}{a+x}} \;dx = (x+a) \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) - \sqrt{ax} + C$

We need to evaluate the definite integral $I = \int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1 + \cos x}}{(1 − \cos x)^{\frac{5}{2}}} \;dx$.

We use the half-angle identities: $1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right)$ and $1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)$.

Substitute these into the integrand:

$\sqrt{1 + \cos x} = \sqrt{2 \cos^2\left(\frac{x}{2}\right)} = \sqrt{2} \left|\cos\left(\frac{x}{2}\right)\right|$

$(1 - \cos x)^{\frac{5}{2}} = (2 \sin^2\left(\frac{x}{2}\right))^{\frac{5}{2}} = 2^{\frac{5}{2}} (\sin^2\left(\frac{x}{2}\right))^{\frac{5}{2}} = 2^{\frac{5}{2}} |\sin\left(\frac{x}{2}\right)|^5$

The limits of integration are from $x = \frac{\pi}{3}$ to $x = \frac{\pi}{2}$.

The corresponding range for $\frac{x}{2}$ is from $\frac{(\pi/3)}{2} = \frac{\pi}{6}$ to $\frac{(\pi/2)}{2} = \frac{\pi}{4}$.

In the interval $\left[\frac{\pi}{6}, \frac{\pi}{4}\right]$, both $\cos\left(\frac{x}{2}\right)$ and $\sin\left(\frac{x}{2}\right)$ are positive.

So, $\left|\cos\left(\frac{x}{2}\right)\right| = \cos\left(\frac{x}{2}\right)$ and $\left|\sin\left(\frac{x}{2}\right)\right| = \sin\left(\frac{x}{2}\right)$.

The integrand becomes:

$\frac{\sqrt{2} \cos\left(\frac{x}{2}\right)}{2^{\frac{5}{2}} \sin^5\left(\frac{x}{2}\right)} = \frac{2^{1/2}}{2^{5/2}} \frac{\cos\left(\frac{x}{2}\right)}{\sin^5\left(\frac{x}{2}\right)} = 2^{1/2 - 5/2} \frac{\cos\left(\frac{x}{2}\right)}{\sin^5\left(\frac{x}{2}\right)}$

$= 2^{-4/2} \frac{\cos\left(\frac{x}{2}\right)}{\sin^5\left(\frac{x}{2}\right)} = 2^{-2} \sin^{-5}\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) = \frac{1}{4} \sin^{-5}\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)$

The integral is $I = \int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{4} \sin^{-5}\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \;dx$.

Use the substitution method.

Let $u = \sin\left(\frac{x}{2}\right)$.

Then $du = \frac{d}{dx}\left(\sin\left(\frac{x}{2}\right)\right) \;dx = \cos\left(\frac{x}{2}\right) \cdot \frac{1}{2} \;dx$.

Change the limits of integration according to the substitution $u = \sin\left(\frac{x}{2}\right)$:

When $x = \frac{\pi}{3}$, $u = \sin\left(\frac{\pi/3}{2}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

When $x = \frac{\pi}{2}$, $u = \sin\left(\frac{\pi/2}{2}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.

Substitute $u$ and $du$ into the integral, and change the limits:

$I = \frac{1}{4} \int\limits_{\frac{1}{2}}^{\frac{\sqrt{2}}{2}} u^{-5} (2 \;du)$

$I = \frac{2}{4} \int\limits_{\frac{1}{2}}^{\frac{\sqrt{2}}{2}} u^{-5} \;du = \frac{1}{2} \int\limits_{\frac{1}{2}}^{\frac{\sqrt{2}}{2}} u^{-5} \;du$

Evaluate the integral using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$ (for $n \neq -1$):

$I = \frac{1}{2} \left[\frac{u^{-5+1}}{-5+1}\right]_{\frac{1}{2}}^{\frac{\sqrt{2}}{2}} = \frac{1}{2} \left[\frac{u^{-4}}{-4}\right]_{\frac{1}{2}}^{\frac{\sqrt{2}}{2}}$

$I = -\frac{1}{8} \left[\frac{1}{u^4}\right]_{\frac{1}{2}}^{\frac{\sqrt{2}}{2}}$

Apply the limits of integration:

$I = -\frac{1}{8} \left(\frac{1}{(\frac{\sqrt{2}}{2})^4} - \frac{1}{(\frac{1}{2})^4}\right)$

Calculate the terms in the parenthesis:

$(\frac{\sqrt{2}}{2})^4 = \left(\frac{(\sqrt{2})^2}{2^2}\right)^2 = \left(\frac{2}{4}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$. So, $\frac{1}{(\frac{\sqrt{2}}{2})^4} = \frac{1}{1/4} = 4$.

$(\frac{1}{2})^4 = \frac{1^4}{2^4} = \frac{1}{16}$. So, $\frac{1}{(\frac{1}{2})^4} = \frac{1}{1/16} = 16$.

Substitute these values back:

$I = -\frac{1}{8} (4 - 16)$

$I = -\frac{1}{8} (-12)$

$I = \frac{12}{8}$

Simplify the fraction:

$I = \frac{\cancel{12}^{3}}{\cancel{8}_{2}}$

$I = \frac{3}{2}$

Thus, the value of the definite integral is $\frac{3}{2}$.

Final Answer:

$\int\limits_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 + \cos x}}{(1 − \cos x)^{\frac{5}{2}}} \;dx = \frac{3}{2}$

We need to evaluate the indefinite integral $I = \int e^{−3x} \cos^3 x \;dx$.

We use the trigonometric identity for $\cos^3 x$. The triple angle formula for cosine is $\cos(3x) = 4\cos^3 x - 3\cos x$.

Rearranging this, we get $\cos^3 x = \frac{1}{4}(\cos(3x) + 3\cos x)$.

Substitute this identity into the integral:

$I = \int e^{−3x} \cdot \frac{1}{4}(\cos(3x) + 3\cos x) \;dx$

$I = \frac{1}{4} \int e^{−3x} (\cos(3x) + 3\cos x) \;dx$

$I = \frac{1}{4} \int e^{−3x} \cos(3x) \;dx + \frac{3}{4} \int e^{−3x} \cos x \;dx$

We use the standard integral formula $\int e^{ax} \cos(bx) \;dx = \frac{e^{ax}}{a^2+b^2}(a \cos(bx) + b \sin(bx)) + C$.

For the first integral, $\int e^{−3x} \cos(3x) \;dx$, we have $a = -3$ and $b = 3$.

$\int e^{−3x} \cos(3x) \;dx = \frac{e^{-3x}}{(-3)^2+3^2}(-3 \cos(3x) + 3 \sin(3x))$

$= \frac{e^{-3x}}{9+9}(-3 \cos(3x) + 3 \sin(3x))$

$= \frac{e^{-3x}}{18}(-3 \cos(3x) + 3 \sin(3x))$

$= \frac{3e^{-3x}}{18}(-\cos(3x) + \sin(3x))$

For the second integral, $\int e^{−3x} \cos x \;dx$, we have $a = -3$ and $b = 1$.

$\int e^{−3x} \cos x \;dx = \frac{e^{-3x}}{(-3)^2+1^2}(-3 \cos x + 1 \sin x)$

$= \frac{e^{-3x}}{9+1}(-3 \cos x + \sin x)$

Substitute the results from (1) and (2) back into the expression for $I$:

$I = \frac{1}{4} \left[\frac{e^{-3x}}{6}(\sin(3x) - \cos(3x))\right] + \frac{3}{4} \left[\frac{e^{-3x}}{10}(\sin x - 3\cos x)\right] + C$

$I = \frac{e^{-3x}}{24}(\sin(3x) - \cos(3x)) + \frac{3e^{-3x}}{40}(\sin x - 3\cos x) + C$

Factor out $e^{-3x}$ and find a common denominator (LCM of 24 and 40 is 120):

$I = e^{-3x} \left[\frac{1}{24}(\sin(3x) - \cos(3x)) + \frac{3}{40}(\sin x - 3\cos x)\right] + C$

$I = e^{-3x} \left[\frac{5}{120}(\sin(3x) - \cos(3x)) + \frac{9}{120}(\sin x - 3\cos x)\right] + C$

$I = \frac{e^{-3x}}{120} [5\sin(3x) - 5\cos(3x) + 9\sin x - 27\cos x] + C$

Thus, the value of the indefinite integral is $\frac{e^{-3x}}{120} (5\sin(3x) - 5\cos(3x) + 9\sin x - 27\cos x) + C$.

Final Answer:

$\int e^{−3x} \cos^3 x \;dx = \frac{e^{-3x}}{120} (5\sin(3x) - 5\cos(3x) + 9\sin x - 27\cos x) + C$

We need to evaluate the indefinite integral $I = \int \sqrt{\tan x} \;dx$.

Following the hint, we use the substitution $\tan x = t^2$. Assume $\tan x \ge 0$, so $t = \sqrt{\tan x}$.

From $\tan x = t^2$, we have $x = \tan^{-1}(t^2)$.

Differentiating $x$ with respect to $t$:

$dx = \frac{d}{dt}(\tan^{-1}(t^2)) \;dt = \frac{1}{1+(t^2)^2} \cdot \frac{d}{dt}(t^2) \;dt = \frac{1}{1+t^4} \cdot 2t \;dt$

$dx = \frac{2t}{1+t^4} \;dt$

Substitute $\sqrt{\tan x} = t$ and $dx = \frac{2t}{1+t^4} \;dt$ into the integral:

$I = \int t \cdot \frac{2t}{1+t^4} \;dt = \int \frac{2t^2}{1+t^4} \;dt$

To evaluate $\int \frac{2t^2}{1+t^4} \;dt$, we can rewrite the integrand. Divide the numerator and denominator by $t^2$ (assuming $t \neq 0$):

$\frac{2t^2}{t^4+1} = \frac{2}{t^2 + \frac{1}{t^2}}$

We can split the numerator $2$ as $(1 + \frac{1}{t^2}) + (1 - \frac{1}{t^2})$:

$\frac{2t^2}{t^4+1} = \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} + \frac{1 - \frac{1}{t^2}}{t^2 + \frac{1}{t^2}}$

So, $I = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \;dt + \int \frac{1 - \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \;dt = I_1 + I_2$

Consider $I_1 = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \;dt$.

The denominator can be written as $t^2 + \frac{1}{t^2} = (t - \frac{1}{t})^2 + 2$.

Let $u = t - \frac{1}{t}$. Then $du = (1 + \frac{1}{t^2}) \;dt$.

$I_1 = \int \frac{du}{u^2 + (\sqrt{2})^2}$

This is a standard integral of the form $\int \frac{du}{u^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right)$.

$I_1 = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t - \frac{1}{t}}{\sqrt{2}}\right) + C_1 = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t^2 - 1}{\sqrt{2}t}\right) + C_1$

Consider $I_2 = \int \frac{1 - \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \;dt$.

The denominator can be written as $t^2 + \frac{1}{t^2} = (t + \frac{1}{t})^2 - 2$.

Let $v = t + \frac{1}{t}$. Then $dv = (1 - \frac{1}{t^2}) \;dt$.

$I_2 = \int \frac{dv}{v^2 - (\sqrt{2})^2}$

This is a standard integral of the form $\int \frac{dv}{v^2-a^2} = \frac{1}{2a} \log\left|\frac{v-a}{v+a}\right|$.

$I_2 = \frac{1}{2\sqrt{2}} \log\left|\frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}}\right| + C_2 = \frac{1}{2\sqrt{2}} \log\left|\frac{t^2 - \sqrt{2}t + 1}{t^2 + \sqrt{2}t + 1}\right| + C_2$

Combine the results for $I_1$ and $I_2$:

$I = I_1 + I_2 = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t^2 - 1}{\sqrt{2}t}\right) + \frac{1}{2\sqrt{2}} \log\left|\frac{t^2 - \sqrt{2}t + 1}{t^2 + \sqrt{2}t + 1}\right| + C$

Substitute back $t = \sqrt{\tan x}$. Then $t^2 = \tan x$.

$I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2}\sqrt{\tan x}}\right) + \frac{1}{2\sqrt{2}} \log\left|\frac{\tan x - \sqrt{2}\sqrt{\tan x} + 1}{\tan x + \sqrt{2}\sqrt{\tan x} + 1}\right| + C$

Rationalize the denominators of the coefficients:

$I = \frac{\sqrt{2}}{2} \tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2\tan x}}\right) + \frac{\sqrt{2}}{4} \log\left|\frac{\tan x - \sqrt{2\tan x} + 1}{\tan x + \sqrt{2\tan x} + 1}\right| + C$

Thus, the value of the indefinite integral is $\frac{\sqrt{2}}{2} \tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2\tan x}}\right) + \frac{\sqrt{2}}{4} \log\left|\frac{\tan x - \sqrt{2\tan x} + 1}{\tan x + \sqrt{2\tan x} + 1}\right| + C$.

Final Answer:

$\int \sqrt{\tan x} \;dx = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2\tan x}}\right) + \frac{1}{2\sqrt{2}} \log\left|\frac{\tan x - \sqrt{2\tan x} + 1}{\tan x + \sqrt{2\tan x} + 1}\right| + C$

We need to evaluate the definite integral $I = \int\limits_0^{\frac{\pi}{2}} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x)^2}$.

Following the hint, we divide the numerator and denominator by $\cos^4 x$:

$I = \int\limits_0^{\frac{\pi}{2}} \frac{\frac{1}{\cos^4 x}}{\frac{(a^2 \cos^2 x + b^2 \sin^2 x)^2}{\cos^4 x}} \;dx$

$I = \int\limits_0^{\frac{\pi}{2}} \frac{\sec^4 x}{\left(\frac{a^2 \cos^2 x}{\cos^2 x} + \frac{b^2 \sin^2 x}{\cos^2 x}\right)^2} \;dx$

$I = \int\limits_0^{\frac{\pi}{2}} \frac{\sec^4 x}{(a^2 + b^2 \tan^2 x)^2} \;dx$

We rewrite $\sec^4 x$ as $\sec^2 x \cdot \sec^2 x = (1+\tan^2 x) \sec^2 x$:

$I = \int\limits_0^{\frac{\pi}{2}} \frac{(1+\tan^2 x) \sec^2 x}{(a^2 + b^2 \tan^2 x)^2} \;dx$

Now, we use the substitution method.

Let $u = \tan x$.

Differentiating with respect to $x$, we get $du = \sec^2 x \;dx$.

We need to change the limits of integration:

When $x = 0$, $u = \tan(0) = 0$.

When $x = \frac{\pi}{2}$, $u = \tan\left(\frac{\pi}{2}\right)$, which approaches $\infty$.

The integral becomes an improper integral with limits from $0$ to $\infty$:

$I = \int\limits_0^{\infty} \frac{1+u^2}{(a^2 + b^2 u^2)^2} \;du$

We use partial fraction decomposition for the integrand $\frac{1+u^2}{(a^2 + b^2 u^2)^2}$. We assume $a \neq 0$ and $b \neq 0$.

Let $y = u^2$. The expression is $\frac{1+y}{(b^2 y + a^2)^2}$. We decompose this as:

$\frac{1+y}{(b^2 y + a^2)^2} = \frac{A}{b^2 y + a^2} + \frac{B}{(b^2 y + a^2)^2}$

$1+y = A(b^2 y + a^2) + B$

$1+y = Ab^2 y + Aa^2 + B$

Comparing coefficients of $y$ and constant terms:

$Ab^2 = 1 \implies A = \frac{1}{b^2}$

$Aa^2 + B = 1 \implies \frac{1}{b^2} a^2 + B = 1 \implies B = 1 - \frac{a^2}{b^2} = \frac{b^2-a^2}{b^2}$

Substituting back $y=u^2$, the decomposition is:

$\frac{1+u^2}{(b^2 u^2 + a^2)^2} = \frac{1/b^2}{b^2 u^2 + a^2} + \frac{(b^2-a^2)/b^2}{(b^2 u^2 + a^2)^2}$

$= \frac{1}{b^2(b^2 u^2 + a^2)} + \frac{b^2-a^2}{b^2(b^2 u^2 + a^2)^2}$

The integral becomes:

$I = \int\limits_0^{\infty} \left( \frac{1}{b^2(b^2 u^2 + a^2)} + \frac{b^2-a^2}{b^2(b^2 u^2 + a^2)^2} \right) \;du$

$I = \frac{1}{b^2} \int\limits_0^{\infty} \frac{du}{b^2 u^2 + a^2} + \frac{b^2-a^2}{b^2} \int\limits_0^{\infty} \frac{du}{(b^2 u^2 + a^2)^2}$

We evaluate the integrals $\int \frac{du}{c^2 u^2 + k^2}$ and $\int \frac{du}{(c^2 u^2 + k^2)^2}$.

$\int \frac{du}{c^2 u^2 + k^2} = \int \frac{du}{c^2(u^2 + (k/c)^2)} = \frac{1}{c^2} \int \frac{du}{u^2 + (k/c)^2} = \frac{1}{c^2} \cdot \frac{1}{k/c} \tan^{-1}\left(\frac{u}{k/c}\right) = \frac{1}{c k} \tan^{-1}\left(\frac{cu}{k}\right)$.

For the first integral term, $c=b$, $k=a$. $\int \frac{du}{b^2 u^2 + a^2} = \frac{1}{ba} \tan^{-1}\left(\frac{bu}{a}\right)$.

$\int\limits_0^{\infty} \frac{du}{b^2 u^2 + a^2} = \left[\frac{1}{ab} \tan^{-1}\left(\frac{bu}{a}\right)\right]_0^{\infty} = \frac{1}{ab} \tan^{-1}(\infty) - \frac{1}{ab} \tan^{-1}(0) = \frac{1}{ab} \cdot \frac{\pi}{2} - 0 = \frac{\pi}{2ab}$.

For the second integral term, $\int \frac{du}{(b^2 u^2 + a^2)^2} = \frac{1}{b^4} \int \frac{du}{(u^2 + (a/b)^2)^2}$. Let $k = a/b$. We use the standard result $\int_0^{\infty} \frac{du}{(u^2+k^2)^2} = \frac{\pi}{4k^3}$.

Substitute $k = a/b$: $\int\limits_0^{\infty} \frac{du}{(u^2 + (a/b)^2)^2} = \frac{\pi}{4(a/b)^3} = \frac{\pi b^3}{4a^3}$.

So, $\int\limits_0^{\infty} \frac{du}{(b^2 u^2 + a^2)^2} = \frac{1}{b^4} \cdot \frac{\pi b^3}{4a^3} = \frac{\pi}{4a^3 b}$.

Substitute these results back into the expression for $I$:

$I = \frac{1}{b^2} \left(\frac{\pi}{2ab}\right) + \frac{b^2-a^2}{b^2} \left(\frac{\pi}{4a^3 b}\right)$

$I = \frac{\pi}{2ab^3} + \frac{(b^2-a^2)\pi}{4a^3 b^3}$

$I = \frac{\pi (2a^2)}{4a^3 b^3} + \frac{(b^2-a^2)\pi}{4a^3 b^3}$

$I = \frac{2\pi a^2 + \pi b^2 - \pi a^2}{4a^3 b^3}$

$I = \frac{\pi a^2 + \pi b^2}{4a^3 b^3}$

$I = \frac{\pi(a^2 + b^2)}{4a^3 b^3}$

This result is valid for $a \neq 0$ and $b \neq 0$.

Thus, the value of the definite integral is $\frac{\pi(a^2+b^2)}{4a^3 b^3}$.

Final Answer:

$\int\limits_0^{\frac{π}{2}} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x)^2} = \frac{\pi(a^2+b^2)}{4a^3 b^3}$, assuming $a \neq 0, b \neq 0$.

We need to evaluate the definite integral:

$\int\limits_0^1 x \log (1+2x) \;dx$

We will solve this integral using the method of Integration by Parts.

The integration by parts formula for definite integrals is:

$\int\limits_a^b u \; dv = \left[ uv \right]_a^b - \int\limits_a^b v \; du$

Let's choose our parts for the integral $\int\limits_0^1 x \log (1+2x) \;dx$:

$u = \log(1+2x)$

$dv = x \;dx$

Now, we find $du$ by differentiating $u$ and $v$ by integrating $dv$:

$du = \frac{d}{dx}(\log(1+2x)) \;dx = \frac{1}{1+2x} \cdot \frac{d}{dx}(1+2x) \;dx = \frac{1}{1+2x} \cdot 2 \;dx = \frac{2}{1+2x} \;dx$

$v = \int x \;dx = \frac{x^2}{2}$

Applying the integration by parts formula with the limits $a=0$ and $b=1$:

$\int\limits_0^1 x \log (1+2x) \;dx = \left[ \frac{x^2}{2} \log(1+2x) \right]_0^1 - \int\limits_0^1 \frac{x^2}{2} \cdot \frac{2}{1+2x} \;dx$

$= \left[ \frac{x^2}{2} \log(1+2x) \right]_0^1 - \int\limits_0^1 \frac{x^2}{1+2x} \;dx$

First, evaluate the boundary term $\left[ \frac{x^2}{2} \log(1+2x) \right]_0^1$:

Substitute the upper limit $x=1$:

$\frac{1^2}{2} \log(1+2(1)) = \frac{1}{2} \log(3)$

Substitute the lower limit $x=0$:

$\frac{0^2}{2} \log(1+2(0)) = 0 \cdot \log(1) = 0$

So, $\left[ \frac{x^2}{2} \log(1+2x) \right]_0^1 = \frac{1}{2} \log(3) - 0 = \frac{1}{2} \log(3)$

Next, we need to evaluate the remaining integral $\int\limits_0^1 \frac{x^2}{1+2x} \;dx$

The integrand $\frac{x^2}{1+2x}$ is a rational function where the degree of the numerator is greater than or equal to the degree of the denominator. We perform polynomial long division or algebraic manipulation to simplify it.

We can write the integrand as Quotient + $\frac{\text{Remainder}}{2x+1}$.

Divide $x^2$ by $2x+1$:

$x^2 = \frac{1}{2}x(2x+1) - \frac{x}{2}$

So, $\frac{x^2}{2x+1} = \frac{\frac{1}{2}x(2x+1) - \frac{x}{2}}{2x+1} = \frac{1}{2}x - \frac{x}{2(2x+1)}$

Now consider the term $\frac{x}{2x+1}$:

$\frac{x}{2x+1} = \frac{\frac{1}{2}(2x)}{2x+1} = \frac{\frac{1}{2}(2x+1 - 1)}{2x+1} = \frac{1}{2} - \frac{1}{2(2x+1)}$

Substitute this back into the expression for $\frac{x^2}{2x+1}$:

$\frac{x^2}{2x+1} = \frac{1}{2}x - \frac{1}{2} \left( \frac{1}{2} - \frac{1}{2(2x+1)} \right)$

$= \frac{1}{2}x - \frac{1}{4} + \frac{1}{4(2x+1)}$

Now, integrate the simplified rational function from $0$ to $1$:

$\int\limits_0^1 \left( \frac{1}{2}x - \frac{1}{4} + \frac{1}{4(2x+1)} \right) \;dx$

$= \left[ \frac{1}{2} \cdot \frac{x^2}{2} - \frac{1}{4}x + \frac{1}{4} \cdot \frac{1}{2} \log|2x+1| \right]_0^1$

$= \left[ \frac{x^2}{4} - \frac{x}{4} + \frac{1}{8} \log|2x+1| \right]_0^1$

Evaluate the definite integral using the limits:

Substitute the upper limit $x=1$:

$\frac{1^2}{4} - \frac{1}{4} + \frac{1}{8} \log|2(1)+1| = \frac{1}{4} - \frac{1}{4} + \frac{1}{8} \log(3) = \frac{1}{8} \log(3)$

Substitute the lower limit $x=0$:

$\frac{0^2}{4} - \frac{0}{4} + \frac{1}{8} \log|2(0)+1| = 0 - 0 + \frac{1}{8} \log(1) = 0$

So, $\int\limits_0^1 \frac{x^2}{1+2x} \;dx = \frac{1}{8} \log(3) - 0 = \frac{1}{8} \log(3)$

Finally, combine the results from the integration by parts formula:

$\int\limits_0^1 x \log (1+2x) \;dx = \left[ \frac{x^2}{2} \log(1+2x) \right]_0^1 - \int\limits_0^1 \frac{x^2}{1+2x} \;dx$

$= \frac{1}{2} \log(3) - \frac{1}{8} \log(3)$

$= \left( \frac{4}{8} - \frac{1}{8} \right) \log(3)$

$= \frac{3}{8} \log(3)$

The final answer is $\frac{3}{8} \log(3)$.

Let the given integral be $I$.

$I = \int\limits_0^π x \log \sin x \;dx$

We use the property of definite integrals: $\int\limits_0^a f(x) \;dx = \int\limits_0^a f(a-x) \;dx$.

Here, $a=π$ and $f(x) = x \log \sin x$.

$I = \int\limits_0^π (π-x) \log \sin(π-x) \;dx$

Since $\sin(π-x) = \sin x$, we have:

$I = \int\limits_0^π (π-x) \log \sin x \;dx$

$I = \int\limits_0^π (π \log \sin x - x \log \sin x) \;dx$

$I = \int\limits_0^π π \log \sin x \;dx - \int\limits_0^π x \log \sin x \;dx$

$I = π \int\limits_0^π \log \sin x \;dx - I$

Adding $I$ to both sides:

$2I = π \int\limits_0^π \log \sin x \;dx$

$I = \frac{π}{2} \int\limits_0^π \log \sin x \;dx$

Now we need to evaluate the integral $J = \int\limits_0^π \log \sin x \;dx$.

We use another property: $\int\limits_0^{2a} f(x) \;dx = 2 \int\limits_0^a f(x) \;dx$ if $f(2a-x) = f(x)$.

Here, $2a=π$, so $a=π/2$. Let $f(x) = \log \sin x$.

$f(π-x) = \log \sin(π-x) = \log \sin x = f(x)$.

So, $J = \int\limits_0^π \log \sin x \;dx = 2 \int\limits_0^{π/2} \log \sin x \;dx$

Substituting this back into the expression for $I$:

$I = \frac{π}{2} \cdot \left( 2 \int\limits_0^{π/2} \log \sin x \;dx \right) = π \int\limits_0^{π/2} \log \sin x \;dx$

The integral $\int\limits_0^{π/2} \log \sin x \;dx$ is a standard integral with the value $-\frac{π}{2} \log 2$. Let's show its derivation.

Let $K = \int\limits_0^{π/2} \log \sin x \;dx$.

Using the property $\int\limits_0^a f(x) \;dx = \int\limits_0^a f(a-x) \;dx$ with $a=π/2$:

$K = \int\limits_0^{π/2} \log \sin(π/2 - x) \;dx = \int\limits_0^{π/2} \log \cos x \;dx$

Adding the two expressions for $K$:

$2K = \int\limits_0^{π/2} \log \sin x \;dx + \int\limits_0^{π/2} \log \cos x \;dx$

$2K = \int\limits_0^{π/2} (\log \sin x + \log \cos x) \;dx$

Using the logarithm property $\log A + \log B = \log (AB)$:

$2K = \int\limits_0^{π/2} \log (\sin x \cos x) \;dx$

Multiply and divide the argument by 2:

$2K = \int\limits_0^{π/2} \log \left( \frac{2 \sin x \cos x}{2} \right) \;dx$

$2K = \int\limits_0^{π/2} \log \left( \frac{\sin 2x}{2} \right) \;dx$

Using the logarithm property $\log (A/B) = \log A - \log B$:

$2K = \int\limits_0^{π/2} (\log \sin 2x - \log 2) \;dx$

$2K = \int\limits_0^{π/2} \log \sin 2x \;dx - \int\limits_0^{π/2} \log 2 \;dx$

$2K = \int\limits_0^{π/2} \log \sin 2x \;dx - [\log 2 \cdot x]_0^{π/2}$

$2K = \int\limits_0^{π/2} \log \sin 2x \;dx - \left( \log 2 \cdot \frac{π}{2} - \log 2 \cdot 0 \right)$

$2K = \int\limits_0^{π/2} \log \sin 2x \;dx - \frac{π}{2} \log 2$

Consider the integral $\int\limits_0^{π/2} \log \sin 2x \;dx$. Let $t = 2x$. Then $dt = 2 \;dx$, so $dx = \frac{1}{2} dt$. The limits of integration change:

When $x=0$, $t = 2(0) = 0$.

When $x=π/2$, $t = 2(π/2) = π$.

So, $\int\limits_0^{π/2} \log \sin 2x \;dx = \int\limits_0^π \log \sin t \cdot \frac{1}{2} dt = \frac{1}{2} \int\limits_0^π \log \sin t \;dt$

Since $\int\limits_0^π \log \sin t \;dt$ is the same as $\int\limits_0^π \log \sin x \;dx$, which we denoted as $J$, this integral is $\frac{1}{2} J = \frac{1}{2} (2K) = K$.

Substitute this back into the equation for $2K$:

$2K = K - \frac{π}{2} \log 2$

$K = -\frac{π}{2} \log 2$

Thus, $\int\limits_0^{π/2} \log \sin x \;dx = -\frac{π}{2} \log 2$.

Now, substitute this value back into the expression for $I$:

$I = π \int\limits_0^{π/2} \log \sin x \;dx = π \left( -\frac{π}{2} \log 2 \right)$

$I = -\frac{π^2}{2} \log 2$

The final answer is $\mathbf{-\frac{π^2}{2} \log 2}$.

Let the given integral be $I$.

$I = \int\limits_{−\frac{π}{4}}^{\frac{π}{4}} \log (\sin x + \cos x) \;dx$

We use the property of definite integrals for symmetric limits: $\int\limits_{-a}^a f(x) \;dx = \int\limits_0^a [f(x) + f(-x)] \;dx$.

Here, $a = \frac{π}{4}$ and $f(x) = \log (\sin x + \cos x)$.

Let's find $f(-x)$:

$f(-x) = \log (\sin(-x) + \cos(-x))$

Using the properties $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$:

$f(-x) = \log (-\sin x + \cos x) = \log (\cos x - \sin x)$

Now, let's find $f(x) + f(-x)$:

$f(x) + f(-x) = \log (\sin x + \cos x) + \log (\cos x - \sin x)$

Using the logarithm property $\log A + \log B = \log (AB)$:

$f(x) + f(-x) = \log [(\sin x + \cos x)(\cos x - \sin x)]$

$f(x) + f(-x) = \log [(\cos x + \sin x)(\cos x - \sin x)]$

Using the algebraic identity $(A+B)(A-B) = A^2 - B^2$ with $A=\cos x$ and $B=\sin x$:

$f(x) + f(-x) = \log (\cos^2 x - \sin^2 x)$

Using the trigonometric identity $\cos^2 x - \sin^2 x = \cos 2x$:

$f(x) + f(-x) = \log (\cos 2x)$

Now, apply the definite integral property:

$I = \int\limits_{−\frac{π}{4}}^{\frac{π}{4}} f(x) \;dx = \int\limits_0^{\frac{π}{4}} [f(x) + f(-x)] \;dx$

$I = \int\limits_0^{\frac{π}{4}} \log (\cos 2x) \;dx$

To evaluate this integral, we use a substitution. Let $t = 2x$.

Then $dt = 2 \;dx$, which means $dx = \frac{1}{2} dt$.

Change the limits of integration:

When $x=0$, $t = 2(0) = 0$.

When $x=\frac{π}{4}$, $t = 2(\frac{π}{4}) = \frac{π}{2}$.

So the integral becomes:

$I = \int\limits_0^{\frac{π}{2}} \log (\cos t) \cdot \frac{1}{2} dt$

$I = \frac{1}{2} \int\limits_0^{\frac{π}{2}} \log (\cos t) \;dt$

Since the variable of integration is a dummy variable, we can write $\int\limits_0^{\frac{π}{2}} \log (\cos t) \;dt = \int\limits_0^{\frac{π}{2}} \log (\cos x) \;dx$.

$I = \frac{1}{2} \int\limits_0^{\frac{π}{2}} \log (\cos x) \;dx$

The integral $\int\limits_0^{\frac{π}{2}} \log (\cos x) \;dx$ is a standard integral, and its value is known to be $-\frac{π}{2} \log 2$. (This is derived similarly to $\int\limits_0^{π/2} \log \sin x \;dx$, using the property $\int\limits_0^a f(x) \;dx = \int\limits_0^a f(a-x) \;dx$ and $\log(\sin x \cos x)$ manipulation as shown in the previous problem).

Substituting this value:

$I = \frac{1}{2} \left( -\frac{π}{2} \log 2 \right)$

$I = -\frac{π}{4} \log 2$

The final answer is $\mathbf{-\frac{π}{4} \log 2}$.

We need to evaluate the integral:

$\int \frac{\cos 2x − \cos 2θ}{\cos x − \cos θ} \;dx$

We use the double angle formula for cosine: $\cos 2A = 2\cos^2 A - 1$.

Apply this to the numerator:

$\cos 2x = 2\cos^2 x - 1$

$\cos 2θ = 2\cos^2 θ - 1$

So, the numerator is:

$\cos 2x - \cos 2θ = (2\cos^2 x - 1) - (2\cos^2 θ - 1)$

$= 2\cos^2 x - 1 - 2\cos^2 θ + 1$

$= 2\cos^2 x - 2\cos^2 θ$

$= 2(\cos^2 x - \cos^2 θ)$

Now, we use the difference of squares formula: $A^2 - B^2 = (A-B)(A+B)$.

$\cos^2 x - \cos^2 θ = (\cos x - \cos θ)(\cos x + \cos θ)$

Substitute this back into the numerator:

Numerator $= 2(\cos x - \cos θ)(\cos x + \cos θ)$

Now, substitute the simplified numerator back into the integral:

$\int \frac{2(\cos x - \cos θ)(\cos x + \cos θ)}{\cos x - \cos θ} \;dx$

Assuming $\cos x \neq \cos θ$, we can cancel the term $(\cos x - \cos θ)$ from the numerator and the denominator:

$\int 2(\cos x + \cos θ) \;dx$

Now, we integrate term by term. Since $\theta$ is a constant, $\cos θ$ is also a constant with respect to $x$.

$\int (2\cos x + 2\cos θ) \;dx$

$= \int 2\cos x \;dx + \int 2\cos θ \;dx$

$= 2 \int \cos x \;dx + 2\cos θ \int 1 \;dx$

We know that $\int \cos x \;dx = \sin x$ and $\int 1 \;dx = x$.

$= 2(\sin x) + 2\cos θ (x) + C$

$= 2\sin x + 2x \cos θ + C$

This can be factored as $2(\sin x + x \cos θ) + C$.

Comparing this result with the given options:

(A) 2(sin x + x cos θ) + C

(B) 2(sin x – x cos θ) + C

(C) 2(sin x + 2x cos θ) + C

(D) 2(sin x – 2x cos θ) + C

The result matches option (A).

The correct option is (A).

Let the given integral be $I$.

$I = \int \frac{dx}{\sin (x−a) \sin(x−b)}$

To solve this integral, we multiply and divide by $\sin(b-a)$. Assume that $\sin(b-a) \neq 0$.

$I = \frac{1}{\sin(b-a)} \int \frac{\sin(b-a)}{\sin(x-a) \sin(x-b)} \;dx$

We can rewrite the term $\sin(b-a)$ in the numerator as $\sin((x-a) - (x-b))$.

$I = \frac{1}{\sin(b-a)} \int \frac{\sin((x-a) - (x-b))}{\sin(x-a) \sin(x-b)} \;dx$

Using the sine subtraction formula, $\sin(A-B) = \sin A \cos B - \cos A \sin B$, where $A = x-a$ and $B = x-b$:

$\sin((x-a) - (x-b)) = \sin(x-a) \cos(x-b) - \cos(x-a) \sin(x-b)$

Substitute this into the numerator of the integral:

$I = \frac{1}{\sin(b-a)} \int \frac{\sin(x-a) \cos(x-b) - \cos(x-a) \sin(x-b)}{\sin(x-a) \sin(x-b)} \;dx$

Now, split the fraction into two terms:

$I = \frac{1}{\sin(b-a)} \int \left( \frac{\sin(x-a) \cos(x-b)}{\sin(x-a) \sin(x-b)} - \frac{\cos(x-a) \sin(x-b)}{\sin(x-a) \sin(x-b)} \right) \;dx$

Cancel the common terms in each fraction:

$I = \frac{1}{\sin(b-a)} \int \left( \frac{\cos(x-b)}{\sin(x-b)} - \frac{\cos(x-a)}{\sin(x-a)} \right) \;dx$

Recognize that $\frac{\cos u}{\sin u} = \cot u$:

$I = \frac{1}{\sin(b-a)} \int (\cot(x-b) - \cot(x-a)) \;dx$

Now, integrate term by term. The integral of $\cot u$ with respect to $u$ is $\log |\sin u| + C$. The integration is with respect to $x$, and the differential $dx$ corresponds directly to $d(x-a)$ and $d(x-b)$.

$\int \cot(x-b) \;dx = \log |\sin(x-b)| + C_1$

$\int \cot(x-a) \;dx = \log |\sin(x-a)| + C_2$

So, the integral becomes:

$I = \frac{1}{\sin(b-a)} [\log |\sin(x-b)| - \log |\sin(x-a)|] + C$

Using the logarithm property $\log A - \log B = \log \left( \frac{A}{B} \right)$:

$I = \frac{1}{\sin(b-a)} \log \left| \frac{\sin(x-b)}{\sin(x-a)} \right| + C$

We can write $\frac{1}{\sin(b-a)}$ as $\text{cosec}(b-a)$.

$I = \text{cosec}(b-a) \log \left| \frac{\sin(x-b)}{\sin(x-a)} \right| + C$

Comparing this result with the given options:

(A) $\sin (b - a) \log \left| \frac{\sin (x − b)}{\sin (x−a)} \right| + C$

(B) $cosec \;(b - a) \log \left| \frac{\sin (x−a)}{\sin (x−b)} \right| + C$

(C) $cosec \;(b - a) \log \left| \frac{\sin (x−b)}{\sin (x−a)} \right| + C$

(D) $\sin (b - a) \log \left| \frac{\sin (x−a)}{\sin (x−b)} \right| + C$

The result matches option (C).

The correct option is (C).

Let the given integral be $I$.

$I = \int \tan^{−1} \sqrt{x} \;dx$

We first use a substitution to simplify the integrand.

Let $u = \sqrt{x}$.

Squaring both sides, we get $u^2 = x$.

Differentiating both sides with respect to $u$, we get $2u \;du = \;dx$.

Substitute $u$ and $dx$ into the integral:

$I = \int \tan^{-1} u \cdot (2u \;du)$

$I = 2 \int u \tan^{-1} u \;du$

Now we evaluate the integral $\int u \tan^{-1} u \;du$ using Integration by Parts.

The formula for integration by parts is $\int P \;dQ = PQ - \int Q \;dP$.

Choose $P = \tan^{-1} u$ and $dQ = u \;du$.

Find $dP$ by differentiating $P$:

$dP = \frac{d}{du}(\tan^{-1} u) \;du = \frac{1}{1+u^2} \;du$

Find $Q$ by integrating $dQ$:

$Q = \int u \;du = \frac{u^2}{2}$

Apply the integration by parts formula:

$\int u \tan^{-1} u \;du = (\tan^{-1} u) \left(\frac{u^2}{2}\right) - \int \left(\frac{u^2}{2}\right) \left(\frac{1}{1+u^2}\right) \;du$

$= \frac{u^2}{2} \tan^{-1} u - \frac{1}{2} \int \frac{u^2}{1+u^2} \;du$

Now, evaluate the remaining integral $\int \frac{u^2}{1+u^2} \;du$.

We can rewrite the integrand by adding and subtracting 1 in the numerator:

$\int \frac{u^2}{1+u^2} \;du = \int \frac{u^2 + 1 - 1}{1+u^2} \;du$

$= \int \left( \frac{u^2+1}{1+u^2} - \frac{1}{1+u^2} \right) \;du$

$= \int \left( 1 - \frac{1}{1+u^2} \right) \;du$

Integrate term by term:

$= \int 1 \;du - \int \frac{1}{1+u^2} \;du$

$= u - \tan^{-1} u + C'$

Substitute this result back into the integration by parts expression for $I$:

$I = 2 \left[ \frac{u^2}{2} \tan^{-1} u - \frac{1}{2} \left( u - \tan^{-1} u \right) \right] + C$

$I = u^2 \tan^{-1} u - (u - \tan^{-1} u) + C$

$I = u^2 \tan^{-1} u - u + \tan^{-1} u + C$

$I = (u^2 + 1) \tan^{-1} u - u + C$

Finally, substitute back $u = \sqrt{x}$:

$I = ((\sqrt{x})^2 + 1) \tan^{-1} \sqrt{x} - \sqrt{x} + C$

$I = (x + 1) \tan^{-1} \sqrt{x} - \sqrt{x} + C$

Comparing this result with the given options:

(A) $(x + 1) \tan^{−1} \sqrt{x} − \sqrt{x} + C$

(B) $x \tan^{−1} \sqrt{x} − \sqrt{x} + C$

(C) $\sqrt{x} − x \tan^{−1} \sqrt{x} + C$

(D) $\sqrt{x} − (x+1) \tan^{−1} \sqrt{x} + C$

The result matches option (A).

The correct option is (A).

Let the given integral be $I$.

$I = \int e^x \left( \frac{1−x}{1+x^2} \right)^2 \;dx$

We expand the term $\left( \frac{1−x}{1+x^2} \right)^2$:

$\left( \frac{1−x}{1+x^2} \right)^2 = \frac{(1-x)^2}{(1+x^2)^2} = \frac{1 - 2x + x^2}{(1+x^2)^2}$

We can rewrite the numerator $1 - 2x + x^2$ as $(1+x^2) - 2x$.

So, the integrand becomes:

$\frac{1 - 2x + x^2}{(1+x^2)^2} = \frac{(1+x^2) - 2x}{(1+x^2)^2}$

Split the fraction into two parts:

$= \frac{1+x^2}{(1+x^2)^2} - \frac{2x}{(1+x^2)^2}$

$= \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2}$

Now the integral is:

$I = \int e^x \left( \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} \right) \;dx$

This integral is in the form $\int e^x (f(x) + f'(x)) \;dx$.

Let $f(x) = \frac{1}{1+x^2}$.

Let's find the derivative $f'(x)$ using the chain rule or quotient rule.

$f(x) = (1+x^2)^{-1}$

$f'(x) = -1 \cdot (1+x^2)^{-2} \cdot \frac{d}{dx}(1+x^2)$

$f'(x) = -1 \cdot (1+x^2)^{-2} \cdot (2x)$

$f'(x) = \frac{-2x}{(1+x^2)^2}$

We can see that the integrand $\frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2}$ is indeed of the form $f(x) + f'(x)$, where $f(x) = \frac{1}{1+x^2}$ and $f'(x) = \frac{-2x}{(1+x^2)^2}$.

The integral $\int e^x (f(x) + f'(x)) \;dx$ is equal to $e^x f(x) + C$.

Using this formula, the integral is:

$I = e^x \cdot f(x) + C$

$I = e^x \cdot \frac{1}{1+x^2} + C$

$I = \frac{e^x}{1+x^2} + C$

Comparing this result with the given options:

(A) $\frac{e^x}{1+x^2} + C$

(B) $\frac{−e^x}{1 + x^2} + C$

(C) $\frac{e^x}{(1+x^2)^2} + C$

(D) $\frac{−e^x}{(1+x^2)^2} + C$

The result matches option (A).

The correct option is (A).

Let the given integral be $I$.

$I = \int \frac{x^9}{(4x^2 + 1)^6} \;dx$

We can rewrite the denominator by factoring out $x^2$ from the term inside the parenthesis:

$(4x^2 + 1)^6 = [x^2(4 + \frac{1}{x^2})]^6 = (x^2)^6 (4 + \frac{1}{x^2})^6 = x^{12} (4 + \frac{1}{x^2})^6$

Substitute this back into the integral:

$I = \int \frac{x^9}{x^{12} (4 + \frac{1}{x^2})^6} \;dx$

$I = \int \frac{1}{x^3 (4 + \frac{1}{x^2})^6} \;dx$

Now, we use a substitution.

Let $u = 4 + \frac{1}{x^2}$.

Then $u = 4 + x^{-2}$.

Differentiate $u$ with respect to $x$ to find $du$:

$\frac{du}{dx} = \frac{d}{dx}(4 + x^{-2}) = 0 - 2x^{-3} = \frac{-2}{x^3}$

So, $du = \frac{-2}{x^3} \;dx$.

This implies $\frac{1}{x^3} \;dx = -\frac{1}{2} \;du$.

Substitute $u$ and $\frac{1}{x^3} \;dx$ into the integral:

$I = \int \frac{1}{(4 + \frac{1}{x^2})^6} \cdot \frac{1}{x^3} \;dx$

$I = \int \frac{1}{u^6} \cdot \left(-\frac{1}{2}\right) \;du$

$I = -\frac{1}{2} \int u^{-6} \;du$

Now, integrate $u^{-6}$ with respect to $u$ using the power rule $\int w^n \;dw = \frac{w^{n+1}}{n+1} + C$ (for $n \neq -1$):

$\int u^{-6} \;du = \frac{u^{-6+1}}{-6+1} + C' = \frac{u^{-5}}{-5} + C'$

$= -\frac{1}{5u^5} + C'$

Substitute this result back into the expression for $I$:

$I = -\frac{1}{2} \left( -\frac{1}{5u^5} \right) + C$

$I = \frac{1}{10u^5} + C$

Finally, substitute back $u = 4 + \frac{1}{x^2}$:

$I = \frac{1}{10 \left( 4 + \frac{1}{x^2} \right)^5} + C$

Using the property $\frac{1}{A^n} = A^{-n}$, we can write the denominator term in the numerator with a negative exponent:

$I = \frac{1}{10} \left( 4 + \frac{1}{x^2} \right)^{-5} + C$

This can also be written as $\frac{1}{10} \left( \frac{1}{x^2} + 4 \right)^{-5} + C$.

Comparing this result with the given options:

(A) $\frac{1}{5x} \left( 4 + \frac{1}{x^2} \right)^{−5} + C$

(B) $\frac{1}{5} \left( 4 + \frac{1}{x^2} \right)^{−5} + C$

(C) $\frac{1}{10x} (1+4)^{−5} + C$ (Incorrect form)

(D) $\frac{1}{10} \left( \frac{1}{x^2} + 4 \right)^{−5} + C$

The result matches option (D).

The correct option is (D).

We need to evaluate the integral $\int \frac{dx}{(x+2) (x^2+1)}$ and compare the result with the given form to find the values of $a$ and $b$.

We use the method of Partial Fraction Decomposition for the integrand.

$\frac{1}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx + C}{x^2+1}$

Multiply both sides by $(x+2)(x^2+1)$:

$1 = A(x^2+1) + (Bx+C)(x+2)$

To find $A$, substitute $x = -2$:

$1 = A((-2)^2+1) + (B(-2)+C)(-2+2)$

$1 = A(4+1) + (-2B+C)(0)$

$1 = 5A$

$A = \frac{1}{5}$

Substitute $A = \frac{1}{5}$ and expand the right side:

$1 = \frac{1}{5}(x^2+1) + (Bx+C)(x+2)$

$1 = \frac{1}{5}x^2 + \frac{1}{5} + Bx^2 + 2Bx + Cx + 2C$

$1 = (\frac{1}{5} + B)x^2 + (2B + C)x + (\frac{1}{5} + 2C)$

Equate the coefficients of the powers of $x$ on both sides:

Coefficient of $x^2$: $0 = \frac{1}{5} + B$

$B = -\frac{1}{5}$

Coefficient of $x$: $0 = 2B + C$

Substitute $B = -\frac{1}{5}$: $0 = 2(-\frac{1}{5}) + C \implies 0 = -\frac{2}{5} + C \implies C = \frac{2}{5}$

Constant term: $1 = \frac{1}{5} + 2C$

Substitute $C = \frac{2}{5}$: $1 = \frac{1}{5} + 2(\frac{2}{5}) = \frac{1}{5} + \frac{4}{5} = \frac{5}{5} = 1$. (This is consistent)

So, the partial fraction decomposition is:

$\frac{1}{(x+2)(x^2+1)} = \frac{1/5}{x+2} + \frac{-\frac{1}{5}x + \frac{2}{5}}{x^2+1}$

$\frac{1}{(x+2)(x^2+1)} = \frac{1}{5(x+2)} + \frac{2-x}{5(x^2+1)}$

Now, we integrate the decomposed expression:

$\int \frac{dx}{(x+2)(x^2+1)} = \int \left( \frac{1}{5(x+2)} + \frac{2-x}{5(x^2+1)} \right) \;dx$

$= \frac{1}{5} \int \frac{1}{x+2} \;dx + \frac{1}{5} \int \frac{2-x}{x^2+1} \;dx$

$= \frac{1}{5} \int \frac{1}{x+2} \;dx + \frac{1}{5} \int \frac{2}{x^2+1} \;dx - \frac{1}{5} \int \frac{x}{x^2+1} \;dx$

Evaluate each integral:

$\int \frac{1}{x+2} \;dx = \log |x+2|$

$\int \frac{2}{x^2+1} \;dx = 2 \tan^{-1} x$

For $\int \frac{x}{x^2+1} \;dx$, let $u = x^2+1$, then $du = 2x \;dx$, so $x \;dx = \frac{1}{2} du$.

$\int \frac{x}{x^2+1} \;dx = \int \frac{1}{u} \cdot \frac{1}{2} \;du = \frac{1}{2} \log |u| = \frac{1}{2} \log |x^2+1|$

Since $x^2+1 > 0$, $|x^2+1| = x^2+1$. So $\int \frac{x}{x^2+1} \;dx = \frac{1}{2} \log (x^2+1)$.

Combine the results:

$\int \frac{dx}{(x+2)(x^2+1)} = \frac{1}{5} \log |x+2| + \frac{1}{5} (2 \tan^{-1} x) - \frac{1}{5} \left( \frac{1}{2} \log (x^2+1) \right) + C$

$= \frac{1}{5} \log |x+2| + \frac{2}{5} \tan^{-1} x - \frac{1}{10} \log (x^2+1) + C$

Rearranging the terms to match the given form:

$= -\frac{1}{10} \log (x^2+1) + \frac{2}{5} \tan^{-1} x + \frac{1}{5} \log |x+2| + C$

The given form is $a \log |1 + x^2| + b \tan^{-1} x + \frac{1}{5} \log |x + 2| + C$.

Since $|1+x^2| = x^2+1$, the given form is $a \log (x^2+1) + b \tan^{-1} x + \frac{1}{5} \log |x + 2| + C$.

Comparing the coefficients:

$a = -\frac{1}{10}$

$b = \frac{2}{5}$

Checking the options, the values $a = -\frac{1}{10}$ and $b = \frac{2}{5}$ correspond to option (C).

The correct option is (C).

We need to evaluate the integral:

$\int \frac{x^3}{x + 1} \;dx$

The integrand is an improper rational function (degree of numerator $>$ degree of denominator), so we perform polynomial long division or algebraic manipulation.

We can rewrite the numerator $x^3$ as $x^3+1-1$.

Using the sum of cubes formula $a^3+b^3 = (a+b)(a^2-ab+b^2)$, we have $x^3+1^3 = (x+1)(x^2-x+1)$.

So, $x^3 = (x+1)(x^2-x+1) - 1$.

Now, substitute this into the integrand:

$\frac{x^3}{x + 1} = \frac{(x+1)(x^2-x+1) - 1}{x + 1}$

Split the fraction:

$= \frac{(x+1)(x^2-x+1)}{x+1} - \frac{1}{x+1}$

Assuming $x+1 \neq 0$, we can cancel the $(x+1)$ term:

$= x^2 - x + 1 - \frac{1}{x+1}$

Now, integrate the simplified expression term by term:

$\int \left( x^2 - x + 1 - \frac{1}{x+1} \right) \;dx$

$= \int x^2 \;dx - \int x \;dx + \int 1 \;dx - \int \frac{1}{x+1} \;dx$

Using the power rule $\int x^n \;dx = \frac{x^{n+1}}{n+1}$ (for $n \neq -1$) and $\int \frac{1}{u} \;du = \log |u|$, we get:

$= \frac{x^{2+1}}{2+1} - \frac{x^{1+1}}{1+1} + x - \log |x+1| + C$

$= \frac{x^3}{3} - \frac{x^2}{2} + x - \log |x+1| + C$

Rearranging the terms to match the format of the options:

$= x - \frac{x^2}{2} + \frac{x^3}{3} - \log |1 + x| + C$

Comparing this result with the given options:

(A) $x + \frac{x^2}{2} + \frac{x^3}{3} - \log |1 - x| + C$

(B) $x + \frac{x^2}{2} − \frac{x^3}{3} − \log |1 - x| + C$

(C) $x - \frac{x^2}{2} - \frac{x^3}{3} - \log | 1 + x| + C$

(D) $x - \frac{x^2}{2} + \frac{x^3}{3} - \log |1 + x| + C$

The result matches option (D).

The correct option is (D).

Let the given integral be $I$.

$I = \int \frac{x + \sin x}{1 + \cos x} \;dx$

We use the half-angle identities for trigonometric functions:

$1 + \cos x = 2 \cos^2 \frac{x}{2}$

$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$

Substitute these identities into the integrand:

$\frac{x + \sin x}{1 + \cos x} = \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}}$

Split the fraction into two parts:

$= \frac{x}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}}$

$= \frac{x}{2} \cdot \frac{1}{\cos^2 \frac{x}{2}} + \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$

$= \frac{x}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2}$

Now, integrate the simplified expression:

$I = \int \left( \frac{x}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) \;dx$

$I = \int \frac{x}{2} \sec^2 \frac{x}{2} \;dx + \int \tan \frac{x}{2} \;dx$

We evaluate the first integral $\int \frac{x}{2} \sec^2 \frac{x}{2} \;dx$ using Integration by Parts. The formula is $\int u \;dv = uv - \int v \;du$.

Let $u = x$ and $dv = \frac{1}{2} \sec^2 \frac{x}{2} \;dx$.

Then $du = \;dx$.

To find $v$, we integrate $dv$: $v = \int \frac{1}{2} \sec^2 \frac{x}{2} \;dx$. Let $w = \frac{x}{2}$, so $dw = \frac{1}{2} \;dx$.

$v = \int \sec^2 w \;dw = \tan w = \tan \frac{x}{2}$.

Applying the integration by parts formula:

$\int \frac{x}{2} \sec^2 \frac{x}{2} \;dx = x \left( \tan \frac{x}{2} \right) - \int \tan \frac{x}{2} \;dx$

$= x \tan \frac{x}{2} - \int \tan \frac{x}{2} \;dx$

Substitute this back into the expression for $I$:

$I = \left( x \tan \frac{x}{2} - \int \tan \frac{x}{2} \;dx \right) + \int \tan \frac{x}{2} \;dx$

$I = x \tan \frac{x}{2} + C$

Comparing this result with the given options:

(A) log |1 + cos x | + C

(B) log |x + sin x| + C

(C) $x - \tan \frac{x}{2} + C$

(D) $x\;.\; \tan \frac{x}{2} + C$

The result matches option (D).

The correct option is (D).

We need to evaluate the integral $\int \frac{x^3}{\sqrt{1 + x^2}} \;dx$ and find the values of $a$ and $b$ by comparing the result with the given form.

Let the integral be $I$.

$I = \int \frac{x^3}{\sqrt{1 + x^2}} \;dx$

We use the substitution method.

Let $u = 1 + x^2$.

Differentiating both sides with respect to $x$:

$du = 2x \;dx$

From the substitution, we also have $x^2 = u - 1$.

We can write $x^3 \;dx = x^2 \cdot (x \;dx)$.

Using $x^2 = u-1$ and $x \;dx = \frac{1}{2} du$, we substitute these into the integral:

$I = \int \frac{(u-1)}{\sqrt{u}} \cdot \frac{1}{2} \;du$

$I = \frac{1}{2} \int \frac{u-1}{\sqrt{u}} \;du$

Split the integrand into two terms:

$I = \frac{1}{2} \int \left( \frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}} \right) \;du$

$I = \frac{1}{2} \int \left( u^{1 - 1/2} - u^{-1/2} \right) \;du$

$I = \frac{1}{2} \int \left( u^{1/2} - u^{-1/2} \right) \;du$

Now, integrate term by term using the power rule $\int w^n \;dw = \frac{w^{n+1}}{n+1}$:

$\int u^{1/2} \;du = \frac{u^{1/2+1}}{1/2+1} = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$

$\int u^{-1/2} \;du = \frac{u^{-1/2+1}}{-1/2+1} = \frac{u^{1/2}}{1/2} = 2 u^{1/2}$

Substitute these back into the expression for $I$:

$I = \frac{1}{2} \left( \frac{2}{3} u^{3/2} - 2 u^{1/2} \right) + C$

$I = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} - \frac{1}{2} \cdot 2 u^{1/2} + C$

$I = \frac{1}{3} u^{3/2} - u^{1/2} + C$

Finally, substitute back $u = 1 + x^2$:

$I = \frac{1}{3} (1+x^2)^{3/2} - (1+x^2)^{1/2} + C$

We can write $(1+x^2)^{1/2}$ as $\sqrt{1+x^2}$.

$I = \frac{1}{3} (1+x^2)^{3/2} - \sqrt{1+x^2} + C$

We are given that the integral is equal to $a (1+x^2 )^{\frac{3}{2}} + b \sqrt{1+x^2} + C$.

Comparing our result with the given form:

$a (1+x^2 )^{\frac{3}{2}} + b \sqrt{1+x^2} + C = \frac{1}{3} (1+x^2)^{3/2} - 1 \cdot \sqrt{1+x^2} + C$

By comparing the coefficients of $(1+x^2)^{3/2}$ and $\sqrt{1+x^2}$, we find:

$a = \frac{1}{3}$

$b = -1$

Comparing these values with the given options:

(A) a = $\frac{1}{3}$ , b = 1

(B) a = $\frac{−1}{3}$ , b = 1

(C) a = $\frac{−1}{3}$, b = -1

(D) a = $\frac{1}{3}$, b = -1

The values $a = \frac{1}{3}$ and $b = -1$ match option (D).

The correct option is (D).

Let the given integral be $I$.

$I = \int\limits_{\frac{−π}{4}}^{\frac{π}{4}} \frac{dx}{1 + \cos 2x}$

We use the trigonometric identity for the denominator:

$1 + \cos 2x = 2 \cos^2 x$

Substitute this into the integrand:

$I = \int\limits_{\frac{−π}{4}}^{\frac{π}{4}} \frac{dx}{2 \cos^2 x}$

$I = \int\limits_{\frac{−π}{4}}^{\frac{π}{4}} \frac{1}{2} \cdot \frac{1}{\cos^2 x} \;dx$

$I = \frac{1}{2} \int\limits_{\frac{−π}{4}}^{\frac{π}{4}} \sec^2 x \;dx$

The integral of $\sec^2 x$ is $\tan x$.

$I = \frac{1}{2} \left[ \tan x \right]_{−\frac{π}{4}}^{\frac{π}{4}}$

Now, we evaluate the definite integral by substituting the limits:

$I = \frac{1}{2} \left( \tan \left( \frac{π}{4} \right) - \tan \left( -\frac{π}{4} \right) \right)$

We know that $\tan \left( \frac{π}{4} \right) = 1$ and $\tan (-θ) = -\tan θ$, so $\tan \left( -\frac{π}{4} \right) = -\tan \left( \frac{π}{4} \right) = -1$.

Substitute these values:

$I = \frac{1}{2} (1 - (-1))$

$I = \frac{1}{2} (1 + 1)$

$I = \frac{1}{2} (2)$

$I = 1$

Comparing this result with the given options:

(A) 1

(B) 2

(C) 3

(D) 4

The result matches option (A).

The correct option is (A).

Let the given integral be $I$.

$I = \int\limits_0^{\frac{π}{2}} \sqrt{1− \sin 2x} \;dx$

We can rewrite the expression under the square root using trigonometric identities:

Using the identity $\sin^2 x + \cos^2 x = 1$ and $\sin 2x = 2 \sin x \cos x$, we have:

$1 - \sin 2x = \cos^2 x + \sin^2 x - 2 \sin x \cos x$

$1 - \sin 2x = (\cos x - \sin x)^2$

Substitute this back into the integral:

$I = \int\limits_0^{\frac{π}{2}} \sqrt{(\cos x - \sin x)^2} \;dx$

$I = \int\limits_0^{\frac{π}{2}} |\cos x - \sin x| \;dx$

We need to consider the sign of $(\cos x - \sin x)$ in the interval $[0, \frac{π}{2}]$.

In the interval $[0, \frac{π}{4}]$, $\cos x \geq \sin x$, so $\cos x - \sin x \geq 0$.

In the interval $[\frac{π}{4}, \frac{π}{2}]$, $\cos x \leq \sin x$, so $\cos x - \sin x \leq 0$, which means $|\cos x - \sin x| = -(\cos x - \sin x) = \sin x - \cos x$.

Split the integral into two parts based on these intervals:

$I = \int\limits_0^{\frac{π}{4}} (\cos x - \sin x) \;dx + \int\limits_{\frac{π}{4}}^{\frac{π}{2}} (\sin x - \cos x) \;dx$

Evaluate the first integral:

$\int (\cos x - \sin x) \;dx = \sin x - (-\cos x) = \sin x + \cos x$

$\int\limits_0^{\frac{π}{4}} (\cos x - \sin x) \;dx = [\sin x + \cos x]_0^{\frac{π}{4}}$

$= (\sin \frac{π}{4} + \cos \frac{π}{4}) - (\sin 0 + \cos 0)$

$= (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) - (0 + 1)$

$= \frac{2\sqrt{2}}{2} - 1 = \sqrt{2} - 1$

Evaluate the second integral:

$\int (\sin x - \cos x) \;dx = -\cos x - \sin x$

$\int\limits_{\frac{π}{4}}^{\frac{π}{2}} (\sin x - \cos x) \;dx = [-\cos x - \sin x]_{\frac{π}{4}}^{\frac{π}{2}}$

$= (-\cos \frac{π}{2} - \sin \frac{π}{2}) - (-\cos \frac{π}{4} - \sin \frac{π}{4})$

$= (-0 - 1) - (-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2})$

$= -1 - (-\frac{2\sqrt{2}}{2}) = -1 - (-\sqrt{2})$

$= -1 + \sqrt{2} = \sqrt{2} - 1$

Add the results of the two integrals to get the value of $I$:

$I = (\sqrt{2} - 1) + (\sqrt{2} - 1)$

$I = \sqrt{2} + \sqrt{2} - 1 - 1$

$I = 2\sqrt{2} - 2$

$I = 2(\sqrt{2} - 1)$

Comparing this result with the given options:

(A) $2\sqrt{2}$

(B) $2 (\sqrt{2} +1)$

(C) 2

(D) $2 (\sqrt{2}−1)$

The result matches option (D).

The correct option is (D).

Let the given integral be $I$.

$I = \int\limits_0^{\frac{π}{2}} \cos x e^{\sin x} \;dx$

We use the substitution method.

Let $u = \sin x$.

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = \cos x$

$du = \cos x \;dx$

Change the limits of integration:

When $x = 0$, $u = \sin(0) = 0$.

When $x = \frac{π}{2}$, $u = \sin\left(\frac{π}{2}\right) = 1$.

Substitute $u$, $du$, and the new limits into the integral:

$I = \int\limits_0^1 e^u \;du$

Now, evaluate the definite integral:

$I = [e^u]_0^1$

$I = e^1 - e^0$

$I = e - 1$

The value of the integral is $e - 1$.

The blank should be filled with $e-1$.

$\int\limits_0^{\frac{π}{2}} \cos x e^{\sin x} \;dx = e-1$.

Let the given integral be $I$.

$I = \int \frac{x + 3}{(x + 4)^2} e^x \;dx$

We want to express the fraction $\frac{x+3}{(x+4)^2}$ in the form $f(x) + f'(x)$ so that we can use the formula $\int e^x (f(x) + f'(x)) \;dx = e^x f(x) + C$.

We can rewrite the numerator $x+3$ as $(x+4) - 1$.

So, the fraction becomes:

$\frac{x+3}{(x+4)^2} = \frac{(x+4) - 1}{(x+4)^2}$

Split the fraction into two parts:

$= \frac{x+4}{(x+4)^2} - \frac{1}{(x+4)^2}$

$= \frac{1}{x+4} - \frac{1}{(x+4)^2}$

Now, the integral is:

$I = \int e^x \left( \frac{1}{x+4} - \frac{1}{(x+4)^2} \right) \;dx$

This integral is in the form $\int e^x (f(x) + f'(x)) \;dx$.

Let $f(x) = \frac{1}{x+4}$.

Let's find the derivative $f'(x)$. Using the power rule, $f(x) = (x+4)^{-1}$.

$f'(x) = -1 \cdot (x+4)^{-2} \cdot \frac{d}{dx}(x+4)$

$f'(x) = -1 \cdot (x+4)^{-2} \cdot 1$

$f'(x) = -\frac{1}{(x+4)^2}$

We can see that the integrand $\frac{1}{x+4} - \frac{1}{(x+4)^2}$ is indeed of the form $f(x) + f'(x)$, where $f(x) = \frac{1}{x+4}$ and $f'(x) = -\frac{1}{(x+4)^2}$.

Using the formula $\int e^x (f(x) + f'(x)) \;dx = e^x f(x) + C$:

$I = e^x \cdot f(x) + C$

$I = e^x \cdot \frac{1}{x+4} + C$

$I = \frac{e^x}{x+4} + C$

The value of the integral is $\frac{e^x}{x+4} + C$.

The blank should be filled with $\frac{e^x}{x+4} + C$.

$\int \frac{x + 3}{(x + 4)^2} e^x \;dx = \frac{e^x}{x+4} + C$.

We are given the equation:

$\int\limits_0^a \frac{1}{1 + 4x^2} \;dx = \frac{π}{8}$

We need to evaluate the definite integral on the left side.

The integral is of the form $\int \frac{1}{1 + (kx)^2} \;dx$. We can use a substitution or recognize the standard integral form $\int \frac{1}{1+u^2} \;du = \tan^{-1} u$.

Let $u = 2x$.

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = 2$

$dx = \frac{1}{2} du$

Change the limits of integration:

When $x = 0$, $u = 2(0) = 0$.

When $x = a$, $u = 2a$.

Substitute $u$, $dx$, and the new limits into the integral:

$\int\limits_0^a \frac{1}{1 + (2x)^2} \;dx = \int\limits_0^{2a} \frac{1}{1 + u^2} \cdot \frac{1}{2} \;du$

$= \frac{1}{2} \int\limits_0^{2a} \frac{1}{1 + u^2} \;du$

Now, evaluate the integral of $\frac{1}{1+u^2}$ which is $\tan^{-1} u$:

$= \frac{1}{2} [\tan^{-1} u]_0^{2a}$

Evaluate using the limits:

$= \frac{1}{2} (\tan^{-1} (2a) - \tan^{-1} (0))$

Since $\tan^{-1} (0) = 0$:

$= \frac{1}{2} \tan^{-1} (2a)$

We are given that this integral is equal to $\frac{π}{8}$.

$\frac{1}{2} \tan^{-1} (2a) = \frac{π}{8}$

Multiply both sides by 2:

$\tan^{-1} (2a) = \frac{π}{4}$

To find $2a$, take the tangent of both sides:

$2a = \tan \left( \frac{π}{4} \right)$

We know that $\tan \left( \frac{π}{4} \right) = 1$.

$2a = 1$

Solve for $a$:

$a = \frac{1}{2}$

The value of $a$ is $\frac{1}{2}$.

The blank should be filled with $\frac{1}{2}$.

If $\int\limits_0^a \frac{1}{1 + 4x^2} \;dx = \frac{π}{8}$, then a = $\frac{1}{2}$.

Let the given integral be $I$.

$I = \int \frac{\sin x}{3 + 4\cos^2 x} \;dx$

We use the substitution method.

Let $u = \cos x$.

Differentiating both sides with respect to $x$:

$du = -\sin x \;dx$

So, $\sin x \;dx = -du$.

Substitute $u$ and $\sin x \;dx$ into the integral:

$I = \int \frac{-du}{3 + 4u^2}$

$I = - \int \frac{du}{3 + 4u^2}$

We can rewrite the denominator to match a standard integral form $\int \frac{1}{a^2 + x^2} \;dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right)$.

$3 + 4u^2 = 3 + (2u)^2$

Here, $a^2 = 3$, so $a = \sqrt{3}$, and we have $(2u)^2$.

$I = - \int \frac{du}{(\sqrt{3})^2 + (2u)^2}$

To integrate $\frac{1}{(\sqrt{3})^2 + (2u)^2} \;du$, let $v = 2u$. Then $dv = 2 \;du$, so $du = \frac{1}{2} dv$.

$I = - \int \frac{\frac{1}{2} dv}{(\sqrt{3})^2 + v^2}$

$I = -\frac{1}{2} \int \frac{dv}{(\sqrt{3})^2 + v^2}$

Now, apply the standard integral formula $\int \frac{1}{a^2 + v^2} \;dv = \frac{1}{a} \tan^{-1} \left( \frac{v}{a} \right)$ with $a = \sqrt{3}$:

$I = -\frac{1}{2} \left( \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{v}{\sqrt{3}} \right) \right) + C$

$I = -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{v}{\sqrt{3}} \right) + C$

Substitute back $v = 2u$:

$I = -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2u}{\sqrt{3}} \right) + C$

Finally, substitute back $u = \cos x$:

$I = -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2\cos x}{\sqrt{3}} \right) + C$

The value of the integral is $-\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2\cos x}{\sqrt{3}} \right) + C$.

The blank should be filled with $-\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2\cos x}{\sqrt{3}} \right) + C$.

$\int \frac{\sin x}{3 + 4\cos^2 x} \;dx = -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2\cos x}{\sqrt{3}} \right) + C$.

Let the given integral be $I$.

$I = \int\limits_{−π}^π \sin^3 x \; \cos^2 x \;dx$

We observe that the limits of integration are symmetric about 0, from $-π$ to $π$.

We need to check if the integrand $f(x) = \sin^3 x \; \cos^2 x$ is an even function or an odd function.

A function $f(x)$ is even if $f(-x) = f(x)$.

A function $f(x)$ is odd if $f(-x) = -f(x)$.

Let's find $f(-x)$:

$f(-x) = \sin^3 (-x) \; \cos^2 (-x)$

Using the trigonometric properties $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$:

$\sin^3 (-x) = (-\sin x)^3 = -\sin^3 x$

$\cos^2 (-x) = (\cos(-x))^2 = (\cos x)^2 = \cos^2 x$

Substitute these back into the expression for $f(-x)$:

$f(-x) = (-\sin^3 x) (\cos^2 x)$

$f(-x) = -\sin^3 x \cos^2 x$

Compare $f(-x)$ with $f(x)$:

$f(-x) = -(\sin^3 x \cos^2 x) = -f(x)$

Since $f(-x) = -f(x)$, the integrand $f(x) = \sin^3 x \; \cos^2 x$ is an odd function.

For a definite integral over a symmetric interval $[-a, a]$, if the integrand $f(x)$ is an odd function, the value of the integral is 0.

The property is: $\int\limits_{-a}^a f(x) \;dx = 0$ if $f(-x) = -f(x)$.

In this case, $a = π$ and $f(x) = \sin^3 x \; \cos^2 x$ is odd.

Therefore, $\int\limits_{−π}^π \sin^3 x \; \cos^2 x \;dx = 0$.

The value of the integral is 0.

The blank should be filled with 0.

The value of $\int\limits_{−π}^π \sin^3 x \; \cos^2 x \;dx$ is 0.