Fundamental integration Questions in English

(A) The anti-derivative $F(x)$ is the integral of $f(x)$ with respect to $x$:
$F(x) = \int (4x^{3} - 6) \, dx$
$F(x) = 4 \int x^{3} \, dx - 6 \int 1 \, dx$
$F(x) = 4 \left( \frac{x^{4}}{4} \right) - 6x + C$
$F(x) = x^{4} - 6x + C$
Given that $F(0) = 3$,we substitute $x = 0$ into the equation:
$3 = (0)^{4} - 6(0) + C$
$3 = 0 - 0 + C$
$C = 3$
Therefore,the required anti-derivative is $F(x) = x^{4} - 6x + 3$.

(A) The anti-derivative of $\sin 2x$ is a function $F(x)$ such that $F'(x) = \sin 2x$.
We know that the derivative of $\cos 2x$ is:
$\frac{d}{dx}(\cos 2x) = -2 \sin 2x$
Dividing both sides by $-2$,we get:
$-\frac{1}{2} \frac{d}{dx}(\cos 2x) = \sin 2x$
Using the linearity property of the derivative,we can write:
$\frac{d}{dx}\left(-\frac{1}{2} \cos 2x\right) = \sin 2x$
Thus,the anti-derivative of $\sin 2x$ is $-\frac{1}{2} \cos 2x$.

(A) The anti-derivative of $\cos 3x$ is a function $F(x)$ such that $F'(x) = \cos 3x$.
We know that the derivative of $\sin 3x$ with respect to $x$ is:
$\frac{d}{dx}(\sin 3x) = 3 \cos 3x$
Dividing both sides by $3$,we get:
$\frac{1}{3} \frac{d}{dx}(\sin 3x) = \cos 3x$
Using the linearity of the derivative,we can write:
$\frac{d}{dx} \left( \frac{1}{3} \sin 3x \right) = \cos 3x$
Thus,the anti-derivative of $\cos 3x$ is $\frac{1}{3} \sin 3x$.

(A) The anti-derivative of $e^{2x}$ is a function $F(x)$ such that $F'(x) = e^{2x}$.
We know that the derivative of $e^{2x}$ with respect to $x$ is:
$\frac{d}{dx}(e^{2x}) = 2e^{2x}$
To isolate $e^{2x}$,we divide both sides by $2$:
$e^{2x} = \frac{1}{2} \frac{d}{dx}(e^{2x})$
Using the linearity property of the derivative,we can write:
$e^{2x} = \frac{d}{dx}\left(\frac{1}{2} e^{2x}\right)$
Thus,by the method of inspection,the anti-derivative of $e^{2x}$ is $\frac{1}{2} e^{2x}$.

(A) The anti-derivative of $\sin 2x - 4e^{3x}$ is a function $F(x)$ such that $F'(x) = \sin 2x - 4e^{3x}$.
We know that $\frac{d}{dx}(\cos 2x) = -2 \sin 2x$,which implies $\frac{d}{dx}(-\frac{1}{2} \cos 2x) = \sin 2x$.
We also know that $\frac{d}{dx}(e^{3x}) = 3e^{3x}$,which implies $\frac{d}{dx}(-\frac{4}{3} e^{3x}) = -4e^{3x}$.
Combining these,we have $\frac{d}{dx}(-\frac{1}{2} \cos 2x - \frac{4}{3} e^{3x}) = \sin 2x - 4e^{3x}$.
Therefore,the anti-derivative is $-\frac{1}{2} \cos 2x - \frac{4}{3} e^{3x}$.

(N/A) To find the integral $\int(4 e^{3 x}+1) d x$,we use the linearity property of integration:
$\int(4 e^{3 x}+1) d x = 4 \int e^{3 x} d x + \int 1 d x$
We know that $\int e^{ax} d x = \frac{e^{ax}}{a} + C$ and $\int 1 d x = x + C$.
Applying these formulas:
$= 4 \left( \frac{e^{3 x}}{3} \right) + x + C$
$= \frac{4}{3} e^{3 x} + x + C$
where $C$ is an arbitrary constant.

Given integral: $\int x^{2}\left(1-\frac{1}{x^{2}}\right) d x$
Multiply $x^{2}$ inside the parentheses:
$= \int \left(x^{2} \cdot 1 - x^{2} \cdot \frac{1}{x^{2}}\right) d x$
$= \int (x^{2} - 1) d x$
Apply the linearity property of integration:
$= \int x^{2} d x - \int 1 d x$
Using the power rule $\int x^{n} d x = \frac{x^{n+1}}{n+1} + C$:
$= \frac{x^{3}}{3} - x + C$
Where $C$ is an arbitrary constant of integration.

To find the integral $\int(ax^{2} + bx + c) dx$,we use the linearity property of integration:
$\int(ax^{2} + bx + c) dx = a \int x^{2} dx + b \int x dx + c \int 1 dx$
Using the power rule $\int x^{n} dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$:
$= a \left(\frac{x^{3}}{3}\right) + b \left(\frac{x^{2}}{2}\right) + c(x) + C$
$= \frac{ax^{3}}{3} + \frac{bx^{2}}{2} + cx + C$
where $C$ is the constant of integration.

To find the integral $\int(2x^{2}+e^{x})dx$,we use the linearity property of integration:
$\int(2x^{2}+e^{x})dx = 2\int x^{2}dx + \int e^{x}dx$
Using the power rule $\int x^{n}dx = \frac{x^{n+1}}{n+1} + C$ and the exponential rule $\int e^{x}dx = e^{x} + C$:
$= 2\left(\frac{x^{3}}{3}\right) + e^{x} + C$
$= \frac{2}{3}x^{3} + e^{x} + C$
where $C$ is an arbitrary constant.

We are given the integral $\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x$.
First,expand the square using the algebraic identity $(a-b)^{2} = a^{2} + b^{2} - 2ab$:
$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} = (\sqrt{x})^{2} + \left(\frac{1}{\sqrt{x}}\right)^{2} - 2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right) = x + \frac{1}{x} - 2$.
Now,substitute this back into the integral:
$\int\left(x+\frac{1}{x}-2\right) d x$.
Using the linearity property of integrals,we can split this into three separate integrals:
$\int x \, d x + \int \frac{1}{x} \, d x - 2 \int 1 \, d x$.
Integrating each term:
$\int x \, d x = \frac{x^{2}}{2}$,
$\int \frac{1}{x} \, d x = \log |x|$,
$-2 \int 1 \, d x = -2x$.
Combining these results and adding the constant of integration $C$:
$\frac{x^{2}}{2} + \log |x| - 2x + C$,where $C$ is an arbitrary constant.

(N/A) To evaluate the integral $\int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x$,we first simplify the integrand by dividing each term in the numerator by the denominator $x^{2}$:
$\int \left( \frac{x^{3}}{x^{2}} + \frac{5x^{2}}{x^{2}} - \frac{4}{x^{2}} \right) d x$
$= \int (x + 5 - 4x^{-2}) d x$
Now,we integrate each term separately using the power rule $\int x^{n} d x = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):
$= \int x d x + \int 5 d x - \int 4x^{-2} d x$
$= \frac{x^{2}}{2} + 5x - 4 \left( \frac{x^{-2+1}}{-2+1} \right) + C$
$= \frac{x^{2}}{2} + 5x - 4 \left( \frac{x^{-1}}{-1} \right) + C$
$= \frac{x^{2}}{2} + 5x + \frac{4}{x} + C$
where $C$ is an arbitrary constant.

Given the integral: $\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$
We can rewrite the integrand by dividing each term in the numerator by $\sqrt{x} = x^{\frac{1}{2}}$:
$= \int \left( \frac{x^{3}}{x^{\frac{1}{2}}} + \frac{3x}{x^{\frac{1}{2}}} + \frac{4}{x^{\frac{1}{2}}} \right) d x$
$= \int \left( x^{3-\frac{1}{2}} + 3x^{1-\frac{1}{2}} + 4x^{-\frac{1}{2}} \right) d x$
$= \int \left( x^{\frac{5}{2}} + 3x^{\frac{1}{2}} + 4x^{-\frac{1}{2}} \right) d x$
Using the power rule $\int x^{n} d x = \frac{x^{n+1}}{n+1} + C$:
$= \frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1} + 3 \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} + 4 \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + C$
$= \frac{x^{\frac{7}{2}}}{\frac{7}{2}} + 3 \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + 4 \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C$
$= \frac{2}{7} x^{\frac{7}{2}} + 2 x^{\frac{3}{2}} + 8 x^{\frac{1}{2}} + C$
$= \frac{2}{7} x^{\frac{7}{2}} + 2 x^{\frac{3}{2}} + 8 \sqrt{x} + C$,where $C$ is an arbitrary constant.

To evaluate the integral $\int \frac{x^{3}-x^{2}+x-1}{x-1} \, dx$,we first simplify the integrand by factorising the numerator.
$\frac{x^{3}-x^{2}+x-1}{x-1} = \frac{x^{2}(x-1) + 1(x-1)}{x-1} = \frac{(x^{2}+1)(x-1)}{x-1}$
For $x \neq 1$,this simplifies to $x^{2}+1$.
Now,we integrate the simplified expression:
$\int (x^{2}+1) \, dx = \int x^{2} \, dx + \int 1 \, dx$
Using the power rule $\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C$,we get:
$= \frac{x^{3}}{3} + x + C$
where $C$ is an arbitrary constant.

To evaluate the integral $\int(1-x) \sqrt{x} \, dx$,we first distribute $\sqrt{x}$ into the parentheses:
$= \int (\sqrt{x} - x \cdot \sqrt{x}) \, dx$
$= \int (x^{\frac{1}{2}} - x^1 \cdot x^{\frac{1}{2}}) \, dx$
$= \int (x^{\frac{1}{2}} - x^{\frac{3}{2}}) \, dx$
Now,apply the power rule for integration,$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$:
$= \int x^{\frac{1}{2}} \, dx - \int x^{\frac{3}{2}} \, dx$
$= \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} - \frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} + C$
$= \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + C$
$= \frac{2}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} + C$
where $C$ is an arbitrary constant.

To find the integral $\int \sqrt{x}(3x^{2} + 2x + 3) dx$,we first distribute $\sqrt{x}$ (which is $x^{1/2}$) into the parenthesis:
$= \int (3x^{2} \cdot x^{1/2} + 2x \cdot x^{1/2} + 3 \cdot x^{1/2}) dx$
$= \int (3x^{5/2} + 2x^{3/2} + 3x^{1/2}) dx$
Now,apply the power rule for integration $\int x^{n} dx = \frac{x^{n+1}}{n+1} + C$:
$= 3 \int x^{5/2} dx + 2 \int x^{3/2} dx + 3 \int x^{1/2} dx$
$= 3 \left( \frac{x^{7/2}}{7/2} \right) + 2 \left( \frac{x^{5/2}}{5/2} \right) + 3 \left( \frac{x^{3/2}}{3/2} \right) + C$
$= 3 \cdot \frac{2}{7} x^{7/2} + 2 \cdot \frac{2}{5} x^{5/2} + 3 \cdot \frac{2}{3} x^{3/2} + C$
$= \frac{6}{7} x^{7/2} + \frac{4}{5} x^{5/2} + 2x^{3/2} + C$
where $C$ is the constant of integration.

(N/A) To find the integral $\int(2x - 3 \cos x + e^x) \, dx$,we use the linearity property of integration:
$= 2 \int x \, dx - 3 \int \cos x \, dx + \int e^x \, dx$
Using the standard integration formulas $\int x^n \, dx = \frac{x^{n+1}}{n+1}$,$\int \cos x \, dx = \sin x$,and $\int e^x \, dx = e^x$:
$= 2 \left( \frac{x^2}{2} \right) - 3(\sin x) + e^x + C$
$= x^2 - 3 \sin x + e^x + C$
where $C$ is an arbitrary constant.

We are given the integral $\int(2x^2 - 3\sin x + 5\sqrt{x}) dx$.
Using the linearity property of integration,we can write:
$\int(2x^2 - 3\sin x + 5\sqrt{x}) dx = 2\int x^2 dx - 3\int \sin x dx + 5\int x^{1/2} dx$.
Applying the power rule $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ and the standard integral $\int \sin x dx = -\cos x + C$:
$= 2(\frac{x^3}{3}) - 3(-\cos x) + 5(\frac{x^{3/2}}{3/2}) + C$.
Simplifying the expression:
$= \frac{2}{3}x^3 + 3\cos x + 5(\frac{2}{3})x^{3/2} + C$.
Final result:
$= \frac{2}{3}x^3 + 3\cos x + \frac{10}{3}x^{3/2} + C$,where $C$ is an arbitrary constant.

(N/A) We are given the integral: $\int \sec x(\sec x+\tan x) \, dx$
First,distribute $\sec x$ inside the parentheses:
$= \int (\sec^2 x + \sec x \tan x) \, dx$
Using the linearity property of integration,we can split this into two separate integrals:
$= \int \sec^2 x \, dx + \int \sec x \tan x \, dx$
We know the standard integrals:
$\int \sec^2 x \, dx = \tan x + C_1$
$\int \sec x \tan x \, dx = \sec x + C_2$
Combining these results,we get:
$= \tan x + \sec x + C$,where $C$ is an arbitrary constant.

We are given the integral $\int \frac{\sec ^{2} x}{\operatorname{cosec}^{2} x} d x$.
Using the trigonometric identities $\sec x = \frac{1}{\cos x}$ and $\operatorname{cosec} x = \frac{1}{\sin x}$,we can rewrite the integrand as:
$\int \frac{\frac{1}{\cos ^{2} x}}{\frac{1}{\sin ^{2} x}} d x = \int \frac{\sin ^{2} x}{\cos ^{2} x} d x$
$= \int \tan ^{2} x d x$
Using the identity $\tan ^{2} x = \sec ^{2} x - 1$,we get:
$= \int (\sec ^{2} x - 1) d x$
$= \int \sec ^{2} x d x - \int 1 d x$
$= \tan x - x + C$,where $C$ is an arbitrary constant.

(A) Given integral: $\int \frac{2-3 \sin x}{\cos ^{2} x} d x$
Split the fraction into two parts:
$= \int \left( \frac{2}{\cos ^{2} x} - \frac{3 \sin x}{\cos ^{2} x} \right) d x$
Using trigonometric identities $\frac{1}{\cos ^{2} x} = \sec ^{2} x$ and $\frac{\sin x}{\cos ^{2} x} = \tan x \sec x$:
$= \int 2 \sec ^{2} x \, d x - 3 \int \tan x \sec x \, d x$
Integrating the terms:
$= 2 \tan x - 3 \sec x + C$
where $C$ is an arbitrary constant.

(B) We need to find the integral: $\int \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) dx$
Rewrite the expression using exponents: $\int \left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right) dx$
Apply the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:
For the first term: $\int x^{\frac{1}{2}} dx = \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} x^{\frac{3}{2}}$
For the second term: $\int x^{-\frac{1}{2}} dx = \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2 x^{\frac{1}{2}}$
Combining these results,we get: $\frac{2}{3} x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + C$,where $C$ is an arbitrary constant.
Hence,the correct answer is $B$.

(C) Given that,$\frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}}$.
To find $f(x)$,we integrate both sides with respect to $x$:
$f(x) = \int \left( 4 x^{3} - 3 x^{-4} \right) dx$
$f(x) = 4 \int x^{3} dx - 3 \int x^{-4} dx$
$f(x) = 4 \left( \frac{x^{4}}{4} \right) - 3 \left( \frac{x^{-3}}{-3} \right) + C$
$f(x) = x^{4} + \frac{1}{x^{3}} + C$
Given $f(2) = 0$,we substitute $x = 2$:
$f(2) = (2)^{4} + \frac{1}{(2)^{3}} + C = 0$
$16 + \frac{1}{8} + C = 0$
$C = -\left( 16 + \frac{1}{8} \right) = -\frac{128+1}{8} = -\frac{129}{8}$
Therefore,$f(x) = x^{4} + \frac{1}{x^{3}} - \frac{129}{8}$.
Thus,the correct option is $C$.

(A) To integrate $\int \sin(mx) \, dx$,we use the method of substitution.
Let $t = mx$. Then,differentiating both sides with respect to $x$,we get $dt = m \, dx$,which implies $dx = \frac{1}{m} \, dt$.
Substituting these into the integral:
$\int \sin(mx) \, dx = \int \sin(t) \cdot \frac{1}{m} \, dt$
$= \frac{1}{m} \int \sin(t) \, dt$
$= \frac{1}{m} (-\cos(t)) + C$
$= -\frac{1}{m} \cos(mx) + C$

(A) The given function is $I = \int \sin (ax+b) \cos (ax+b) \, dx$.
We know that $\sin 2\theta = 2 \sin \theta \cos \theta$,so $\sin \theta \cos \theta = \frac{\sin 2\theta}{2}$.
Substituting $\theta = ax+b$,we get:
$I = \int \frac{\sin 2(ax+b)}{2} \, dx = \frac{1}{2} \int \sin(2ax + 2b) \, dx$.
Using the integration formula $\int \sin(kx+c) \, dx = -\frac{\cos(kx+c)}{k} + C$:
$I = \frac{1}{2} \left( -\frac{\cos(2ax+2b)}{2a} \right) + C$.
$I = -\frac{\cos 2(ax+b)}{4a} + C$,where $C$ is an arbitrary constant.

(A) Let $I = \int e^{2x+3} dx$.
Substitute $t = 2x+3$.
Then,$dt = 2 dx$,which implies $dx = \frac{1}{2} dt$.
Substituting these into the integral:
$I = \int e^t \cdot \frac{1}{2} dt$
$I = \frac{1}{2} \int e^t dt$
$I = \frac{1}{2} e^t + C$
Substituting back $t = 2x+3$:
$I = \frac{1}{2} e^{2x+3} + C$,where $C$ is an arbitrary constant.

(N/A) We need to evaluate the integral $\int \tan ^{2}(2 x-3) \, dx$.
Using the trigonometric identity $\tan ^{2} \theta = \sec ^{2} \theta - 1$,we can rewrite the integral as:
$\int \tan ^{2}(2 x-3) \, dx = \int (\sec ^{2}(2 x-3) - 1) \, dx$
Now,we split the integral into two parts:
$= \int \sec ^{2}(2 x-3) \, dx - \int 1 \, dx$
Let $t = 2x - 3$. Then $dt = 2 \, dx$,which implies $dx = \frac{1}{2} \, dt$.
Substituting this into the integral:
$= \frac{1}{2} \int \sec ^{2} t \, dt - x + C$
Since $\int \sec ^{2} t \, dt = \tan t + C$,we have:
$= \frac{1}{2} \tan t - x + C$
Substituting back $t = 2x - 3$:
$= \frac{1}{2} \tan (2 x-3) - x + C$
where $C$ is an arbitrary constant.

(A) Let $I = \int \frac{d x}{\sin ^{2} x \cos ^{2} x}$.
We know that $1 = \sin ^{2} x + \cos ^{2} x$.
Substituting this in the integral,we get:
$I = \int \frac{\sin ^{2} x + \cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$
$I = \int \frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x} d x + \int \frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$
$I = \int \frac{1}{\cos ^{2} x} d x + \int \frac{1}{\sin ^{2} x} d x$
$I = \int \sec ^{2} x d x + \int \csc ^{2} x d x$
Using the standard integration formulas $\int \sec ^{2} x d x = \tan x + C_1$ and $\int \csc ^{2} x d x = -\cot x + C_2$:
$I = \tan x - \cot x + C$.
Hence,the correct option is $A$.

(A) We use the trigonometric identity $\cos 2x = 2 \cos^{2} x - 1$,which implies $\cos^{2} x = \frac{1 + \cos 2x}{2}$.
Substituting this into the integral,we get:
$\int \cos^{2} x \, dx = \int \frac{1 + \cos 2x}{2} \, dx$
$= \frac{1}{2} \int (1 + \cos 2x) \, dx$
$= \frac{1}{2} \left( \int 1 \, dx + \int \cos 2x \, dx \right)$
$= \frac{1}{2} \left( x + \frac{\sin 2x}{2} \right) + C$
$= \frac{x}{2} + \frac{1}{4} \sin 2x + C$

(B) To evaluate the integral $\int \sin 2x \cos 3x \, dx$,we use the trigonometric identity: $\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$.
Here,$A = 2x$ and $B = 3x$.
Substituting these into the identity,we get:
$\sin 2x \cos 3x = \frac{1}{2} [\sin(2x+3x) + \sin(2x-3x)] = \frac{1}{2} [\sin 5x + \sin(-x)]$.
Since $\sin(-x) = -\sin x$,the expression becomes:
$\sin 2x \cos 3x = \frac{1}{2} [\sin 5x - \sin x]$.
Now,integrate both sides:
$\int \sin 2x \cos 3x \, dx = \frac{1}{2} \int (\sin 5x - \sin x) \, dx$.
$= \frac{1}{2} [\int \sin 5x \, dx - \int \sin x \, dx]$.
$= \frac{1}{2} [-\frac{1}{5} \cos 5x - (-\cos x)] + C$.
$= -\frac{1}{10} \cos 5x + \frac{1}{2} \cos x + C$.

We use the trigonometric identity $\sin 3x = 3 \sin x - 4 \sin^{3} x$.
Rearranging this,we get $\sin^{3} x = \frac{3 \sin x - \sin 3x}{4}$.
Now,integrate both sides:
$\int \sin^{3} x \, dx = \int \frac{3 \sin x - \sin 3x}{4} \, dx$
$= \frac{3}{4} \int \sin x \, dx - \frac{1}{4} \int \sin 3x \, dx$
$= \frac{3}{4} (-\cos x) - \frac{1}{4} \left(-\frac{\cos 3x}{3}\right) + C$
$= -\frac{3}{4} \cos x + \frac{1}{12} \cos 3x + C$.
Alternatively,using substitution:
$\int \sin^{3} x \, dx = \int \sin^{2} x \cdot \sin x \, dx = \int (1 - \cos^{2} x) \sin x \, dx$.
Let $t = \cos x$,then $dt = -\sin x \, dx$,or $\sin x \, dx = -dt$.
Substituting these into the integral:
$\int (1 - t^{2}) (-dt) = \int (t^{2} - 1) \, dt = \frac{t^{3}}{3} - t + C$.
Substituting back $t = \cos x$:
$= \frac{1}{3} \cos^{3} x - \cos x + C$.

(A) We know that $\sin ^{2} \theta = \frac{1 - \cos 2\theta}{2}$.
Substituting $\theta = 2x + 5$,we get:
$\sin ^{2}(2 x+5) = \frac{1 - \cos 2(2 x+5)}{2} = \frac{1 - \cos (4 x+10)}{2}$.
Now,we integrate the expression:
$\int \sin ^{2}(2 x+5) dx = \int \frac{1 - \cos (4 x+10)}{2} dx$
$= \frac{1}{2} \int 1 dx - \frac{1}{2} \int \cos (4 x+10) dx$
$= \frac{1}{2} x - \frac{1}{2} \left( \frac{\sin (4 x+10)}{4} \right) + C$
$= \frac{1}{2} x - \frac{1}{8} \sin (4 x+10) + C$.

We use the trigonometric identity: $\sin A \cos B = \frac{1}{2} \{\sin(A+B) + \sin(A-B)\}$.
Applying this to the integral:
$\int \sin 3x \cos 4x \, dx = \int \frac{1}{2} \{\sin(3x+4x) + \sin(3x-4x)\} \, dx$
$= \frac{1}{2} \int \{\sin 7x + \sin(-x)\} \, dx$
Since $\sin(-x) = -\sin x$,we have:
$= \frac{1}{2} \int (\sin 7x - \sin x) \, dx$
$= \frac{1}{2} \int \sin 7x \, dx - \frac{1}{2} \int \sin x \, dx$
$= \frac{1}{2} \left( \frac{-\cos 7x}{7} \right) - \frac{1}{2} (-\cos x) + C$
$= -\frac{\cos 7x}{14} + \frac{\cos x}{2} + C$,where $C$ is an arbitrary constant.

We use the trigonometric identity $\cos A \cos B = \frac{1}{2} \{\cos(A+B) + \cos(A-B)\}$.
First,consider the integral $I = \int \cos 2x \cos 4x \cos 6x \, dx$.
Using the identity for $\cos 4x \cos 6x$:
$I = \int \cos 2x \left[ \frac{1}{2} (\cos(4x+6x) + \cos(4x-6x)) \right] dx$
$I = \frac{1}{2} \int \cos 2x (\cos 10x + \cos(-2x)) \, dx$
$I = \frac{1}{2} \int (\cos 2x \cos 10x + \cos^2 2x) \, dx$
Using $\cos^2 2x = \frac{1 + \cos 4x}{2}$ and the identity for $\cos 2x \cos 10x$:
$I = \frac{1}{2} \int \left[ \frac{1}{2} (\cos 12x + \cos(-8x)) + \frac{1 + \cos 4x}{2} \right] dx$
$I = \frac{1}{4} \int (\cos 12x + \cos 8x + 1 + \cos 4x) \, dx$
Integrating term by term:
$I = \frac{1}{4} \left[ \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + x + \frac{\sin 4x}{4} \right] + C$
$I = \frac{\sin 12x}{48} + \frac{\sin 8x}{32} + \frac{x}{4} + \frac{\sin 4x}{16} + C$

We use the trigonometric identity $\sin A \sin B = \frac{1}{2} \{\cos(A-B) - \cos(A+B)\}$.
First,consider $\sin 2x \sin 3x = \frac{1}{2} \{\cos(2x-3x) - \cos(2x+3x)\} = \frac{1}{2} \{\cos(-x) - \cos(5x)\} = \frac{1}{2} \{\cos x - \cos 5x\}$.
Now,the integral becomes $\int \sin x \cdot \frac{1}{2} \{\cos x - \cos 5x\} \, dx = \frac{1}{2} \int (\sin x \cos x - \sin x \cos 5x) \, dx$.
Using $\sin x \cos x = \frac{\sin 2x}{2}$ and $\sin A \cos B = \frac{1}{2} \{\sin(A+B) + \sin(A-B)\}$:
$= \frac{1}{2} \int \frac{\sin 2x}{2} \, dx - \frac{1}{2} \int \frac{1}{2} \{\sin(x+5x) + \sin(x-5x)\} \, dx$
$= \frac{1}{4} \int \sin 2x \, dx - \frac{1}{4} \int (\sin 6x - \sin 4x) \, dx$
$= \frac{1}{4} \left( \frac{-\cos 2x}{2} \right) - \frac{1}{4} \left( \frac{-\cos 6x}{6} + \frac{\cos 4x}{4} \right) + C$
$= -\frac{\cos 2x}{8} + \frac{\cos 6x}{24} - \frac{\cos 4x}{16} + C$.

We use the trigonometric identity: $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$.
Applying this to the integral:
$\int \sin 4x \sin 8x \, dx = \int \frac{1}{2} [\cos(4x-8x) - \cos(4x+8x)] \, dx$
$= \frac{1}{2} \int [\cos(-4x) - \cos(12x)] \, dx$
Since $\cos(-\theta) = \cos(\theta)$,we have:
$= \frac{1}{2} \int (\cos 4x - \cos 12x) \, dx$
Integrating term by term:
$= \frac{1}{2} [\frac{\sin 4x}{4} - \frac{\sin 12x}{12}] + C$
$= \frac{\sin 4x}{8} - \frac{\sin 12x}{24} + C$,where $C$ is an arbitrary constant.

We need to evaluate the integral $I = \int \frac{1-\cos x}{1+\cos x} dx$.
Using the trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$,we have:
$\frac{1-\cos x}{1+\cos x} = \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \tan^2 \frac{x}{2}$.
Using the identity $\tan^2 \theta = \sec^2 \theta - 1$,we can write:
$\tan^2 \frac{x}{2} = \sec^2 \frac{x}{2} - 1$.
Now,integrate the expression:
$I = \int (\sec^2 \frac{x}{2} - 1) dx = \int \sec^2 \frac{x}{2} dx - \int 1 dx$.
The integral of $\sec^2(ax)$ is $\frac{1}{a} \tan(ax) + C$.
Therefore,$I = \frac{\tan(x/2)}{1/2} - x + C = 2 \tan \frac{x}{2} - x + C$,where $C$ is an arbitrary constant.

We need to evaluate the integral $I = \int \frac{\cos x}{1+\cos x} dx$.
First,we simplify the integrand:
$\frac{\cos x}{1+\cos x} = \frac{\cos x + 1 - 1}{1+\cos x} = 1 - \frac{1}{1+\cos x}$.
Using the identity $1+\cos x = 2\cos^2 \frac{x}{2}$,we get:
$1 - \frac{1}{2\cos^2 \frac{x}{2}} = 1 - \frac{1}{2}\sec^2 \frac{x}{2}$.
Now,integrate term by term:
$I = \int (1 - \frac{1}{2}\sec^2 \frac{x}{2}) dx = \int 1 dx - \frac{1}{2} \int \sec^2 \frac{x}{2} dx$.
$= x - \frac{1}{2} \cdot \frac{\tan \frac{x}{2}}{\frac{1}{2}} + C$.
$= x - \tan \frac{x}{2} + C$,where $C$ is an arbitrary constant.

We know that $\cos^{2} \theta = \frac{1+\cos 2\theta}{2}$.
$\cos^{4} 2x = (\cos^{2} 2x)^{2} = \left(\frac{1+\cos 4x}{2}\right)^{2}$
$= \frac{1}{4} (1 + 2\cos 4x + \cos^{2} 4x)$
$= \frac{1}{4} \left[1 + 2\cos 4x + \frac{1+\cos 8x}{2}\right]$
$= \frac{1}{4} \left[1 + 2\cos 4x + \frac{1}{2} + \frac{\cos 8x}{2}\right]$
$= \frac{1}{4} \left[\frac{3}{2} + 2\cos 4x + \frac{\cos 8x}{2}\right] = \frac{3}{8} + \frac{1}{2}\cos 4x + \frac{1}{8}\cos 8x$
Now,integrating with respect to $x$:
$\int \cos^{4} 2x \, dx = \int \left(\frac{3}{8} + \frac{1}{2}\cos 4x + \frac{1}{8}\cos 8x\right) \, dx$
$= \frac{3}{8}x + \frac{1}{2} \cdot \frac{\sin 4x}{4} + \frac{1}{8} \cdot \frac{\sin 8x}{8} + C$
$= \frac{3}{8}x + \frac{\sin 4x}{8} + \frac{\sin 8x}{64} + C$,where $C$ is an arbitrary constant.

We are given the integral $I = \int \frac{\sin ^{2} x}{1+\cos x} dx$.
Using the trigonometric identities $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos ^{2} \frac{x}{2}$,we can simplify the integrand:
$\frac{\sin ^{2} x}{1+\cos x} = \frac{(2 \sin \frac{x}{2} \cos \frac{x}{2})^{2}}{2 \cos ^{2} \frac{x}{2}}$
$= \frac{4 \sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}$
$= 2 \sin ^{2} \frac{x}{2}$
Using the identity $2 \sin ^{2} \theta = 1 - \cos 2\theta$,we have $2 \sin ^{2} \frac{x}{2} = 1 - \cos x$.
Therefore,$I = \int (1 - \cos x) dx$
$= x - \sin x + C$,where $C$ is an arbitrary constant.

Consider the expression $\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}$.
Using the formula $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$,we have:
$\frac{-2 \sin(x+\alpha) \sin(x-\alpha)}{-2 \sin(\frac{x+\alpha}{2}) \sin(\frac{x-\alpha}{2})} = \frac{\sin(x+\alpha) \sin(x-\alpha)}{\sin(\frac{x+\alpha}{2}) \sin(\frac{x-\alpha}{2})}$
Using $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$= \frac{[2 \sin(\frac{x+\alpha}{2}) \cos(\frac{x+\alpha}{2})] [2 \sin(\frac{x-\alpha}{2}) \cos(\frac{x-\alpha}{2})]}{\sin(\frac{x+\alpha}{2}) \sin(\frac{x-\alpha}{2})}$
$= 4 \cos(\frac{x+\alpha}{2}) \cos(\frac{x-\alpha}{2})$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$= 2[\cos(\frac{x+\alpha}{2} + \frac{x-\alpha}{2}) + \cos(\frac{x+\alpha}{2} - \frac{x-\alpha}{2})]$
$= 2[\cos x + \cos \alpha] = 2 \cos x + 2 \cos \alpha$
Now,integrating with respect to $x$:
$\int (2 \cos x + 2 \cos \alpha) dx = 2 \sin x + 2x \cos \alpha + C$,where $C$ is an arbitrary constant.

(N/A) To find the integral $I = \int \tan ^{4} x \, dx$,we can simplify the integrand as follows:
$\tan ^{4} x = \tan ^{2} x \cdot \tan ^{2} x$
Using the identity $\tan ^{2} x = \sec ^{2} x - 1$,we get:
$\tan ^{4} x = (\sec ^{2} x - 1) \tan ^{2} x = \sec ^{2} x \tan ^{2} x - \tan ^{2} x$
Substituting $\tan ^{2} x = \sec ^{2} x - 1$ again:
$\tan ^{4} x = \sec ^{2} x \tan ^{2} x - (\sec ^{2} x - 1) = \sec ^{2} x \tan ^{2} x - \sec ^{2} x + 1$
Now,integrating term by term:
$\int \tan ^{4} x \, dx = \int \sec ^{2} x \tan ^{2} x \, dx - \int \sec ^{2} x \, dx + \int 1 \, dx$
For the first integral,let $u = \tan x$,then $du = \sec ^{2} x \, dx$. Thus,$\int \sec ^{2} x \tan ^{2} x \, dx = \int u^{2} \, du = \frac{u^{3}}{3} = \frac{\tan ^{3} x}{3}$.
Substituting this back:
$\int \tan ^{4} x \, dx = \frac{\tan ^{3} x}{3} - \tan x + x + C$,where $C$ is the constant of integration.

(A) We are given the integral $I = \int \frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} dx$.
First,simplify the integrand:
$\frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} = \frac{\sin ^{3} x}{\sin ^{2} x \cos ^{2} x} + \frac{\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}$
$= \frac{\sin x}{\cos ^{2} x} + \frac{\cos x}{\sin ^{2} x}$
$= \tan x \sec x + \cot x \csc x$
Now,integrate each term:
$\int (\tan x \sec x + \cot x \csc x) dx = \int \tan x \sec x dx + \int \cot x \csc x dx$
$= \sec x - \csc x + C$,where $C$ is an arbitrary constant.

Given the integral: $\int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} dx$
Using the trigonometric identity $\cos 2x = 1 - 2\sin^2 x$,we can rewrite the numerator:
$\cos 2x + 2\sin^2 x = (1 - 2\sin^2 x) + 2\sin^2 x = 1$
Substituting this into the integral:
$\int \frac{1}{\cos^2 x} dx$
Since $\frac{1}{\cos^2 x} = \sec^2 x$:
$\int \sec^2 x dx = \tan x + C$
where $C$ is an arbitrary constant.

(A) We know that $\sin ^{-1}(\cos x) = \sin ^{-1}(\sin(\frac{\pi}{2}-x))$.
Since $\sin ^{-1}(\sin \theta) = \theta$ for $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $\sin ^{-1}(\cos x) = \frac{\pi}{2}-x$.
Now,we integrate the function:
$\int \sin ^{-1}(\cos x) \, dx = \int (\frac{\pi}{2}-x) \, dx$
$= \int \frac{\pi}{2} \, dx - \int x \, dx$
$= \frac{\pi}{2}x - \frac{x^{2}}{2} + C_{1}$.

(C) We are given the integral $I = \int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$.
Split the fraction into two separate terms:
$I = \int \left( \frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x} - \frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} \right) d x$.
Simplify each term:
$I = \int \left( \frac{1}{\cos ^{2} x} - \frac{1}{\sin ^{2} x} \right) d x$.
Using trigonometric identities $\frac{1}{\cos ^{2} x} = \sec ^{2} x$ and $\frac{1}{\sin ^{2} x} = \operatorname{cosec} ^{2} x$:
$I = \int (\sec ^{2} x - \operatorname{cosec} ^{2} x) d x$.
Integrating term by term:
$\int \sec ^{2} x d x = \tan x$ and $\int \operatorname{cosec} ^{2} x d x = -\cot x$.
Therefore,$I = \tan x - (-\cot x) + C = \tan x + \cot x + C$.
Hence,the correct option is $C$.

(A) We know that the standard integral formula is $\int \frac{dx}{x^{2}-a^{2}} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C$.
Given integral is $I = \int \frac{dx}{x^{2}-16}$.
We can write $16$ as $4^{2}$,so $I = \int \frac{dx}{x^{2}-4^{2}}$.
Here,$a = 4$.
Applying the formula,we get $I = \frac{1}{2(4)} \log \left| \frac{x-4}{x+4} \right| + C$.
Therefore,$I = \frac{1}{8} \log \left| \frac{x-4}{x+4} \right| + C$.

(A) We have $x^{2}-6 x+13 = x^{2}-6 x+9-9+13 = (x-3)^{2}+4 = (x-3)^{2}+2^{2}$.
So,$\int \frac{d x}{x^{2}-6 x+13} = \int \frac{d x}{(x-3)^{2}+2^{2}}$.
Let $x-3 = t$. Then $dx = dt$.
Using the standard integral formula $\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C$,we get:
$\int \frac{dt}{t^{2}+2^{2}} = \frac{1}{2} \tan^{-1} \left( \frac{t}{2} \right) + C$.
Substituting $t = x-3$ back,we get:
$= \frac{1}{2} \tan^{-1} \left( \frac{x-3}{2} \right) + C$.

(A) The given integral is of the form $\int \frac{d x}{a x^{2}+b x+c}$. We write the denominator of the integrand as:
$3 x^{2}+13 x-10 = 3\left(x^{2}+\frac{13}{3}x-\frac{10}{3}\right)$
$= 3\left[\left(x+\frac{13}{6}\right)^{2}-\left(\frac{13}{6}\right)^{2}-\frac{10}{3}\right]$
$= 3\left[\left(x+\frac{13}{6}\right)^{2}-\frac{169}{36}-\frac{120}{36}\right] = 3\left[\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}\right]$
Thus,$\int \frac{d x}{3 x^{2}+13 x-10} = \frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}}$
Using the formula $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + C$,we get:
$= \frac{1}{3} \times \frac{1}{2 \times \frac{17}{6}} \log \left|\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right| + C$
$= \frac{1}{17} \log \left|\frac{x-\frac{4}{6}}{x+\frac{30}{6}}\right| + C = \frac{1}{17} \log \left|\frac{x-\frac{2}{3}}{x+5}\right| + C$
$= \frac{1}{17} \log \left|\frac{3x-2}{3(x+5)}\right| + C = \frac{1}{17} \log \left|\frac{3x-2}{x+5}\right| + C'$

We have $\int \frac{dx}{\sqrt{5x^{2}-2x}} = \int \frac{dx}{\sqrt{5(x^{2}-\frac{2x}{5})}}$.
$= \frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{(x-\frac{1}{5})^{2}-(\frac{1}{5})^{2}}}$ (completing the square).
Let $t = x - \frac{1}{5}$,then $dx = dt$.
Therefore,$\int \frac{dx}{\sqrt{5x^{2}-2x}} = \frac{1}{\sqrt{5}} \int \frac{dt}{\sqrt{t^{2}-(\frac{1}{5})^{2}}}$.
Using the standard formula $\int \frac{dx}{\sqrt{x^{2}-a^{2}}} = \log |x + \sqrt{x^{2}-a^{2}}| + C$,we get:
$= \frac{1}{\sqrt{5}} \log |t + \sqrt{t^{2}-(\frac{1}{5})^{2}}| + C$.
Substituting $t = x - \frac{1}{5}$ back,we get:
$= \frac{1}{\sqrt{5}} \log |x - \frac{1}{5} + \sqrt{x^{2}-\frac{2x}{5}}| + C$.

Let $2x = t$.
Then,$2 dx = dt$,which implies $dx = \frac{1}{2} dt$.
Substituting these into the integral:
$\int \frac{1}{\sqrt{1+4x^2}} dx = \int \frac{1}{\sqrt{1+t^2}} \cdot \frac{1}{2} dt = \frac{1}{2} \int \frac{1}{\sqrt{1+t^2}} dt$.
Using the standard integral formula $\int \frac{1}{\sqrt{x^2+a^2}} dx = \log |x + \sqrt{x^2+a^2}| + C$:
$= \frac{1}{2} \log |t + \sqrt{t^2+1}| + C$.
Substituting $t = 2x$ back into the expression:
$= \frac{1}{2} \log |2x + \sqrt{4x^2+1}| + C$,where $C$ is an arbitrary constant.