Fundamental integration Questions in English

(A) Let $2-x=t$.
Then,$-dx = dt$,which implies $dx = -dt$.
Substituting these into the integral:
$\int \frac{1}{\sqrt{(2-x)^{2}+1}} dx = -\int \frac{1}{\sqrt{t^{2}+1}} dt$.
Using the standard integral formula $\int \frac{1}{\sqrt{x^{2}+a^{2}}} dx = \log |x + \sqrt{x^{2}+a^{2}}| + C$,we get:
$= -\log |t + \sqrt{t^{2}+1}| + C$.
Substituting $t = 2-x$ back into the expression:
$= -\log |2-x + \sqrt{(2-x)^{2}+1}| + C$.
Since $-\log|u| = \log|1/u|$,this can be written as:
$= \log \left| \frac{1}{(2-x) + \sqrt{x^{2}-4x+5}} \right| + C$,where $C$ is an arbitrary constant.

(A) Let $5x = t$.
Then,$5 dx = dt$,which implies $dx = \frac{1}{5} dt$.
Substituting these into the integral:
$\int \frac{1}{\sqrt{9-25 x^{2}}} dx = \int \frac{1}{\sqrt{3^{2}-(5x)^{2}}} dx$
$= \frac{1}{5} \int \frac{1}{\sqrt{3^{2}-t^{2}}} dt$
Using the standard integral formula $\int \frac{1}{\sqrt{a^{2}-x^{2}}} dx = \sin^{-1}(\frac{x}{a}) + C$:
$= \frac{1}{5} \sin^{-1}(\frac{t}{3}) + C$
Substituting $t = 5x$ back:
$= \frac{1}{5} \sin^{-1}(\frac{5x}{3}) + C$,where $C$ is an arbitrary constant.

(N/A) To integrate $\int \frac{1}{\sqrt{x^{2}+2x+2}} dx$,we first complete the square for the quadratic expression in the denominator.
$x^{2}+2x+2 = (x^{2}+2x+1)+1 = (x+1)^{2}+1^{2}$.
Now,the integral becomes $\int \frac{1}{\sqrt{(x+1)^{2}+1^{2}}} dx$.
Let $t = x+1$,then $dt = dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{t^{2}+a^{2}}} dt = \log |t + \sqrt{t^{2}+a^{2}}| + C$,where $a=1$:
$\int \frac{1}{\sqrt{t^{2}+1^{2}}} dt = \log |t + \sqrt{t^{2}+1}| + C$.
Substituting $t = x+1$ back into the expression:
$= \log |(x+1) + \sqrt{(x+1)^{2}+1}| + C$
$= \log |(x+1) + \sqrt{x^{2}+2x+2}| + C$,where $C$ is an arbitrary constant.

(N/A) To integrate $\int \frac{1}{\sqrt{7-6x-x^{2}}} dx$,we first complete the square for the quadratic expression $7-6x-x^{2}$.
$7-6x-x^{2} = 7 - (x^{2} + 6x)$
$= 7 - (x^{2} + 6x + 9 - 9)$
$= 7 - ((x+3)^{2} - 9)$
$= 7 + 9 - (x+3)^{2}$
$= 16 - (x+3)^{2}$
$= (4)^{2} - (x+3)^{2}$
Now,the integral becomes:
$\int \frac{1}{\sqrt{(4)^{2} - (x+3)^{2}}} dx$
Let $u = x+3$,then $du = dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{a^{2}-u^{2}}} du = \sin^{-1}(\frac{u}{a}) + C$,we get:
$= \sin^{-1}(\frac{x+3}{4}) + C$,where $C$ is the constant of integration.

(C) To evaluate the integral $\int \frac{d x}{x^{2}+2 x+2}$,we first complete the square in the denominator.
$x^{2}+2 x+2 = (x^{2}+2 x+1) + 1 = (x+1)^{2} + 1^{2}$.
Now,the integral becomes $\int \frac{d x}{(x+1)^{2} + 1^{2}}$.
Using the standard integration formula $\int \frac{d u}{u^{2} + a^{2}} = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + C$,where $u = x+1$ and $a = 1$,we get:
$\int \frac{d x}{(x+1)^{2} + 1^{2}} = \tan^{-1}(x+1) + C$.
Thus,the correct option is $C$.

(D) To evaluate the integral $I = \int \frac{dx}{\sqrt{9x-4x^{2}}}$,we complete the square inside the square root.
$I = \int \frac{dx}{\sqrt{-4(x^{2}-\frac{9}{4}x)}}$
$I = \int \frac{dx}{\sqrt{-4(x^{2}-\frac{9}{4}x + \frac{81}{64} - \frac{81}{64})}}$
$I = \int \frac{dx}{\sqrt{-4[(x-\frac{9}{8})^{2} - (\frac{9}{8})^{2}]}}$
$I = \int \frac{dx}{2\sqrt{(\frac{9}{8})^{2} - (x-\frac{9}{8})^{2}}}$
Using the standard integral formula $\int \frac{dy}{\sqrt{a^{2}-y^{2}}} = \sin^{-1}(\frac{y}{a}) + C$,where $a = \frac{9}{8}$ and $y = x - \frac{9}{8}$:
$I = \frac{1}{2} \sin^{-1}\left(\frac{x-\frac{9}{8}}{\frac{9}{8}}\right) + C$
$I = \frac{1}{2} \sin^{-1}\left(\frac{8x-9}{9}\right) + C$
Thus,the correct option is $D$.

(N/A) Let $I = \int \sqrt{4-x^{2}} \, dx = \int \sqrt{(2)^{2}-(x)^{2}} \, dx$.
We use the standard integration formula: $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right) + C$.
Here,$a = 2$.
Substituting $a = 2$ into the formula:
$I = \frac{x}{2} \sqrt{4-x^{2}} + \frac{4}{2} \sin^{-1} \left(\frac{x}{2}\right) + C$.
Simplifying the expression:
$I = \frac{x}{2} \sqrt{4-x^{2}} + 2 \sin^{-1} \left(\frac{x}{2}\right) + C$,where $C$ is an arbitrary constant.

Let $I = \int \sqrt{1-4x^2} dx = \int \sqrt{(1)^2 - (2x)^2} dx$.
Substitute $2x = t$,then $2 dx = dt$,which implies $dx = \frac{1}{2} dt$.
Substituting these into the integral:
$I = \frac{1}{2} \int \sqrt{1^2 - t^2} dt$.
Using the standard formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) + C$:
$I = \frac{1}{2} \left[ \frac{t}{2} \sqrt{1 - t^2} + \frac{1^2}{2} \sin^{-1} \left( \frac{t}{1} \right) \right] + C$.
Simplifying the expression:
$I = \frac{t}{4} \sqrt{1 - t^2} + \frac{1}{4} \sin^{-1} (t) + C$.
Substituting $t = 2x$ back into the expression:
$I = \frac{2x}{4} \sqrt{1 - (2x)^2} + \frac{1}{4} \sin^{-1} (2x) + C$.
$I = \frac{x}{2} \sqrt{1 - 4x^2} + \frac{1}{4} \sin^{-1} (2x) + C$,where $C$ is the constant of integration.

Let $I = \int \sqrt{1-4x-x^{2}} dx$.
To integrate this,we complete the square inside the square root:
$1-4x-x^{2} = 1 - (x^{2} + 4x) = 1 - (x^{2} + 4x + 4 - 4) = 1 - ((x+2)^{2} - 4) = 5 - (x+2)^{2}$.
So,$I = \int \sqrt{(\sqrt{5})^{2} - (x+2)^{2}} dx$.
Using the standard integral formula $\int \sqrt{a^{2} - t^{2}} dt = \frac{t}{2} \sqrt{a^{2} - t^{2}} + \frac{a^{2}}{2} \sin^{-1}(\frac{t}{a}) + C$,where $t = x+2$ and $a = \sqrt{5}$:
$I = \frac{x+2}{2} \sqrt{5 - (x+2)^{2}} + \frac{5}{2} \sin^{-1}(\frac{x+2}{\sqrt{5}}) + C$.
Simplifying the term under the square root back to the original form:
$I = \frac{x+2}{2} \sqrt{1-4x-x^{2}} + \frac{5}{2} \sin^{-1}(\frac{x+2}{\sqrt{5}}) + C$,where $C$ is an arbitrary constant.

Let $I = \int \sqrt{1+\frac{x^{2}}{9}} \, dx$.
We can rewrite the integrand as:
$I = \int \sqrt{\frac{9+x^{2}}{9}} \, dx = \frac{1}{3} \int \sqrt{9+x^{2}} \, dx = \frac{1}{3} \int \sqrt{3^{2}+x^{2}} \, dx$.
Using the standard integration formula $\int \sqrt{x^{2}+a^{2}} \, dx = \frac{x}{2} \sqrt{x^{2}+a^{2}} + \frac{a^{2}}{2} \ln |x + \sqrt{x^{2}+a^{2}}| + C$,where $a = 3$:
$I = \frac{1}{3} \left[ \frac{x}{2} \sqrt{x^{2}+3^{2}} + \frac{3^{2}}{2} \ln |x + \sqrt{x^{2}+3^{2}}| \right] + C$.
$I = \frac{1}{3} \left[ \frac{x}{2} \sqrt{x^{2}+9} + \frac{9}{2} \ln |x + \sqrt{x^{2}+9}| \right] + C$.
$I = \frac{x}{6} \sqrt{x^{2}+9} + \frac{3}{2} \ln |x + \sqrt{x^{2}+9}| + C$,where $C$ is an arbitrary constant.

(D) We use the standard integration formula for $\int \sqrt{x^{2}+a^{2}} \, dx$:
$\int \sqrt{x^{2}+a^{2}} \, dx = \frac{x}{2} \sqrt{x^{2}+a^{2}} + \frac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}| + C$
Comparing $\int \sqrt{1+x^{2}} \, dx$ with the standard form,we have $a = 1$.
Substituting $a = 1$ into the formula:
$\int \sqrt{1+x^{2}} \, dx = \frac{x}{2} \sqrt{1+x^{2}} + \frac{1^{2}}{2} \log |x+\sqrt{1+x^{2}}| + C$
$= \frac{x}{2} \sqrt{1+x^{2}} + \frac{1}{2} \log |x+\sqrt{1+x^{2}}| + C$
Thus,the correct option is $D$.

To integrate $\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} dx$,we first rationalize the denominator:
$\frac{1}{\sqrt{x+a}+\sqrt{x+b}} = \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}$
$= \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} = \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b}$
Now,integrate the expression:
$\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} dx = \frac{1}{a-b} \int (\sqrt{x+a}-\sqrt{x+b}) dx$
$= \frac{1}{a-b} \left[ \int (x+a)^{\frac{1}{2}} dx - \int (x+b)^{\frac{1}{2}} dx \right]$
$= \frac{1}{a-b} \left[ \frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}} - \frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}} \right] + C$
$= \frac{2}{3(a-b)} \left[ (x+a)^{\frac{3}{2}} - (x+b)^{\frac{3}{2}} \right] + C$

Given the integral $I = \int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} dx$.
Using the property $a \log b = \log b^a$ and $e^{\log x} = x$,we simplify the integrand:
$\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} = \frac{e^{\log x^5} - e^{\log x^4}}{e^{\log x^3} - e^{\log x^2}} = \frac{x^5 - x^4}{x^3 - x^2}$
Factor out the terms in the numerator and denominator:
$= \frac{x^4(x - 1)}{x^2(x - 1)}$
For $x \neq 1$ and $x \neq 0$,we can cancel $(x - 1)$ and $x^2$:
$= x^2$
Now,integrate the simplified function:
$\int x^2 dx = \frac{x^3}{3} + C$,where $C$ is the constant of integration.

(D) Let $I = \int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} \, dx$.
We know that $\sin ^{8} x - \cos ^{8} x = (\sin ^{4} x - \cos ^{4} x)(\sin ^{4} x + \cos ^{4} x) = (\sin ^{2} x - \cos ^{2} x)(\sin ^{2} x + \cos ^{2} x)(\sin ^{4} x + \cos ^{4} x)$.
Since $\sin ^{2} x + \cos ^{2} x = 1$,we have $\sin ^{8} x - \cos ^{8} x = (\sin ^{2} x - \cos ^{2} x)(\sin ^{4} x + \cos ^{4} x)$.
Also,note that $\sin ^{4} x + \cos ^{4} x = (\sin ^{2} x + \cos ^{2} x)^{2} - 2 \sin ^{2} x \cos ^{2} x = 1 - 2 \sin ^{2} x \cos ^{2} x$.
Substituting these into the integrand:
$\frac{(\sin ^{2} x - \cos ^{2} x)(1 - 2 \sin ^{2} x \cos ^{2} x)}{1 - 2 \sin ^{2} x \cos ^{2} x} = \sin ^{2} x - \cos ^{2} x = -(\cos ^{2} x - \sin ^{2} x) = -\cos 2x$.
Therefore,$I = \int -\cos 2x \, dx = -\frac{\sin 2x}{2} + C$.

(N/A) Let $I=\int \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}} d x$.
Since $\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}=\frac{\pi}{2}$,we have $\cos ^{-1} \sqrt{x}=\frac{\pi}{2}-\sin ^{-1} \sqrt{x}$.
Substituting this into the integral:
$I=\int \frac{\sin ^{-1} \sqrt{x}-(\frac{\pi}{2}-\sin ^{-1} \sqrt{x})}{\frac{\pi}{2}} d x = \frac{2}{\pi} \int (2 \sin ^{-1} \sqrt{x}-\frac{\pi}{2}) d x = \frac{4}{\pi} \int \sin ^{-1} \sqrt{x} d x - x$.
Let $I_1 = \int \sin ^{-1} \sqrt{x} d x$. Put $\sqrt{x}=t \Rightarrow x=t^2 \Rightarrow dx=2t dt$.
$I_1 = \int \sin ^{-1} t \cdot 2t dt = 2 [\frac{t^2}{2} \sin ^{-1} t - \int \frac{t^2}{2\sqrt{1-t^2}} dt] = t^2 \sin ^{-1} t - \int \frac{t^2}{\sqrt{1-t^2}} dt$.
Using $\int \frac{t^2}{\sqrt{1-t^2}} dt = \int \frac{-(1-t^2)+1}{\sqrt{1-t^2}} dt = -\int \sqrt{1-t^2} dt + \int \frac{1}{\sqrt{1-t^2}} dt = -[\frac{t}{2}\sqrt{1-t^2} + \frac{1}{2}\sin ^{-1} t] + \sin ^{-1} t = -\frac{t}{2}\sqrt{1-t^2} + \frac{1}{2}\sin ^{-1} t$.
Thus,$I_1 = t^2 \sin ^{-1} t + \frac{t}{2}\sqrt{1-t^2} - \frac{1}{2}\sin ^{-1} t = (t^2 - \frac{1}{2}) \sin ^{-1} t + \frac{t}{2}\sqrt{1-t^2}$.
Substituting back $t=\sqrt{x}$:
$I = \frac{4}{\pi} [(x - \frac{1}{2}) \sin ^{-1} \sqrt{x} + \frac{\sqrt{x}}{2}\sqrt{1-x}] - x + C = \frac{2(2x-1)}{\pi} \sin ^{-1} \sqrt{x} + \frac{2}{\pi} \sqrt{x-x^2} - x + C$.

(A) Let $I = \int \tan ^{-1}(\sec x+\tan x) d x$.
We know that $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$,so $\sec x + \tan x = \frac{1+\sin x}{\cos x}$.
Using half-angle formulas,$1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = (\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})$.
Thus,$\frac{1+\sin x}{\cos x} = \frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})} = \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}$.
Dividing numerator and denominator by $\cos \frac{x}{2}$,we get $\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}} = \tan(\frac{\pi}{4} + \frac{x}{2})$.
Therefore,$I = \int \tan^{-1}(\tan(\frac{\pi}{4} + \frac{x}{2})) d x = \int (\frac{\pi}{4} + \frac{x}{2}) d x$.
Integrating term by term,we get $I = \frac{\pi x}{4} + \frac{x^2}{4} + c$.

(C) Given $f^{\prime}(x)=x-\frac{5}{x^5}$.
Integrating both sides with respect to $x$,we get:
$f(x) = \int \left(x - 5x^{-5}\right) dx$
$f(x) = \frac{x^2}{2} - 5 \left( \frac{x^{-4}}{-4} \right) + c$
$f(x) = \frac{x^2}{2} + \frac{5}{4x^4} + c$
Given $f(1) = 4$,substitute $x=1$:
$4 = \frac{1^2}{2} + \frac{5}{4(1)^4} + c$
$4 = \frac{1}{2} + \frac{5}{4} + c$
$4 = \frac{2+5}{4} + c$
$4 = \frac{7}{4} + c$
$c = 4 - \frac{7}{4} = \frac{16-7}{4} = \frac{9}{4}$
Thus,$f(x) = \frac{x^2}{2} + \frac{5}{4x^4} + \frac{9}{4}$.

(A) Given velocity $v = 6t - \frac{t^2}{6}$.
We know that $v = \frac{ds}{dt}$,therefore $ds = v \, dt$.
Integrating both sides: $\int ds = \int (6t - \frac{t^2}{6}) dt$.
$s = 6 \frac{t^2}{2} - \frac{1}{6} \frac{t^3}{3} + C = 3t^2 - \frac{t^3}{18} + C$.
Given that $s = 0$ at $t = 0$,substituting these values gives $0 = 3(0)^2 - \frac{(0)^3}{18} + C$,so $C = 0$.
Thus,the displacement equation is $s = 3t^2 - \frac{t^3}{18}$.
To find the distance travelled in $3 \text{ s}$,we calculate $s$ at $t = 3$:
$s(3) = 3(3)^2 - \frac{(3)^3}{18} = 3(9) - \frac{27}{18} = 27 - \frac{3}{2} = \frac{54 - 3}{2} = \frac{51}{2} \text{ units}$.

(C) Given,$v = \frac{ds}{dt} = 6t - \frac{t^2}{6}$.
Integrating both sides with respect to $t$,we get:
$s = \int (6t - \frac{t^2}{6}) dt = 3t^2 - \frac{t^3}{18} + C$.
Given that $s = 0$ at $t = 0$,we substitute these values to find the constant $C$:
$0 = 3(0)^2 - \frac{(0)^3}{18} + C \implies C = 0$.
Thus,the displacement function is $s(t) = 3t^2 - \frac{t^3}{18}$.
To find the distance traveled in $3 \ s$,we calculate $s(3)$:
$s(3) = 3(3)^2 - \frac{(3)^3}{18} = 3(9) - \frac{27}{18} = 27 - \frac{3}{2} = \frac{54 - 3}{2} = \frac{51}{2}$.

(C) Given the integral $I = \int \frac{e^{2030 \log x}-e^{2029 \log x}}{e^{2028 \log x}-e^{2027 \log x}} \,d x$.
Using the property $e^{n \log x} = e^{\log x^n} = x^n$,we can rewrite the integrand:
$I = \int \frac{x^{2030} - x^{2029}}{x^{2028} - x^{2027}} \,d x$.
Factor out the highest common powers from the numerator and denominator:
$I = \int \frac{x^{2029}(x - 1)}{x^{2027}(x - 1)} \,d x$.
Assuming $x \neq 1$,we can cancel the term $(x - 1)$:
$I = \int \frac{x^{2029}}{x^{2027}} \,d x = \int x^{2029 - 2027} \,d x = \int x^2 \,d x$.
Integrating $x^2$ with respect to $x$,we get:
$I = \frac{x^3}{3} + c$,where $c$ is the constant of integration.

(C) We know that $\cos 2\theta = 2\cos^2 \theta - 1$.
Substituting this into the integral,we get:
$\int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} dx$
$= \int \frac{2\cos^2 x - 2\cos^2 \alpha}{\cos x - \cos \alpha} dx$
$= 2 \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} dx$
$= 2 \int (\cos x + \cos \alpha) dx$
$= 2 (\sin x + x \cos \alpha) + c$
$= 2 \sin x + 2x \cos \alpha + c$.

(B) We have the integral $I = \int \frac{dx}{\sin^2 x \cos^2 x}$.
Since $1 = \sin^2 x + \cos^2 x$,we can write the integral as:
$I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx$
$I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) dx$
$I = \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) dx$
$I = \int (\sec^2 x + \csc^2 x) dx$
Using the standard integrals $\int \sec^2 x dx = \tan x$ and $\int \csc^2 x dx = -\cot x$,we get:
$I = \tan x - \cot x + c$.

(B) Let $I = \int \frac{x^2+1}{(x+1)^2} dx$.
We can rewrite the numerator as $x^2+1 = (x^2+2x+1) - 2x = (x+1)^2 - 2x$.
So,$I = \int \frac{(x+1)^2 - 2x}{(x+1)^2} dx = \int 1 dx - \int \frac{2x}{(x+1)^2} dx$.
For the second integral,write $2x = 2(x+1) - 2$.
$I = x - \int \frac{2(x+1)-2}{(x+1)^2} dx = x - 2 \int \frac{1}{x+1} dx + 2 \int \frac{1}{(x+1)^2} dx$.
Integrating these terms,we get $I = x - 2 \log |x+1| + 2 \left( -\frac{1}{x+1} \right) + c$.
$I = x - 2 \log |x+1| - \frac{2}{x+1} + c$.

(A) Let $I = \int \tan ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) d x$
Using the identities $1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ and $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$:
$I = \int \tan ^{-1}\left(\sqrt{\frac{(\cos \frac{x}{2} - \sin \frac{x}{2})^2}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}}\right) d x$
$I = \int \tan ^{-1}\left(\frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}}\right) d x$
Dividing numerator and denominator by $\cos \frac{x}{2}$:
$I = \int \tan ^{-1}\left(\frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}}\right) d x$
Since $\tan(\frac{\pi}{4} - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta}$:
$I = \int \tan ^{-1}\left(\tan \left(\frac{\pi}{4} - \frac{x}{2}\right)\right) d x$
$I = \int \left(\frac{\pi}{4} - \frac{x}{2}\right) d x$
$I = \frac{\pi}{4} x - \frac{x^2}{4} + c$

(B) We know that $x^4+x^2+1 = (x^2+1)^2 - x^2 = (x^2+x+1)(x^2-x+1)$.
Therefore,the integral becomes:
$I = \int \frac{(x^2+x+1)(x^2-x+1)}{x^2-x+1} \,d x$
$I = \int (x^2+x+1) \,d x$
Integrating term by term:
$I = \frac{x^3}{3} + \frac{x^2}{2} + x + c$
Thus,the correct option is $B$.

(B) Given that $f^{\prime}(x)=4 x^3-3 x^{-4}$.
Integrating both sides with respect to $x$:
$f(x) = \int (4x^3 - 3x^{-4}) dx$
$f(x) = 4 \cdot \frac{x^4}{4} - 3 \cdot \frac{x^{-3}}{-3} + c$
$f(x) = x^4 + \frac{1}{x^3} + c$
Given that $f(2) = 0$,substitute $x = 2$:
$0 = (2)^4 + \frac{1}{2^3} + c$
$0 = 16 + \frac{1}{8} + c$
$0 = \frac{128+1}{8} + c$
$c = -\frac{129}{8}$
Therefore,$f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}$.

(C) First, perform polynomial long division on the integrand $\frac{x^3-7x+6}{x^2+3x}$.
Dividing $x^3-7x+6$ by $x^2+3x$ gives a quotient of $(x-3)$ and a remainder of $(2x+6)$.
Thus, $\frac{x^3-7x+6}{x^2+3x} = x-3 + \frac{2x+6}{x^2+3x}$.
Simplify the remainder term: $\frac{2x+6}{x^2+3x} = \frac{2(x+3)}{x(x+3)} = \frac{2}{x}$ for $x \neq -3$.
Now, integrate the expression: $\int (x-3+\frac{2}{x}) dx$.
$= \int x dx - \int 3 dx + \int \frac{2}{x} dx$.
$= \frac{x^2}{2} - 3x + 2 \log |x| + c$.

(C) We have $I = \int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we write $\frac{5x}{2} = 2x + \frac{x}{2}$.
$I = \int \frac{\sin(2x + \frac{x}{2})}{\sin \frac{x}{2}} dx = \int \frac{\sin 2x \cos \frac{x}{2} + \cos 2x \sin \frac{x}{2}}{\sin \frac{x}{2}} dx$.
$I = \int (\sin 2x \cot \frac{x}{2} + \cos 2x) dx$.
Alternatively,using the sum-to-product identity or expansion:
$\frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} = \frac{\sin(2x + \frac{x}{2})}{\sin \frac{x}{2}} = \frac{\sin 2x \cos \frac{x}{2} + \cos 2x \sin \frac{x}{2}}{\sin \frac{x}{2}} = \cos 2x + \sin 2x \cot \frac{x}{2}$.
Using $\sin 2x = 2 \sin x \cos x$ and $\cot \frac{x}{2} = \frac{1+\cos x}{\sin x}$:
$\sin 2x \cot \frac{x}{2} = 2 \sin x \cos x \cdot \frac{1+\cos x}{\sin x} = 2 \cos x(1+\cos x) = 2 \cos x + 2 \cos^2 x$.
Since $2 \cos^2 x = 1 + \cos 2x$,we have:
$\cos 2x + 2 \cos x + 1 + \cos 2x = 1 + 2 \cos x + 2 \cos 2x$.
Integrating: $\int (1 + 2 \cos x + 2 \cos 2x) dx = x + 2 \sin x + \sin 2x + C$.

(A) We know that the algebraic identity for the difference of two $n$-th powers is given by:
$(x-a)(x^{n-1}+x^{n-2}a+\ldots+a^{n-1}) = x^n - a^n$.
Substituting this into the integral,we get:
$\int(x^n - a^n)dx$.
Integrating term by term with respect to $x$:
$\int x^n dx - \int a^n dx = \frac{x^{n+1}}{n+1} - a^n x + C$.

(C) Let $I = \int [1+2 \tan^2 x + 2 \tan x \sec x]^{1/2} dx$.
Since $1 + \tan^2 x = \sec^2 x$,we can rewrite the expression inside the square root as:
$I = \int [\sec^2 x + \tan^2 x + 2 \sec x \tan x]^{1/2} dx$.
This is a perfect square:
$I = \int [(\sec x + \tan x)^2]^{1/2} dx = \int (\sec x + \tan x) dx$.
Integrating term by term:
$I = \int \sec x dx + \int \tan x dx$.
$I = \log |\sec x + \tan x| + \log |\sec x| + c$.
Using the property $\log a + \log b = \log(ab)$:
$I = \log |\sec x(\sec x + \tan x)| + c$.

(A) To evaluate the integral $I = \int \frac{1}{\cos x+\sqrt{3} \sin x} dx$,we multiply and divide the denominator by $2$:
$I = \int \frac{1}{2(\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x)} dx$
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we can write $\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x = \sin(x + \frac{\pi}{6})$:
$I = \frac{1}{2} \int \frac{1}{\sin(x + \frac{\pi}{6})} dx$
$I = \frac{1}{2} \int \csc(x + \frac{\pi}{6}) dx$
Using the standard integral formula $\int \csc \theta d\theta = \log |\tan(\frac{\theta}{2})| + C$:
$I = \frac{1}{2} \log |\tan(\frac{x}{2} + \frac{\pi}{12})| + C$

(A) To evaluate the integral $I = \int \frac{dx}{\sqrt{5+4x-x^2}}$,we first complete the square for the quadratic expression inside the square root.
$5+4x-x^2 = -(x^2-4x-5) = -((x-2)^2 - 4 - 5) = -( (x-2)^2 - 9 ) = 9 - (x-2)^2$.
Substituting this back into the integral,we get:
$I = \int \frac{dx}{\sqrt{3^2 - (x-2)^2}}$.
Using the standard integration formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + c$,we obtain:
$I = \sin^{-1}\left(\frac{x-2}{3}\right) + c$.

(A) Given $f^{\prime}(x)=k(\cos x+\sin x)$.
Integrating both sides with respect to $x$,we get:
$f(x) = \int k(\cos x + \sin x) dx = k(\sin x - \cos x) + C$.
Using the condition $f(0) = 9$:
$f(0) = k(\sin 0 - \cos 0) + C = k(0 - 1) + C = -k + C = 9$ ...$(1)$.
Using the condition $f\left(\frac{\pi}{2}\right) = 15$:
$f\left(\frac{\pi}{2}\right) = k\left(\sin \frac{\pi}{2} - \cos \frac{\pi}{2}\right) + C = k(1 - 0) + C = k + C = 15$ ...$(2)$.
Adding equations $(1)$ and $(2)$:
$(-k + C) + (k + C) = 9 + 15 \Rightarrow 2C = 24 \Rightarrow C = 12$.
Substituting $C = 12$ into equation $(2)$:
$k + 12 = 15 \Rightarrow k = 3$.
Thus,$f(x) = 3(\sin x - \cos x) + 12$.

(B) We are given the integral $I = \int \frac{dx}{\cos 2x + \sin^2 x}$.
Using the trigonometric identity $\cos 2x = 1 - 2\sin^2 x$,we substitute this into the denominator:
$I = \int \frac{dx}{1 - 2\sin^2 x + \sin^2 x}$
$I = \int \frac{dx}{1 - \sin^2 x}$
Using the identity $1 - \sin^2 x = \cos^2 x$,we get:
$I = \int \frac{dx}{\cos^2 x}$
$I = \int \sec^2 x \, dx$
The integral of $\sec^2 x$ is $\tan x + c$.
Therefore,$I = \tan x + c$.

(D) We are given the integral $I = \int \frac{dx}{\cos 2x - \cos^2 x}$.
Using the trigonometric identity $\cos 2x = 2\cos^2 x - 1$,we substitute this into the denominator:
$I = \int \frac{dx}{(2\cos^2 x - 1) - \cos^2 x} = \int \frac{dx}{\cos^2 x - 1}$.
We know that $\cos^2 x - 1 = - (1 - \cos^2 x) = -\sin^2 x$.
Substituting this back into the integral:
$I = \int \frac{dx}{-\sin^2 x} = -\int \csc^2 x \, dx$.
Since the derivative of $\cot x$ is $-\csc^2 x$,the integral of $-\csc^2 x$ is $\cot x + c$.
Therefore,$I = \cot x + c$.

(A) To evaluate the integral $\int \frac{d x}{x^{2}+4 x+13}$,we complete the square in the denominator:
$x^{2}+4 x+13 = (x^{2}+4 x+4) + 9 = (x+2)^{2} + 3^{2}$.
Now,the integral becomes $\int \frac{d x}{(x+2)^{2} + 3^{2}}$.
Using the standard formula $\int \frac{d x}{x^{2}+a^{2}} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + c$,where $a=3$ and the variable is $(x+2)$:
$\int \frac{d x}{(x+2)^{2} + 3^{2}} = \frac{1}{3} \tan^{-1}\left(\frac{x+2}{3}\right) + c$.
Comparing this with the given options,the correct option is $A$.

(D) $I = \int \frac{1}{\sin x \cdot \cos^2 x} \, dx$
$I = \int \frac{\sin^2 x + \cos^2 x}{\sin x \cdot \cos^2 x} \, dx$
$I = \int \frac{\sin^2 x}{\sin x \cdot \cos^2 x} \, dx + \int \frac{\cos^2 x}{\sin x \cdot \cos^2 x} \, dx$
$I = \int \frac{\sin x}{\cos^2 x} \, dx + \int \frac{1}{\sin x} \, dx$
$I = \int \tan x \sec x \, dx + \int \operatorname{cosec} x \, dx$
$I = \sec x + \ln |\operatorname{cosec} x - \cot x| + c$

(C) We know that $\int \frac{1}{x^{2}+a^{2}} d x = \frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c$.
Given integral is $I = \int \frac{1}{16 x^{2}+9} d x$.
Taking $16$ as a common factor from the denominator:
$I = \frac{1}{16} \int \frac{1}{x^{2}+\frac{9}{16}} d x = \frac{1}{16} \int \frac{1}{x^{2}+\left(\frac{3}{4}\right)^{2}} d x$.
Using the formula with $a = \frac{3}{4}$:
$I = \frac{1}{16} \times \frac{1}{3/4} \tan ^{-1}\left(\frac{x}{3/4}\right)+c$.
$I = \frac{1}{16} \times \frac{4}{3} \tan ^{-1}\left(\frac{4 x}{3}\right)+c$.
$I = \frac{1}{12} \tan ^{-1}\left(\frac{4 x}{3}\right)+c$.

(B) Given that $f(x) = x$ and $g(x) = \sin x$.
First,we find the composite function $f(g(x))$:
$f(g(x)) = f(\sin x) = \sin x$.
Now,we evaluate the integral:
$\int f(g(x)) \, dx = \int \sin x \, dx$.
Using the standard integral formula $\int \sin x \, dx = -\cos x + c$,we get:
$\int f(g(x)) \, dx = -\cos x + c$.

(D) Let $I = \int \cos \left(\frac{x}{16}\right) \cdot \cos \left(\frac{x}{8}\right) \cdot \cos \left(\frac{x}{4}\right) \cdot \sin \left(\frac{x}{16}\right) dx$.
Using the identity $\sin(2\theta) = 2\sin\theta \cos\theta$,we have $\sin \left(\frac{x}{16}\right) \cos \left(\frac{x}{16}\right) = \frac{1}{2} \sin \left(\frac{x}{8}\right)$.
Substituting this into the integral:
$I = \int \frac{1}{2} \sin \left(\frac{x}{8}\right) \cos \left(\frac{x}{8}\right) \cos \left(\frac{x}{4}\right) dx$.
Using the identity again,$\sin \left(\frac{x}{8}\right) \cos \left(\frac{x}{8}\right) = \frac{1}{2} \sin \left(\frac{x}{4}\right)$.
So,$I = \int \frac{1}{2} \cdot \frac{1}{2} \sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{4}\right) dx = \frac{1}{4} \int \sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{4}\right) dx$.
Using the identity one more time,$\sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{4}\right) = \frac{1}{2} \sin \left(\frac{x}{2}\right)$.
Thus,$I = \frac{1}{4} \int \frac{1}{2} \sin \left(\frac{x}{2}\right) dx = \frac{1}{8} \int \sin \left(\frac{x}{2}\right) dx$.
Integrating $\sin \left(\frac{x}{2}\right)$,we get $-2 \cos \left(\frac{x}{2}\right)$.
$I = \frac{1}{8} \cdot (-2 \cos \left(\frac{x}{2}\right)) + c = -\frac{1}{4} \cos \left(\frac{x}{2}\right) + c$.

(C) We have the integral $I = \int \frac{\sin 7 x}{\cos 9 x \cos 2 x} \,d x$.
Since $7x = 9x - 2x$, we can write $\sin 7x = \sin(9x - 2x)$.
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$, we get:
$I = \int \frac{\sin 9x \cos 2x - \cos 9x \sin 2x}{\cos 9x \cos 2x} \,d x$.
Splitting the fraction:
$I = \int \left( \frac{\sin 9x \cos 2x}{\cos 9x \cos 2x} - \frac{\cos 9x \sin 2x}{\cos 9x \cos 2x} \right) \,d x$.
$I = \int (\tan 9x - \tan 2x) \,d x$.
Integrating term by term:
$I = \int \tan 9x \,d x - \int \tan 2x \,d x$.
Using the formula $\int \tan(ax) \,d x = \frac{1}{a} \ln |\sec(ax)| + c$:
$I = \frac{1}{9} \ln |\sec 9x| - \frac{1}{2} \ln |\sec 2x| + c$.
Thus, the correct option is $C$.

(C) Let $I = \int \frac{dx}{\cos x(1+\cos x)}$.
We can write the integrand as $\frac{1}{\cos x(1+\cos x)} = \frac{1+\cos x - \cos x}{\cos x(1+\cos x)} = \frac{1}{\cos x} - \frac{1}{1+\cos x}$.
Now,$I = \int \sec x \, dx - \int \frac{dx}{1+\cos x}$.
We know that $\int \sec x \, dx = \log |\sec x + \tan x| + c_1$.
For the second part,use the identity $1+\cos x = 2 \cos^2 \left(\frac{x}{2}\right)$.
So,$\int \frac{dx}{1+\cos x} = \int \frac{dx}{2 \cos^2 \left(\frac{x}{2}\right)} = \frac{1}{2} \int \sec^2 \left(\frac{x}{2}\right) dx$.
Integrating this,we get $\frac{1}{2} \cdot \frac{\tan(x/2)}{1/2} = \tan \left(\frac{x}{2}\right) + c_2$.
Combining these,$I = \log |\sec x + \tan x| - \tan \left(\frac{x}{2}\right) + c$.

(B) Let $I = \int (1 - \cos x) \operatorname{cosec}^2 x \, dx$
Using the identity $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we have:
$I = \int \frac{2 \sin^2 \frac{x}{2}}{\sin^2 x} \, dx$
$I = \int \frac{2 \sin^2 \frac{x}{2}}{(2 \sin \frac{x}{2} \cos \frac{x}{2})^2} \, dx$
$I = \int \frac{2 \sin^2 \frac{x}{2}}{4 \sin^2 \frac{x}{2} \cos^2 \frac{x}{2}} \, dx$
$I = \frac{1}{2} \int \frac{1}{\cos^2 \frac{x}{2}} \, dx$
$I = \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx$
Integrating $\sec^2 \frac{x}{2}$,we get:
$I = \frac{1}{2} \cdot \frac{\tan \frac{x}{2}}{1/2} + c$
$I = \tan \frac{x}{2} + c$

(D) Given integral $I = \int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} \,d x$.
Using the identity $1+\cos \theta = 2 \cos^2 \frac{\theta}{2}$, we have $\cos 8x + 1 = 2 \cos^2 4x$.
The denominator is $\frac{\cos 2x}{\sin 2x} - \frac{\sin 2x}{\cos 2x} = \frac{\cos^2 2x - \sin^2 2x}{\sin 2x \cos 2x} = \frac{\cos 4x}{\frac{1}{2} \sin 4x} = \frac{2 \cos 4x}{\sin 4x}$.
Substituting these into the integral:
$I = \int \frac{2 \cos^2 4x}{\frac{2 \cos 4x}{\sin 4x}} \,d x = \int \cos 4x \sin 4x \,d x$.
Multiply and divide by $2$:
$I = \frac{1}{2} \int 2 \sin 4x \cos 4x \,d x = \frac{1}{2} \int \sin 8x \,d x$.
Integrating $\sin 8x$ gives $\frac{-\cos 8x}{8}$.
So, $I = \frac{1}{2} \left( \frac{-\cos 8x}{8} \right) + c = \frac{-\cos 8x}{16} + c$.
Comparing with $A \cos 8x + c$, we get $A = \frac{-1}{16}$.

(A) To evaluate the integral $I = \int \sqrt{\frac{1+x}{1-x}} \, dx$,we multiply the numerator and denominator inside the square root by $(1+x)$:
$I = \int \sqrt{\frac{(1+x)(1+x)}{(1-x)(1+x)}} \, dx = \int \sqrt{\frac{(1+x)^2}{1-x^2}} \, dx$
$I = \int \frac{1+x}{\sqrt{1-x^2}} \, dx = \int \frac{1}{\sqrt{1-x^2}} \, dx + \int \frac{x}{\sqrt{1-x^2}} \, dx$
For the first part,$\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1} x$.
For the second part,let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$:
$\int \frac{x}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int u^{-1/2} \, du = -\frac{1}{2} \cdot 2u^{1/2} = -\sqrt{1-x^2}$.
Combining these,we get $I = \sin^{-1} x - \sqrt{1-x^2} + C$.

(B) We have the integral $I = \int \frac{5(x^6+1)}{x^2+1} \, dx$.
Using the algebraic identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$,where $a=x^2$ and $b=1$,we can write $x^6+1 = (x^2)^3+1^3 = (x^2+1)(x^4-x^2+1)$.
Substituting this into the integral:
$I = \int \frac{5(x^2+1)(x^4-x^2+1)}{x^2+1} \, dx$
$I = 5 \int (x^4-x^2+1) \, dx$
Integrating term by term:
$I = 5 \left( \frac{x^5}{5} - \frac{x^3}{3} + x \right) + C$
$I = x^5 - \frac{5x^3}{3} + 5x + C$.

(A) Let $I = \int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} dx$.
Using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$,we can factor the numerator:
$\sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x) = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)(\sin^4 x + \cos^4 x)$.
Since $\sin^2 x + \cos^2 x = 1$,this simplifies to $(\sin^2 x - \cos^2 x)(\sin^4 x + \cos^4 x)$.
We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - 2 \sin^2 x \cos^2 x$.
Substituting this back into the integral:
$I = \int \frac{(\sin^2 x - \cos^2 x)(1 - 2 \sin^2 x \cos^2 x)}{1 - 2 \sin^2 x \cos^2 x} dx$.
$I = \int (\sin^2 x - \cos^2 x) dx$.
Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we have $\sin^2 x - \cos^2 x = -\cos 2x$.
$I = \int -\cos 2x dx = -\frac{1}{2} \sin 2x + C$.

(C) We have the integral $I = \int \frac{\sin 4x}{\sin x} \, dx$.
Using the identity $\sin 4x = 2 \sin 2x \cos 2x = 4 \sin x \cos x \cos 2x$,we get:
$I = \int \frac{4 \sin x \cos x \cos 2x}{\sin x} \, dx = 4 \int \cos x \cos 2x \, dx$.
Using the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have $2 \cos 2x \cos x = \cos(2x+x) + \cos(2x-x) = \cos 3x + \cos x$.
Thus,$I = 2 \int (\cos 3x + \cos x) \, dx$.
Integrating term by term,we get:
$I = 2 \left( \frac{\sin 3x}{3} + \sin x \right) + C = \frac{2}{3} \sin 3x + 2 \sin x + C$.

(B) Let $I = \int \frac{2 x^3-1}{x^4+x} \,d x$.
We can rewrite the denominator as $x(x^3+1)$.
So,$I = \int \frac{2 x^3-1}{x(x^3+1)} \,d x$.
Using partial fractions or observation,we can write $\frac{2x^3-1}{x(x^3+1)} = \frac{A}{x} + \frac{B x^2}{x^3+1}$.
By inspection,$\frac{2x^3-1}{x(x^3+1)} = \frac{-1}{x} + \frac{3x^2}{x^3+1}$.
Integrating both terms:
$I = \int \left( \frac{3x^2}{x^3+1} - \frac{1}{x} \right) \,d x$.
$I = \log|x^3+1| - \log|x| + C$.
$I = \log \left| \frac{x^3+1}{x} \right| + C$.

(A) Let $I = \int x^{x}(1+\log x) dx$.
Consider the substitution $u = x^{x}$.
Taking the natural logarithm on both sides,$\log u = x \log x$.
Differentiating with respect to $x$,$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.
Thus,$\frac{du}{dx} = u(1 + \log x) = x^{x}(1 + \log x)$.
Therefore,$du = x^{x}(1 + \log x) dx$.
Substituting this into the integral,$I = \int du = u + c$.
Replacing $u$ with $x^{x}$,we get $I = x^{x} + c$.
Comparing this with the given expression $k x^{x} + c$,we find $k = 1$.
Since $\log_{e} e = 1$,the correct option is $A$.