Fundamental integration Questions in English

(D) We know that $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + C$.
Given integral is $I = \int \frac{dx}{\sqrt{4^2 - (3x)^2}}$.
Using the substitution $u = 3x$,we have $du = 3dx$,so $dx = \frac{du}{3}$.
$I = \int \frac{du/3}{\sqrt{4^2 - u^2}} = \frac{1}{3} \int \frac{du}{\sqrt{4^2 - u^2}}$.
$I = \frac{1}{3} \sin^{-1}(\frac{u}{4}) + C = \frac{1}{3} \sin^{-1}(\frac{3x}{4}) + C$.
Comparing this with $A \sin^{-1}(Bx) + C$,we get $A = \frac{1}{3}$ and $B = \frac{3}{4}$.
Therefore,$A + B = \frac{1}{3} + \frac{3}{4} = \frac{4 + 9}{12} = \frac{13}{12}$.

(A) We are given the integral $\int \frac{1}{\sqrt{9-16 x^2}} d x$.
Rewrite the denominator as $\sqrt{3^2-(4 x)^2}$.
Using the standard formula $\int \frac{1}{\sqrt{a^2-u^2}} d u = \sin^{-1}(\frac{u}{a}) + c$,where $u = 4x$ and $du = 4 dx$,we get:
$\int \frac{1}{\sqrt{3^2-(4 x)^2}} d x = \frac{1}{4} \int \frac{1}{\sqrt{3^2-(4 x)^2}} d(4x) = \frac{1}{4} \sin^{-1}(\frac{4x}{3}) + c$.
Comparing this with $\alpha \sin^{-1}(\beta x) + c$,we identify $\alpha = \frac{1}{4}$ and $\beta = \frac{4}{3}$.
Now,calculate $\alpha + \frac{1}{\beta} = \frac{1}{4} + \frac{1}{4/3} = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$.

(A) Given the integral equation: $\int \frac{f(x)}{\log (\sin x)} d x = \log [\log \sin x] + c$.
To find $f(x)$,we differentiate both sides with respect to $x$:
$\frac{d}{d x} \left( \int \frac{f(x)}{\log (\sin x)} d x \right) = \frac{d}{d x} (\log [\log \sin x] + c)$.
Using the Fundamental Theorem of Calculus on the left side,we get:
$\frac{f(x)}{\log (\sin x)} = \frac{d}{d x} (\log [\log \sin x])$.
Applying the chain rule on the right side:
$\frac{d}{d x} (\log [\log \sin x]) = \frac{1}{\log \sin x} \cdot \frac{d}{d x} (\log \sin x)$.
Since $\frac{d}{d x} (\log \sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$,we have:
$\frac{f(x)}{\log \sin x} = \frac{1}{\log \sin x} \cdot \cot x$.
Comparing both sides,we obtain $f(x) = \cot x$.

(D) Let $I = \int \frac{dx}{\sqrt{8+2x-x^2}}$.
First,complete the square for the quadratic expression inside the square root:
$8+2x-x^2 = -(x^2-2x-8) = -(x^2-2x+1-9) = -( (x-1)^2 - 9 ) = 9 - (x-1)^2$.
Now,substitute this back into the integral:
$I = \int \frac{dx}{\sqrt{9-(x-1)^2}} = \int \frac{dx}{\sqrt{3^2-(x-1)^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\frac{x}{a}\right)+c$,we get:
$I = \sin^{-1}\left(\frac{x-1}{3}\right)+c$.

(B) Let $I = \int \frac{1}{7-6 x-x^2} dx$.
First,complete the square in the denominator: $7-6x-x^2 = 7 - (x^2+6x) = 7 - (x^2+6x+9-9) = 16 - (x+3)^2$.
So,$I = \int \frac{1}{16-(x+3)^2} dx$.
Using the standard integral formula $\int \frac{1}{a^2-x^2} dx = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + c$,where $a=4$ and $x$ is replaced by $(x+3)$:
$I = \frac{1}{2(4)} \log \left| \frac{4+(x+3)}{4-(x+3)} \right| + c$.
$I = \frac{1}{8} \log \left| \frac{7+x}{1-x} \right| + c$.

(A) We have $I = \int \tan^4 x dx = \int \tan^2 x (\sec^2 x - 1) dx$.
$I = \int \tan^2 x \sec^2 x dx - \int \tan^2 x dx$.
$I = \int \tan^2 x \sec^2 x dx - \int (\sec^2 x - 1) dx$.
$I = \frac{\tan^3 x}{3} - \tan x + x + k$.
Comparing this with $a \tan^3 x + b \tan x + c x + k$,we get $a = \frac{1}{3}$,$b = -1$,and $c = 1$.
Therefore,$a - b + c = \frac{1}{3} - (-1) + 1 = \frac{1}{3} + 1 + 1 = \frac{1}{3} + 2 = \frac{7}{3}$.

(A) Given $f\left(\frac{x-4}{x-2}\right)=2x+1$.
Let $y = \frac{x-4}{x-2}$.
Then $y = \frac{x-2-2}{x-2} = 1 - \frac{2}{x-2}$.
So,$\frac{2}{x-2} = 1-y$,which implies $x-2 = \frac{2}{1-y} = \frac{-2}{y-1}$.
Thus,$x = 2 - \frac{2}{y-1} = \frac{2y-2-2}{y-1} = \frac{2y-4}{y-1}$.
Now,$f(y) = 2x+1 = 2\left(\frac{2y-4}{y-1}\right) + 1 = \frac{4y-8+y-1}{y-1} = \frac{5y-9}{y-1} = \frac{5(y-1)-4}{y-1} = 5 - \frac{4}{y-1}$.
Therefore,$f(x) = 5 - \frac{4}{x-1}$.
Now,$\int f(x) dx = \int \left(5 - \frac{4}{x-1}\right) dx = 5x - 4 \log |x-1| + c$.

(D) To evaluate the integral $I = \int \frac{d x}{7+6 x-x^2}$,we first complete the square for the quadratic expression in the denominator:
$7+6 x-x^2 = -(x^2-6 x-7) = -(x^2-6 x+9-16) = 16-(x-3)^2$.
Thus,the integral becomes $I = \int \frac{d x}{4^2-(x-3)^2}$.
Using the standard formula $\int \frac{d x}{a^2-x^2} = \frac{1}{2 a} \log \left| \frac{a+x}{a-x} \right| + c$,where $a=4$ and $x$ is replaced by $(x-3)$:
$I = \frac{1}{2(4)} \log \left| \frac{4+(x-3)}{4-(x-3)} \right| + c$.
Simplifying the expression inside the logarithm:
$I = \frac{1}{8} \log \left| \frac{x+1}{7-x} \right| + c$.

(B) Given $f(x) = \frac{x}{x+1}$.
We find $F(x) = (fof)(x) = f(f(x)) = f\left(\frac{x}{x+1}\right) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{x}{x + (x+1)} = \frac{x}{2x+1}$.
Now,we evaluate the integral $\int F(x) \, dx = \int \frac{x}{2x+1} \, dx$.
To integrate,we manipulate the numerator: $\frac{x}{2x+1} = \frac{1}{2} \left( \frac{2x+1-1}{2x+1} \right) = \frac{1}{2} \left( 1 - \frac{1}{2x+1} \right) = \frac{1}{2} - \frac{1}{2(2x+1)}$.
Integrating term by term: $\int \left( \frac{1}{2} - \frac{1}{2(2x+1)} \right) \, dx = \frac{1}{2}x - \frac{1}{2} \cdot \frac{1}{2} \log |2x+1| + c = \frac{x}{2} - \frac{1}{4} \log |2x+1| + c$.

(C) Let $I = \int \operatorname{cosec}(x-a) \cdot \operatorname{cosec} x \, dx$.
We can write this as $I = \int \frac{1}{\sin (x-a) \sin x} \, dx$.
Multiply and divide by $\sin a$:
$I = \frac{1}{\sin a} \int \frac{\sin a}{\sin (x-a) \sin x} \, dx$.
Express $a$ as $x - (x-a)$:
$I = \frac{1}{\sin a} \int \frac{\sin (x - (x-a))}{\sin (x-a) \sin x} \, dx$.
Using the formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin a} \int \frac{\sin x \cos (x-a) - \cos x \sin (x-a)}{\sin (x-a) \sin x} \, dx$.
$I = \frac{1}{\sin a} \int \left( \frac{\sin x \cos (x-a)}{\sin (x-a) \sin x} - \frac{\cos x \sin (x-a)}{\sin (x-a) \sin x} \right) \, dx$.
$I = \frac{1}{\sin a} \int (\cot (x-a) - \cot x) \, dx$.
Integrating,we get:
$I = \frac{1}{\sin a} [\log |\sin (x-a)| - \log |\sin x|] + c$.
Using the property $\log m - \log n = \log(\frac{m}{n})$:
$I = \frac{1}{\sin a} \log \left| \frac{\sin (x-a)}{\sin x} \right| + c$.
Since $\frac{1}{\sin x} = \operatorname{cosec} x$,this is equivalent to:
$I = \frac{1}{\sin a} \log |\sin (x-a) \cdot \operatorname{cosec} x| + c$.

(D) Given the integral $I = \int \tan^{-1} \left( \sqrt{\frac{1 - \sin x}{1 + \sin x}} \right) dx$.
Using the identities $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ and $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$,we have:
$I = \int \tan^{-1} \sqrt{\frac{(\cos \frac{x}{2} - \sin \frac{x}{2})^2}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}} dx$
$I = \int \tan^{-1} \left( \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}} \right) dx$
Dividing numerator and denominator by $\cos \frac{x}{2}$,we get:
$I = \int \tan^{-1} \left( \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} \right) dx$
Using the formula $\tan(\frac{\pi}{4} - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta}$:
$I = \int \tan^{-1} \left( \tan \left( \frac{\pi}{4} - \frac{x}{2} \right) \right) dx$
$I = \int \left( \frac{\pi}{4} - \frac{x}{2} \right) dx$
Integrating term by term:
$I = \frac{\pi}{4} x - \frac{x^2}{4} + C$.

(A) Let $I = \int \frac{\sin 2x}{\sin^2 x \cos^2 x} dx$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get:
$I = \int \frac{2 \sin x \cos x}{\sin^2 x \cos^2 x} dx$
$I = 2 \int \frac{1}{\sin x \cos x} dx$
Multiply numerator and denominator by $2$:
$I = 2 \int \frac{2}{2 \sin x \cos x} dx = 4 \int \frac{1}{\sin 2x} dx$
$I = 4 \int \operatorname{cosec} 2x dx$
Using the formula $\int \operatorname{cosec} u du = \log |\tan(u/2)| + c$,we get:
$I = 4 \times \frac{1}{2} \log |\tan x| + c = 2 \log |\tan x| + c$
Using the property $n \log a = \log a^n$,we get:
$I = \log |\tan^2 x| + c$.

(A) Let $I = \int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x$.
Substitute $x = (x - \frac{\pi}{4}) + \frac{\pi}{4}$,so $I = \int \frac{\sin \left((x-\frac{\pi}{4}) + \frac{\pi}{4}\right)}{\sin \left(x-\frac{\pi}{4}\right)} d x$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $I = \int \frac{\sin (x-\frac{\pi}{4}) \cos \frac{\pi}{4} + \cos (x-\frac{\pi}{4}) \sin \frac{\pi}{4}}{\sin (x-\frac{\pi}{4})} d x$.
Since $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we have $I = \int \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cot (x-\frac{\pi}{4}) \right) d x$.
Integrating term by term,$I = \frac{1}{\sqrt{2}} \int 1 d x + \frac{1}{\sqrt{2}} \int \cot (x-\frac{\pi}{4}) d x$.
The integral of $\cot u$ is $\log |\sin u|$,so $I = \frac{1}{\sqrt{2}} [x + \log |\sin (x-\frac{\pi}{4})|] + c$.

(C) Given integral $I = \int \frac{dx}{32-2x^2} = \frac{1}{2} \int \frac{dx}{16-x^2}$.
Using the formula $\int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + c$,where $a=4$:
$I = \frac{1}{2} \left[ \frac{1}{2(4)} \log \left| \frac{4+x}{4-x} \right| \right] + c$
$I = \frac{1}{16} [ \log |4+x| - \log |4-x| ] + c$
$I = -\frac{1}{16} \log |4-x| + \frac{1}{16} \log |4+x| + c$.
Comparing this with $A \log(4-x) + B \log(4+x) + c$,we get $A = -\frac{1}{16}$ and $B = \frac{1}{16}$.

(B) To solve the integral $I = \int \frac{1}{\sqrt{4x-x^2}} dx$,we first complete the square for the quadratic expression inside the square root.
$4x - x^2 = -(x^2 - 4x) = -(x^2 - 4x + 4 - 4) = -( (x-2)^2 - 4 ) = 4 - (x-2)^2$.
Substituting this back into the integral,we get:
$I = \int \frac{1}{\sqrt{4 - (x-2)^2}} dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}(\frac{u}{a}) + c$,where $a = 2$ and $u = x-2$:
$I = \sin^{-1}\left(\frac{x-2}{2}\right) + c$.
Thus,the correct option is $B$.

(C) We know that $x^5 + 1 = (x+1)(x^4 - x^3 + x^2 - x + 1)$.
Therefore,the integral becomes $\int \frac{(x+1)(x^4 - x^3 + x^2 - x + 1)}{x+1} \, dx = \int (x^4 - x^3 + x^2 - x + 1) \, dx$.
Integrating term by term,we get $\frac{x^5}{5} - \frac{x^4}{4} + \frac{x^3}{3} - \frac{x^2}{2} + x + c$.
This can be written in summation notation as $\sum_{n=1}^5 (-1)^{n+1} \cdot \frac{x^n}{n} + c$.
Comparing this with the given options,the correct option is $C$.

(B) We are given the integral $I = \int \frac{\cos 3x}{\sin x} dx$.
Using the trigonometric identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we can also write $\cos 3x = \cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x$.
Alternatively,use $\cos 3x = 1 - 4 \sin^2 x$ is incorrect,rather $\cos 3x = \cos(2x+x) = \cos 2x \cos x - \sin 2x \sin x = (1 - 2 \sin^2 x) \cos x - (2 \sin x \cos x) \sin x = \cos x - 2 \sin^2 x \cos x - 2 \sin^2 x \cos x = \cos x - 4 \sin^2 x \cos x$.
Thus,$\frac{\cos 3x}{\sin x} = \frac{\cos x - 4 \sin^2 x \cos x}{\sin x} = \cot x - 4 \sin x \cos x = \cot x - 2 \sin 2x$.
Now,integrate: $\int (\cot x - 2 \sin 2x) dx = \int \cot x dx - 2 \int \sin 2x dx$.
$= \log |\sin x| - 2 (-\frac{\cos 2x}{2}) + C = \log |\sin x| + \cos 2x + C$.
Comparing this with $p \cos 2x + q \log |\sin x| + C$,we get $p = 1$ and $q = 1$.
Therefore,$p + q = 1 + 1 = 2$.

(A) We know that $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$.
Substituting these into the integral:
$\int \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} d x = \int \tan^2 \frac{x}{2} d x$.
Using the identity $\tan^2 \theta = \sec^2 \theta - 1$:
$\int (\sec^2 \frac{x}{2} - 1) d x$.
Integrating term by term:
$= \int \sec^2 \frac{x}{2} d x - \int 1 d x$.
$= 2 \tan \frac{x}{2} - x + C$.

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

(B) To evaluate the integral $I = \int \frac{d x}{x^2+2 x+5}$,we first complete the square in the denominator.
$x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 2^2$.
Now,the integral becomes $I = \int \frac{d x}{(x+1)^2 + 2^2}$.
Using the standard integration formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,where $a = 2$ and the variable is $(x+1)$:
$I = \frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) + C$.
Comparing this with the given options,the correct option is $B$.

(A) To evaluate the integral $I = \int \frac{1}{e^x+1} dx$,we can rewrite the integrand as follows:
$I = \int \frac{1}{e^x(1 + e^{-x})} dx = \int \frac{e^{-x}}{1 + e^{-x}} dx$.
Let $u = 1 + e^{-x}$. Then $du = -e^{-x} dx$,which implies $e^{-x} dx = -du$.
Substituting these into the integral,we get:
$I = \int \frac{-du}{u} = -\log |u| + C = -\log |1 + e^{-x}| + C$.
We can simplify this as:
$I = -\log \left| \frac{e^x + 1}{e^x} \right| + C = -(\log |e^x + 1| - \log |e^x|) + C = -\log |e^x + 1| + x + C$.
Alternatively,using the form $I = \int \frac{1}{e^x+1} dx = \int \frac{e^x+1-e^x}{e^x+1} dx = \int (1 - \frac{e^x}{e^x+1}) dx = x - \log |e^x+1| + C$.
Since $x = \log(e^x)$,we have $I = \log(e^x) - \log |e^x+1| + C = \log \left| \frac{e^x}{e^x+1} \right| + C$.
Thus,the correct option is $A$.

(B) We have the integral $I = \int \frac{\operatorname{cosec}^2 x}{\sec ^2 x} \, dx$.
Using the trigonometric identities $\operatorname{cosec}^2 x = \frac{1}{\sin^2 x}$ and $\sec^2 x = \frac{1}{\cos^2 x}$,we get:
$I = \int \frac{\cos^2 x}{\sin^2 x} \, dx = \int \cot^2 x \, dx$.
Using the identity $\cot^2 x = \operatorname{cosec}^2 x - 1$,we have:
$I = \int (\operatorname{cosec}^2 x - 1) \, dx$.
Integrating term by term,we get:
$I = \int \operatorname{cosec}^2 x \, dx - \int 1 \, dx = -\cot x - x + C$.
Thus,the correct option is $B$.

(B) To solve the integral $I = \int \frac{1}{\sqrt{2x-x^2}} dx$,we first complete the square for the expression inside the square root:
$2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = -( (x-1)^2 - 1 ) = 1 - (x-1)^2$.
Substituting this back into the integral,we get:
$I = \int \frac{1}{\sqrt{1 - (x-1)^2}} dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}(\frac{u}{a}) + C$,where $u = x-1$ and $a = 1$:
$I = \sin^{-1}(\frac{x-1}{1}) + C = \sin^{-1}(x-1) + C$.
Therefore,the correct option is $B$.

(A) Given $f^{\prime}(x) = 3 \sin x - 4 \sin^3 x$.
Using the trigonometric identity $\sin(3x) = 3 \sin x - 4 \sin^3 x$,we can rewrite the derivative as:
$f^{\prime}(x) = \sin(3x)$.
To find $f(x)$,we integrate $f^{\prime}(x)$ with respect to $x$:
$f(x) = \int \sin(3x) \, dx = -\frac{\cos(3x)}{3} + c$.
Given the initial condition $f(0) = \frac{1}{3}$,we substitute $x = 0$ into the equation:
$f(0) = -\frac{\cos(0)}{3} + c = \frac{1}{3}$.
Since $\cos(0) = 1$,we have:
$-\frac{1}{3} + c = \frac{1}{3}$.
Solving for $c$:
$c = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$.

(C) Given the rate of digging $\frac{dA}{dt} = \frac{2}{\sqrt{t}}$.
Integrating both sides with respect to $t$:
$A = \int \frac{2}{\sqrt{t}} dt = 2 \int t^{-1/2} dt = 2 \cdot \frac{t^{1/2}}{1/2} + C = 4\sqrt{t} + C$.
At $t = 0$,the area dug $A = 0$,so $C = 0$.
Thus,$A = 4\sqrt{t}$.
To find the time $t$ to dig $40$ square metres,set $A = 40$:
$40 = 4\sqrt{t} \Rightarrow \sqrt{t} = 10$.
Squaring both sides,$t = 10^2 = 100$ minutes.

(A) We are given the integral $ I = \int \frac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta} dx $.
Using the trigonometric identity $\cos 2A = 2\cos^2 A - 1$,we can rewrite the numerator:
$\cos 2x - \cos 2\theta = (2\cos^2 x - 1) - (2\cos^2 \theta - 1) = 2\cos^2 x - 2\cos^2 \theta = 2(\cos^2 x - \cos^2 \theta)$.
Now,substitute this into the integral:
$I = \int \frac{2(\cos^2 x - \cos^2 \theta)}{\cos x - \cos \theta} dx$.
Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$:
$I = \int \frac{2(\cos x - \cos \theta)(\cos x + \cos \theta)}{\cos x - \cos \theta} dx$.
Canceling the common term $(\cos x - \cos \theta)$:
$I = 2 \int (\cos x + \cos \theta) dx$.
Integrating with respect to $x$ (noting that $\cos \theta$ is a constant):
$I = 2(\sin x + x \cos \theta) + C$.

(C) To evaluate the integral $ I = \int \frac{1}{1+e^{x}} d x $:
Multiply the numerator and denominator by $ e^{-x} $:
$ I = \int \frac{e^{-x}}{e^{-x}(1+e^{x})} d x $
$ I = \int \frac{e^{-x}}{e^{-x}+1} d x $
Let $ u = e^{-x} + 1 $. Then $ du = -e^{-x} d x $,which implies $ e^{-x} d x = -du $.
Substituting these into the integral:
$ I = \int \frac{-du}{u} $
$ I = -\ln |u| + C $
$ I = -\ln |e^{-x} + 1| + C $
Simplify the expression inside the logarithm:
$ I = -\ln \left| \frac{1}{e^{x}} + 1 \right| + C $
$ I = -\ln \left| \frac{1+e^{x}}{e^{x}} \right| + C $
Using the property $ -\ln(a/b) = \ln(b/a) $:
$ I = \ln \left| \frac{e^{x}}{1+e^{x}} \right| + C $
Thus,the correct option is $ C $.

(B) Given that $\int f(x) dx = g(x)$.
This implies that $f(x) = g'(x)$.
We need to evaluate the integral $\int f(x) g(x) dx$.
Substituting $f(x) = g'(x)$,we get $\int g'(x) g(x) dx$.
Let $g(x) = u$,then $g'(x) dx = du$.
The integral becomes $\int u du = \frac{u^2}{2} + C$.
Substituting back $u = g(x)$,we get $\frac{1}{2} g^2(x) + C$.

(B) We have $I = \int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx$.
Using the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we multiply the numerator and denominator by $2 \cos \frac{x}{2}$:
$I = \int \frac{2 \sin \frac{5x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} dx = \int \frac{\sin(3x) + \sin(2x)}{\sin x} dx$.
Using the formulas $\sin 3x = 3 \sin x - 4 \sin^3 x$ and $\sin 2x = 2 \sin x \cos x$:
$I = \int \frac{3 \sin x - 4 \sin^3 x + 2 \sin x \cos x}{\sin x} dx = \int (3 - 4 \sin^2 x + 2 \cos x) dx$.
Using $4 \sin^2 x = 2(1 - \cos 2x)$:
$I = \int (3 - 2(1 - \cos 2x) + 2 \cos x) dx = \int (3 - 2 + 2 \cos 2x + 2 \cos x) dx$.
$I = \int (1 + 2 \cos 2x + 2 \cos x) dx = x + \sin 2x + 2 \sin x + C$.

(B) Given integral: $I = \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx$
Using the identity $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$I = \int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} dx$
$I = \int \frac{2\cos^2 x - 2\cos^2 \alpha}{\cos x - \cos \alpha} dx$
$I = 2 \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} dx$
$I = 2 \int (\cos x + \cos \alpha) dx$
Integrating with respect to $x$:
$I = 2(\sin x + x \cos \alpha) + c$

(A) Let $I = \int \frac{\sec x}{\sec x+\tan x} dx$.
Multiplying the numerator and denominator by $(\sec x - \tan x)$,we get:
$I = \int \frac{\sec x(\sec x - \tan x)}{\sec^2 x - \tan^2 x} dx$.
Since $\sec^2 x - \tan^2 x = 1$,the integral simplifies to:
$I = \int (\sec^2 x - \sec x \tan x) dx$.
Integrating term by term,we obtain:
$I = \tan x - \sec x + C$.

(D) We know that $1 + \cos 2\theta = 2 \cos^2 \theta$.
Applying this identity for $\theta = 4x$,we get $1 + \cos 8x = 2 \cos^2 4x$.
Substituting this into the integral:
$I = \int \frac{1}{2 \cos^2 4x} dx$
$I = \frac{1}{2} \int \sec^2 4x dx$
Using the standard integral $\int \sec^2(ax) dx = \frac{\tan(ax)}{a} + c$,we get:
$I = \frac{1}{2} \cdot \frac{\tan 4x}{4} + c$
$I = \frac{\tan 4x}{8} + c$

(C) Let $I = \int \sqrt{5-2x+x^2} dx$.
Completing the square,we have $5-2x+x^2 = (x-1)^2 + 4 = (x-1)^2 + 2^2$.
Using the standard integral formula $\int \sqrt{x^2+a^2} dx = \frac{x}{2} \sqrt{x^2+a^2} + \frac{a^2}{2} \log |x + \sqrt{x^2+a^2}| + C$,where $x$ is replaced by $(x-1)$ and $a=2$:
$I = \frac{x-1}{2} \sqrt{(x-1)^2 + 2^2} + \frac{2^2}{2} \log |(x-1) + \sqrt{(x-1)^2 + 2^2}| + C$.
$I = \frac{x-1}{2} \sqrt{5-2x+x^2} + 2 \log |(x-1) + \sqrt{5-2x+x^2}| + C$.

(B) Given the integral: $ I = \int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} dx $
Using the logarithmic property $ e^{k \log x} = x^k $,we can simplify the expression:
$ I = \int \frac{x^6 - x^5}{x^4 - x^3} dx $
Factor out the common terms in the numerator and denominator:
$ I = \int \frac{x^5(x - 1)}{x^3(x - 1)} dx $
Assuming $ x \neq 1 $,we can cancel the term $ (x - 1) $:
$ I = \int \frac{x^5}{x^3} dx = \int x^2 dx $
Integrating $ x^2 $ with respect to $ x $ gives:
$ I = \frac{x^3}{3} + c $

(C) Let $I = \int \operatorname{cosec}(x-a) \operatorname{cosec} x \, dx$.
Multiply and divide by $\sin a$:
$I = \int \frac{\sin a}{\sin a \sin(x-a) \sin x} \, dx$.
Since $\sin a = \sin(x - (x-a))$,we have:
$I = \frac{1}{\sin a} \int \frac{\sin(x - (x-a))}{\sin(x-a) \sin x} \, dx$.
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin a} \int \frac{\sin x \cos(x-a) - \cos x \sin(x-a)}{\sin(x-a) \sin x} \, dx$.
$I = \frac{1}{\sin a} \int \left( \frac{\cos(x-a)}{\sin(x-a)} - \frac{\cos x}{\sin x} \right) \, dx$.
$I = \frac{1}{\sin a} \int (\cot(x-a) - \cot x) \, dx$.
Integrating,we get:
$I = \frac{1}{\sin a} [\log |\sin(x-a)| - \log |\sin x|] + c$.
$I = \frac{1}{\sin a} \log \left| \frac{\sin(x-a)}{\sin x} \right| + c$.
This can be written as $\frac{1}{\sin a} \log |\sin(x-a) \operatorname{cosec} x| + c$.

(C) Let $I = \int \frac{x^{2}+1}{x^{2}-1} d x$.
We can rewrite the numerator as $x^{2}-1+2$.
So,$I = \int \frac{x^{2}-1+2}{x^{2}-1} d x$.
$I = \int \left( \frac{x^{2}-1}{x^{2}-1} + \frac{2}{x^{2}-1} \right) d x$.
$I = \int 1 d x + 2 \int \frac{1}{x^{2}-1} d x$.
Using the standard integral formula $\int \frac{1}{x^{2}-a^{2}} d x = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + c$,where $a=1$:
$I = x + 2 \cdot \frac{1}{2(1)} \log \left| \frac{x-1}{x+1} \right| + c$.
$I = x + \log \left| \frac{x-1}{x+1} \right| + c$.

(B) Given the integral $ I = \int \frac{\sin ^{2} x}{1+\cos x} d x $.
Using the trigonometric identity $ \sin ^{2} x = 1 - \cos ^{2} x $,we get:
$ I = \int \frac{1 - \cos ^{2} x}{1 + \cos x} d x $.
Since $ 1 - \cos ^{2} x = (1 - \cos x)(1 + \cos x) $,the expression becomes:
$ I = \int \frac{(1 - \cos x)(1 + \cos x)}{1 + \cos x} d x $.
Canceling the common term $ (1 + \cos x) $,we have:
$ I = \int (1 - \cos x) d x $.
Integrating term by term,we get:
$ I = x - \sin x + C $.

(B) Given,$f^{\prime}(x)=a \sin x+b \cos x$ and $f^{\prime}(0)=4$.
Substituting $x=0$ in the derivative: $f^{\prime}(0)=a \sin(0)+b \cos(0) = b = 4$.
Now,integrate $f^{\prime}(x)$ to find $f(x)$:
$f(x) = \int (a \sin x + 4 \cos x) dx = -a \cos x + 4 \sin x + C$.
Using $f(0)=3$: $f(0) = -a \cos(0) + 4 \sin(0) + C = -a + C = 3 \Rightarrow C = a+3$.
Using $f\left(\frac{\pi}{2}\right)=5$: $f\left(\frac{\pi}{2}\right) = -a \cos\left(\frac{\pi}{2}\right) + 4 \sin\left(\frac{\pi}{2}\right) + C = 0 + 4(1) + C = 4+C = 5 \Rightarrow C = 1$.
Substituting $C=1$ into $C=a+3$: $1 = a+3 \Rightarrow a = -2$.
Thus,$f(x) = -(-2) \cos x + 4 \sin x + 1 = 2 \cos x + 4 \sin x + 1$.

(C) We are given the integral $\int \frac{1}{1 + \sin x} dx$.
Multiply the numerator and denominator by $(1 - \sin x)$:
$\int \frac{1 - \sin x}{1 - \sin^2 x} dx = \int \frac{1 - \sin x}{\cos^2 x} dx = \int (\sec^2 x - \sec x \tan x) dx = \tan x - \sec x + c$.
Alternatively,using the half-angle identity $\sin x = \cos(\frac{\pi}{2} - x)$:
$\int \frac{1}{1 + \cos(\frac{\pi}{2} - x)} dx = \int \frac{1}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} dx = \frac{1}{2} \int \sec^2(\frac{\pi}{4} - \frac{x}{2}) dx$.
Integrating this,we get $\frac{1}{2} \cdot \frac{\tan(\frac{\pi}{4} - \frac{x}{2})}{-\frac{1}{2}} + c = -\tan(\frac{\pi}{4} - \frac{x}{2}) + c = \tan(\frac{x}{2} - \frac{\pi}{4}) + c$.
Comparing this with $\tan(f(x)) + c$,we have $f(x) = \frac{x}{2} - \frac{\pi}{4}$.
Thus,$f'(x) = \frac{d}{dx}(\frac{x}{2} - \frac{\pi}{4}) = \frac{1}{2}$.
Therefore,$f'(0) = \frac{1}{2}$.

(C) We know that $1 - \sin 2x = \cos^2 x + \sin^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2$.
Thus,$\int \sqrt{1 - \sin 2x} \, dx = \int |\cos x - \sin x| \, dx$.
Given $x \notin [2n\pi - \frac{\pi}{4}, 2n\pi + \frac{3\pi}{4}]$,the expression $\cos x - \sin x$ is negative in this interval.
Therefore,$|\cos x - \sin x| = -(\cos x - \sin x) = \sin x - \cos x$.
Integrating this,we get $\int (\sin x - \cos x) \, dx = -\cos x - \sin x + c$.

(B) Given the integral: $\int \frac{1}{1-\cos x} dx = \int \frac{1}{2 \sin^2 \frac{x}{2}} dx$
$= \frac{1}{2} \int \operatorname{cosec}^2 \frac{x}{2} dx$
$= \frac{1}{2} \left(-2 \cot \frac{x}{2}\right) + c = -\cot \frac{x}{2} + c$
We know that $-\cot \theta = \tan \left(\theta + \frac{\pi}{2}\right)$ or $\tan \left(\theta - \frac{\pi}{2}\right)$.
Using $-\cot \frac{x}{2} = \tan \left(\frac{x}{2} - \frac{\pi}{2}\right)$,we compare this with $\tan \left(\frac{x}{\alpha} + \beta\right)$.
Thus,$\alpha = 2$ and $\beta = -\frac{\pi}{2}$.
Calculating the required value: $\frac{\pi \alpha}{4} - \beta = \frac{\pi(2)}{4} - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$.

(B) Let $I = \int \frac{x^5}{x^2+1} \, dx$.
We can rewrite the integrand as:
$\frac{x^5}{x^2+1} = \frac{x^3(x^2+1) - x^3}{x^2+1} = x^3 - \frac{x^3}{x^2+1}$.
Further,$\frac{x^3}{x^2+1} = \frac{x(x^2+1) - x}{x^2+1} = x - \frac{x}{x^2+1}$.
So,$\frac{x^5}{x^2+1} = x^3 - x + \frac{x}{x^2+1}$.
Now,integrating term by term:
$I = \int (x^3 - x + \frac{x}{x^2+1}) \, dx = \frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{2} \int \frac{2x}{x^2+1} \, dx$.
Using the substitution $u = x^2+1$,$du = 2x \, dx$,we get:
$I = \frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{2} \log(x^2+1) + c$.

(B) We know that the Taylor series expansion for the exponential function is $e^u = \sum_{r=0}^{\infty} \frac{u^r}{r!}$.
Substituting $u = 3x$,we get $\sum_{r=0}^{\infty} \frac{(3x)^r}{r!} = \sum_{r=0}^{\infty} \frac{x^r 3^r}{r!} = e^{3x}$.
Now,we need to evaluate the integral $\int \left(\sum_{r=0}^{\infty} \frac{x^r 3^r}{r!}\right) dx$.
Substituting the series with $e^{3x}$,the integral becomes $\int e^{3x} dx$.
Using the integration formula $\int e^{ax} dx = \frac{e^{ax}}{a} + c$,we get $\int e^{3x} dx = \frac{e^{3x}}{3} + c$.

(C) We have the integral $I = \int \frac{\sin 7x}{\sin 2x \sin 5x} dx$.
Using the identity $\sin 7x = \sin(5x + 2x)$,we can write:
$I = \int \frac{\sin(5x + 2x)}{\sin 2x \sin 5x} dx$.
Applying the formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$:
$I = \int \frac{\sin 5x \cos 2x + \cos 5x \sin 2x}{\sin 2x \sin 5x} dx$.
Dividing each term in the numerator by the denominator:
$I = \int \frac{\sin 5x \cos 2x}{\sin 2x \sin 5x} dx + \int \frac{\cos 5x \sin 2x}{\sin 2x \sin 5x} dx$.
$I = \int \cot 2x dx + \int \cot 5x dx$.
Using the standard integral $\int \cot(ax) dx = \frac{1}{a} \log |\sin(ax)| + c$:
$I = \frac{1}{2} \log |\sin 2x| + \frac{1}{5} \log |\sin 5x| + c$.

(C) Given $f(x) = \int \frac{dx}{x^2+(\sqrt{2})^2}$.
Using the standard integral formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$f(x) = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{x}{\sqrt{2}}) + C$.
Given $f(\sqrt{2}) = 0$,substitute $x = \sqrt{2}$:
$0 = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{\sqrt{2}}{\sqrt{2}}) + C$
$0 = \frac{1}{\sqrt{2}} \tan^{-1}(1) + C$
$0 = \frac{1}{\sqrt{2}} (\frac{\pi}{4}) + C$
$C = -\frac{\pi}{4\sqrt{2}}$.
Now,find $f(0)$:
$f(0) = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{0}{\sqrt{2}}) - \frac{\pi}{4\sqrt{2}}$
$f(0) = \frac{1}{\sqrt{2}} (0) - \frac{\pi}{4\sqrt{2}} = -\frac{\pi}{4\sqrt{2}}$.

(B) We are given the integral $I = \int 3^{-\log _9 x^2} d x$.
Using the property of logarithms $n \log_b a = \log_b a^n$,we have $-\log _9 x^2 = \log _9 (x^2)^{-1} = \log _9 (\frac{1}{x^2})$.
Substituting this into the integral,we get $I = \int 3^{\log _9 (\frac{1}{x^2})} d x$.
Using the property $a^{\log _b c} = c^{\log _b a}$,we rewrite $3^{\log _9 (\frac{1}{x^2})}$ as $(\frac{1}{x^2})^{\log _9 3}$.
Since $\log _9 3 = \log _{3^2} 3 = \frac{1}{2} \log _3 3 = \frac{1}{2}$,the expression becomes $(\frac{1}{x^2})^{\frac{1}{2}} = \frac{1}{x}$.
Thus,$I = \int \frac{1}{x} d x = \log |x| + C$.

(A) Let $I = \int \frac{3x+2}{4x^2+4x+5} dx$.
We express the numerator as a derivative of the denominator:
$3x+2 = \frac{3}{8}(8x+4) + (2 - \frac{12}{8}) = \frac{3}{8}(8x+4) + \frac{1}{2}$.
Substituting this into the integral:
$I = \int \frac{\frac{3}{8}(8x+4) + \frac{1}{2}}{4x^2+4x+5} dx = \frac{3}{8} \int \frac{8x+4}{4x^2+4x+5} dx + \frac{1}{2} \int \frac{1}{(2x+1)^2 + 2^2} dx$.
For the first part,let $u = 4x^2+4x+5$,then $du = (8x+4)dx$.
$I = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{2} \int \frac{1}{(2x+1)^2 + 2^2} dx$.
Using the formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \tan^{-1}(\frac{2x+1}{2}) + C$.
$I = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{8} \tan^{-1}(x+\frac{1}{2}) + C$.
Comparing with the given form,$A = \frac{3}{8}$ and $B = \frac{1}{8}$.

(B) Given $f(x)=1+2^x \log 2$.
Since $g(x)$ is an antiderivative of $f(x)$,we have $g(x) = \int f(x) dx$.
$g(x) = \int (1+2^x \log 2) dx = x + \frac{2^x \log 2}{\log 2} + c = x + 2^x + c$.
Thus,$y = x + 2^x + c$ ... $(i)$.
The curve passes through $\left(-1, \frac{1}{2}\right)$,so $\frac{1}{2} = -1 + 2^{-1} + c$.
$\frac{1}{2} = -1 + \frac{1}{2} + c \implies c = 1$.
Substituting $c=1$ in $(i)$,we get $y = x + 2^x + 1$.
To find the intersection with the $Y$-axis,set $x=0$:
$y = 0 + 2^0 + 1 = 1 + 1 = 2$.
Therefore,the curve meets the $Y$-axis at $(0,2)$.

(B) Let $I = \int \frac{1 + \tan x \tan(x + a)}{\tan x \tan(x + a)} dx$.
We know that $\tan a = \tan((x + a) - x) = \frac{\tan(x + a) - \tan x}{1 + \tan x \tan(x + a)}$.
Therefore,$1 + \tan x \tan(x + a) = \frac{\tan(x + a) - \tan x}{\tan a}$.
Substituting this into the integral,we get:
$I = \int \frac{\tan(x + a) - \tan x}{\tan a \tan x \tan(x + a)} dx$.
$I = \frac{1}{\tan a} \int \left( \frac{\tan(x + a)}{\tan x \tan(x + a)} - \frac{\tan x}{\tan x \tan(x + a)} \right) dx$.
$I = \cot a \int \left( \frac{1}{\tan x} - \frac{1}{\tan(x + a)} \right) dx$.
$I = \cot a \int (\cot x - \cot(x + a)) dx$.
Integrating,we get:
$I = \cot a (\log |\sin x| - \log |\sin(x + a)|) + C$.
Thus,the correct option is $B$.

(A) Let $I = \int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} \, dx$.
Since $1 = \operatorname{cosec}^2 x - \cot^2 x$,we have:
$I = \int \sqrt{\operatorname{cosec}^2 x - \cot^2 x + 2 \cot^2 x + 2 \cot x \operatorname{cosec} x} \, dx$
$I = \int \sqrt{\operatorname{cosec}^2 x + \cot^2 x + 2 \cot x \operatorname{cosec} x} \, dx$
$I = \int \sqrt{(\operatorname{cosec} x + \cot x)^2} \, dx$
$I = \int (\operatorname{cosec} x + \cot x) \, dx$
Using the standard integrals $\int \operatorname{cosec} x \, dx = \log |\operatorname{cosec} x - \cot x|$ and $\int \cot x \, dx = \log |\sin x|$:
$I = \log |\operatorname{cosec} x - \cot x| + \log |\sin x| + C$
$I = \log \left| \frac{1 - \cos x}{\sin x} \cdot \sin x \right| + C$
$I = \log |1 - \cos x| + C$
Using the identity $1 - \cos x = 2 \sin^2 \frac{x}{2}$:
$I = \log |2 \sin^2 \frac{x}{2}| + C$
$I = \log 2 + 2 \log |\sin \frac{x}{2}| + C$
Since $\log 2$ is a constant,we can absorb it into $C$:
$I = 2 \log |\sin \frac{x}{2}| + C$