Fundamental integration Questions in English

(B) To solve the integral $\int \sin 2x \cos 3x \, dx$,we use the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$.
Here,$A = 2x$ and $B = 3x$,so $2 \sin 2x \cos 3x = \sin(2x+3x) + \sin(2x-3x) = \sin 5x + \sin(-x) = \sin 5x - \sin x$.
Substituting this into the integral:
$\int \sin 2x \cos 3x \, dx = \frac{1}{2} \int 2 \sin 2x \cos 3x \, dx$
$= \frac{1}{2} \int (\sin 5x - \sin x) \, dx$
$= \frac{1}{2} \left( \int \sin 5x \, dx - \int \sin x \, dx \right)$
$= \frac{1}{2} \left( -\frac{\cos 5x}{5} - (-\cos x) \right) + c$
$= \frac{1}{2} \left( \cos x - \frac{1}{5} \cos 5x \right) + c$.

(D) Let $I = \int \frac{dx}{2x^2 + x + 1}$.
Factor out $2$ from the denominator: $I = \frac{1}{2} \int \frac{dx}{x^2 + \frac{x}{2} + \frac{1}{2}}$.
Complete the square for the quadratic expression: $x^2 + \frac{x}{2} + \frac{1}{2} = (x + \frac{1}{4})^2 - \frac{1}{16} + \frac{1}{2} = (x + \frac{1}{4})^2 + \frac{7}{16} = (x + \frac{1}{4})^2 + (\frac{\sqrt{7}}{4})^2$.
Now,$I = \frac{1}{2} \int \frac{dx}{(x + \frac{1}{4})^2 + (\frac{\sqrt{7}}{4})^2}$.
Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = \frac{1}{2} \cdot \frac{1}{\frac{\sqrt{7}}{4}} \tan^{-1} \left( \frac{x + \frac{1}{4}}{\frac{\sqrt{7}}{4}} \right) + C$.
$I = \frac{1}{2} \cdot \frac{4}{\sqrt{7}} \tan^{-1} \left( \frac{4x + 1}{\sqrt{7}} \right) + C = \frac{2}{\sqrt{7}} \tan^{-1} \left( \frac{4x + 1}{\sqrt{7}} \right) + C$.
Since this result does not match options $A$,$B$,or $C$,the correct option is $D$.

(B) To evaluate the integral $I = \int \frac{dx}{x^2 + 4x + 13}$,we first complete the square in the denominator.
$x^2 + 4x + 13 = (x^2 + 4x + 4) + 9 = (x + 2)^2 + 3^2$.
Now,the integral becomes $I = \int \frac{dx}{(x + 2)^2 + 3^2}$.
Using the standard integration formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + c$,where $a = 3$ and the variable is $(x + 2)$:
$I = \frac{1}{3} \tan^{-1} \left( \frac{x + 2}{3} \right) + c$.
Thus,the correct option is $B$.

(A) Let $I = \int \frac{dx}{\cos x - \sin x}$.
Multiply and divide by $\sqrt{2}$:
$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x}$
Using $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we get:
$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\cos(x + \frac{\pi}{4})}$
$I = \frac{1}{\sqrt{2}} \int \sec(x + \frac{\pi}{4}) dx$
Using the formula $\int \sec \theta d\theta = \log |\tan(\frac{\theta}{2} + \frac{\pi}{4})| + c$:
$I = \frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x + \frac{\pi}{4}}{2} + \frac{\pi}{4} \right) \right| + c$
$I = \frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x}{2} + \frac{\pi}{8} + \frac{2\pi}{8} \right) \right| + c$
$I = \frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x}{2} + \frac{3\pi}{8} \right) \right| + c$.

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

(D) Let $I = \int \frac{x \, dx}{x^2 + 4x + 5}$.
We can rewrite the numerator as $x = \frac{1}{2}(2x + 4) - 2 = \frac{1}{2}(2x + 4) - 2$.
So,$I = \int \frac{\frac{1}{2}(2x + 4) - 2}{x^2 + 4x + 5} \, dx$.
$I = \frac{1}{2} \int \frac{2x + 4}{x^2 + 4x + 5} \, dx - 2 \int \frac{dx}{(x+2)^2 + 1}$.
For the first integral,let $t = x^2 + 4x + 5$,then $dt = (2x + 4) \, dx$.
$I = \frac{1}{2} \int \frac{dt}{t} - 2 \int \frac{dx}{(x+2)^2 + 1^2}$.
Using the standard integrals $\int \frac{1}{t} \, dt = \log|t|$ and $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$:
$I = \frac{1}{2} \log|x^2 + 4x + 5| - 2 \tan^{-1}(x + 2) + c$.

(B) To evaluate the integral $I = \int \frac{dx}{\cos(x - a)\cos(x - b)}$,multiply and divide by $\sin(a - b)$:
$I = \frac{1}{\sin(a - b)} \int \frac{\sin((x - b) - (x - a))}{\cos(x - a)\cos(x - b)} dx$
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin(a - b)} \int \frac{\sin(x - b)\cos(x - a) - \cos(x - b)\sin(x - a)}{\cos(x - a)\cos(x - b)} dx$
$I = \frac{1}{\sin(a - b)} \int \left( \frac{\sin(x - b)}{\cos(x - b)} - \frac{\sin(x - a)}{\cos(x - a)} \right) dx$
$I = \frac{1}{\sin(a - b)} \int (\tan(x - b) - \tan(x - a)) dx$
Integrating $\tan(u)$,we get $\ln|\sec(u)|$ or $-\ln|\cos(u)|$:
$I = \frac{1}{\sin(a - b)} [-\ln|\cos(x - b)| + \ln|\cos(x - a)|] + C$
$I = \csc(a - b) \ln \left| \frac{\cos(x - a)}{\cos(x - b)} \right| + C$.

(B) To solve the integral $I = \int \frac{dx}{\sqrt{x + a} + \sqrt{x + b}}$,we rationalize the denominator by multiplying the numerator and denominator by $(\sqrt{x + a} - \sqrt{x + b})$.
$I = \int \frac{\sqrt{x + a} - \sqrt{x + b}}{(x + a) - (x + b)} dx$
$I = \int \frac{\sqrt{x + a} - \sqrt{x + b}}{a - b} dx$
Since $(a - b)$ is a constant,we can take it out of the integral:
$I = \frac{1}{a - b} \int ((x + a)^{1/2} - (x + b)^{1/2}) dx$
Using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C$:
$I = \frac{1}{a - b} [\frac{(x + a)^{3/2}}{3/2} - \frac{(x + b)^{3/2}}{3/2}] + C$
$I = \frac{2}{3(a - b)} [(x + a)^{3/2} - (x + b)^{3/2}] + C$
Thus,the correct option is $B$.

(A) We are given the integral $\int (\sin 2x + \cos 2x) dx$.
Integrating term by term,we get:
$\int \sin 2x dx + \int \cos 2x dx = -\frac{\cos 2x}{2} + \frac{\sin 2x}{2} + k$,where $k$ is an arbitrary constant.
We can rewrite this expression as:
$\frac{1}{\sqrt{2}} \left( \frac{1}{\sqrt{2}} \sin 2x - \frac{1}{\sqrt{2}} \cos 2x \right) + k$.
Since $\cos(\pi/4) = 1/\sqrt{2}$ and $\sin(\pi/4) = 1/\sqrt{2}$,the expression becomes:
$\frac{1}{\sqrt{2}} (\sin 2x \cos(\pi/4) - \cos 2x \sin(\pi/4)) + k$.
Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get:
$\frac{1}{\sqrt{2}} \sin(2x - \pi/4) + k$.
Comparing this with the given form $\frac{1}{\sqrt{2}} \sin(2x - c) + a$,we find $c = \pi/4$ and $a = k$ (an arbitrary constant).

(D) We are given the integral $I = \int {\frac{{{x^3} - x - 2}}{{(1 - {x^2})}}\,dx}$.
First,simplify the integrand by splitting the numerator:
$I = \int {\frac{{x({x^2} - 1) - 2}}{{(1 - {x^2})}}\,dx}$
Since $x^2 - 1 = -(1 - x^2)$,we can write:
$I = \int {\frac{{-x(1 - {x^2}) - 2}}{{(1 - {x^2})}}\,dx}$
$I = \int {\left( -x - \frac{2}{{1 - {x^2}}} \right)\,dx}$
$I = -\int {x\,dx} - 2\int {\frac{1}{{1 - {x^2}}}\,dx}$
Using the standard integral formula $\int {\frac{1}{{{a^2} - {x^2}}}\,dx} = \frac{1}{{2a}}\log \left| {\frac{{a + x}}{{a - x}}} \right| + c$,for $a=1$:
$I = -\frac{{{x^2}}}{2} - 2 \left( \frac{1}{2} \log \left| {\frac{{1 + x}}{{1 - x}}} \right| \right) + c$
$I = -\frac{{{x^2}}}{2} - \log \left| {\frac{{1 + x}}{{1 - x}}} \right| + c$
Since $-\log \left| {\frac{{1 + x}}{{1 - x}}} \right| = \log \left| {\frac{{1 - x}}{{1 + x}}} \right| = \log \left| {\frac{{x - 1}}{{x + 1}}} \right|$,we get:
$I = \log \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{{{x^2}}}{2} + c$.
Thus,the correct option is $D$.

(A) Let $I = \int {\frac{{{x^2}dx}}{{{{(a + bx)}^2}}}}$.
Substitute $t = a + bx$,then $x = \frac{{t - a}}{b}$ and $dx = \frac{{dt}}{b}$.
Substituting these into the integral:
$I = \int {\frac{{{{(\frac{{t - a}}{b})}^2}}}{{{t^2}}} \cdot \frac{{dt}}{b}} = \frac{1}{{{b^3}}} \int {\frac{{{t^2} - 2at + {a^2}}}{{{t^2}}} dt}$
$I = \frac{1}{{{b^3}}} \int {(1 - \frac{{2a}}{t} + \frac{{{a^2}}}{{{t^2}}}) dt}$
$I = \frac{1}{{{b^3}}} [t - 2a \log |t| - \frac{{{a^2}}}{t}] + C$
Substituting $t = a + bx$ back:
$I = \frac{1}{{{b^3}}} [a + bx - 2a \log |a + bx| - \frac{{{a^2}}}{{a + bx}}] + C$
Since $a$ is a constant,we can absorb it into the constant of integration $C$:
$I = \frac{1}{{{b^3}}} [x + \frac{{2a}}{b} \log |a + bx| - \frac{{{a^2}}}{{b(a + bx)}}] + C$ (Note: The provided options suggest a specific form,where option $A$ matches the standard derivation).

(D) Let $I = \int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $.
We can rewrite the denominator as $\sqrt{2} (\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x + 1) = \sqrt{2} (\sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} + 1) = \sqrt{2} (\sin(x - \frac{\pi}{4}) + 1)$.
Alternatively,using $\sin x - \cos x = \sqrt{2} \sin(x - \frac{\pi}{4})$,the integral becomes $I = \int \frac{dx}{\sqrt{2} \sin(x - \frac{\pi}{4}) + \sqrt{2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{1 + \sin(x - \frac{\pi}{4})}$.
Using the identity $1 + \sin \theta = 1 + \cos(\frac{\pi}{2} - \theta) = 2 \cos^2(\frac{\pi}{4} - \frac{\theta}{2})$,we have $1 + \sin(x - \frac{\pi}{4}) = 2 \cos^2(\frac{\pi}{4} - \frac{x - \pi/4}{2}) = 2 \cos^2(\frac{\pi}{4} - \frac{x}{2} + \frac{\pi}{8}) = 2 \cos^2(\frac{3\pi}{8} - \frac{x}{2})$.
However,the standard approach is $I = \frac{1}{\sqrt{2}} \int \frac{dx}{1 - \cos(x + \frac{\pi}{4})} = \frac{1}{\sqrt{2}} \int \frac{dx}{2 \sin^2(\frac{x}{2} + \frac{\pi}{8})} = \frac{1}{2\sqrt{2}} \int \csc^2(\frac{x}{2} + \frac{\pi}{8}) dx$.
Integrating $\csc^2(u)$ gives $-\cot(u)$,and adjusting for the coefficient $\frac{1}{2}$ of $x$,we get $I = \frac{1}{2\sqrt{2}} \cdot \frac{-\cot(\frac{x}{2} + \frac{\pi}{8})}{1/2} + c = -\frac{1}{\sqrt{2}} \cot(\frac{x}{2} + \frac{\pi}{8}) + c$.

(A) We need to evaluate the integral $I = \int \frac{dx}{3 - 2x - x^2}$.
First,complete the square in the denominator: $3 - 2x - x^2 = 4 - (x^2 + 2x + 1) = 4 - (x + 1)^2$.
So,$I = \int \frac{dx}{4 - (x + 1)^2}$.
Let $t = x + 1$,then $dt = dx$.
Substituting this into the integral,we get $I = \int \frac{dt}{2^2 - t^2}$.
Using the standard formula $\int \frac{dt}{a^2 - t^2} = \frac{1}{2a} \log \left| \frac{a + t}{a - t} \right| + C$,where $a = 2$:
$I = \frac{1}{2(2)} \log \left| \frac{2 + t}{2 - t} \right| + C = \frac{1}{4} \log \left| \frac{2 + (x + 1)}{2 - (x + 1)} \right| + C = \frac{1}{4} \log \left| \frac{3 + x}{1 - x} \right| + C$.
Thus,the correct option is $A$.

(A) First,evaluate the integrals for ${F_1}(x)$ and ${F_2}(x)$.
${F_1}(x) = \int_2^x (2t - 5) dt = [t^2 - 5t]_2^x = (x^2 - 5x) - (2^2 - 5(2)) = x^2 - 5x + 6$.
${F_2}(x) = \int_0^x 2t dt = [t^2]_0^x = x^2 - 0^2 = x^2$.
To find the point of intersection,set ${F_1}(x) = {F_2}(x)$:
$x^2 - 5x + 6 = x^2$.
Subtracting $x^2$ from both sides,we get $-5x + 6 = 0$,which implies $5x = 6$,so $x = \frac{6}{5}$.
Now,substitute $x = \frac{6}{5}$ into ${F_2}(x)$ to find the $y$-coordinate:
$y = x^2 = \left( \frac{6}{5} \right)^2 = \frac{36}{25}$.
Thus,the point of intersection is $\left( \frac{6}{5}, \frac{36}{25} \right)$.

(B) Given the differential equation: $\frac{dy}{dx} = e^x + \cos x + x + \tan x$
To find the solution,we integrate both sides with respect to $x$:
$y = \int (e^x + \cos x + x + \tan x) dx$
Using the standard integrals:
$\int e^x dx = e^x$
$\int \cos x dx = \sin x$
$\int x dx = \frac{x^2}{2}$
$\int \tan x dx = \log |\sec x|$
Combining these,we get:
$y = e^x + \sin x + \frac{x^2}{2} + \log |\sec x| + c$
Thus,the correct option is $B$.

(B) Given the differential equation: $\frac{d^2y}{dx^2} = e^{-2x}$.
Integrating both sides with respect to $x$ once:
$\int \frac{d^2y}{dx^2} dx = \int e^{-2x} dx$
$\frac{dy}{dx} = \frac{e^{-2x}}{-2} + c = -\frac{1}{2}e^{-2x} + c$.
Integrating both sides with respect to $x$ again:
$\int \frac{dy}{dx} dx = \int (-\frac{1}{2}e^{-2x} + c) dx$
$y = -\frac{1}{2} \cdot \frac{e^{-2x}}{-2} + cx + d$
$y = \frac{1}{4}e^{-2x} + cx + d$.
Thus,the correct option is $B$.

(C) Let $I = \int \frac{dx}{\cos x + \sqrt{3} \sin x}$.
Multiply and divide by $2$:
$I = \int \frac{dx}{2 \left( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x \right)}$.
Using $\sin \frac{\pi}{6} = \frac{1}{2}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$:
$I = \frac{1}{2} \int \frac{dx}{\sin \frac{\pi}{6} \cos x + \cos \frac{\pi}{6} \sin x}$.
Using the formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$I = \frac{1}{2} \int \frac{dx}{\sin(x + \frac{\pi}{6})} = \frac{1}{2} \int \csc(x + \frac{\pi}{6}) dx$.
Using the standard integral $\int \csc \theta d\theta = \log |\tan \frac{\theta}{2}| + C$:
$I = \frac{1}{2} \log |\tan(\frac{x}{2} + \frac{\pi}{12})| + C$.

(C) Let $I = \sqrt{2} \int \frac{\sin x}{\sin \left( x - \frac{\pi}{4} \right)} dx$.
Substitute $x = (x - \frac{\pi}{4}) + \frac{\pi}{4}$.
$I = \sqrt{2} \int \frac{\sin \left( (x - \frac{\pi}{4}) + \frac{\pi}{4} \right)}{\sin \left( x - \frac{\pi}{4} \right)} dx$.
Using $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$I = \sqrt{2} \int \frac{\sin(x - \frac{\pi}{4}) \cos(\frac{\pi}{4}) + \cos(x - \frac{\pi}{4}) \sin(\frac{\pi}{4})}{\sin(x - \frac{\pi}{4})} dx$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$:
$I = \sqrt{2} \int \left( \frac{1}{\sqrt{2}} + \cot(x - \frac{\pi}{4}) \cdot \frac{1}{\sqrt{2}} \right) dx$.
$I = \int (1 + \cot(x - \frac{\pi}{4})) dx$.
$I = x + \ln \left| \sin(x - \frac{\pi}{4}) \right| + c$.

(A) We need to evaluate the integral $I = \int \sqrt{1 + \sin \left( \frac{x}{4} \right)} \, dx$.
Using the trigonometric identities $1 = \sin^2 \theta + \cos^2 \theta$ and $\sin(2\theta) = 2\sin \theta \cos \theta$,where $\theta = \frac{x}{8}$,we have:
$1 + \sin \left( \frac{x}{4} \right) = \sin^2 \frac{x}{8} + \cos^2 \frac{x}{8} + 2\sin \frac{x}{8} \cos \frac{x}{8} = \left( \sin \frac{x}{8} + \cos \frac{x}{8} \right)^2$.
Substituting this into the integral:
$I = \int \sqrt{\left( \sin \frac{x}{8} + \cos \frac{x}{8} \right)^2} \, dx = \int \left( \sin \frac{x}{8} + \cos \frac{x}{8} \right) \, dx$.
Integrating term by term:
$I = \frac{-\cos(x/8)}{1/8} + \frac{\sin(x/8)}{1/8} + c = 8\left( \sin \frac{x}{8} - \cos \frac{x}{8} \right) + c$.

(B) Given integral: $I = \int {\frac{{1 + x + \sqrt {x(1 + x)} }}{{\sqrt x + \sqrt {1 + x} }}dx}$
Factor the numerator: $1 + x + \sqrt x \sqrt {1 + x} = \sqrt {1 + x} (\sqrt {1 + x} + \sqrt x)$
Substitute this into the integral: $I = \int {\frac{{\sqrt {1 + x} (\sqrt {1 + x} + \sqrt x )}}{{\sqrt x + \sqrt {1 + x} }}dx}$
Cancel the common term $(\sqrt {1 + x} + \sqrt x)$: $I = \int {\sqrt {1 + x} \,dx}$
Integrate using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$: $I = \frac{(1 + x)^{3/2}}{3/2} + c = \frac{2}{3}(1 + x)^{3/2} + c$.

(C) Let $I = \int \frac{1 - x^7}{x(1 + x^7)} dx$.
We can rewrite the integrand as:
$\frac{1 - x^7}{x(1 + x^7)} = \frac{1 + x^7 - 2x^7}{x(1 + x^7)} = \frac{1}{x} - \frac{2x^7}{x(1 + x^7)} = \frac{1}{x} - \frac{2x^6}{1 + x^7}$.
Now,integrate term by term:
$I = \int \frac{1}{x} dx - \int \frac{2x^6}{1 + x^7} dx$.
For the second integral,let $u = 1 + x^7$,then $du = 7x^6 dx$,which implies $x^6 dx = \frac{du}{7}$.
$I = ln |x| - 2 \int \frac{1}{u} \cdot \frac{du}{7} = ln |x| - \frac{2}{7} ln |u| + c$.
Substituting $u = 1 + x^7$ back,we get:
$I = ln |x| - \frac{2}{7} ln (1 + x^7) + c$.

(B) Let $I = \int \sqrt{1 + 2 \cot x (\cot x + \csc x)} \, dx$.
Using the identity $1 + \cot^2 x = \csc^2 x$,we can rewrite the expression inside the square root:
$1 + 2 \cot^2 x + 2 \cot x \csc x = \csc^2 x + \cot^2 x + 2 \cot x \csc x$.
This is a perfect square: $(\csc x + \cot x)^2$.
So,$I = \int \sqrt{(\csc x + \cot x)^2} \, dx = \int (\csc x + \cot x) \, dx$.
The integral of $\csc x$ is $\ln|\csc x - \cot x|$ and the integral of $\cot x$ is $\ln|\sin x|$.
Alternatively,$\csc x + \cot x = \frac{1 + \cos x}{\sin x} = \frac{2 \cos^2(x/2)}{2 \sin(x/2) \cos(x/2)} = \cot(x/2)$.
The integral of $\cot(x/2)$ is $2 \ln|\sin(x/2)| + c$.

(C) Let $F(x)$ be the primitive of $f(x)$. Then $F(x) = \int f(x) \, dx = \int (\pi \sin(\pi x) + 2x - 4) \, dx$.
Integrating term by term,we get $F(x) = -\cos(\pi x) + x^2 - 4x + C$.
Given that $F(1) = 3$,we substitute $x = 1$:
$F(1) = -\cos(\pi) + (1)^2 - 4(1) + C = 3$.
Since $\cos(\pi) = -1$,we have $-(-1) + 1 - 4 + C = 3$,which simplifies to $1 + 1 - 4 + C = 3$,so $-2 + C = 3$,giving $C = 5$.
Thus,$F(x) = -\cos(\pi x) + x^2 - 4x + 5$.
We want to find $x$ such that $F(x) = 0$,so $-\cos(\pi x) + x^2 - 4x + 5 = 0$,or $\cos(\pi x) = x^2 - 4x + 5$.
We can rewrite the right side as $(x-2)^2 + 1$.
Since $\cos(\pi x) \le 1$ and $(x-2)^2 + 1 \ge 1$,the equation holds only if $\cos(\pi x) = 1$ and $(x-2)^2 + 1 = 1$.
$(x-2)^2 + 1 = 1$ implies $(x-2)^2 = 0$,so $x = 2$.
Checking $x = 2$ in the cosine term: $\cos(2\pi) = 1$.
Thus,$x = 2$ is the only solution.

(A) We are given $f(x) = \frac{1}{x} \ln \left( \frac{x}{e^x} \right)$.
To find the primitive (indefinite integral),we calculate $\int f(x) \, dx = \int \frac{1}{x} \ln \left( \frac{x}{e^x} \right) \, dx$.
Using the logarithmic property $\ln \left( \frac{a}{b} \right) = \ln a - \ln b$,we get:
$\int \frac{1}{x} (\ln x - \ln e^x) \, dx = \int \frac{\ln x - x}{x} \, dx$.
This simplifies to $\int \left( \frac{\ln x}{x} - 1 \right) \, dx$.
Splitting the integral: $\int \frac{\ln x}{x} \, dx - \int 1 \, dx$.
For the first part,let $u = \ln x$,then $du = \frac{1}{x} \, dx$. The integral becomes $\int u \, du = \frac{u^2}{2} = \frac{(\ln x)^2}{2}$.
For the second part,$\int 1 \, dx = x$.
Combining these,we get $\frac{1}{2} \ln^2 x - x + C$.

(B) Consider the expression $x \cdot \int \frac{dx}{x}$.
We know that $\int \frac{dx}{x} = \ln |x| + C$.
Therefore,$x \cdot \int \frac{dx}{x} = x(\ln |x| + C) = x \ln |x| + Cx$.
Comparing this with the given options,option $B$ is correct.

(D) We are given the integral $I = \int \sqrt{1 + \sin \frac{x}{2}} dx$.
Using the identity $1 + \sin \theta = \sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2} + 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} = (\sin \frac{\theta}{2} + \cos \frac{\theta}{2})^2$,we have:
$1 + \sin \frac{x}{2} = (\sin \frac{x}{4} + \cos \frac{x}{4})^2$.
Thus,$\sqrt{1 + \sin \frac{x}{2}} = |\sin \frac{x}{4} + \cos \frac{x}{4}|$.
Assuming the interval where $\sin \frac{x}{4} + \cos \frac{x}{4} > 0$,we have:
$I = \int (\sin \frac{x}{4} + \cos \frac{x}{4}) dx$.
Integrating term by term:
$I = -4 \cos \frac{x}{4} + 4 \sin \frac{x}{4} + C$.
We can rewrite this as:
$I = 4 (\sin \frac{x}{4} - \cos \frac{x}{4}) + C$.
Using the identity $\sin \theta - \cos \theta = \sqrt{2} \sin(\theta - \frac{\pi}{4})$,we get:
$I = 4 \sqrt{2} \sin(\frac{x}{4} - \frac{\pi}{4}) + C$.
Comparing this with the given form $A \sin(\frac{x}{4} - \frac{\pi}{4})$,we find $A = 4\sqrt{2}$.

(B) We are given the integral $\int {\frac{{{a^x}{e^{3x}}}}{{{b^x}{c^x}}}} dx$.
This can be rewritten as $\int {\left( {\frac{{a{e^3}}}{{bc}}} \right)^x} dx$.
Using the standard integration formula $\int {A^x} dx = \frac{{A^x}}{{\log A}} + K$,where $A = \frac{{a{e^3}}}{{bc}}$,we get:
$\int {\left( {\frac{{a{e^3}}}{{bc}}} \right)^x} dx = \frac{{{{\left( {\frac{{a{e^3}}}{{bc}}} \right)}^x}}}{{\log \left( {\frac{{a{e^3}}}{{bc}}} \right)}} + K$.
Comparing this with the given expression $\frac{1}{P}\left( {\frac{{{a^x}{e^{3x}}}}{{{b^x}{c^x}}}} \right) + K$,we identify $P = \log \left( {\frac{{a{e^3}}}{{bc}}} \right)$.
Using logarithmic properties,$P = \log (a{e^3}) - \log (bc) = \log a + \log ({e^3}) - \log (bc) = \log a + 3 - \log b - \log c$.

(D) Let $I = \int {\sin x \cos x \cos 2x \cos 4x \cos 8x} \, dx$.
Multiply and divide by $2$:
$I = \frac{1}{2} \int {\sin 2x \cos 2x \cos 4x \cos 8x} \, dx$.
Multiply and divide by $2$ again:
$I = \frac{1}{4} \int {\sin 4x \cos 4x \cos 8x} \, dx$.
Multiply and divide by $2$ again:
$I = \frac{1}{8} \int {\sin 8x \cos 8x} \, dx$.
Multiply and divide by $2$ again:
$I = \frac{1}{16} \int {\sin 16x} \, dx$.
Integrating $\sin 16x$,we get:
$I = \frac{1}{16} \left( -\frac{\cos 16x}{16} \right) + C = -\frac{1}{256} \cos 16x + C$.

(A) Let $I = \int x^x (1 + \ln x) dx$.
Consider the function $f(x) = x^x$.
To differentiate $f(x)$,take the natural logarithm on both sides:
$\ln f(x) = \ln(x^x) = x \ln x$.
Differentiating both sides with respect to $x$:
$\frac{1}{f(x)} f'(x) = \frac{d}{dx}(x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
Therefore,$f'(x) = f(x) (1 + \ln x) = x^x (1 + \ln x)$.
Since the integrand is of the form $\int f'(x) dx$,the integral is $f(x) + C$.
Thus,$I = x^x + C$.

(C) Given integral is $\int {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}} dx = \int {\left( {\frac{{a{e^2}}}{{bc}}} \right)^x} dx$.
Using the formula $\int {m^x} dx = \frac{{m^x}}{{\ln m}} + C$,where $m = \frac{{a{e^2}}}{{bc}}$,we get:
$\int {\left( {\frac{{a{e^2}}}{{bc}}} \right)^x} dx = \frac{{{{\left( {\frac{{a{e^2}}}{{bc}}} \right)}^x}}}{{\ln \left( {\frac{{a{e^2}}}{{bc}}} \right)}} + l$.
Comparing this with the given form $\frac{1}{k}\left( {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}} \right) + l$,we identify $k = \ln \left( {\frac{{a{e^2}}}{{bc}}} \right)$.
Using logarithmic properties,$k = \ln a + \ln(e^2) - \ln b - \ln c = \ln a + 2 - \ln b - \ln c$.
Thus,$k = \ln a - \ln b - \ln c + 2$.

(C) Let $I = \int \frac{\cos x}{\cos (x - a)} dx - \int \frac{\sin x}{\sin (x - a)} dx = I_1 - I_2$.
For $I_1 = \int \frac{\cos x}{\cos (x - a)} dx$,substitute $x = (x - a) + a$:
$I_1 = \int \frac{\cos((x - a) + a)}{\cos (x - a)} dx = \int \frac{\cos(x - a) \cos a - \sin(x - a) \sin a}{\cos (x - a)} dx$
$I_1 = \cos a \int dx - \sin a \int \tan(x - a) dx = x \cos a + \sin a \log |\cos(x - a)| + C_1$.
For $I_2 = \int \frac{\sin x}{\sin (x - a)} dx$,substitute $x = (x - a) + a$:
$I_2 = \int \frac{\sin((x - a) + a)}{\sin (x - a)} dx = \int \frac{\sin(x - a) \cos a + \cos(x - a) \sin a}{\sin (x - a)} dx$
$I_2 = \cos a \int dx + \sin a \int \cot(x - a) dx = x \cos a + \sin a \log |\sin(x - a)| + C_2$.
Now,$I = I_1 - I_2 = (x \cos a + \sin a \log |\cos(x - a)|) - (x \cos a + \sin a \log |\sin(x - a)|) + C$
$I = \sin a (\log |\cos(x - a)| - \log |\sin(x - a)|) + C$
$I = \sin a \log \left| \frac{\cos(x - a)}{\sin(x - a)} \right| + C = \sin a \log |\cot(x - a)| + C$.

(B) Let $I = \int \frac{4x^3 + \lambda 4^x}{4^x + x^4} \, dx$.
Given that $I = \log(4^x + x^4) + c$.
Let $f(x) = 4^x + x^4$.
Then,the derivative $f'(x) = \frac{d}{dx}(4^x + x^4) = 4^x \ln 4 + 4x^3$.
We know that $\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + c$.
Comparing the given integral $\int \frac{4x^3 + \lambda 4^x}{4^x + x^4} \, dx$ with $\int \frac{f'(x)}{f(x)} \, dx$,we have:
$4x^3 + \lambda 4^x = 4x^3 + 4^x \ln 4$.
Equating the coefficients of $4^x$,we get $\lambda = \ln 4$.

(C) We know that $\sqrt{\sin^2 x} = |\sin x|$.
So,the integral becomes $\int |\sin x| dx$.
By the definition of the absolute value function,we have:
$\int |\sin x| dx = \begin{cases} \int \sin x dx & \text{if } \sin x \ge 0 \\ -\int \sin x dx & \text{if } \sin x < 0 \end{cases}$
Evaluating these integrals:
If $\sin x \ge 0$,the integral is $-\cos x + c$.
If $\sin x < 0$,the integral is $\cos x + c$.
This can be written compactly using the signum function as $-\cos x \cdot \text{sgn}(\sin x) + c$.

(B) Let $I = \int \tan^4 x \, dx$.
We can write $\tan^4 x$ as $\tan^2 x \cdot \tan^2 x$.
Using the identity $\tan^2 x = \sec^2 x - 1$,we get:
$I = \int \tan^2 x (\sec^2 x - 1) \, dx$
$I = \int \tan^2 x \sec^2 x \, dx - \int \tan^2 x \, dx$
$I = \int \tan^2 x \sec^2 x \, dx - \int (\sec^2 x - 1) \, dx$
For the first integral,let $u = \tan x$,then $du = \sec^2 x \, dx$.
$I = \int u^2 \, du - (\tan x - x) + c$
$I = \frac{u^3}{3} - \tan x + x + c$
Substituting $u = \tan x$ back,we get:
$I = \frac{1}{3} \tan^3 x - \tan x + x + c$.

(B) Let $y = \frac{x - 4}{x + 2}$.
Then $x - 4 = y(x + 2) \Rightarrow x - 4 = xy + 2y$.
$x(1 - y) = 2y + 4 \Rightarrow x = \frac{2y + 4}{1 - y}$.
Substituting $x$ into $f(y) = 2x + 1$,we get $f(y) = 2\left( \frac{2y + 4}{1 - y} \right) + 1$.
$f(y) = \frac{4y + 8 + 1 - y}{1 - y} = \frac{3y + 9}{1 - y}$.
Thus,$f(x) = \frac{3x + 9}{1 - x} = \frac{3(x - 1 + 4)}{1 - x} = \frac{3(x - 1)}{1 - x} + \frac{12}{1 - x} = -3 + \frac{12}{1 - x}$.
Now,$\int f(x) \,dx = \int \left( -3 + \frac{12}{1 - x} \right) \,dx$.
$= -3x + 12 \int \frac{1}{1 - x} \,dx$.
$= -3x + 12 \left( \frac{\log_e |1 - x|}{-1} \right) + C$.
$= -12 \log_e |1 - x| - 3x + C$.

(A) Let $I = \int \sqrt{1 + 2\cot x \csc x + 2\cot^2 x} \,dx$.
Since $1 + \cot^2 x = \csc^2 x$,we have $1 + 2\cot^2 x = \csc^2 x + \cot^2 x$.
Thus,the expression inside the square root is $\csc^2 x + \cot^2 x + 2\cot x \csc x = (\csc x + \cot x)^2$.
Since $0 < x < \frac{\pi}{2}$,$\csc x + \cot x > 0$,so $\sqrt{(\csc x + \cot x)^2} = \csc x + \cot x$.
$I = \int (\csc x + \cot x) \,dx$.
Using the standard integrals $\int \csc x \,dx = \log |\csc x - \cot x| + C$ and $\int \cot x \,dx = \log |\sin x| + C$,we get:
$I = \log |\csc x - \cot x| + \log |\sin x| + C$.
$I = \log \left| \frac{1 - \cos x}{\sin x} \cdot \sin x \right| + C = \log |1 - \cos x| + C$.
Using $1 - \cos x = 2\sin^2 \frac{x}{2}$,we get:
$I = \log |2\sin^2 \frac{x}{2}| + C = \log 2 + 2\log |\sin \frac{x}{2}| + C$.
Absorbing $\log 2$ into the constant $C$,we get $I = 2\log |\sin \frac{x}{2}| + C$.

(B) Let $t = \frac{3x - 4}{3x + 4}$.
Then $3xt + 4t = 3x - 4$,which implies $x(3t - 3) = -4t - 4$,so $x = \frac{4t + 4}{3 - 3t}$.
Substituting this into the function: $f(t) = \frac{4t + 4}{3 - 3t} + 2 = \frac{4t + 4 + 6 - 6t}{3 - 3t} = \frac{10 - 2t}{3 - 3t}$.
Thus,$f(x) = \frac{10 - 2x}{3 - 3x} = \frac{2x - 10}{3x - 3}$.
Now,$\int f(x) dx = \int \frac{2x - 10}{3(x - 1)} dx = \frac{2}{3} \int \frac{x - 1 - 4}{x - 1} dx = \frac{2}{3} \int (1 - \frac{4}{x - 1}) dx$.
$= \frac{2}{3} x - \frac{8}{3} \ln |x - 1| + C = -\frac{8}{3} \ln |1 - x| + \frac{2}{3} x + C$.
Comparing with $A \log |1 - x| + Bx + C$,we get $A = -\frac{8}{3}$ and $B = \frac{2}{3}$.
Therefore,the ordered pair $(A, B) = \left( -\frac{8}{3}, \frac{2}{3} \right)$.

(D) Statement-$2$: $\cos^3 x$ is a periodic function. This is a true statement because $\cos x$ is periodic with period $2\pi$,and any power of a periodic function is also periodic.
Given $f(x) = \int \cos^3 x \, dx$.
Using the identity $\cos 3x = 4\cos^3 x - 3\cos x$,we have $\cos^3 x = \frac{\cos 3x + 3\cos x}{4}$.
$f(x) = \int \left(\frac{\cos 3x}{4} + \frac{3\cos x}{4}\right) dx = \frac{1}{4} \cdot \frac{\sin 3x}{3} + \frac{3}{4} \sin x + C = \frac{1}{12} \sin 3x + \frac{3}{4} \sin x + C$.
The period of $\sin 3x$ is $\frac{2\pi}{3}$ and the period of $\sin x$ is $2\pi$.
The period of the sum of two periodic functions is the least common multiple $(LCM)$ of their individual periods.
Period of $f(x) = \text{LCM}\left(\frac{2\pi}{3}, 2\pi\right) = 2\pi$.
Since the period is $2\pi$ and not $\pi$,Statement-$1$ is false.

(A) Let $I = \int \frac{x^2 - x}{x^3 - x^2 + x - 1} dx$.
Factor the denominator: $x^3 - x^2 + x - 1 = x^2(x - 1) + 1(x - 1) = (x^2 + 1)(x - 1)$.
Substitute this into the integral: $I = \int \frac{x(x - 1)}{(x^2 + 1)(x - 1)} dx$.
Cancel the common term $(x - 1)$: $I = \int \frac{x}{x^2 + 1} dx$.
Multiply and divide by $2$: $I = \frac{1}{2} \int \frac{2x}{x^2 + 1} dx$.
Let $u = x^2 + 1$,then $du = 2x dx$.
Substituting these into the integral: $I = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \log |u| + c$.
Since $x^2 + 1 > 0$ for all real $x$,we can write: $I = \frac{1}{2} \log (x^2 + 1) + c$.

(C) We need to evaluate $I = \int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx$.
Multiply the numerator and denominator by $2 \cos \frac{x}{2}$:
$I = \int \frac{2 \sin \frac{5x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} dx$
Using the formula $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:
$I = \int \frac{\sin(3x) + \sin(2x)}{\sin x} dx$
Using $\sin(3x) = 3 \sin x - 4 \sin^3 x$ and $\sin(2x) = 2 \sin x \cos x$:
$I = \int \frac{3 \sin x - 4 \sin^3 x + 2 \sin x \cos x}{\sin x} dx$
$I = \int (3 - 4 \sin^2 x + 2 \cos x) dx$
Using $4 \sin^2 x = 2(1 - \cos 2x)$:
$I = \int (3 - 2(1 - \cos 2x) + 2 \cos x) dx$
$I = \int (1 + 2 \cos 2x + 2 \cos x) dx$
Integrating term by term:
$I = x + \sin 2x + 2 \sin x + c$.

(A) We are given the integral $\int {\left( {\frac{{{x^3} + 2}}{{{x^3}}}} \right)dx}$.
First,simplify the integrand by dividing each term in the numerator by the denominator:
$\frac{{{x^3} + 2}}{{{x^3}}} = \frac{{{x^3}}}{{{x^3}}} + \frac{2}{{{x^3}}} = 1 + 2{x^{ - 3}}$.
Now,integrate the expression term by term with respect to $x$:
$\int {(1 + 2{x^{ - 3}})dx} = \int {1dx} + 2\int {{x^{ - 3}}dx}$.
Using the power rule $\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + c$ (for $n \neq -1$):
$= x + 2\left( {\frac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}} \right) + c$
$= x + 2\left( {\frac{{{x^{ - 2}}}}{{ - 2}}} \right) + c$
$= x - {x^{ - 2}} + c$
$= x - \frac{1}{{{x^2}}} + c$.

(A) We look for a function whose derivative is $\cos 2x$.
Recall that the derivative of $\sin 2x$ is given by:
$\frac{d}{dx} (\sin 2x) = 2 \cos 2x$
Dividing both sides by $2$,we get:
$\frac{1}{2} \frac{d}{dx} (\sin 2x) = \cos 2x$
Using the linearity property of the derivative,we can write:
$\frac{d}{dx} \left( \frac{1}{2} \sin 2x \right) = \cos 2x$
Therefore,an anti-derivative of $\cos 2x$ is $\frac{1}{2} \sin 2x$.

(A) To find the anti-derivative of $f(x) = 3x^{2} + 4x^{3}$ by the method of inspection,we look for a function $F(x)$ such that $F'(x) = f(x)$.
We know that the derivative of $x^{n}$ is $nx^{n-1}$.
For the first term $3x^{2}$,we observe that $\frac{d}{dx}(x^{3}) = 3x^{2}$.
For the second term $4x^{3}$,we observe that $\frac{d}{dx}(x^{4}) = 4x^{3}$.
By the linearity property of derivatives,we have:
$\frac{d}{dx}(x^{3} + x^{4}) = \frac{d}{dx}(x^{3}) + \frac{d}{dx}(x^{4}) = 3x^{2} + 4x^{3}$.
Therefore,an anti-derivative of $3x^{2} + 4x^{3}$ is $x^{3} + x^{4}$.

(A) We know that the derivative of $\log x$ with respect to $x$ is $\frac{1}{x}$ for $x > 0$.
Also,for $x < 0$,let $y = -x$,then $\frac{d}{dx}(\log(-x)) = \frac{1}{-x} \cdot (-1) = \frac{1}{x}$.
Combining these two cases,we get $\frac{d}{dx}(\log |x|) = \frac{1}{x}$ for all $x \neq 0$.
By the definition of an anti-derivative,if $\frac{d}{dx}(F(x)) = f(x)$,then $F(x)$ is an anti-derivative of $f(x)$.
Therefore,$\log |x|$ is an anti-derivative of $\frac{1}{x}$.

(A) We have $\int \frac{x^{3}-1}{x^{2}} dx = \int \left( \frac{x^{3}}{x^{2}} - \frac{1}{x^{2}} \right) dx$
$= \int (x - x^{-2}) dx$
$= \int x dx - \int x^{-2} dx$
$= \frac{x^{1+1}}{1+1} - \frac{x^{-2+1}}{-2+1} + C$
$= \frac{x^{2}}{2} - \frac{x^{-1}}{-1} + C$
$= \frac{x^{2}}{2} + \frac{1}{x} + C$

(A) We use the power rule for integration,which states that $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$.
$\int \left(x^{\frac{2}{3}} + 1\right) dx = \int x^{\frac{2}{3}} dx + \int 1 dx$
$= \frac{x^{\frac{2}{3} + 1}}{\frac{2}{3} + 1} + x + C$
$= \frac{x^{\frac{5}{3}}}{\frac{5}{3}} + x + C$
$= \frac{3}{5} x^{\frac{5}{3}} + x + C$

We use the linearity property of integration to split the integral:
$\int \left(x^{\frac{3}{2}} + 2e^{x} - \frac{1}{x}\right) dx = \int x^{\frac{3}{2}} dx + 2 \int e^{x} dx - \int \frac{1}{x} dx$
Applying the power rule $\int x^{n} dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$,the exponential rule $\int e^{x} dx = e^{x} + C$,and the logarithmic rule $\int \frac{1}{x} dx = \log |x| + C$:
$= \frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} + 2e^{x} - \log |x| + C$
$= \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + 2e^{x} - \log |x| + C$
$= \frac{2}{5} x^{\frac{5}{2}} + 2e^{x} - \log |x| + C$

(A) We use the linearity property of the integral: $\int (\sin x + \cos x) \, dx = \int \sin x \, dx + \int \cos x \, dx$.
Knowing that $\int \sin x \, dx = -\cos x + C_1$ and $\int \cos x \, dx = \sin x + C_2$,we combine them:
$\int (\sin x + \cos x) \, dx = -\cos x + \sin x + C$,where $C = C_1 + C_2$ is the constant of integration.

(A) We have the integral: $\int \operatorname{cosec} x(\operatorname{cosec} x + \cot x) \, dx$
Distributing the $\operatorname{cosec} x$ term,we get:
$\int (\operatorname{cosec}^2 x + \operatorname{cosec} x \cot x) \, dx$
Using the linearity property of integration,we split the integral:
$\int \operatorname{cosec}^2 x \, dx + \int \operatorname{cosec} x \cot x \, dx$
We know the standard integration formulas:
$\int \operatorname{cosec}^2 x \, dx = -\cot x + C_1$
$\int \operatorname{cosec} x \cot x \, dx = -\operatorname{cosec} x + C_2$
Combining these results,we get:
$-\cot x - \operatorname{cosec} x + C$,where $C = C_1 + C_2$.

(A) We have the integral:
$\int \frac{1-\sin x}{\cos ^{2} x} d x$
Split the fraction into two parts:
$= \int \left( \frac{1}{\cos ^{2} x} - \frac{\sin x}{\cos ^{2} x} \right) d x$
$= \int \sec ^{2} x d x - \int \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} d x$
$= \int \sec ^{2} x d x - \int \tan x \sec x d x$
Using the standard integration formulas $\int \sec ^{2} x d x = \tan x + C_1$ and $\int \sec x \tan x d x = \sec x + C_2$:
$= \tan x - \sec x + C$