Fundamental integration Questions in English

(D) Given the integral: $I = \int \sqrt{1 - \sin 2x} \, dx$
Using the trigonometric identity $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,we have:
$1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2$
Thus,$\sqrt{1 - \sin 2x} = \sqrt{(\cos x - \sin x)^2} = |\cos x - \sin x|$
Since $x \in (0, \pi/4)$,we know that $\cos x > \sin x$,so $|\cos x - \sin x| = \cos x - \sin x$
Now,integrating the expression:
$I = \int (\cos x - \sin x) \, dx$
$I = \sin x - (-\cos x) + c$
$I = \sin x + \cos x + c$

(C) We are given the integral $I = \int {\frac{{1 + {{\cos }^2}x}}{{{{\sin }^2}x}}} \,dx$.
Splitting the fraction,we get $I = \int {(\frac{1}{{{{\sin }^2}x}} + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}})} \,dx$.
Using trigonometric identities,$\frac{1}{{{{\sin }^2}x}} = \csc^2 x$ and $\frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = \cot^2 x$.
So,$I = \int {(\csc^2 x + \cot^2 x)} \,dx$.
Since $\cot^2 x = \csc^2 x - 1$,we substitute this into the integral:
$I = \int {(\csc^2 x + \csc^2 x - 1)} \,dx = \int {(2\csc^2 x - 1)} \,dx$.
Integrating term by term,we get $I = 2 \int \csc^2 x \,dx - \int 1 \,dx$.
Since $\int \csc^2 x \,dx = -\cot x$,the result is $I = -2\cot x - x + c$.

(C) We know that $\sin^{-1}(\cos x) = \sin^{-1}(\sin(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x$.
Therefore,the integral becomes:
$\int (\frac{\pi}{2} - x) \, dx$
$= \int \frac{\pi}{2} \, dx - \int x \, dx$
$= \frac{\pi}{2}x - \frac{x^2}{2} + C$
$= \frac{\pi x - x^2}{2} + C$.

(D) We have the integral $I = \int \frac{dx}{\tan x + \cot x}$.
First,express $\tan x$ and $\cot x$ in terms of $\sin x$ and $\cos x$:
$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$.
Substituting this into the integral:
$I = \int \sin x \cos x \, dx$.
Multiply and divide by $2$:
$I = \frac{1}{2} \int 2 \sin x \cos x \, dx = \frac{1}{2} \int \sin 2x \, dx$.
Integrating $\sin 2x$ gives $-\frac{\cos 2x}{2}$:
$I = \frac{1}{2} \left( -\frac{\cos 2x}{2} \right) + c = -\frac{\cos 2x}{4} + c$.

(C) We know that $e^{a \log x} = e^{\log x^a} = x^a$ and $e^{x \log a} = e^{\log a^x} = a^x$.
Therefore,the integral becomes $\int (x^a + a^x) dx$.
Using the standard integration formulas $\int x^n dx = \frac{x^{n+1}}{n+1} + c$ and $\int a^x dx = \frac{a^x}{\log a} + c$,we get:
$\int (x^a + a^x) dx = \frac{x^{a+1}}{a+1} + \frac{a^x}{\log a} + c$.

(C) Given that $f'(x) = x^2 + 5$.
To find $f(x)$,we integrate $f'(x)$ with respect to $x$:
$f(x) = \int (x^2 + 5) dx = \frac{x^3}{3} + 5x + C$.
We are given the condition $f(0) = -1$.
Substituting $x = 0$ into the expression for $f(x)$:
$f(0) = \frac{0^3}{3} + 5(0) + C = -1$.
This gives $C = -1$.
Therefore,the function is $f(x) = \frac{x^3}{3} + 5x - 1$.

(C) We know that $1 - \cos 2x = 2\sin^2 x$ and $1 + \cos 2x = 2\cos^2 x$.
Substituting these into the integral:
$\int {{\tan ^{ - 1}}\sqrt {\frac{{2\sin^2 x}}{{2\cos^2 x}}} } \,dx = \int {{\tan ^{ - 1}}\sqrt {\tan^2 x} } \,dx$
$= \int {{\tan ^{ - 1}}(\tan x)\,dx} = \int x \,dx$
$= \frac{{{x^2}}}{2} + c$.

(A) To solve the integral $I = \int \frac{dx}{\sqrt{x} + \sqrt{x - 2}}$,we rationalize the denominator by multiplying the numerator and denominator by $(\sqrt{x} - \sqrt{x - 2})$.
$I = \int \frac{\sqrt{x} - \sqrt{x - 2}}{(\sqrt{x} + \sqrt{x - 2})(\sqrt{x} - \sqrt{x - 2})} dx$
$I = \int \frac{\sqrt{x} - \sqrt{x - 2}}{x - (x - 2)} dx$
$I = \int \frac{\sqrt{x} - \sqrt{x - 2}}{2} dx$
$I = \frac{1}{2} \int (x^{1/2} - (x - 2)^{1/2}) dx$
Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$,we get:
$I = \frac{1}{2} \left[ \frac{x^{3/2}}{3/2} - \frac{(x - 2)^{3/2}}{3/2} \right] + c$
$I = \frac{1}{2} \cdot \frac{2}{3} [x^{3/2} - (x - 2)^{3/2}] + c$
$I = \frac{1}{3} [x^{3/2} - (x - 2)^{3/2}] + c$

(D) We have the integral $I = \int {\frac{{\cos x - 1}}{{\cos x + 1}}\,dx}$.
Using the trigonometric identities $\cos x = 1 - 2\sin^2 \frac{x}{2}$ and $\cos x = 2\cos^2 \frac{x}{2} - 1$,we get:
$I = \int {\frac{{(1 - 2\sin^2 \frac{x}{2}) - 1}}{{(2\cos^2 \frac{x}{2} - 1) + 1}}\,dx}$
$I = \int {\frac{{ - 2\sin^2 \frac{x}{2}}}{{2\cos^2 \frac{x}{2}}}\,dx} = - \int {\tan^2 \frac{x}{2}\,dx}$
Using the identity $\tan^2 \theta = \sec^2 \theta - 1$,we have:
$I = - \int {(\sec^2 \frac{x}{2} - 1)\,dx} = \int {(1 - \sec^2 \frac{x}{2})\,dx}$
Integrating term by term,we get:
$I = x - 2\tan \frac{x}{2} + c$.

(D) To solve the integral $I = \int \frac{dx}{1 - \sin x}$,we multiply the numerator and denominator by $(1 + \sin x)$:
$I = \int \frac{1 + \sin x}{(1 - \sin x)(1 + \sin x)} dx$
$I = \int \frac{1 + \sin x}{1 - \sin^2 x} dx$
Using the identity $1 - \sin^2 x = \cos^2 x$,we get:
$I = \int \frac{1 + \sin x}{\cos^2 x} dx$
$I = \int (\sec^2 x + \sec x \tan x) dx$
Integrating term by term:
$I = \tan x + \sec x + c$.

(D) Given $\int (\sin 2x - \cos 2x) \,dx = \frac{1}{\sqrt{2}} \sin(2x - a) + b$.
Integrating the left side: $\int \sin 2x \,dx - \int \cos 2x \,dx = -\frac{1}{2} \cos 2x - \frac{1}{2} \sin 2x + C$.
So,$-\frac{1}{2} (\sin 2x + \cos 2x) + C = \frac{1}{\sqrt{2}} \sin(2x - a) + b$.
Multiply both sides by $-\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin 2x + \frac{1}{\sqrt{2}} \cos 2x - \sqrt{2}C = -\sin(2x - a) - \sqrt{2}b$.
Using $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we have $\sin(2x + \frac{\pi}{4}) = \frac{1}{\sqrt{2}} \sin 2x + \frac{1}{\sqrt{2}} \cos 2x$.
Thus,$\sin(2x + \frac{\pi}{4}) = -\sin(2x - a) - \sqrt{2}b + \sqrt{2}C$.
Since $-\sin \theta = \sin(\theta - \pi)$,we have $\sin(2x + \frac{\pi}{4}) = \sin(2x - a - \pi) - \sqrt{2}b + \sqrt{2}C$.
Comparing the arguments: $2x + \frac{\pi}{4} = 2x - a - \pi \Rightarrow a = -\pi - \frac{\pi}{4} = -\frac{5\pi}{4}$.
Also,$b$ is an arbitrary constant.

(B) The given expression inside the integral is the Taylor series expansion for the exponential function $e^x$.
We know that $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$.
Substituting this into the integral,we get:
$\int {\left( {1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \dots} \right) dx} = \int {{e^x} dx}$.
The integral of $e^x$ with respect to $x$ is $e^x + c$.
Therefore,the correct option is $B$.

(D) We are given the integral $I = \int {\frac{{\cot x \tan x}}{{{{\sec }^2}x - 1}}} \;dx$.
Since $\cot x \tan x = 1$ and ${{\sec }^2}x - 1 = {\tan ^2}x$,the integral becomes:
$I = \int {\frac{1}{{{{\tan }^2}x}}} \;dx = \int {{{\cot }^2}x} \;dx$.
Using the trigonometric identity ${{\cot }^2}x = \csc^2 x - 1$,we get:
$I = \int {(\csc^2 x - 1)} \;dx$.
Integrating term by term,we have:
$I = - \cot x - x + c$.

(A) We need to evaluate the integral $I = \int (\sec x + \tan x)^2 dx$.
Expanding the integrand using the identity $(a+b)^2 = a^2 + b^2 + 2ab$:
$I = \int (\sec^2 x + \tan^2 x + 2\sec x \tan x) dx$.
Using the trigonometric identity $\tan^2 x = \sec^2 x - 1$:
$I = \int (\sec^2 x + (\sec^2 x - 1) + 2\sec x \tan x) dx$.
$I = \int (2\sec^2 x - 1 + 2\sec x \tan x) dx$.
Integrating term by term:
$I = 2 \int \sec^2 x dx - \int 1 dx + 2 \int \sec x \tan x dx$.
$I = 2\tan x - x + 2\sec x + c$.
$I = 2(\sec x + \tan x) - x + c$.

(A) We know that for all $x \in \mathbb{R}$,the identity holds: ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$.
Substituting this into the integral,we get:
$\int {{x^{51}}({{\tan }^{ - 1}}x + {\cot ^{ - 1}}x)} \,dx = \int {{x^{51}} \cdot \frac{\pi }{2}} \,dx$.
Taking the constant $\frac{\pi }{2}$ outside the integral:
$= \frac{\pi }{2} \int {{x^{51}}} \,dx$.
Using the power rule $\int {{x^n}} \,dx = \frac{{{x^{n + 1}}}}{{n + 1}} + c$:
$= \frac{\pi }{2} \cdot \frac{{{x^{52}}}}{{52}} + c = \frac{{\pi {x^{52}}}}{{104}} + c$.
Since $\frac{\pi }{2} = {\tan ^{ - 1}}x + {\cot ^{ - 1}}x$,we can rewrite the result as:
$= \frac{{{x^{52}}}}{{52}} \cdot \frac{\pi }{2} + c = \frac{{{x^{52}}}}{{52}}({\tan ^{ - 1}}x + {\cot ^{ - 1}}x) + c$.

(B) We know that the integral of $\sin x$ with respect to $x$ is $-\cos x + c$.
Applying this to the given integral:
$\int 5 \sin x \, dx = 5 \int \sin x \, dx$
$= 5(-\cos x) + c$
$= -5 \cos x + c$.

(B) We have the integral $I = \int \frac{\tan x}{\sec x + \tan x} \, dx$.
Multiplying the numerator and denominator by $(\sec x - \tan x)$,we get:
$I = \int \frac{\tan x(\sec x - \tan x)}{(\sec x + \tan x)(\sec x - \tan x)} \, dx$
Since $(\sec^2 x - \tan^2 x) = 1$,the expression simplifies to:
$I = \int (\sec x \tan x - \tan^2 x) \, dx$
Using the identity $\tan^2 x = \sec^2 x - 1$,we have:
$I = \int (\sec x \tan x - (\sec^2 x - 1)) \, dx$
$I = \int \sec x \tan x \, dx - \int \sec^2 x \, dx + \int 1 \, dx$
Integrating each term,we get:
$I = \sec x - \tan x + x + c$.

(C) To solve the integral $\int \frac{dx}{\sin^2 x \cos^2 x}$,we use the trigonometric identity $1 = \sin^2 x + \cos^2 x$.
Substituting this into the numerator,we get:
$\int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx$
Splitting the fraction,we have:
$\int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) dx$
This simplifies to:
$\int (\sec^2 x + \csc^2 x) dx$
Integrating term by term,we get:
$\int \sec^2 x dx + \int \csc^2 x dx = \tan x - \cot x + c$.

(B) We expand the integrand using the identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ or $a^3 + 3a^2b + 3ab^2 + b^3$.
$\int (x + \frac{1}{x})^3 dx = \int (x^3 + 3x^2(\frac{1}{x}) + 3x(\frac{1}{x^2}) + \frac{1}{x^3}) dx$
$= \int (x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}) dx$
Integrating term by term:
$= \frac{x^4}{4} + \frac{3x^2}{2} + 3\log|x| + \frac{x^{-2}}{-2} + c$
$= \frac{x^4}{4} + \frac{3x^2}{2} + 3\log|x| - \frac{1}{2x^2} + c$

(C) We know that $1 = \sin^2 \frac{x}{4} + \cos^2 \frac{x}{4}$ and $\sin \frac{x}{2} = 2 \sin \frac{x}{4} \cos \frac{x}{4}$.
Substituting these into the integral,we get:
$\int \sqrt{\sin^2 \frac{x}{4} + \cos^2 \frac{x}{4} + 2 \sin \frac{x}{4} \cos \frac{x}{4}} \, dx$
$= \int \sqrt{(\sin \frac{x}{4} + \cos \frac{x}{4})^2} \, dx$
$= \int (\sin \frac{x}{4} + \cos \frac{x}{4}) \, dx$
$= -4 \cos \frac{x}{4} + 4 \sin \frac{x}{4} + c$
$= 4(\sin \frac{x}{4} - \cos \frac{x}{4}) + c$.

(C) Given integral: $I = \int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx$
Since $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$,
Substituting this into the integral:
$I = \int \frac{\sin x + \cos x}{\sqrt{(\sin x + \cos x)^2}} \, dx$
$I = \int \frac{\sin x + \cos x}{\sin x + \cos x} \, dx$
$I = \int 1 \, dx = x + c$
Thus,the correct option is $C$.

(A) To evaluate the integral $I = \int \frac{x - 1}{(x + 1)^2} \, dx$,we rewrite the numerator in terms of the denominator:
$x - 1 = (x + 1) - 2$.
Substituting this into the integral:
$I = \int \frac{(x + 1) - 2}{(x + 1)^2} \, dx$
$I = \int \left( \frac{x + 1}{(x + 1)^2} - \frac{2}{(x + 1)^2} \right) \, dx$
$I = \int \frac{1}{x + 1} \, dx - 2 \int (x + 1)^{-2} \, dx$
Using the standard integration formulas $\int \frac{1}{u} \, du = \log |u| + C$ and $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$:
$I = \log |x + 1| - 2 \left( \frac{(x + 1)^{-1}}{-1} \right) + C$
$I = \log |x + 1| + \frac{2}{x + 1} + C$.

(A) We have the integral $I = \int \frac{\text{cosec}\theta - \cot\theta}{\text{cosec}\theta + \cot\theta} \, d\theta$.
Multiply the numerator and denominator by $(\text{cosec}\theta - \cot\theta)$:
$I = \int \frac{(\text{cosec}\theta - \cot\theta)^2}{\text{cosec}^2\theta - \cot^2\theta} \, d\theta$.
Since $\text{cosec}^2\theta - \cot^2\theta = 1$,we get:
$I = \int (\text{cosec}\theta - \cot\theta)^2 \, d\theta = \int (\text{cosec}^2\theta + \cot^2\theta - 2\text{cosec}\theta \cot\theta) \, d\theta$.
Using the identity $\cot^2\theta = \text{cosec}^2\theta - 1$:
$I = \int (\text{cosec}^2\theta + \text{cosec}^2\theta - 1 - 2\text{cosec}\theta \cot\theta) \, d\theta = \int (2\text{cosec}^2\theta - 1 - 2\text{cosec}\theta \cot\theta) \, d\theta$.
Integrating term by term:
$I = 2(-\cot\theta) - \theta - 2(-\text{cosec}\theta) + c = 2\text{cosec}\theta - 2\cot\theta - \theta + c$.

(B) The given series is $1 + 2x + 3x^2 + 4x^3 + \dots = \sum_{n=1}^{\infty} n x^{n-1}$.
This is the derivative of the geometric series $\sum_{n=0}^{\infty} x^n = (1 - x)^{-1}$ for $|x| < 1$.
Thus,$1 + 2x + 3x^2 + 4x^3 + \dots = \frac{d}{dx} (1 - x)^{-1} = (1 - x)^{-2}$.
Now,we integrate this expression:
$\int (1 - x)^{-2} \, dx = \frac{(1 - x)^{-1}}{-1 \cdot (-1)} + c = (1 - x)^{-1} + c$.

(C) Given that $\int (\cos x - \sin x) \, dx = \sqrt{2} \sin (x + \alpha) + c$.
Integrating the left side: $\int (\cos x - \sin x) \, dx = \sin x + \cos x + c$.
Now,equate this to the right side: $\sin x + \cos x + c = \sqrt{2} \sin (x + \alpha) + c$.
Divide and multiply the left side by $\sqrt{2}$: $\sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right) + c = \sqrt{2} \sin (x + \alpha) + c$.
Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we know that $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
So,$\sqrt{2} \left( \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} \right) + c = \sqrt{2} \sin (x + \alpha) + c$.
This simplifies to $\sqrt{2} \sin (x + \frac{\pi}{4}) + c = \sqrt{2} \sin (x + \alpha) + c$.
Comparing both sides,we get $\alpha = \frac{\pi}{4}$.

(D) We are given the integral $\int {\frac{{3{x^3} - 2\sqrt x }}{x}} dx$.
First,divide each term in the numerator by the denominator $x$:
$\int {(\frac{{3{x^3}}}{x} - \frac{{2\sqrt x }}{x})} dx = \int {(3{x^2} - 2{x^{1/2 - 1}})} dx$
$= \int {(3{x^2} - 2{x^{ - 1/2}})} dx$.
Now,apply the power rule for integration $\int {x^n} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + c$:
$= 3 \cdot \frac{{{x^3}}}{3} - 2 \cdot \frac{{{x^{ - 1/2 + 1}}}}{{ - 1/2 + 1}} + c$
$= {x^3} - 2 \cdot \frac{{{x^{1/2}}}}{{1/2}} + c$
$= {x^3} - 4{x^{1/2}} + c$
$= {x^3} - 4\sqrt x + c$.

(D) We have the integral $I = \int \frac{dx}{1 + \sin x}$.
Multiply the numerator and denominator by $(1 - \sin x)$:
$I = \int \frac{1 - \sin x}{1 - \sin^2 x} dx = \int \frac{1 - \sin x}{\cos^2 x} dx$
$I = \int (\sec^2 x - \sec x \tan x) dx$
$I = \tan x - \sec x + C$
Using half-angle formulas $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$:
$I = \int \frac{dx}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2} = \int \sec^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) \cdot \frac{1}{2} dx$ is not the direct path,let's use:
$I = \int \frac{dx}{1 + \cos(\frac{\pi}{2} - x)} = \int \frac{dx}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = \frac{1}{2} \int \sec^2(\frac{\pi}{4} - \frac{x}{2}) dx$
$I = \frac{1}{2} \cdot \frac{\tan(\frac{\pi}{4} - \frac{x}{2})}{-1/2} + C = -\tan(\frac{\pi}{4} - \frac{x}{2}) + C$
$I = \tan(\frac{x}{2} - \frac{\pi}{4}) + C$
Comparing this with $\tan(\frac{x}{2} + a) + b$,we get $a = -\frac{\pi}{4}$ and $b$ is an arbitrary constant.

(C) To evaluate the integral $I = \int \frac{dx}{\sin x + \cos x}$,we multiply and divide by $\sqrt{2}$:
$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x}$
Using the identity $\sin(x + \frac{\pi}{4}) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x$,we get:
$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\sin(x + \frac{\pi}{4})}$
$I = \frac{1}{\sqrt{2}} \int \csc(x + \frac{\pi}{4}) dx$
Using the standard integral formula $\int \csc \theta d\theta = \log |\tan(\frac{\theta}{2})| + c$,we have:
$I = \frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x + \frac{\pi}{4}}{2} \right) \right| + c$
$I = \frac{1}{\sqrt{2}} \log \tan \left( \frac{x}{2} + \frac{\pi}{8} \right) + c$

(B) The given integral is a standard integral form.
By the standard formula of integration,we know that:
$\int \frac{1}{x\sqrt{x^2 - 1}} \, dx = \sec^{-1}x + c$
Therefore,the correct option is $B$.

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

(D) We are given the integral $I = \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \, dx$.
Dividing each term in the numerator by the denominator,we get:
$I = \int \left( \frac{\sin^3 x}{\sin^2 x \cos^2 x} + \frac{\cos^3 x}{\sin^2 x \cos^2 x} \right) \, dx$
$I = \int \left( \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} \right) \, dx$
Using trigonometric identities $\frac{\sin x}{\cos^2 x} = \sec x \tan x$ and $\frac{\cos x}{\sin^2 x} = \csc x \cot x$,we have:
$I = \int (\sec x \tan x + \csc x \cot x) \, dx$
Integrating term by term:
$\int \sec x \tan x \, dx = \sec x$ and $\int \csc x \cot x \, dx = -\csc x$.
Therefore,$I = \sec x - \csc x + c$.

(A) We are given the integral: $\int {\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)^2} dx$
Expanding the square using the identity $(a-b)^2 = a^2 + b^2 - 2ab$:
$= \int {\left( {\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}} \right)} dx$
Using the trigonometric identities $\cos^2 \theta + \sin^2 \theta = 1$ and $2\sin \theta \cos \theta = \sin(2\theta)$:
$= \int {(1 - \sin x)} dx$
Integrating term by term:
$= \int 1 dx - \int \sin x dx$
$= x - (-\cos x) + c$
$= x + \cos x + c$

(A) To solve the integral $\int \frac{1}{x^2} (2x + 1)^3 dx$,we first expand the binomial $(2x + 1)^3$.
Using the formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we get $(2x + 1)^3 = (2x)^3 + 3(2x)^2(1) + 3(2x)(1)^2 + (1)^3 = 8x^3 + 12x^2 + 6x + 1$.
Now,substitute this into the integral: $\int \frac{8x^3 + 12x^2 + 6x + 1}{x^2} dx$.
Divide each term by $x^2$: $\int (8x + 12 + \frac{6}{x} + \frac{1}{x^2}) dx$.
Integrate term by term: $\int 8x dx + \int 12 dx + \int \frac{6}{x} dx + \int x^{-2} dx$.
This results in $8 \cdot \frac{x^2}{2} + 12x + 6\log|x| + \frac{x^{-1}}{-1} + c$.
Simplifying,we get $4x^2 + 12x + 6\log|x| - \frac{1}{x} + c$.

(B) We are given 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,we get:
$I = \int \frac{5(x^2 + 1)(x^4 - x^2 + 1)}{x^2 + 1} dx$
$I = \int 5(x^4 - x^2 + 1) dx$
Integrating term by term:
$I = 5 \left( \frac{x^5}{5} - \frac{x^3}{3} + x \right) + c$
$I = x^5 - \frac{5}{3}x^3 + 5x + c$.
Thus,the correct option is $B$.

(D) To solve the integral $\int {\frac{{a{x^{ - 2}} + b{x^{ - 1}} + c}}{{{x^{ - 3}}}}} \,dx$,first simplify the integrand by dividing each term in the numerator by ${x^{ - 3}}$:
$\frac{{a{x^{ - 2}} + b{x^{ - 1}} + c}}{{{x^{ - 3}}}} = a{x^{ - 2 - ( - 3)}} + b{x^{ - 1 - ( - 3)}} + c{x^{ - ( - 3)}} = a{x^1} + b{x^2} + c{x^3}$.
Now,integrate the expression with respect to $x$:
$\int {(ax + b{x^2} + c{x^3})} \,dx = a\int x \,dx + b\int {{x^2}} \,dx + c\int {{x^3}} \,dx$.
Using the power rule $\int {{x^n}} \,dx = \frac{{{x^{n + 1}}}}{{n + 1}} + k$,we get:
$a(\frac{{{x^2}}}{2}) + b(\frac{{{x^3}}}{3}) + c(\frac{{{x^4}}}{4}) + k = \frac{1}{2}a{x^2} + \frac{1}{3}b{x^3} + \frac{1}{4}c{x^4} + k$.
Thus,the correct option is $D$.

(B) To solve the integral $\int \frac{dx}{\sin x + \sqrt{3} \cos x}$,we multiply and divide by $2$:
$\int \frac{dx}{\sin x + \sqrt{3} \cos x} = \frac{1}{2} \int \frac{dx}{\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x}$
Using the formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we can write $\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x = \sin(x + \frac{\pi}{3})$:
$= \frac{1}{2} \int \frac{dx}{\sin(x + \frac{\pi}{3})} = \frac{1}{2} \int \csc(x + \frac{\pi}{3}) dx$
Using the standard integral $\int \csc \theta d\theta = \log \tan(\frac{\theta}{2}) + c$:
$= \frac{1}{2} \log \tan \left( \frac{x + \frac{\pi}{3}}{2} \right) + c = \frac{1}{2} \log \tan \left( \frac{x}{2} + \frac{\pi}{6} \right) + c$.

(D) Given integral: $I = \int {\frac{{{x^2} + x - 6}}{{(x - 2)(x - 1)}}dx}$
Factorize the numerator: ${x^2} + x - 6 = (x + 3)(x - 2)$
Substitute this into the integral: $I = \int {\frac{{(x + 3)(x - 2)}}{{(x - 2)(x - 1)}}dx} = \int {\frac{{x + 3}}{{x - 1}}dx}$
Rewrite the numerator: $x + 3 = (x - 1) + 4$
So,$I = \int {\frac{{(x - 1) + 4}}{{x - 1}}dx} = \int {\left( {1 + \frac{4}{{x - 1}}} \right)dx}$
Integrating term by term: $I = \int {1dx} + 4\int {\frac{1}{{x - 1}}dx} = x + 4\log |x - 1| + c$

(B) We know the trigonometric identity $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$.
Substituting $\theta = 2x$,we get $\cos 6x = 4\cos^3 2x - 3\cos 2x$.
Therefore,the integral becomes:
$\int \frac{dx}{4\cos^3 2x - 3\cos 2x} = \int \frac{dx}{\cos 6x} = \int \sec 6x \, dx$.
Using the standard integral formula $\int \sec u \, du = \log |\sec u + \tan u| + c$,we have:
$\int \sec 6x \, dx = \frac{1}{6} \log |\sec 6x + \tan 6x| + c$.

(A) We know that $\sin 3x = 3\sin x - 4\sin^3 x$.
Substituting this into the integral:
$\int \frac{3\sin x - 4\sin^3 x}{\sin x} \, dx = \int (3 - 4\sin^2 x) \, dx$
Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$:
$\int (3 - 4(\frac{1 - \cos 2x}{2})) \, dx = \int (3 - 2(1 - \cos 2x)) \, dx$
$= \int (3 - 2 + 2\cos 2x) \, dx = \int (1 + 2\cos 2x) \, dx$
$= x + 2(\frac{\sin 2x}{2}) + c = x + \sin 2x + c$.

(D) Given the equation: $\int \frac{f(x) \, dx}{\log \sin x} = \log \log \sin x$.
To find $f(x)$,we differentiate both sides with respect to $x$ using the Fundamental Theorem of Calculus and the Chain Rule.
Let $y = \log \sin x$. Then $\frac{dy}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x$.
By differentiating the right side: $\frac{d}{dx} (\log(\log \sin x)) = \frac{1}{\log \sin x} \cdot \frac{d}{dx}(\log \sin x) = \frac{1}{\log \sin x} \cdot \cot x$.
Equating the derivatives: $\frac{f(x)}{\log \sin x} = \frac{\cot x}{\log \sin x}$.
Therefore,$f(x) = \cot x$.

(A) We are given the integral $I = \int \frac{\sin x + \csc x}{\tan x} \, dx$.
First,simplify the integrand:
$\frac{\sin x + \csc x}{\tan x} = \frac{\sin x}{\tan x} + \frac{\csc x}{\tan x}$
$= \frac{\sin x}{(\sin x / \cos x)} + \frac{(1 / \sin x)}{(\sin x / \cos x)}$
$= \cos x + \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = \cos x + \csc x \cot x$.
Now,integrate term by term:
$I = \int (\cos x + \csc x \cot x) \, dx$
$= \int \cos x \, dx + \int \csc x \cot x \, dx$
$= \sin x - \csc x + c$.

(D) We need to evaluate the integral $I = \int (\tan x - \cot x)^2 \, dx$.
Expanding the square,we get:
$I = \int (\tan^2 x + \cot^2 x - 2 \tan x \cot x) \, dx$
Since $\tan x \cot x = 1$,the expression becomes:
$I = \int (\tan^2 x + \cot^2 x - 2) \, dx$
Using the trigonometric identities $\tan^2 x = \sec^2 x - 1$ and $\cot^2 x = \csc^2 x - 1$,we substitute:
$I = \int ((\sec^2 x - 1) + (\csc^2 x - 1) - 2) \, dx$
$I = \int (\sec^2 x + \csc^2 x - 4) \, dx$
Integrating term by term:
$I = \int \sec^2 x \, dx + \int \csc^2 x \, dx - \int 4 \, dx$
$I = \tan x - \cot x - 4x + c$.

(C) Given integral is $I = \int {\{ 1 + 2\tan^2 x + 2\tan x \sec x \}^{1/2}} dx$.
Since $1 + \tan^2 x = \sec^2 x$,we can rewrite the expression inside the square root as:
$1 + 2\tan^2 x + 2\tan x \sec x = \sec^2 x + \tan^2 x + 2\tan x \sec x = (\sec x + \tan x)^2$.
Substituting this back into the integral:
$I = \int (\sec x + \tan x) dx$.
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 $\log a + \log b = \log(ab)$:
$I = \log |\sec x(\sec x + \tan x)| + c$.
Thus,the correct option is $C$.

(A) To evaluate the integral $\int \frac{2x}{(2x + 1)^2} dx$,we rewrite the numerator as $(2x + 1 - 1)$.
$\int \frac{2x + 1 - 1}{(2x + 1)^2} dx = \int \left( \frac{2x + 1}{(2x + 1)^2} - \frac{1}{(2x + 1)^2} \right) dx$
$= \int \frac{1}{2x + 1} dx - \int (2x + 1)^{-2} dx$
Using the standard integration formulas $\int \frac{1}{ax+b} dx = \frac{1}{a}\log|ax+b| + c$ and $\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c$:
$= \frac{1}{2}\log|2x + 1| - \frac{(2x + 1)^{-1}}{2(-1)} + c$
$= \frac{1}{2}\log(2x + 1) + \frac{1}{2(2x + 1)} + c$.

(A) Given integral: $I = \int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx$
Split the fraction: $I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} - \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) dx$
Simplify the terms: $I = \int \left( \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} \right) dx$
Use trigonometric identities: $I = \int (\sec^2 x - \csc^2 x) dx$
Integrate: $I = \tan x - (-\cot x) + c$
Final result: $I = \tan x + \cot x + c$

(B) We need to evaluate the integral $\int (3 \csc^2 x + 2 \sin 3x) \, dx$.
Using the linearity property of integration,we can write this as $\int 3 \csc^2 x \, dx + \int 2 \sin 3x \, dx$.
The integral of $\csc^2 x$ is $-\cot x$ and the integral of $\sin(ax)$ is $-\frac{1}{a} \cos(ax)$.
Thus,$\int 3 \csc^2 x \, dx = -3 \cot x$ and $\int 2 \sin 3x \, dx = 2 \left( -\frac{1}{3} \cos 3x \right) = -\frac{2}{3} \cos 3x$.
Combining these,we get $-3 \cot x - \frac{2}{3} \cos 3x + c$.
Factoring out the negative sign,we get $-(3 \cot x + \frac{2}{3} \cos 3x) + c$.

(A) Given $f'(x) = \frac{1}{x} + x$.
To find $f(x)$,we integrate $f'(x)$ with respect to $x$:
$f(x) = \int (\frac{1}{x} + x) dx = \log|x| + \frac{x^2}{2} + C$.
We are given $f(1) = \frac{5}{2}$. Substituting $x = 1$ into the equation:
$f(1) = \log(1) + \frac{1^2}{2} + C = \frac{5}{2}$.
Since $\log(1) = 0$,we have $0 + \frac{1}{2} + C = \frac{5}{2}$.
$C = \frac{5}{2} - \frac{1}{2} = \frac{4}{2} = 2$.
Substituting the value of $C$ back into the expression for $f(x)$,we get:
$f(x) = \log x + \frac{x^2}{2} + 2$.

(D) To solve the integral $\int \frac{dx}{\sqrt{1 + x} + \sqrt{x}}$,we rationalize the denominator by multiplying the numerator and denominator by $(\sqrt{1 + x} - \sqrt{x})$.
$\int \frac{1}{\sqrt{1 + x} + \sqrt{x}} \times \frac{\sqrt{1 + x} - \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}} dx$
$= \int \frac{\sqrt{1 + x} - \sqrt{x}}{(1 + x) - x} dx$
$= \int (\sqrt{1 + x} - \sqrt{x}) dx$
$= \int (1 + x)^{1/2} dx - \int x^{1/2} dx$
$= \frac{(1 + x)^{3/2}}{3/2} - \frac{x^{3/2}}{3/2} + c$
$= \frac{2}{3}(1 + x)^{3/2} - \frac{2}{3}x^{3/2} + c$.

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

(C) We know that $\tan \left( \frac{\pi}{4} - x \right) = \frac{\tan(\pi/4) - \tan x}{1 + \tan(\pi/4) \tan x} = \frac{1 - \tan x}{1 + \tan x}$.
Therefore,the integral becomes $\int \tan \left( \frac{\pi}{4} - x \right) \, dx$.
Using the formula $\int \tan \theta \, d\theta = \log |\sec \theta| + c$ or $-\log |\cos \theta| + c$,we get:
$\int \tan \left( \frac{\pi}{4} - x \right) \, dx = -\log \left| \cos \left( \frac{\pi}{4} - x \right) \right| \cdot (-1) + c = \log \left| \sec \left( \frac{\pi}{4} - x \right) \right| + c$.
Alternatively,using $\cos \left( \frac{\pi}{4} - x \right) = \sin \left( \frac{\pi}{2} - \left( \frac{\pi}{4} - x \right) \right) = \sin \left( \frac{\pi}{4} + x \right)$,we have:
$-\log \left| \cos \left( \frac{\pi}{4} - x \right) \right| + c = \log \left| \frac{1}{\cos \left( \frac{\pi}{4} - x \right)} \right| + c = \log \left| \sec \left( \frac{\pi}{4} - x \right) \right| + c$.
Wait,checking the options: $\log \sin \left( \frac{\pi}{4} + x \right) + c$ is equivalent to $-\log \cos \left( \frac{\pi}{4} - x \right) + c$. Since $\cos(\frac{\pi}{4}-x) = \sin(\frac{\pi}{2} - (\frac{\pi}{4}-x)) = \sin(\frac{\pi}{4}+x)$,the result is $\log \sin(\frac{\pi}{4}+x) + c$.