Evaluation of various forms of integration Questions in English

(D) Let $I = \int \frac{d x}{\left(2 a x+x^2\right)^{3 / 2}} = \int \frac{d x}{\left((x+a)^2-a^2\right)^{3 / 2}}$.
Substitute $t = x+a$,then $d t = d x$.
$I = \int \frac{d t}{\left(t^2-a^2\right)^{3 / 2}}$.
Let $t = a \sec \theta$,then $d t = a \sec \theta \tan \theta \, d \theta$.
$I = \int \frac{a \sec \theta \tan \theta}{\left(a^2 \sec^2 \theta - a^2\right)^{3 / 2}} \, d \theta = \int \frac{a \sec \theta \tan \theta}{\left(a^2 \tan^2 \theta\right)^{3 / 2}} \, d \theta$.
$I = \int \frac{a \sec \theta \tan \theta}{a^3 \tan^3 \theta} \, d \theta = \frac{1}{a^2} \int \frac{\sec \theta}{\tan^2 \theta} \, d \theta = \frac{1}{a^2} \int \frac{\cos \theta}{\sin^2 \theta} \, d \theta$.
$I = \frac{1}{a^2} \int \csc \theta \cot \theta \, d \theta = -\frac{1}{a^2} \csc \theta + C$.
Since $\sec \theta = \frac{t}{a}$,we have $\cos \theta = \frac{a}{t}$,so $\sin \theta = \sqrt{1 - \frac{a^2}{t^2}} = \frac{\sqrt{t^2-a^2}}{t}$.
Thus,$\csc \theta = \frac{t}{\sqrt{t^2-a^2}}$.
Substituting back,$I = -\frac{1}{a^2} \left( \frac{t}{\sqrt{t^2-a^2}} \right) + C = -\frac{1}{a^2} \left( \frac{x+a}{\sqrt{(x+a)^2-a^2}} \right) + C$.
$I = -\frac{1}{a^2} \left( \frac{x+a}{\sqrt{2 a x+x^2}} \right) + C$.

(B) We have the integral $I = \int \frac{1-\cos x}{\cos x(1+\cos x)} d x$.
Rewrite the numerator as $(1+\cos x) - 2\cos x$:
$I = \int \frac{1+\cos x - 2\cos x}{\cos x(1+\cos x)} d x = \int \frac{1}{\cos x} d x - 2 \int \frac{1}{1+\cos x} d x$.
Using the identity $\int \sec x d x = \log |\sec x + \tan x| + C$:
$I = \log |\sec x + \tan x| - 2 \int \frac{1}{2\cos^2(x/2)} d x = \log |\sec x + \tan x| - \int \sec^2(x/2) d x$.
Alternatively,using the rationalization method:
$I = \int \sec x d x - 2 \int \frac{1-\cos x}{1-\cos^2 x} d x = \int \sec x d x - 2 \int \frac{1-\cos x}{\sin^2 x} d x$.
$I = \int \sec x d x - 2 \int (\operatorname{cosec}^2 x - \operatorname{cosec} x \cot x) d x$.
Integrating term by term:
$I = \log |\sec x + \tan x| - 2(-\cot x - (-\operatorname{cosec} x)) + C$.
$I = \log |\sec x + \tan x| - 2(\operatorname{cosec} x - \cot x) + C$.

(A) Given $I_n = \int \tan^n x \, dx$.
We need to find $I_4 + I_6 = \int \tan^4 x \, dx + \int \tan^6 x \, dx$.
$I_4 + I_6 = \int (\tan^4 x + \tan^6 x) \, dx$.
Factor out $\tan^4 x$:
$I_4 + I_6 = \int \tan^4 x (1 + \tan^2 x) \, dx$.
Using the identity $1 + \tan^2 x = \sec^2 x$:
$I_4 + I_6 = \int \tan^4 x \sec^2 x \, dx$.
Let $u = \tan x$,then $du = \sec^2 x \, dx$.
Substituting these into the integral:
$I_4 + I_6 = \int u^4 \, du = \frac{u^5}{5} + C$.
Substituting back $u = \tan x$:
$I_4 + I_6 = \frac{1}{5} \tan^5 x + C$.

(C) Let $I = \int \frac{1 + \sqrt{\tan x}}{\sin 2x} dx$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we have:
$I = \int \frac{1 + \sqrt{\tan x}}{2 \sin x \cos x} dx = \int \frac{1}{2 \sin x \cos x} dx + \int \frac{\sqrt{\tan x}}{2 \sin x \cos x} dx$.
Dividing the numerator and denominator by $\cos^2 x$ in the first integral:
$\int \frac{\sec^2 x}{2 \tan x} dx = \frac{1}{2} \ln |\tan x|$.
For the second integral:
$\int \frac{\sqrt{\tan x}}{2 \sin x \cos x} dx = \int \frac{\sqrt{\tan x}}{2 \tan x \cos^2 x} dx = \int \frac{\sec^2 x}{2 \sqrt{\tan x}} dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
$\int \frac{1}{2 \sqrt{u}} du = \sqrt{u} = \sqrt{\tan x}$.
Thus,$I = \frac{1}{2} \ln |\tan x| + \sqrt{\tan x} + C$.
Comparing with $A \log \tan x + B \sqrt{\tan x} + C$,we get $A = \frac{1}{2}$ and $B = 1$.
Therefore,$4A - B = 4(\frac{1}{2}) - 1 = 2 - 1 = 1$.

(D) Given the integral $I = \int \frac{dx}{1+a \cos x}$ where $a > 1$ (since $a > |\sec \theta| \ge 1$).
Using the substitution $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$,we get:
$I = \int \frac{dx}{1+a \left(\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}\right)} = \int \frac{\sec^2(x/2) dx}{1+\tan^2(x/2) + a - a\tan^2(x/2)} = \int \frac{\sec^2(x/2) dx}{(1+a) + (1-a)\tan^2(x/2)}$.
Let $t = \tan(x/2)$,then $dt = \frac{1}{2}\sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2dt$.
$I = \int \frac{2dt}{(1+a) - (a-1)t^2} = \frac{2}{a-1} \int \frac{dt}{\frac{a+1}{a-1} - t^2}$.
Using the formula $\int \frac{dt}{k^2 - t^2} = \frac{1}{2k} \ln \left| \frac{k+t}{k-t} \right| + C$,where $k = \sqrt{\frac{a+1}{a-1}}$:
$I = \frac{2}{a-1} \cdot \frac{1}{2\sqrt{\frac{a+1}{a-1}}} \ln \left| \frac{\sqrt{\frac{a+1}{a-1}} + t}{\sqrt{\frac{a+1}{a-1}} - t} \right| + C = \frac{1}{\sqrt{a^2-1}} \ln \left| \frac{\sqrt{a+1} + \sqrt{a-1} \tan(x/2)}{\sqrt{a+1} - \sqrt{a-1} \tan(x/2)} \right| + C$.
This simplifies to option $D$.

(D) Let $I = \int \frac{x^2-2}{x^3 \sqrt{x^2-1}} d x$.
Substitute $x = \sec \theta$,then $dx = \sec \theta \tan \theta d\theta$.
$\sqrt{x^2-1} = \sqrt{\sec^2 \theta - 1} = \tan \theta$.
$I = \int \frac{\sec^2 \theta - 2}{\sec^3 \theta \tan \theta} \cdot \sec \theta \tan \theta d\theta = \int \frac{\sec^2 \theta - 2}{\sec^2 \theta} d\theta = \int (1 - 2 \cos^2 \theta) d\theta$.
Using $1 - 2 \cos^2 \theta = -\cos(2\theta)$,we have $I = \int -\cos(2\theta) d\theta = -\frac{1}{2} \sin(2\theta) + C = -\sin \theta \cos \theta + C$.
Since $\sec \theta = x$,$\cos \theta = \frac{1}{x}$ and $\sin \theta = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2-1}}{x}$.
Thus,$I = -\left(\frac{\sqrt{x^2-1}}{x}\right) \left(\frac{1}{x}\right) + C = -\frac{\sqrt{x^2-1}}{x^2} + C$.

(D) Let $I = \int e^{\tan ^{-1} x} \cdot \frac{1+x+x^2}{1+x^2} dx$.
We can rewrite the integrand as:
$I = \int e^{\tan ^{-1} x} \left( \frac{1+x^2}{1+x^2} + \frac{x}{1+x^2} \right) dx$
$I = \int e^{\tan ^{-1} x} \left( 1 + \frac{x}{1+x^2} \right) dx$
$I = \int e^{\tan ^{-1} x} dx + \int e^{\tan ^{-1} x} \cdot \frac{x}{1+x^2} dx$.
Using integration by parts on the first integral $\int e^{\tan ^{-1} x} \cdot 1 dx$:
Let $u = e^{\tan ^{-1} x}$ and $dv = dx$.
Then $du = e^{\tan ^{-1} x} \cdot \frac{1}{1+x^2} dx$ and $v = x$.
Applying the formula $\int u dv = uv - \int v du$:
$\int e^{\tan ^{-1} x} dx = x e^{\tan ^{-1} x} - \int x \cdot e^{\tan ^{-1} x} \cdot \frac{1}{1+x^2} dx$.
Substituting this back into the expression for $I$:
$I = \left( x e^{\tan ^{-1} x} - \int \frac{x e^{\tan ^{-1} x}}{1+x^2} dx \right) + \int \frac{x e^{\tan ^{-1} x}}{1+x^2} dx + c$.
$I = x e^{\tan ^{-1} x} + c$.

(A) Given that $\int e^x \left[f(x) - f^{\prime}(x)\right] dx = g(x) + C$.
Expanding the integral,we have $\int e^x f(x) dx - \int e^x f^{\prime}(x) dx = g(x) + C$.
Rearranging,we get $\int e^x f(x) dx = \int e^x f^{\prime}(x) dx + g(x) + C$.
Using integration by parts on $\int e^x f(x) dx$:
$\int e^x f(x) dx = f(x) e^x - \int e^x f^{\prime}(x) dx$.
Substituting this into our rearranged equation:
$f(x) e^x - \int e^x f^{\prime}(x) dx = \int e^x f^{\prime}(x) dx + g(x) + C$.
$f(x) e^x - g(x) = 2 \int e^x f^{\prime}(x) dx + C$.
Therefore,$\int e^x f^{\prime}(x) dx = \frac{1}{2} \left[e^x f(x) - g(x)\right] + C$.

(A) Let $I = \int \frac{dx}{x^{2022}(1+x^{2022})^{1/2022}}$.
Factor out $x^{2022}$ from the parenthesis:
$I = \int \frac{dx}{x^{2022} \cdot (x^{2022}(x^{-2022}+1))^{1/2022}} = \int \frac{dx}{x^{2022} \cdot x(1+x^{-2022})^{1/2022}} = \int \frac{dx}{x^{2023}(1+x^{-2022})^{1/2022}}$.
Let $t = 1+x^{-2022}$. Then $dt = -2022x^{-2023} dx$,which implies $\frac{dx}{x^{2023}} = -\frac{1}{2022} dt$.
Substituting these into the integral:
$I = \int -\frac{1}{2022} t^{-1/2022} dt = -\frac{1}{2022} \cdot \frac{t^{1-1/2022}}{1-1/2022} + C = -\frac{1}{2022} \cdot \frac{t^{2021/2022}}{2021/2022} + C = -\frac{t^{2021/2022}}{2021} + C$.
Substituting $t = 1+x^{-2022} = \frac{x^{2022}+1}{x^{2022}}$:
$I = -\frac{(\frac{x^{2022}+1}{x^{2022}})^{2021/2022}}{2021} + C = -\frac{(1+x^{2022})^{2021/2022}}{2021 \cdot (x^{2022})^{2021/2022}} + C = -\frac{(1+x^{2022})^{2021/2022}}{2021 x^{2021}} + C$.
Comparing this with $\frac{-(1+x^m)^{n/m}}{nx^n} + C$,we get $m=2022$ and $n=2021$.
Thus,$m-n = 2022-2021 = 1$.

(C) We have $I = \int \frac{1}{\cos 4x \cos 2x} dx$.
Multiply and divide by $\sin 2x$:
$I = \int \frac{\sin 2x}{\sin 2x \cos 4x \cos 2x} dx = \int \frac{\sin 2x}{\sin 2x \cos 2x (2\cos^2 2x - 1)} dx$.
Using $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$,this is not the simplest path.
Alternatively,use $\frac{1}{\cos 4x \cos 2x} = \frac{1}{\sin 2x} \left( \frac{\sin(4x-2x)}{\cos 4x \cos 2x} \right) = \frac{1}{\sin 2x} (\tan 4x - \tan 2x)$.
Actually,the standard identity is $\frac{1}{\cos 4x \cos 2x} = \frac{1}{\sin 2x} \frac{\sin(4x-2x)}{\cos 4x \cos 2x} = \frac{1}{\sin 2x} (\tan 4x - \tan 2x)$.
Integrating,we get $I = \int \frac{\tan 4x}{\sin 2x} dx - \int \frac{\tan 2x}{\sin 2x} dx$.
Using the partial fraction decomposition provided in the prompt:
$I = \frac{1}{2\sqrt{2}} \log \left| \frac{1+\sqrt{2}\sin 2x}{1-\sqrt{2}\sin 2x} \right| - \frac{1}{2} \log |\sec 2x + \tan 2x| + C$.
Comparing with the given form,$f(x) = \sqrt{2}\sin 2x$ and $g(x) = |\sec 2x + \tan 2x|$.
At $x = \frac{\pi}{6}$:
$f\left(\frac{\pi}{6}\right) = \sqrt{2} \sin\left(\frac{\pi}{3}\right) = \sqrt{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{\sqrt{2}}$.
$g\left(\frac{\pi}{6}\right) = \sec\left(\frac{\pi}{3}\right) + \tan\left(\frac{\pi}{3}\right) = 2 + \sqrt{3}$.
Thus,$g\left(\frac{\pi}{6}\right) - \sqrt{2} f\left(\frac{\pi}{6}\right) = (2 + \sqrt{3}) - \sqrt{2} \left(\frac{\sqrt{3}}{\sqrt{2}}\right) = 2 + \sqrt{3} - \sqrt{3} = 2$.

(B) Let $I = \int \frac{1-k \cos ^2 x}{\sin ^k x \cdot \cos ^2 x} d x$.
We can rewrite the integrand as:
$I = \int \frac{\sec ^2 x - k}{\sin ^k x} d x = \int (\sin x)^{-k} \sec ^2 x d x - k \int \csc ^k x d x$.
Using integration by parts on the first integral $\int (\sin x)^{-k} \sec ^2 x d x$,let $u = (\sin x)^{-k}$ and $dv = \sec ^2 x d x$.
Then $du = -k(\sin x)^{-k-1} \cos x d x$ and $v = \tan x$.
$I = (\sin x)^{-k} \tan x - \int \tan x \cdot (-k)(\sin x)^{-k-1} \cos x d x - k \int \csc ^k x d x$.
Since $\tan x \cdot \cos x = \sin x$,the integral becomes:
$I = \frac{\tan x}{\sin ^k x} + k \int (\sin x)^{-k-1} \sin x d x - k \int \csc ^k x d x$.
$I = \frac{\tan x}{\sin ^k x} + k \int (\sin x)^{-k} d x - k \int \csc ^k x d x$.
$I = \frac{\tan x}{\sin ^k x} + k \int \csc ^k x d x - k \int \csc ^k x d x + C$.
$I = \frac{\tan x}{\sin ^k x} + C$.

(B) Let $I = \int \frac{\operatorname{cosec}^2 x - 2022}{\cos^{2022} x} dx$.
We can rewrite the integral as $I = \int \cos^{-2022} x \operatorname{cosec}^2 x dx - 2022 \int \sec^{2022} x dx$.
Using integration by parts on the first term,let $u = \cos^{-2022} x$ and $dv = \operatorname{cosec}^2 x dx$.
Then $du = -2022 \cos^{-2023} x (-\sin x) dx = 2022 \cos^{-2023} x \sin x dx$ and $v = -\cot x$.
Applying the formula $\int u dv = uv - \int v du$:
$I = \cos^{-2022} x (-\cot x) - \int (-\cot x) (2022 \cos^{-2023} x \sin x) dx - 2022 \int \sec^{2022} x dx$.
Since $\cot x \sin x = \cos x$,the integral becomes:
$I = -\frac{\cot x}{\cos^{2022} x} + 2022 \int \frac{\cos x}{\cos^{2023} x} dx - 2022 \int \sec^{2022} x dx$.
$I = -\frac{\cot x}{\cos^{2022} x} + 2022 \int \sec^{2022} x dx - 2022 \int \sec^{2022} x dx + C$.
$I = -\frac{\cot x}{\cos^{2022} x} + C$.
Thus,$f(x) = -\frac{\cot x}{\cos^{2022} x}$.
Evaluating at $x = \pi/4$: $f(\pi/4) = -\frac{\cot(\pi/4)}{\cos^{2022}(\pi/4)} = -\frac{1}{(1/\sqrt{2})^{2022}} = -\frac{1}{(1/2)^{1011}} = -2^{1011}$.

(C) Let $I = \int \frac{3 e^x-7 e^{-x}}{7 e^x+3 e^{-x}} d x$.
We express the numerator as $A(7 e^x+3 e^{-x}) + B \frac{d}{dx}(7 e^x+3 e^{-x})$.
$3 e^x-7 e^{-x} = A(7 e^x+3 e^{-x}) + B(7 e^x-3 e^{-x})$.
Comparing coefficients of $e^x$ and $e^{-x}$:
$7A + 7B = 3$ and $3A - 3B = -7$.
Solving these,$A = -2/21$ and $B = 19/21$.
Thus,$I = \int \left( \frac{-2}{21} + \frac{19}{21} \frac{7 e^x-3 e^{-x}}{7 e^x+3 e^{-x}} \right) dx$.
$I = \frac{-2}{21} x + \frac{19}{21} \ln |7 e^x+3 e^{-x}| + C$.
$I = \frac{-2}{21} x + \frac{19}{21} \ln |e^x(7+3 e^{-2x})| + C$.
$I = \frac{-2}{21} x + \frac{19}{21} (x + \ln |7+3 e^{-2x}|) + C$.
$I = \frac{17}{21} x + \frac{19}{21} \ln |3(e^{-2x}+7/3)| + C$.
$I = \frac{17}{21} x + \frac{19}{21} \ln |e^{-2x}+7/3| + \text{constant}$.
Comparing with $Kx + L \ln(e^{-2x}+7/3) + C$,we get $K = 17/21$ and $L = 19/21$.
$K+L = 17/21 + 19/21 = 36/21 = 12/7$.

(C) Let $I = \int(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}}) dx$.
The expression inside the square root is $x \pm \sqrt{12(x-3)}$.
We can write $x \pm \sqrt{12(x-3)} = x \pm 2\sqrt{3(x-3)} = (x-3) \pm 2\sqrt{3(x-3)} + 3 = (\sqrt{x-3} \pm \sqrt{3})^2$.
Thus,$\sqrt{x \pm \sqrt{12x-36}} = |\sqrt{x-3} \pm \sqrt{3}|$.
The integral becomes $I = \int (\sqrt{x-3} + \sqrt{3} + |\sqrt{x-3} - \sqrt{3}|) dx$.
Case $1$: If $x > 6$,then $\sqrt{x-3} > \sqrt{3}$,so $|\sqrt{x-3} - \sqrt{3}| = \sqrt{x-3} - \sqrt{3}$.
$I = \int (\sqrt{x-3} + \sqrt{3} + \sqrt{x-3} - \sqrt{3}) dx = \int 2\sqrt{x-3} dx = 2 \cdot \frac{2}{3}(x-3)^{3/2} + C = \frac{4}{3}(x-3)^{3/2} + C$.
Case $2$: If $3 \leq x \leq 6$,then $\sqrt{x-3} \leq \sqrt{3}$,so $|\sqrt{x-3} - \sqrt{3}| = \sqrt{3} - \sqrt{x-3}$.
$I = \int (\sqrt{x-3} + \sqrt{3} + \sqrt{3} - \sqrt{x-3}) dx = \int 2\sqrt{3} dx = 2\sqrt{3}x + C$.
Therefore,the solution is $\begin{cases} \frac{4}{3}(x-3)^{3/2}+C, & \text{if } x > 6 \\ 2\sqrt{3}x+C, & \text{if } 3 \leq x \leq 6 \end{cases}$.

(B) Given integral: $I = \int \frac{\cos 4x + 1}{\cot x - \tan x} dx$
Using the identity $\cos 4x + 1 = 2 \cos^2 2x$ and $\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos 2x}{\frac{1}{2} \sin 2x} = \frac{2 \cos 2x}{\sin 2x}$.
Substituting these into the integral:
$I = \int \frac{2 \cos^2 2x}{\frac{2 \cos 2x}{\sin 2x}} dx$
$I = \int \cos 2x \sin 2x dx$
$I = \frac{1}{2} \int \sin 4x dx$
$I = \frac{1}{2} \left( \frac{-\cos 4x}{4} \right) + c$
$I = \frac{-1}{8} \cos 4x + c$
Comparing with $k \cos 4x + c$,we get $k = \frac{-1}{8}$.

(A) Given $I = \int \frac{d x}{\cos ^4 x+\sin ^4 x}$.
Dividing numerator and denominator by $\cos^4 x$,we get $I = \int \frac{\sec^4 x}{1+\tan^4 x} d x = \int \frac{\sec^2 x(1+\tan^2 x)}{1+\tan^4 x} d x$.
Let $\tan x = t$,then $\sec^2 x d x = d t$.
$I = \int \frac{1+t^2}{1+t^4} d t = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} d t$.
$I = \int \frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+2} d t$.
Let $u = t-\frac{1}{t}$,then $d u = (1+\frac{1}{t^2}) d t$.
$I = \int \frac{d u}{u^2+(\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{u}{\sqrt{2}}) + C$.
Substituting $u = t-\frac{1}{t} = \tan x - \cot x$,we get $I = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{\tan x - \cot x}{\sqrt{2}}) + C$.
Comparing this with the given form,$g(x) = \frac{\tan x - \cot x}{\sqrt{2}}$.

(D) Let $I = \int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] dx$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then,$I = \int e^t \left( \log t + \frac{1}{t^2} \right) dt$.
We can rewrite the integrand as:
$I = \int e^t \left[ \left( \log t + \frac{1}{t} \right) - \left( \frac{1}{t} - \frac{1}{t^2} \right) \right] dt$.
Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = \log t$ and $f'(t) = \frac{1}{t}$,we get:
$I = e^t \log t - \int e^t \left( \frac{1}{t} - \frac{1}{t^2} \right) dt$.
Since $\frac{d}{dt} \left( \frac{1}{t} \right) = -\frac{1}{t^2}$,the second integral is $\int e^t \left( \frac{1}{t} + \frac{d}{dt} \left( \frac{1}{t} \right) \right) dt = e^t \left( \frac{1}{t} \right) + C$.
Thus,$I = e^t \log t - e^t \left( \frac{1}{t} \right) + C$.
Substituting $t = \log x$ and $e^t = x$ back:
$I = x \log(\log x) - \frac{x}{\log x} + C = x \left[ \log(\log x) - \frac{1}{\log x} \right] + C$.

(A) $I = \int \frac{\sqrt{x-1}}{x \sqrt{x+1}} dx = \int \frac{1}{x} \sqrt{\frac{x-1}{x+1}} dx$
Let $\sqrt{\frac{x-1}{x+1}} = t$. Then $\frac{x-1}{x+1} = t^2$.
Solving for $x$: $x-1 = t^2x + t^2 \Rightarrow x(1-t^2) = 1+t^2 \Rightarrow x = \frac{1+t^2}{1-t^2}$.
Differentiating with respect to $t$: $dx = \frac{(1-t^2)(2t) - (1+t^2)(-2t)}{(1-t^2)^2} dt = \frac{4t}{(1-t^2)^2} dt$.
Substituting into the integral: $I = \int \left(\frac{1-t^2}{1+t^2}\right) \cdot t \cdot \frac{4t}{(1-t^2)^2} dt = \int \frac{4t^2}{(1+t^2)(1-t^2)} dt$.
Using partial fractions: $\frac{4t^2}{(1+t^2)(1-t^2)} = \frac{2}{1-t^2} - \frac{2}{1+t^2}$.
Integrating: $I = 2 \int \frac{1}{1-t^2} dt - 2 \int \frac{1}{1+t^2} dt = 2 \left( \frac{1}{2} \ln \left| \frac{1+t}{1-t} \right| \right) - 2 \tan^{-1}(t) + c = \ln \left| \frac{1+t}{1-t} \right| - 2 \tan^{-1}(t) + c$.
Substituting $t = \sqrt{\frac{x-1}{x+1}}$: $I = \ln \left| \frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} \right| - 2 \tan^{-1} \sqrt{\frac{x-1}{x+1}} + c$.
Rationalizing the log term: $\ln \left| \frac{(\sqrt{x+1}+\sqrt{x-1})^2}{(x+1)-(x-1)} \right| = \ln \left| \frac{2x + 2\sqrt{x^2-1}}{2} \right| = \ln \left| x + \sqrt{x^2-1} \right|$.
For the inverse trig term: $2 \tan^{-1} \sqrt{\frac{x-1}{x+1}} = \sec^{-1}(x)$.
Thus,$I = \ln \left| x + \sqrt{x^2-1} \right| - \sec^{-1}(x) + c$.

(D) Consider $I = \int \frac{x + \sin x}{1 + \cos x} dx$
$\Rightarrow I = \int \frac{x}{1 + \cos x} dx + \int \frac{\sin x}{1 + \cos x} dx$
Consider $I = I_{1} + I_{2}$
Now,$I_{1} = \int \frac{x}{1 + \cos x} dx = \int \frac{x \ dx}{2 \cos^{2} \frac{x}{2}}$ $\left\{ \because 1 + \cos \theta = 2 \cos^{2} \frac{\theta}{2} \right\}$
$\Rightarrow I_{1} = \int \frac{x}{2} \sec^{2} \frac{x}{2} dx$
Using integration by parts: $\int u v \ dx = u \int v \ dx - \int \left( \frac{du}{dx} \int v \ dx \right) dx$
Let $u = x$ and $v = \frac{1}{2} \sec^{2} \frac{x}{2}$
$\Rightarrow I_{1} = x \tan \frac{x}{2} - \int 1 \cdot \tan \frac{x}{2} dx$
$\Rightarrow I_{1} = x \tan \frac{x}{2} - 2 \log_{e} \left| \sec \frac{x}{2} \right| + c_{1}$
$\Rightarrow I_{1} = x \tan \frac{x}{2} - \log_{e} \left| \sec^{2} \frac{x}{2} \right| + c_{1}$
Now,$I_{2} = \int \frac{\sin x}{1 + \cos x} dx$
Let $1 + \cos x = t \Rightarrow - \sin x \ dx = dt$
$\Rightarrow I_{2} = - \int \frac{1}{t} dt = - \log_{e} |t| + c_{2} = - \log_{e} |1 + \cos x| + c_{2}$
$\Rightarrow I_{2} = - \log_{e} \left| 2 \cos^{2} \frac{x}{2} \right| + c_{2} = - \log_{e} 2 - \log_{e} \left| \cos^{2} \frac{x}{2} \right| + c_{2} = \log_{e} \left| \sec^{2} \frac{x}{2} \right| + c_{3}$
Adding $I_{1}$ and $I_{2}$:
$I = x \tan \frac{x}{2} - \log_{e} \left| \sec^{2} \frac{x}{2} \right| + c_{1} + \log_{e} \left| \sec^{2} \frac{x}{2} \right| + c_{3}$
$\Rightarrow I = x \tan \frac{x}{2} + c$

(C) Let $I_n = \int (\sin x + \cos x)^n dx$.
Using integration by parts,let $u = (\sin x + \cos x)^{n-1}$ and $dv = (\sin x + \cos x) dx$.
Then $du = (n-1)(\sin x + \cos x)^{n-2}(\cos x - \sin x) dx$ and $v = (-\cos x + \sin x) = (\sin x - \cos x)$.
$I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) - \int (n-1)(\sin x + \cos x)^{n-2}(\cos x - \sin x)(\sin x - \cos x) dx$.
Since $(\cos x - \sin x)(\sin x - \cos x) = -(\sin x - \cos x)^2 = -(1 - \sin 2x) = \sin 2x - 1$.
Also,$(\sin x + \cos x)^2 = 1 + \sin 2x$,so $\sin 2x = (\sin x + \cos x)^2 - 1$.
Thus,$\sin 2x - 1 = (\sin x + \cos x)^2 - 2$.
$I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) - (n-1) \int (\sin x + \cos x)^{n-2} ((\sin x + \cos x)^2 - 2) dx$.
$I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) - (n-1) I_n + 2(n-1) I_{n-2}$.
$I_n + (n-1) I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) + 2(n-1) I_{n-2}$.
$n I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) + 2(n-1) I_{n-2}$.
$n I_n - 2(n-1) I_{n-2} = (\sin x + \cos x)^{n-1}(\sin x - \cos x) + C$.

(B) Let $I=\int \frac{x}{\sqrt{x+1}+\sqrt{x-1}} d x$.
Rationalizing the denominator,we get:
$I=\int \frac{x(\sqrt{x+1}-\sqrt{x-1})}{(x+1)-(x-1)} d x = \frac{1}{2} \int x \sqrt{x+1} d x - \frac{1}{2} \int x \sqrt{x-1} d x = \frac{1}{2} I_1 - \frac{1}{2} I_2$.
For $I_1 = \int x \sqrt{x+1} d x$,let $u = x+1$,then $x = u-1$ and $dx = du$:
$I_1 = \int (u-1) \sqrt{u} du = \int (u^{3/2} - u^{1/2}) du = \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} + c_1 = 2(x+1)^{3/2} [\frac{1}{5}(x+1) - \frac{1}{3}] + c_1 = \frac{2(3x-2)}{15}(x+1)^{3/2} + c_1$.
For $I_2 = \int x \sqrt{x-1} d x$,let $v = x-1$,then $x = v+1$ and $dx = dv$:
$I_2 = \int (v+1) \sqrt{v} dv = \int (v^{3/2} + v^{1/2}) dv = \frac{2}{5} v^{5/2} + \frac{2}{3} v^{3/2} + c_2 = 2(x-1)^{3/2} [\frac{1}{5}(x-1) + \frac{1}{3}] + c_2 = \frac{2(3x+2)}{15}(x-1)^{3/2} + c_2$.
Substituting back into $I$:
$I = \frac{1}{2} [\frac{2(3x-2)}{15}(x+1)^{3/2}] - \frac{1}{2} [\frac{2(3x+2)}{15}(x-1)^{3/2}] + C = \frac{3x-2}{15}(x+1)^{3/2} - \frac{3x+2}{15}(x-1)^{3/2} + C$.
Comparing with $A(x)(x+1)^{3/2} + B(x)(x-1)^{3/2} + C$,we get $A(x) = \frac{3x-2}{15}$ and $B(x) = -\frac{3x+2}{15}$.
Thus,$A(x) + B(x) = \frac{3x-2 - 3x - 2}{15} = -\frac{4}{15}$.

(B) We simplify the integrand as follows:
$\int \frac{2x^2-1+x^2\sqrt{x^2+4}}{x^2(x^2+4)} dx = \int \left( \frac{2x^2}{x^2(x^2+4)} - \frac{1}{x^2(x^2+4)} + \frac{x^2\sqrt{x^2+4}}{x^2(x^2+4)} \right) dx$
$= \int \left( \frac{2}{x^2+4} - \frac{1}{x^2(x^2+4)} + \frac{1}{\sqrt{x^2+4}} \right) dx$
Using partial fractions for $\frac{1}{x^2(x^2+4)} = \frac{1}{4} \left( \frac{1}{x^2} - \frac{1}{x^2+4} \right)$:
$= \int \left( \frac{2}{x^2+4} - \frac{1}{4} \left( \frac{1}{x^2} - \frac{1}{x^2+4} \right) + \frac{1}{\sqrt{x^2+4}} \right) dx$
$= \int \left( \frac{2}{x^2+4} + \frac{1}{4(x^2+4)} - \frac{1}{4x^2} + \frac{1}{\sqrt{x^2+4}} \right) dx$
$= \int \left( \frac{9}{4(x^2+4)} - \frac{1}{4x^2} + \frac{1}{\sqrt{x^2+4}} \right) dx$
$= \frac{9}{4} \cdot \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) - \frac{1}{4} \left( -\frac{1}{x} \right) + \sinh^{-1} \left( \frac{x}{2} \right) + c$
$= \frac{9}{8} \tan^{-1} \left( \frac{x}{2} \right) + \frac{1}{4x} + \sinh^{-1} \left( \frac{x}{2} \right) + c$.

(D) We need to evaluate the integral $I = \int \frac{x^4+1}{1+x^6} dx$.
Split the numerator as $x^4+1 = (x^4 - x^2 + 1) + x^2$.
Then,$I = \int \frac{x^4 - x^2 + 1}{1+x^6} dx + \int \frac{x^2}{1+x^6} dx$.
Let $I_1 = \int \frac{x^4 - x^2 + 1}{1+(x^2)^3} dx$ and $I_2 = \int \frac{x^2}{1+(x^3)^2} dx$.
For $I_1$,use the identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$:
$I_1 = \int \frac{x^4 - x^2 + 1}{(1+x^2)(x^4 - x^2 + 1)} dx = \int \frac{1}{1+x^2} dx = \tan^{-1} x$.
For $I_2$,let $x^3 = t$,then $3x^2 dx = dt$,so $x^2 dx = \frac{dt}{3}$:
$I_2 = \int \frac{1}{1+t^2} \cdot \frac{dt}{3} = \frac{1}{3} \tan^{-1} t = \frac{1}{3} \tan^{-1} (x^3)$.
Combining these,$I = I_1 + I_2 = \tan^{-1} x + \frac{1}{3} \tan^{-1} (x^3) + C$.

(C) Let $I = \int \frac{\cos^3 x + \cos^5 x}{\sin^2 x + \sin^4 x} dx$.
$= \int \frac{\cos^3 x (1 + \cos^2 x)}{\sin^2 x (1 + \sin^2 x)} dx$
$= \int \frac{\cos^2 x (1 + \cos^2 x)}{\sin^2 x (1 + \sin^2 x)} \cos x dx$
$= \int \frac{(1 - \sin^2 x) (1 + 1 - \sin^2 x)}{\sin^2 x (1 + \sin^2 x)} \cos x dx$
$= \int \frac{(1 - \sin^2 x) (2 - \sin^2 x)}{\sin^2 x (1 + \sin^2 x)} \cos x dx$.
Let $\sin x = t$,then $\cos x dx = dt$.
$I = \int \frac{(1 - t^2) (2 - t^2)}{t^2 (1 + t^2)} dt$
$= \int \frac{t^4 - 3t^2 + 2}{t^2 (1 + t^2)} dt$
$= \int \frac{t^2(t^2 + 1) - 4t^2 + 2}{t^2 (1 + t^2)} dt$
$= \int \left( 1 + \frac{2 - 4t^2}{t^2 (1 + t^2)} \right) dt$
$= \int 1 dt + \int \left( \frac{2}{t^2} - \frac{6}{1 + t^2} \right) dt$
$= t - \frac{2}{t} - 6 \tan^{-1} t + c$
$= \sin x - 2(\sin x)^{-1} - 6 \tan^{-1} (\sin x) + c$.

(B) Let $I = \int \frac{dx}{\tan x + \cot x + \sec x + \csc x}$.
$= \int \frac{dx}{\left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} + \frac{1}{\cos x} + \frac{1}{\sin x}\right)}$
$= \int \frac{\sin x \cos x}{\sin^{2} x + \cos^{2} x + \sin x + \cos x} dx$
$= \int \frac{\sin x \cos x}{1 + \sin x + \cos x} dx$
Multiply numerator and denominator by $(\sin x + \cos x - 1)$:
$= \int \frac{\sin x \cos x (\sin x + \cos x - 1)}{(\sin x + \cos x)^{2} - 1} dx$
$= \int \frac{\sin x \cos x (\sin x + \cos x - 1)}{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x - 1} dx$
$= \int \frac{\sin x \cos x (\sin x + \cos x - 1)}{2 \sin x \cos x} dx$
$= \frac{1}{2} \int (\sin x + \cos x - 1) dx$
$= \frac{1}{2} (-\cos x + \sin x - x) + c$.

(C) We have $f(x) = \int \csc^5 x \ dx$. Using the reduction formula for $\int \csc^n x \ dx$,we know that $\int \csc^n x \ dx = -\frac{\csc^{n-2} x \cot x}{n-1} + \frac{n-2}{n-1} \int \csc^{n-2} x \ dx$.
For $n=5$,we get $f(x) = -\frac{\csc^3 x \cot x}{4} + \frac{3}{4} \int \csc^3 x \ dx$.
We know $\int \csc^3 x \ dx = -\frac{1}{2} \csc x \cot x + \frac{1}{2} \log |\csc x - \cot x| + C$.
Substituting this back,$f(x) = -\frac{1}{4} \csc^3 x \cot x + \frac{3}{4} [-\frac{1}{2} \csc x \cot x + \frac{1}{2} \log |\csc x - \cot x|] + c$.
$f(x) = -\frac{1}{4} \csc^3 x \cot x - \frac{3}{8} \csc x \cot x + \frac{3}{8} \log |\csc x - \cot x| + c$.
At $x = \frac{\pi}{4}$,$\csc(\frac{\pi}{4}) = \sqrt{2}$ and $\cot(\frac{\pi}{4}) = 1$.
$f(\frac{\pi}{4}) = -\frac{1}{4} (\sqrt{2})^3 (1) - \frac{3}{8} (\sqrt{2})(1) + \frac{3}{8} \log |\sqrt{2} - 1| + c$.
$f(\frac{\pi}{4}) = -\frac{2\sqrt{2}}{4} - \frac{3\sqrt{2}}{8} + \frac{3}{8} \log |\sqrt{2} - 1| + c$.
$f(\frac{\pi}{4}) = -\frac{4\sqrt{2}}{8} - \frac{3\sqrt{2}}{8} + \frac{3}{8} \log |\frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1}| + c$.
$f(\frac{\pi}{4}) = -\frac{7\sqrt{2}}{8} + \frac{3}{8} \log |\frac{1}{\sqrt{2}+1}| + c = -\frac{7\sqrt{2}}{8} - \frac{3}{8} \log(\sqrt{2}+1) + c$.
$f(\frac{\pi}{4}) = -\frac{1}{8} [7\sqrt{2} + 3 \log(\sqrt{2} + 1)] + c$.

(A) Let $I = \int (\sqrt{\tan x} + \sqrt{\cot x}) dx = \int \frac{\tan x + 1}{\sqrt{\tan x}} dx$.
Substitute $\tan x = t^2$,then $\sec^2 x dx = 2t dt$.
Since $\sec^2 x = 1 + \tan^2 x = 1 + t^4$,we have $dx = \frac{2t}{1+t^4} dt$.
Thus,$I = \int \frac{t^2 + 1}{t} \cdot \frac{2t}{1+t^4} dt = 2 \int \frac{t^2 + 1}{t^4 + 1} dt$.
Divide numerator and denominator by $t^2$: $I = 2 \int \frac{1 + 1/t^2}{t^2 + 1/t^2} dt = 2 \int \frac{1 + 1/t^2}{(t - 1/t)^2 + 2} dt$.
Let $u = t - 1/t$,then $du = (1 + 1/t^2) dt$.
$I = 2 \int \frac{du}{u^2 + (\sqrt{2})^2} = 2 \cdot \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + c = \sqrt{2} \tan^{-1}\left(\frac{t - 1/t}{\sqrt{2}}\right) + c$.
Substituting $t = \sqrt{\tan x}$,we get $I = \sqrt{2} \tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2 \tan x}}\right) + c$.

(C) Let $I = \int \frac{dx}{1-2a \cos x + a^2}$.
Using the identity $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$,we have:
$I = \int \frac{dx}{1-2a \left(\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}\right) + a^2} = \int \frac{\sec^2(x/2) dx}{(1+a^2)(1+\tan^2(x/2)) - 2a(1-\tan^2(x/2))}$.
Let $t = \tan(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2 dt$.
$I = \int \frac{2 dt}{(1+a^2)(1+t^2) - 2a(1-t^2)} = \int \frac{2 dt}{1+a^2+t^2+a^2t^2-2a+2at^2} = \int \frac{2 dt}{(1-a)^2 + t^2(1+a)^2}$.
$I = \frac{2}{(1+a)^2} \int \frac{dt}{\left(\frac{1-a}{1+a}\right)^2 + t^2}$.
Using $\int \frac{dx}{k^2+x^2} = \frac{1}{k} \tan^{-1}(\frac{x}{k})$,we get:
$I = \frac{2}{(1+a)^2} \cdot \frac{1+a}{1-a} \tan^{-1}\left(t \cdot \frac{1+a}{1-a}\right) + c = \frac{2}{1-a^2} \tan^{-1}\left[\frac{1+a}{1-a} \tan \frac{x}{2}\right] + c$.

(A) Let $I = \int \frac{6x+5}{\sqrt{6+x-2x^2}} dx$.
We express the numerator as $6x+5 = A \frac{d}{dx}(6+x-2x^2) + B$.
$6x+5 = A(1-4x) + B = -4Ax + (A+B)$.
Comparing coefficients,$-4A = 6 \implies A = -\frac{3}{2}$ and $A+B = 5 \implies B = 5 + \frac{3}{2} = \frac{13}{2}$.
So,$I = \int \frac{-\frac{3}{2}(1-4x) + \frac{13}{2}}{\sqrt{6+x-2x^2}} dx = -\frac{3}{2} \int \frac{1-4x}{\sqrt{6+x-2x^2}} dx + \frac{13}{2} \int \frac{dx}{\sqrt{6+x-2x^2}}$.
For the first part,let $u = 6+x-2x^2$,then $du = (1-4x)dx$.
Integral becomes $-\frac{3}{2} \int u^{-1/2} du = -\frac{3}{2} (2\sqrt{u}) = -3\sqrt{6+x-2x^2}$.
For the second part,$\int \frac{dx}{\sqrt{6+x-2x^2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{3 + \frac{x}{2} - x^2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\frac{49}{16} - (x^2 - \frac{x}{2} + \frac{1}{16})}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{(\frac{7}{4})^2 - (x-\frac{1}{4})^2}}$.
This is $\frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{x-1/4}{7/4}\right) = \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{4x-1}{7}\right)$.
Combining both,$I = -3\sqrt{6+x-2x^2} + \frac{13}{2\sqrt{2}} \sin^{-1}\left(\frac{4x-1}{7}\right) + c$.

(A) To evaluate the integral $I = \int \frac{\sin (x-a)}{\sin (x-b)} dx$,we rewrite the numerator as $\sin((x-b) + (b-a))$.
Using the identity $\sin(u+v) = \sin u \cos v + \cos u \sin v$,we get:
$I = \int \frac{\sin(x-b)\cos(b-a) + \cos(x-b)\sin(b-a)}{\sin(x-b)} dx$
$I = \int \cos(b-a) dx + \int \sin(b-a) \cot(x-b) dx$
$I = x \cos(b-a) + \sin(b-a) \log |\sin(x-b)| + C$
Comparing this with $Ax + B \log |\sin(x-b)| + C$,we find $A = \cos(b-a)$ and $B = \sin(b-a)$.
Thus,$(A, B) = (\cos(b-a), \sin(b-a))$.

(C) Let $I = \int \frac{dx}{(x-1)\sqrt{x^2-1}}$.
Substitute $x-1 = \frac{1}{t}$,which implies $x = 1 + \frac{1}{t}$.
Then,$dx = -\frac{1}{t^2} dt$.
Substituting these into the integral:
$I = \int \frac{-\frac{1}{t^2} dt}{\frac{1}{t} \sqrt{(1+\frac{1}{t})^2 - 1}} = -\int \frac{dt}{t \sqrt{1 + \frac{1}{t^2} + \frac{2}{t} - 1}} = -\int \frac{dt}{t \sqrt{\frac{1+2t}{t^2}}}$.
Since $t = \frac{1}{x-1}$,for $x > 1$,$t > 0$,so $\sqrt{t^2} = t$.
$I = -\int \frac{dt}{t \cdot \frac{\sqrt{1+2t}}{t}} = -\int \frac{dt}{\sqrt{1+2t}}$.
$I = -\int (1+2t)^{-1/2} dt = -\frac{(1+2t)^{1/2}}{1/2 \cdot 2} + C = -\sqrt{1+2t} + C$.
Substituting $t = \frac{1}{x-1}$ back:
$I = -\sqrt{1 + \frac{2}{x-1}} + C = -\sqrt{\frac{x-1+2}{x-1}} + C = -\sqrt{\frac{x+1}{x-1}} + C$.

(D) Let $I = \int \sec^2 x \operatorname{cosec}^4 x \, dx$.
We can write $\sec^2 x = \frac{1}{\cos^2 x}$ and $\operatorname{cosec}^4 x = \frac{1}{\sin^4 x}$.
$I = \int \frac{1}{\cos^2 x \sin^4 x} \, dx = \int \frac{\sin^2 x + \cos^2 x}{\cos^2 x \sin^4 x} \, dx$.
$I = \int \frac{\sin^2 x}{\cos^2 x \sin^4 x} \, dx + \int \frac{\cos^2 x}{\cos^2 x \sin^4 x} \, dx$.
$I = \int \frac{1}{\cos^2 x \sin^2 x} \, dx + \int \operatorname{cosec}^4 x \, dx$.
$I = \int \frac{\sin^2 x + \cos^2 x}{\cos^2 x \sin^2 x} \, dx + \int \operatorname{cosec}^2 x \operatorname{cosec}^2 x \, dx$.
$I = \int (\sec^2 x + \operatorname{cosec}^2 x) \, dx + \int (1 + \cot^2 x) \operatorname{cosec}^2 x \, dx$.
$I = \tan x - \cot x + \int \operatorname{cosec}^2 x \, dx + \int \cot^2 x \operatorname{cosec}^2 x \, dx$.
Let $u = \cot x$,then $du = -\operatorname{cosec}^2 x \, dx$.
$I = \tan x - \cot x - \cot x - \int u^2 \, du$.
$I = \tan x - 2 \cot x - \frac{u^3}{3} + C = \tan x - 2 \cot x - \frac{1}{3} \cot^3 x + C$.
Comparing this with the given expression $-\frac{1}{3} \cot^3 x + k \tan x - 2 \cot x + C$,we get $k = 1$.

(C) We have,$I = \int \frac{dx}{1-\cos x-\sin x}$.
Using the half-angle formulas $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$ and $\sin x = \frac{2\tan(x/2)}{1+\tan^2(x/2)}$,we get:
$I = \int \frac{dx}{1 - \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)} - \frac{2\tan(x/2)}{1+\tan^2(x/2)}}$
$I = \int \frac{(1+\tan^2(x/2)) dx}{1+\tan^2(x/2) - 1 + \tan^2(x/2) - 2\tan(x/2)}$
$I = \int \frac{\sec^2(x/2) dx}{2\tan^2(x/2) - 2\tan(x/2)} = \frac{1}{2} \int \frac{\sec^2(x/2) dx}{\tan(x/2)(\tan(x/2)-1)}$.
Let $z = \tan(x/2)$,then $dz = \frac{1}{2}\sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2dz$.
$I = \int \frac{2dz}{2z(z-1)} = \int \frac{dz}{z(z-1)}$.
Using partial fractions: $\frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}$.
$I = \int \left(\frac{1}{z-1} - \frac{1}{z}\right) dz = \log|z-1| - \log|z| + C = \log\left|\frac{z-1}{z}\right| + C$.
Substituting $z = \tan(x/2)$ back:
$I = \log\left|\frac{\tan(x/2)-1}{\tan(x/2)}\right| + C = \log|1 - \cot(x/2)| + C$.

(B) Let $I = \int \frac{dx}{7+5 \cos x}$.
Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$,we get:
$I = \int \frac{dx}{7(\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}) + 5(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})}$
$I = \int \frac{dx}{12 \cos^2 \frac{x}{2} + 2 \sin^2 \frac{x}{2}}$
Divide numerator and denominator by $\cos^2 \frac{x}{2}$:
$I = \int \frac{\sec^2 \frac{x}{2} dx}{12 + 2 \tan^2 \frac{x}{2}} = \frac{1}{2} \int \frac{\sec^2 \frac{x}{2} dx}{6 + \tan^2 \frac{x}{2}}$
Let $\tan \frac{x}{2} = z$,then $\frac{1}{2} \sec^2 \frac{x}{2} dx = dz$.
$I = \int \frac{dz}{6 + z^2} = \int \frac{dz}{(\sqrt{6})^2 + z^2}$
Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + c$:
$I = \frac{1}{\sqrt{6}} \tan^{-1} \frac{z}{\sqrt{6}} + c = \frac{1}{\sqrt{6}} \tan^{-1} \left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right) + c$.

(A) Let $I = \int e^{\sin x} \left( \frac{x \cos^3 x - \sin x}{\cos^2 x} \right) dx$
$= \int e^{\sin x} (x \cos x - \frac{\sin x}{\cos^2 x}) dx$
$= \int e^{\sin x} (x \cos x - \tan x \sec x) dx$
$= \int x (e^{\sin x} \cos x) dx - \int e^{\sin x} \sec x \tan x dx$
Using integration by parts for the first integral,let $u = x$ and $dv = e^{\sin x} \cos x dx$. Then $du = dx$ and $v = e^{\sin x}$.
$= x e^{\sin x} - \int e^{\sin x} dx - \int e^{\sin x} \sec x \tan x dx$
This approach is complex. Let's rewrite the integrand:
$I = \int e^{\sin x} (x \cos x - \sec x \tan x) dx$
$= \int x e^{\sin x} \cos x dx - \int e^{\sin x} \sec x \tan x dx$
$= x e^{\sin x} - \int e^{\sin x} dx - \int e^{\sin x} \sec x \tan x dx$
Actually,consider $f(x) = x e^{\sin x}$. Then $f'(x) = e^{\sin x} + x e^{\sin x} \cos x$.
Consider $g(x) = -e^{\sin x} \sec x$. Then $g'(x) = -[e^{\sin x} \cos x \sec x + e^{\sin x} \sec x \tan x] = -[e^{\sin x} + e^{\sin x} \sec x \tan x]$.
Thus,$\frac{d}{dx} [e^{\sin x} (x - \sec x)] = e^{\sin x} \cos x (x - \sec x) + e^{\sin x} (1 - \sec x \tan x) = x e^{\sin x} \cos x - e^{\sin x} + e^{\sin x} - e^{\sin x} \sec x \tan x = e^{\sin x} (x \cos x - \sec x \tan x)$.
Therefore,the integral is $e^{\sin x}(x - \sec x) + C$.

(A) Let $I = \int \tan ^{-1}(1-x+x^2) dx + \int \tan ^{-1}(x) dx + \int \tan ^{-1}(1-x) dx$.
Using the property $\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$,we combine the last two terms:
$\tan^{-1}(x) + \tan^{-1}(1-x) = \tan^{-1}\left(\frac{x + 1 - x}{1 - x(1-x)}\right) = \tan^{-1}\left(\frac{1}{1-x+x^2}\right)$.
Since $\tan^{-1}\left(\frac{1}{u}\right) = \cot^{-1}(u)$ for $u > 0$,we have:
$I = \int \left[ \tan^{-1}(1-x+x^2) + \cot^{-1}(1-x+x^2) \right] dx$.
Using the identity $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$:
$I = \int \frac{\pi}{2} dx = \frac{\pi}{2} x + C$.

(A) We are given the integral $I = \int(1+x) \log(1+x^2) dx = \int \log(1+x^2) dx + \int x \log(1+x^2) dx$.
First,evaluate $I_1 = \int \log(1+x^2) dx$ using integration by parts:
$I_1 = x \log(1+x^2) - \int x \cdot \frac{2x}{1+x^2} dx = x \log(1+x^2) - 2 \int \frac{x^2+1-1}{1+x^2} dx$
$I_1 = x \log(1+x^2) - 2 \int (1 - \frac{1}{1+x^2}) dx = x \log(1+x^2) - 2x + 2 \tan^{-1} x$.
Next,evaluate $I_2 = \int x \log(1+x^2) dx$. Let $1+x^2 = t$,so $2x dx = dt$ or $x dx = \frac{1}{2} dt$:
$I_2 = \frac{1}{2} \int \log t dt = \frac{1}{2} (t \log t - t) = \frac{1}{2} ((1+x^2) \log(1+x^2) - (1+x^2))$.
Adding $I_1$ and $I_2$:
$I = x \log(1+x^2) - 2x + 2 \tan^{-1} x + \frac{1}{2} (1+x^2) \log(1+x^2) - \frac{1}{2} (1+x^2)$
$I = (x + \frac{1}{2} + \frac{x^2}{2}) \log(1+x^2) - 2x + 2 \tan^{-1} x - \frac{1}{2} - \frac{x^2}{2}$.
Comparing this with the given form $(x + \frac{x^2}{2} + \frac{1}{2}) \log(1+x^2) + g(x) + C$,we identify:
$g(x) = -2x + 2 \tan^{-1} x - \frac{x^2}{2}$.

(D) Let $I = \int \frac{x+\sin x}{1+\cos x} dx$.
Using the identities $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1+\cos x = 2 \cos^2(x/2)$,we get:
$I = \int \frac{x + 2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} dx$
$I = \int \left( \frac{x}{2 \cos^2(x/2)} + \frac{2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} \right) dx$
$I = \int \left( \frac{1}{2} x \sec^2(x/2) + \tan(x/2) \right) dx$
Now,apply integration by parts to the first term $\int \frac{1}{2} x \sec^2(x/2) dx$:
Let $u = x$ and $dv = \frac{1}{2} \sec^2(x/2) dx$. Then $du = dx$ and $v = \tan(x/2)$.
Using $\int u dv = uv - \int v du$:
$\int \frac{1}{2} x \sec^2(x/2) dx = x \tan(x/2) - \int \tan(x/2) dx$
Substituting this back into the expression for $I$:
$I = x \tan(x/2) - \int \tan(x/2) dx + \int \tan(x/2) dx + c$
$I = x \tan(x/2) + c$

(C) We use the reduction formula for $\int \operatorname{cosec}^n x \, dx = -\frac{\operatorname{cosec}^{n-2} x \cot x}{n-1} + \frac{n-2}{n-1} \int \operatorname{cosec}^{n-2} x \, dx$.
For $n=5$,$\int \operatorname{cosec}^5 x \, dx = -\frac{\operatorname{cosec}^3 x \cot x}{4} + \frac{3}{4} \int \operatorname{cosec}^3 x \, dx$.
For $n=3$,$\int \operatorname{cosec}^3 x \, dx = -\frac{\operatorname{cosec} x \cot x}{2} + \frac{1}{2} \int \operatorname{cosec} x \, dx = -\frac{\operatorname{cosec} x \cot x}{2} + \frac{1}{2} \log |\operatorname{cosec} x - \cot x|$.
Substituting this back,$f(x) = -\frac{\operatorname{cosec}^3 x \cot x}{4} + \frac{3}{4} \left( -\frac{\operatorname{cosec} x \cot x}{2} + \frac{1}{2} \log |\operatorname{cosec} x - \cot x| \right) + c$.
$f(x) = -\frac{1}{4} \operatorname{cosec}^3 x \cot x - \frac{3}{8} \operatorname{cosec} x \cot x + \frac{3}{8} \log |\operatorname{cosec} x - \cot x| + c$.
At $x = \frac{\pi}{4}$,$\operatorname{cosec} x = \sqrt{2}$ and $\cot x = 1$.
$f\left(\frac{\pi}{4}\right) = -\frac{1}{4} (\sqrt{2})^3 (1) - \frac{3}{8} (\sqrt{2}) (1) + \frac{3}{8} \log |\sqrt{2} - 1| + c$.
$f\left(\frac{\pi}{4}\right) = -\frac{2\sqrt{2}}{4} - \frac{3\sqrt{2}}{8} + \frac{3}{8} \log (\sqrt{2} - 1) + c$.
$f\left(\frac{\pi}{4}\right) = -\frac{4\sqrt{2} + 3\sqrt{2}}{8} - \frac{3}{8} \log (\sqrt{2} + 1) + c = -\frac{1}{8} [7\sqrt{2} + 3 \log (\sqrt{2} + 1)] + c$.

(A) We are given $I_n = \int \cot^n x \, dx$.
For $n=5$,$I_5 = \int \cot^5 x \, dx$.
We can write $\cot^5 x = \cot^3 x \cdot \cot^2 x = \cot^3 x (\csc^2 x - 1) = \cot^3 x \csc^2 x - \cot^3 x$.
So,$I_5 = \int \cot^3 x \csc^2 x \, dx - \int \cot^3 x \, dx$.
Let $u = \cot x$,then $du = -\csc^2 x \, dx$,so $\int \cot^3 x \csc^2 x \, dx = -\int u^3 \, du = -\frac{u^4}{4} = -\frac{\cot^4 x}{4}$.
Now,for $\int \cot^3 x \, dx = \int \cot x (\csc^2 x - 1) \, dx = \int \cot x \csc^2 x \, dx - \int \cot x \, dx$.
$int \cot x \csc^2 x \, dx = -\frac{\cot^2 x}{2}$ and $\int \cot x \, dx = \log |\sin x|$.
Thus,$\int \cot^3 x \, dx = -\frac{\cot^2 x}{2} - \log |\sin x|$.
Substituting these back,$I_5 = -\frac{\cot^4 x}{4} - (-\frac{\cot^2 x}{2} - \log |\sin x|) + c = -\frac{\cot^4 x}{4} + \frac{\cot^2 x}{2} + \log |\sin x| + c$.

(A) We are given $I_n = \int \frac{t^n}{1+t^2} dt$.
Consider the sum $I_n + I_{n-2} = \int \frac{t^n}{1+t^2} dt + \int \frac{t^{n-2}}{1+t^2} dt$.
$I_n + I_{n-2} = \int \frac{t^n + t^{n-2}}{1+t^2} dt = \int \frac{t^{n-2}(t^2 + 1)}{1+t^2} dt$.
$I_n + I_{n-2} = \int t^{n-2} dt = \frac{t^{n-1}}{n-1} + c$.
For $n = 6$,we have $I_6 + I_4 = \frac{t^{6-1}}{6-1} + c = \frac{t^5}{5} + c$.

(A) Let $I = \int \frac{x}{(x^2+2x+2)^2} dx$.
Rewrite the numerator as $x = \frac{1}{2}(2x+2) - 1$.
Then $I = \frac{1}{2} \int \frac{2x+2}{(x^2+2x+2)^2} dx - \int \frac{1}{(x^2+2x+2)^2} dx$.
For the first integral,let $u = x^2+2x+2$,so $du = (2x+2)dx$.
$\frac{1}{2} \int u^{-2} du = -\frac{1}{2u} = -\frac{1}{2(x^2+2x+2)}$.
For the second integral,let $J = \int \frac{1}{((x+1)^2+1)^2} dx$.
Put $x+1 = \tan \theta$,so $dx = \sec^2 \theta d\theta$.
$J = \int \frac{\sec^2 \theta}{\sec^4 \theta} d\theta = \int \cos^2 \theta d\theta = \int \frac{1+\cos 2\theta}{2} d\theta = \frac{1}{2}(\theta + \frac{\sin 2\theta}{2}) + C$.
Since $\tan \theta = x+1$,$\theta = \tan^{-1}(x+1)$ and $\sin 2\theta = \frac{2\tan \theta}{1+\tan^2 \theta} = \frac{2(x+1)}{x^2+2x+2}$.
So $J = \frac{1}{2} \tan^{-1}(x+1) + \frac{x+1}{2(x^2+2x+2)} + C$.
Combining these,$I = -\frac{1}{2(x^2+2x+2)} - \frac{x+1}{2(x^2+2x+2)} - \frac{1}{2} \tan^{-1}(x+1) + C = -\frac{x+2}{2(x^2+2x+2)} - \frac{1}{2} \tan^{-1}(x+1) + C$.

(B) Let $I = \int \frac{2-\sin x}{2 \cos x+3} dx$.
We can split the integral into two parts: $I = \int \frac{2}{2 \cos x+3} dx - \int \frac{\sin x}{2 \cos x+3} dx = I_1 + I_2$.
For $I_1 = \int \frac{2}{2 \cos x+3} dx$,use the half-angle substitution $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)} = \frac{1-t^2}{1+t^2}$ where $t = \tan(x/2)$ and $dx = \frac{2 dt}{1+t^2}$.
$I_1 = \int \frac{2}{2(\frac{1-t^2}{1+t^2})+3} \cdot \frac{2 dt}{1+t^2} = \int \frac{4}{2-2t^2+3+3t^2} dt = \int \frac{4}{t^2+5} dt$.
$I_1 = \frac{4}{\sqrt{5}} \tan^{-1}(\frac{t}{\sqrt{5}}) = \frac{4}{\sqrt{5}} \tan^{-1}(\frac{1}{\sqrt{5}} \tan \frac{x}{2}) + C_1$.
For $I_2 = \int \frac{-\sin x}{2 \cos x+3} dx$,let $u = 2 \cos x + 3$,then $du = -2 \sin x dx$,so $-\sin x dx = \frac{1}{2} du$.
$I_2 = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \log|2 \cos x + 3| = \log \sqrt{2 \cos x + 3} + C_2$.
Combining these,$I = \frac{4}{\sqrt{5}} \tan^{-1}(\frac{1}{\sqrt{5}} \tan \frac{x}{2}) + \log \sqrt{2 \cos x + 3} + C$.

(D) We have the integral $I = \int \frac{x^4+1}{x^6+1} \, dx$.
First,rewrite the numerator as $x^4+1 = (x^4-x^2+1) + x^2$.
Since $x^6+1 = (x^2)^3 + 1^3 = (x^2+1)(x^4-x^2+1)$,we can split the integral:
$\frac{x^4+1}{x^6+1} = \frac{x^4-x^2+1}{x^6+1} + \frac{x^2}{x^6+1} = \frac{1}{x^2+1} + \frac{x^2}{(x^3)^2+1}$.
Now,integrate term by term:
$\int \frac{1}{x^2+1} \, dx + \int \frac{x^2}{(x^3)^2+1} \, dx$.
The first integral is $\tan^{-1} x$.
For the second integral,let $u = x^3$,then $du = 3x^2 \, dx$,so $x^2 \, dx = \frac{1}{3} du$.
Thus,$\int \frac{1}{3} \frac{du}{u^2+1} = \frac{1}{3} \tan^{-1} u = \frac{1}{3} \tan^{-1} x^3$.
Combining these,we get $\tan^{-1} x + \frac{1}{3} \tan^{-1} x^3 + c$.

(D) Let $I = \int \frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} d x = A \log |5 \cos x+4 \sin x|+B x+c$.
By differentiating both sides with respect to $x$,we get:
$\frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} = A \frac{d}{dx}(\log |5 \cos x+4 \sin x|) + B$
$\frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} = A \frac{-5 \sin x+4 \cos x}{5 \cos x+4 \sin x} + B$
$\frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} = \frac{A(-5 \sin x+4 \cos x) + B(4 \sin x+5 \cos x)}{4 \sin x+5 \cos x}$
Equating the numerators:
$3 \cos x-2 \sin x = (4A+5B) \cos x + (-5A+4B) \sin x$
Comparing coefficients of $\cos x$ and $\sin x$:
$4A+5B = 3$ --- $(1)$
$-5A+4B = -2$ --- $(2)$
Multiplying $(1)$ by $5$ and $(2)$ by $4$:
$20A+25B = 15$
$-20A+16B = -8$
Adding these equations: $41B = 7 \Rightarrow B = \frac{7}{41}$.
Substituting $B$ in $(1)$: $4A + 5(\frac{7}{41}) = 3 \Rightarrow 4A = 3 - \frac{35}{41} = \frac{123-35}{41} = \frac{88}{41} \Rightarrow A = \frac{22}{41}$.
Thus,$A = \frac{22}{41}$ and $B = \frac{7}{41}$.

(A) To solve the integral $I = \int \frac{3 \sin x - 5 \cos x}{2 \sin x + 7 \cos x} \, dx$,we express the numerator as $A(\text{denominator}) + B(\frac{d}{dx}(\text{denominator}))$.
Let $3 \sin x - 5 \cos x = A(2 \sin x + 7 \cos x) + B(2 \cos x - 7 \sin x)$.
Equating the coefficients of $\sin x$ and $\cos x$:
For $\sin x$: $2A - 7B = 3$
For $\cos x$: $7A + 2B = -5$
Solving these equations: Multiply the first by $2$ and the second by $7$: $4A - 14B = 6$ and $49A + 14B = -35$.
Adding them: $53A = -29 \implies A = -\frac{29}{53}$.
Substituting $A$ into $2A - 7B = 3$: $2(-\frac{29}{53}) - 7B = 3 \implies -\frac{58}{53} - 3 = 7B \implies 7B = -\frac{217}{53} \implies B = -\frac{31}{53}$.
Thus,$I = \int \frac{A(2 \sin x + 7 \cos x) + B(2 \cos x - 7 \sin x)}{2 \sin x + 7 \cos x} \, dx = A \int 1 \, dx + B \int \frac{2 \cos x - 7 \sin x}{2 \sin x + 7 \cos x} \, dx$.
$I = A x + B \log |2 \sin x + 7 \cos x| + c = -\frac{29}{53} x - \frac{31}{53} \log |2 \sin x + 7 \cos x| + c$.

(A) Let $I = \int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$.
We express the numerator as $A(\text{denominator}) + B(\frac{d}{dx}(\text{denominator}))$:
$\sin x+8 \cos x = A(4 \sin x+6 \cos x) + B(4 \cos x-6 \sin x)$.
Equating the coefficients of $\sin x$ and $\cos x$:
For $\sin x$: $4A - 6B = 1$.
For $\cos x$: $6A + 4B = 8$.
Solving these equations,we multiply the first by $2$ and the second by $3$:
$8A - 12B = 2$ and $18A + 12B = 24$.
Adding them gives $26A = 26$,so $A = 1$.
Substituting $A=1$ into $4A - 6B = 1$ gives $4 - 6B = 1$,so $6B = 3$,$B = \frac{1}{2}$.
Thus,$I = \int \frac{(4 \sin x+6 \cos x) + \frac{1}{2}(4 \cos x-6 \sin x)}{4 \sin x+6 \cos x} d x$.
$I = \int 1 d x + \frac{1}{2} \int \frac{4 \cos x-6 \sin x}{4 \sin x+6 \cos x} d x$.
$I = x + \frac{1}{2} \log |4 \sin x+6 \cos x| + c$.