Integration by substitution Questions in English

(A) Let $I = \int \frac{d x}{\sqrt[3]{\sin ^{11} x \cdot \cos x}}$.
Divide the numerator and denominator by $\cos^{\frac{12}{3}} x = \cos^4 x$:
$I = \int \frac{\sec^4 x}{\tan^{\frac{11}{3}} x} d x$.
Using $\sec^4 x = (1 + \tan^2 x) \sec^2 x$,we get:
$I = \int \frac{(1 + \tan^2 x) \sec^2 x}{\tan^{\frac{11}{3}} x} d x$.
Let $\tan x = t$,so $\sec^2 x d x = d t$.
$I = \int \frac{1 + t^2}{t^{\frac{11}{3}}} d t = \int (t^{-\frac{11}{3}} + t^{-\frac{5}{3}}) d t$.
Integrating term by term:
$I = \frac{t^{-\frac{8}{3}}}{-\frac{8}{3}} + \frac{t^{-\frac{2}{3}}}{-\frac{2}{3}} + c = -\left(\frac{3}{8} t^{-\frac{8}{3}} + \frac{3}{2} t^{-\frac{2}{3}}\right) + c$.
Substituting $t = \tan x$:
$I = -\left(\frac{3}{8} \tan^{-\frac{8}{3}} x + \frac{3}{2} \tan^{-\frac{2}{3}} x\right) + c$.
Comparing with the given form,we get $f(x) = \tan^{-\frac{8}{3}} x$ and $g(x) = \tan^{-\frac{2}{3}} x$.

(A) Let $I = \int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} dx$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$,but it is more convenient to write $8 - \sin 2x = 9 - (1 + \sin 2x) = 9 - (\sin x + \cos x)^2$.
So,$I = \int \frac{\cos x - \sin x}{\sqrt{9 - (\sin x + \cos x)^2}} dx$.
Let $t = \sin x + \cos x$.
Then $dt = (\cos x - \sin x) dx$.
Substituting these into the integral,we get $I = \int \frac{dt}{\sqrt{3^2 - t^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + c$,we have $I = \sin^{-1}(\frac{t}{3}) + c$.
Substituting $t = \sin x + \cos x$ back,we get $I = \sin^{-1}\left(\frac{\sin x + \cos x}{3}\right) + c$.
Comparing this with the given expression $a \sin^{-1}\left(\frac{\sin x + \cos x}{b}\right) + c$,we find $a = 1$ and $b = 3$.
Thus,the ordered pair $(a, b)$ is $(1, 3)$.

(C) Let $I = \int \sqrt{e^x-1} \, dx$.
Substitute $e^x - 1 = t^2$,which implies $e^x = t^2 + 1$.
Differentiating both sides,$e^x \, dx = 2t \, dt$.
Thus,$dx = \frac{2t}{t^2+1} \, dt$.
Substituting these into the integral:
$I = \int t \cdot \frac{2t}{t^2+1} \, dt = \int \frac{2t^2}{t^2+1} \, dt$.
Rewrite the numerator:
$I = \int \frac{2(t^2+1) - 2}{t^2+1} \, dt = \int \left( 2 - \frac{2}{t^2+1} \right) \, dt$.
Integrating term by term:
$I = 2t - 2 \tan^{-1}(t) + c$.
Substituting $t = \sqrt{e^x-1}$ back:
$I = 2 \sqrt{e^x-1} - 2 \tan^{-1} \sqrt{e^x-1} + c$.

(B) Let $I = \int \frac{\sqrt{x}}{x+1} \,dx$.
Substitute $\sqrt{x} = t$, which implies $x = t^2$ and $dx = 2t \,dt$.
Substituting these into the integral:
$I = \int \frac{t}{t^2+1} (2t) \,dt = \int \frac{2t^2}{t^2+1} \,dt$.
Rewrite the numerator:
$I = 2 \int \frac{t^2+1-1}{t^2+1} \,dt = 2 \int \left(1 - \frac{1}{t^2+1}\right) \,dt$.
Integrating term by term:
$I = 2 \left( \int 1 \,dt - \int \frac{1}{t^2+1} \,dt \right) = 2(t - \tan^{-1} t) + c$.
Substituting $t = \sqrt{x}$ back:
$I = 2(\sqrt{x} - \tan^{-1} \sqrt{x}) + c$.

(C) Let $I = \int(2x+4)\sqrt{x-1} \, dx$.
Substitute $x-1 = t^2$,which implies $x = t^2+1$ and $dx = 2t \, dt$.
Substituting these into the integral:
$I = \int(2(t^2+1)+4) \cdot t \cdot (2t) \, dt$
$I = \int(2t^2+2+4) \cdot 2t^2 \, dt$
$I = \int(2t^2+6) \cdot 2t^2 \, dt$
$I = \int(4t^4 + 12t^2) \, dt$
Integrating term by term:
$I = 4 \int t^4 \, dt + 12 \int t^2 \, dt$
$I = 4 \cdot \frac{t^5}{5} + 12 \cdot \frac{t^3}{3} + c$
$I = \frac{4}{5}t^5 + 4t^3 + c$
Substituting back $t = (x-1)^{\frac{1}{2}}$:
$I = \frac{4}{5}(x-1)^{\frac{5}{2}} + 4(x-1)^{\frac{3}{2}} + c$
Comparing this with the given form $a(x-1)^{\frac{5}{2}} + b(x-1)^{\frac{3}{2}} + c$,we get $a = \frac{4}{5}$ and $b = 4$.
Therefore,$a+b = \frac{4}{5} + 4 = \frac{4+20}{5} = \frac{24}{5}$.

(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.
Factor out $x^4$ from the parenthesis: $I = \int \frac{dx}{x^2(x^4(1 + x^{-4}))^{3/4}} = \int \frac{dx}{x^2 \cdot x^3(1 + x^{-4})^{3/4}} = \int \frac{dx}{x^5(1 + x^{-4})^{3/4}}$.
Let $t = 1 + x^{-4}$.
Then $dt = -4x^{-5} dx$,which implies $x^{-5} dx = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + c = -t^{1/4} + c$.
Substituting back $t = 1 + x^{-4} = \frac{x^4+1}{x^4}$:
$I = -(\frac{x^4+1}{x^4})^{1/4} + c$.

(B) Let $I = \int \frac{dx}{\sqrt{e^x-1}}$.
Substitute $\sqrt{e^x-1} = t$.
Then $e^x - 1 = t^2$,which implies $e^x = t^2 + 1$.
Differentiating both sides,we get $e^x dx = 2t dt$.
Thus,$dx = \frac{2t}{e^x} dt = \frac{2t}{t^2+1} dt$.
Substituting these into the integral:
$I = \int \frac{1}{t} \cdot \frac{2t}{t^2+1} dt = 2 \int \frac{1}{t^2+1} dt$.
Integrating,we get $I = 2 \tan^{-1}(t) + c$.
Substituting $t = \sqrt{e^x-1}$ back,we get $I = 2 \tan^{-1}(\sqrt{e^x-1}) + c$.
Comparing this with the given expression $2 \tan^{-1}(f(x)) + c$,we find $f(x) = \sqrt{e^x-1}$.

(A) Let $I = \int \frac{\cos ^3 x}{\sin ^2 x+\sin x} \,d x$.
Using the identity $\cos^2 x = 1 - \sin^2 x$, we have:
$I = \int \frac{(1-\sin^2 x) \cos x}{\sin x(1+\sin x)} \,d x$.
Since $1-\sin^2 x = (1-\sin x)(1+\sin x)$, the expression simplifies to:
$I = \int \frac{(1-\sin x)(1+\sin x) \cos x}{\sin x(1+\sin x)} \,d x = \int \frac{1-\sin x}{\sin x} \cos x \,d x$.
Substitute $t = \sin x$, then $dt = \cos x \,d x$.
$I = \int \left(\frac{1}{t} - 1\right) dt$.
Integrating with respect to $t$:
$I = \log |t| - t + C$.
Substituting back $t = \sin x$:
$I = \log |\sin x| - \sin x + C$.

(D) Let $I = \int(7x-2)\sqrt{3x+2} \, dx$.
Substitute $3x+2 = t$,so $x = \frac{t-2}{3}$ and $dx = \frac{1}{3} dt$.
Substituting these into the integral:
$I = \int \left[7\left(\frac{t-2}{3}\right) - 2\right] \sqrt{t} \cdot \frac{1}{3} dt$
$I = \frac{1}{3} \int \left(\frac{7t - 14 - 6}{3}\right) \sqrt{t} \, dt$
$I = \frac{1}{9} \int (7t - 20) t^{\frac{1}{2}} \, dt$
$I = \frac{7}{9} \int t^{\frac{3}{2}} \, dt - \frac{20}{9} \int t^{\frac{1}{2}} \, dt$
$I = \frac{7}{9} \cdot \frac{t^{\frac{5}{2}}}{\frac{5}{2}} - \frac{20}{9} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + c$
$I = \frac{14}{45} t^{\frac{5}{2}} - \frac{40}{27} t^{\frac{3}{2}} + c$
Substituting $t = 3x+2$ back:
$I = \frac{14}{45}(3x+2)^{\frac{5}{2}} - \frac{40}{27}(3x+2)^{\frac{3}{2}} + c$
Comparing this with the given form,we get $A = \frac{14}{45}$ and $B = \frac{-40}{27}$.

(A) We have the integral $I = \int \frac{dx}{1+3 \sin^2 x}$.
Dividing the numerator and denominator by $\cos^2 x$,we get:
$I = \int \frac{\sec^2 x dx}{\sec^2 x + 3 \tan^2 x} = \int \frac{\sec^2 x dx}{1 + \tan^2 x + 3 \tan^2 x} = \int \frac{\sec^2 x dx}{1 + 4 \tan^2 x}$.
Let $t = \tan x$,then $dt = \sec^2 x dx$.
The integral becomes:
$I = \int \frac{dt}{1 + 4t^2} = \int \frac{dt}{1 + (2t)^2}$.
Using the formula $\int \frac{du}{a^2 + u^2} = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + c$,where $u = 2t$ and $du = 2dt$:
$I = \frac{1}{2} \int \frac{2dt}{1 + (2t)^2} = \frac{1}{2} \tan^{-1}(2t) + c$.
Substituting $t = \tan x$ back,we get:
$I = \frac{1}{2} \tan^{-1}(2 \tan x) + c$.
Comparing this with the given expression $\frac{1}{2} \tan^{-1}(f(x)) + c$,we find $f(x) = 2 \tan x$.

(A) Let $I = \int(\sqrt{\tan x} + \sqrt{\cot x}) dx$.
We can write this as $I = \int \left(\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}}\right) dx$.
Simplifying the integrand,we get $I = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} dx$.
Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$.
Squaring both sides of $t = \sin x - \cos x$,we get $t^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - 2 \sin x \cos x$.
Thus,$2 \sin x \cos x = 1 - t^2$,which implies $\sin x \cos x = \frac{1 - t^2}{2}$.
Substituting these into the integral,we get $I = \int \frac{dt}{\sqrt{\frac{1 - t^2}{2}}} = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}}$.
Integrating,we obtain $I = \sqrt{2} \sin^{-1}(t) + c$.
Substituting $t = \sin x - \cos x$ back,we get $I = \sqrt{2} \sin^{-1}(\sin x - \cos x) + c$.

(D) Let $I = \int \frac{\sin x}{3+4 \cos ^2 x} \,dx$.
Substitute $\cos x = t$, then $-\sin x \,dx = dt$, which implies $\sin x \,dx = -dt$.
Substituting these into the integral, we get $I = \int \frac{-dt}{3+4 t^2} = -\int \frac{dt}{(\sqrt{3})^2 + (2t)^2}$.
Using the standard integral formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$, we have:
$I = -\frac{1}{\sqrt{3}} \cdot \frac{1}{2} \tan^{-1}(\frac{2t}{\sqrt{3}}) + C = -\frac{1}{2\sqrt{3}} \tan^{-1}(\frac{2 \cos x}{\sqrt{3}}) + C$.
Comparing this with $A \tan^{-1}(B \cos x) + C$, we identify $A = -\frac{1}{2\sqrt{3}}$ and $B = \frac{2}{\sqrt{3}}$.
Therefore, $A + B = -\frac{1}{2\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{-1 + 4}{2\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.

(C) Given $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.
We can rewrite the integral by taking $x^4$ common from the bracket:
$I = \int \frac{dx}{x^2 \left[x^4(1 + \frac{1}{x^4})\right]^{3/4}} = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{3/4}}$.
Let $t = 1 + \frac{1}{x^4}$. Then $dt = -\frac{4}{x^5} dx$,which implies $\frac{dx}{x^5} = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + c = -t^{1/4} + c$.
Substituting back $t = 1 + \frac{1}{x^4} = \frac{x^4+1}{x^4}$:
$I = -(1 + \frac{1}{x^4})^{1/4} + c = -\left(\frac{x^4+1}{x^4}\right)^{1/4} + c = -\frac{(x^4+1)^{1/4}}{x} + c$.

(D) Let $I = \int \frac{dx}{\sin x + \cos x}$.
Using the substitution $\tan \frac{x}{2} = t$,we have $dx = \frac{2}{1+t^2} dt$,$\sin x = \frac{2t}{1+t^2}$,and $\cos x = \frac{1-t^2}{1+t^2}$.
Substituting these into the integral:
$I = \int \frac{\frac{2}{1+t^2}}{\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}} dt = \int \frac{2}{2t + 1 - t^2} dt = -2 \int \frac{1}{t^2 - 2t - 1} dt$.
Completing the square in the denominator:
$t^2 - 2t - 1 = (t-1)^2 - 2 = (t-1)^2 - (\sqrt{2})^2$.
Using the formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + c$:
$I = -2 \times \frac{1}{2\sqrt{2}} \log \left| \frac{t-1-\sqrt{2}}{t-1+\sqrt{2}} \right| + c = -\frac{1}{\sqrt{2}} \log \left| \frac{\tan \frac{x}{2} - 1 - \sqrt{2}}{\tan \frac{x}{2} - 1 + \sqrt{2}} \right| + c$.
Simplifying the expression:
$I = -\frac{1}{\sqrt{2}} \log \left| \frac{\tan \frac{x}{2} - (\sqrt{2} + 1)}{\tan \frac{x}{2} + \sqrt{2} - 1} \right| + c$.

(C) Let $I = \int \frac{(x^2-1) dx}{x^3 \sqrt{2x^4-2x^2+1}}$.
Divide the numerator and denominator inside the square root by $x^4$:
$I = \int \frac{(x^2-1) dx}{x^3 \cdot x^2 \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} = \int \frac{(\frac{1}{x^3} - \frac{1}{x^5}) dx}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}$.
Let $t = 2-\frac{2}{x^2}+\frac{1}{x^4}$.
Then $dt = (\frac{4}{x^3} - \frac{4}{x^5}) dx$,which implies $(\frac{1}{x^3} - \frac{1}{x^5}) dx = \frac{dt}{4}$.
Substituting these into the integral:
$I = \int \frac{dt/4}{\sqrt{t}} = \frac{1}{4} \int t^{-1/2} dt = \frac{1}{4} \cdot \frac{t^{1/2}}{1/2} + c = \frac{1}{2} \sqrt{t} + c$.
Substituting $t$ back,we get $I = \frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}} + c$.

(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{\frac{3}{4}}}$.
We can rewrite the integral as $I = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{\frac{3}{4}}} = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{\frac{3}{4}}}$.
Let $t = 1 + \frac{1}{x^4}$. Then $dt = -\frac{4}{x^5} dx$,which implies $\frac{dx}{x^5} = -\frac{1}{4} dt$.
Substituting these into the integral,we get:
$I = -\frac{1}{4} \int t^{-\frac{3}{4}} dt$.
Integrating with respect to $t$,we have:
$I = -\frac{1}{4} \cdot \frac{t^{\frac{1}{4}}}{\frac{1}{4}} + c = -t^{\frac{1}{4}} + c$.
Substituting back $t = 1 + \frac{1}{x^4}$,we get:
$I = -(1 + \frac{1}{x^4})^{\frac{1}{4}} + c = -\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}} + c$.

(D) Let $I = \int \frac{\log (\cot x)}{\sin 2 x} \,d x$.
Put $\log (\cot x) = t$.
Differentiating both sides with respect to $x$:
$\frac{1}{\cot x} \cdot (-\csc^2 x) \,d x = dt$.
$\Rightarrow \frac{\sin x}{\cos x} \cdot (-\frac{1}{\sin^2 x}) \,d x = dt$.
$\Rightarrow \frac{-1}{\sin x \cos x} \,d x = dt$.
Since $\sin 2x = 2 \sin x \cos x$, we have $\sin x \cos x = \frac{\sin 2x}{2}$.
Substituting this: $\frac{-1}{\frac{\sin 2x}{2}} \,d x = dt \Rightarrow \frac{-2}{\sin 2x} \,d x = dt \Rightarrow \frac{d x}{\sin 2x} = \frac{-dt}{2}$.
Now, substitute these into the integral:
$I = \int t \cdot (\frac{-dt}{2}) = -\frac{1}{2} \int t \,dt$.
$I = -\frac{1}{2} \cdot \frac{t^2}{2} + c = -\frac{1}{4} t^2 + c$.
Substituting $t = \log (\cot x)$ back:
$I = -\frac{1}{4} [\log (\cot x)]^2 + c$.

(C) Let $I = \int \frac{dx}{x \sqrt{1-x^3}}$.
Multiply the numerator and denominator by $x^2$:
$I = \int \frac{x^2 dx}{x^3 \sqrt{1-x^3}}$.
Let $1-x^3 = t^2$,then $-3x^2 dx = 2t dt$,so $x^2 dx = -\frac{2}{3} t dt$.
Also,$x^3 = 1-t^2$.
Substituting these into the integral:
$I = \int \frac{-\frac{2}{3} t dt}{(1-t^2) t} = -\frac{2}{3} \int \frac{dt}{1-t^2} = \frac{2}{3} \int \frac{dt}{t^2-1}$.
Using the formula $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + c$:
$I = \frac{2}{3} \times \frac{1}{2} \log \left|\frac{t-1}{t+1}\right| + c = \frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right| + c$.
Comparing this with the given expression,we find $k = \frac{1}{3}$.

(B) Let $I = \int \frac{e^x(1+x)}{\cos^2(e^x \cdot x)} dx$.
Substitute $t = e^x \cdot x$.
Then,by the product rule,$dt = (e^x \cdot x + e^x \cdot 1) dx = e^x(x+1) dx$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{\cos^2 t} = \int \sec^2 t dt$.
The integral of $\sec^2 t$ is $\tan t + c$.
Therefore,$I = \tan(x \cdot e^x) + c$.

(A) Let $I = \int x \sqrt{\frac{2 \sin \left(x^2+1\right)-\sin 2\left(x^2+1\right)}{2 \sin \left(x^2+1\right)+\sin 2\left(x^2+1\right)}} \, dx$
Using $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$I = \int x \sqrt{\frac{2 \sin \left(x^2+1\right) - 2 \sin \left(x^2+1\right) \cos \left(x^2+1\right)}{2 \sin \left(x^2+1\right) + 2 \sin \left(x^2+1\right) \cos \left(x^2+1\right)}} \, dx$
$I = \int x \sqrt{\frac{1 - \cos \left(x^2+1\right)}{1 + \cos \left(x^2+1\right)}} \, dx$
Using $1 - \cos \theta = 2 \sin^2(\theta/2)$ and $1 + \cos \theta = 2 \cos^2(\theta/2)$:
$I = \int x \sqrt{\frac{2 \sin^2 \left(\frac{x^2+1}{2}\right)}{2 \cos^2 \left(\frac{x^2+1}{2}\right)}} \, dx$
$I = \int x \tan \left(\frac{x^2+1}{2}\right) \, dx$
Let $t = \frac{x^2+1}{2}$,then $dt = x \, dx$.
$I = \int \tan t \, dt = \log |\sec t| + c$
$I = \log \left| \sec \left(\frac{x^2+1}{2}\right) \right| + c$

(B) Let $I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^5 x + \cos ^3 x \sin ^2 x + \sin ^3 x \cos ^2 x + \cos ^5 x)^2} \,dx$.
Factor the denominator: $\sin ^5 x + \sin ^3 x \cos ^2 x + \cos ^3 x \sin ^2 x + \cos ^5 x = \sin ^3 x(\sin ^2 x + \cos ^2 x) + \cos ^3 x(\sin ^2 x + \cos ^2 x) = \sin ^3 x + \cos ^3 x$.
Thus, $I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^3 x + \cos ^3 x)^2} \,dx$.
Divide the numerator and denominator by $\cos ^6 x$: $I = \int \frac{\tan ^2 x \sec ^2 x}{(1 + \tan ^3 x)^2} \,dx$.
Let $t = 1 + \tan ^3 x$. Then $dt = 3 \tan ^2 x \sec ^2 x \,dx$, which implies $\tan ^2 x \sec ^2 x \,dx = \frac{1}{3} dt$.
Substituting these into the integral: $I = \frac{1}{3} \int \frac{1}{t^2} \,dt = \frac{1}{3} (-t^{-1}) + c = -\frac{1}{3t} + c$.
Substituting $t$ back: $I = \frac{-1}{3(1 + \tan ^3 x)} + c$.

(A) Let $I = \int \frac{\operatorname{cosec} x \, dx}{\cos^2(1 + \log \tan \frac{x}{2})}$.
Let $t = 1 + \log(\tan \frac{x}{2})$.
Differentiating both sides with respect to $x$,we get:
$dt = \frac{1}{\tan \frac{x}{2}} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} \, dx$.
Using the identity $\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$ and $\sec^2 \frac{x}{2} = \frac{1}{\cos^2 \frac{x}{2}}$,we have:
$dt = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \cdot \frac{1}{\cos^2 \frac{x}{2}} \cdot \frac{1}{2} \, dx = \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \, dx$.
Since $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get $dt = \frac{1}{\sin x} \, dx = \operatorname{cosec} x \, dx$.
Substituting these into the integral:
$I = \int \frac{1}{\cos^2 t} \, dt = \int \sec^2 t \, dt$.
Integrating,we get $I = \tan t + c$.
Substituting back the value of $t$,we get $I = \tan(1 + \log(\tan \frac{x}{2})) + c$.

(B) Let $I = \int \left( \frac{\tan \left( \frac{1}{x} \right)}{x} \right)^2 \, dx$.
Substitute $t = \frac{1}{x}$,then $dt = -\frac{1}{x^2} \, dx$,which implies $dx = -x^2 \, dt = -\frac{1}{t^2} \, dt$.
Substituting these into the integral:
$I = \int \tan^2(t) \cdot \frac{1}{x^2} \cdot (-x^2 \, dt) = -\int \tan^2(t) \, dt$.
Using the identity $\tan^2(t) = \sec^2(t) - 1$:
$I = -\int (\sec^2(t) - 1) \, dt = \int (1 - \sec^2(t)) \, dt$.
Integrating term by term:
$I = t - \tan(t) + c$.
Substituting back $t = \frac{1}{x}$:
$I = \frac{1}{x} - \tan \left( \frac{1}{x} \right) + c$.

(A) Let $I = \int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \,d x$.
Since $\sin 2x = 2 \sin x \cos x$,we have $I = \int \frac{1}{\cos ^3 x \sqrt{2 \sin x \cos x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{1}{\cos ^3 x \sqrt{\sin x} \sqrt{\cos x}} \,d x$.
This simplifies to $I = \frac{1}{\sqrt{2}} \int \frac{1}{\cos^{3.5} x \sqrt{\sin x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{\sec^{3.5} x}{\sqrt{\tan x}} \,d x$.
Wait,let us rewrite the integrand as $\frac{\sec^2 x \cdot \sec^2 x}{\sqrt{2 \sin x \cos x}} = \frac{\sec^2 x \cdot \sec^2 x}{\sqrt{2 \tan x \cos^2 x}} = \frac{\sec^4 x}{\sqrt{2 \tan x}}$.
So,$I = \frac{1}{\sqrt{2}} \int \frac{\sec^4 x}{\sqrt{\tan x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{(1+\tan^2 x) \sec^2 x}{\sqrt{\tan x}} \,d x$.
Let $\tan x = t$,then $\sec^2 x \,d x = dt$.
$I = \frac{1}{\sqrt{2}} \int \frac{1+t^2}{\sqrt{t}} \,dt = \frac{1}{\sqrt{2}} \int (t^{-1/2} + t^{3/2}) \,dt$.
$I = \frac{1}{\sqrt{2}} \left( \frac{t^{1/2}}{1/2} + \frac{t^{5/2}}{5/2} \right) + c = \frac{1}{\sqrt{2}} \left( 2\sqrt{t} + \frac{2}{5} t^{5/2} \right) + c$.
$I = \sqrt{2} \sqrt{t} + \frac{\sqrt{2}}{5} t^{5/2} + c = \sqrt{2} \left( \sqrt{\tan x} + \frac{1}{5} (\tan x)^{5/2} \right) + c$.

(B) Substitute $x = \tan \theta$,then $dx = \sec^2 \theta \, d\theta$.
Then,$f(x) = \int \frac{\tan^2 \theta \sec^2 \theta \, d\theta}{\sec^2 \theta(1+\sec \theta)} = \int \frac{\tan^2 \theta \, d\theta}{1+\sec \theta}$.
Since $\tan^2 \theta = \sec^2 \theta - 1$,we have $\int \frac{\sec^2 \theta - 1}{1+\sec \theta} \, d\theta = \int (\sec \theta - 1) \, d\theta$.
Integrating gives $f(x) = \log |\sec \theta + \tan \theta| - \theta + c$.
Substituting back $x = \tan \theta$ and $\sec \theta = \sqrt{1+x^2}$,we get $f(x) = \log |x + \sqrt{1+x^2}| - \tan^{-1} x + c$.
Given $f(0) = 0$,we have $\log |0 + 1| - \tan^{-1}(0) + c = 0$,which implies $c = 0$.
Thus,$f(x) = \log |x + \sqrt{1+x^2}| - \tan^{-1} x$.
Evaluating at $x=1$,$f(1) = \log |1 + \sqrt{2}| - \tan^{-1}(1) = \log (1+\sqrt{2}) - \frac{\pi}{4}$.

(A) Let $I = \int \frac{2+\cos \frac{x}{2}}{x+\sin \frac{x}{2}} \,d x$.
Substitute $t = x + \sin \frac{x}{2}$.
Differentiating both sides with respect to $x$,we get:
$\frac{dt}{dx} = 1 + \cos \frac{x}{2} \cdot \frac{1}{2} = 1 + \frac{1}{2} \cos \frac{x}{2} = \frac{2 + \cos \frac{x}{2}}{2}$.
Therefore,$(2 + \cos \frac{x}{2}) dx = 2 dt$.
Substituting these into the integral:
$I = \int \frac{2}{t} dt = 2 \int \frac{1}{t} dt$.
Integrating,we get:
$I = 2 \log |t| + c$.
Substituting back $t = x + \sin \frac{x}{2}$:
$I = 2 \log \left| x + \sin \frac{x}{2} \right| + c$.

(B) Let $t = \sin^2 x + \cos^3 x$.
Then,differentiating with respect to $x$,we get:
$dt = (2 \sin x \cos x + 3 \cos^2 x (-\sin x)) \, dx$
$dt = (\sin 2x - 3 \cos^2 x \sin x) \, dx$
$dt = \sin 2x (1 - \frac{3}{2} \cos x) \, dx$
Substituting this into the integral:
$\int \frac{1}{e^t} \, dt = \int e^{-t} \, dt$
$= -e^{-t} + c$
$= -e^{-(\sin^2 x + \cos^3 x)} + c$.

(A) Let $I = \int \frac{dx}{\cos x \sqrt{\cos 2x}}$.
We know that $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$.
So,$I = \int \frac{dx}{\cos x \sqrt{\frac{1-\tan^2 x}{1+\tan^2 x}}} = \int \frac{dx}{\cos x \frac{\sqrt{1-\tan^2 x}}{\sec x}}$.
Since $\frac{1}{\cos x} = \sec x$,we have $I = \int \frac{\sec x \cdot \sec x}{\sqrt{1-\tan^2 x}} \, dx = \int \frac{\sec^2 x}{\sqrt{1-\tan^2 x}} \, dx$.
Let $\tan x = t$,then $\sec^2 x \, dx = dt$.
Substituting these into the integral,we get $I = \int \frac{dt}{\sqrt{1-t^2}}$.
Using the standard integral formula $\int \frac{dt}{\sqrt{1-t^2}} = \sin^{-1}(t) + C$,we get $I = \sin^{-1}(\tan x) + C$.

(D) Let $I = \int \frac{3 x^{13}+2 x^{11}}{\left(2 x^4+3 x^2+1\right)^4} d x$.
Divide the numerator and denominator by $x^{16}$ inside the integral:
$I = \int \frac{3 x^{-3}+2 x^{-5}}{\left(2+3 x^{-2}+x^{-4}\right)^4} d x$.
Let $u = 2+3 x^{-2}+x^{-4}$.
Then $du = (-6 x^{-3}-4 x^{-5}) d x = -2(3 x^{-3}+2 x^{-5}) d x$.
So,$(3 x^{-3}+2 x^{-5}) d x = -\frac{1}{2} du$.
Substituting these into the integral:
$I = \int \frac{-1/2}{u^4} du = -\frac{1}{2} \int u^{-4} du = -\frac{1}{2} \left( \frac{u^{-3}}{-3} \right) + C = \frac{1}{6 u^3} + C$.
Substituting $u$ back:
$I = \frac{1}{6(2+3 x^{-2}+x^{-4})^3} + C = \frac{1}{6(\frac{2x^4+3x^2+1}{x^4})^3} + C = \frac{x^{12}}{6(2 x^4+3 x^2+1)^3} + C$.

(B) Given $f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 1 + 2x^7)^2} dx$.
Divide the numerator and denominator by $x^{14}$ inside the integral:
$f(x) = \int \frac{5x^{-6} + 7x^{-8}}{(x^{-5} + x^{-7} + 2)^2} dx$.
Let $t = x^{-5} + x^{-7} + 2$. Then $dt = (-5x^{-6} - 7x^{-8}) dx$,which implies $-(5x^{-6} + 7x^{-8}) dx = dt$.
Substituting this into the integral:
$f(x) = -\int \frac{dt}{t^2} = \frac{1}{t} + C = \frac{1}{x^{-5} + x^{-7} + 2} + C$.
Simplifying the expression:
$f(x) = \frac{x^7}{1 + x^2 + 2x^7} + C$.
Given $f(0) = 0$,we have $0 = \frac{0}{1} + C$,so $C = 0$.
Thus,$f(x) = \frac{x^7}{x^2 + 1 + 2x^7}$.
Evaluating at $x = 1$:
$f(1) = \frac{1^7}{1^2 + 1 + 2(1)^7} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$.

(A) Given $f(x) = \sqrt{\tan x}$ and $g(x) = \sin x \cdot \cos x$.
We need to evaluate the integral $I = \int \frac{f(x)}{g(x)} dx$.
$I = \int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} dx$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sqrt{\tan x}}{\frac{\sin x \cdot \cos x}{\cos^2 x}} \cdot \frac{1}{\cos^2 x} dx = \int \frac{\sqrt{\tan x}}{\tan x} \cdot \sec^2 x dx$.
$I = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
$I = \int \frac{1}{\sqrt{u}} du = \int u^{-1/2} du = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C$.
Substituting back $u = \tan x$,we get $I = 2\sqrt{\tan x} + C$.

(C) Let $I = \int(3-x) \sqrt{4-x} \, dx$.
Substitute $u = 4-x$,then $du = -dx$,which means $dx = -du$.
Also,$x = 4-u$,so $3-x = 3-(4-u) = u-1$.
Substituting these into the integral:
$I = \int (u-1) \sqrt{u} (-du) = \int (1-u) \sqrt{u} \, du$
$I = \int (u^{1/2} - u^{3/2}) \, du$
$I = \frac{u^{3/2}}{3/2} - \frac{u^{5/2}}{5/2} + C$
$I = \frac{2}{3} u^{3/2} - \frac{2}{5} u^{5/2} + C$
Substituting $u = 4-x$ back:
$I = \frac{2}{3} (4-x)^{3/2} - \frac{2}{5} (4-x)^{5/2} + C$.

(A) Let $I = \int \frac{x}{\sqrt{1-2 x^4}} \, dx$.
Substitute $u = \sqrt{2} x^2$.
Then $du = 2 \sqrt{2} x \, dx$,which implies $x \, dx = \frac{du}{2 \sqrt{2}}$.
Substituting these into the integral:
$I = \int \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{2 \sqrt{2}} = \frac{1}{2 \sqrt{2}} \int \frac{du}{\sqrt{1-u^2}}$.
Using the standard integral formula $\int \frac{1}{\sqrt{1-u^2}} \, du = \sin^{-1}(u) + C$,we get:
$I = \frac{1}{2 \sqrt{2}} \sin^{-1}(u) + C = \frac{1}{2 \sqrt{2}} \sin^{-1}(\sqrt{2} x^2) + C$.

(B) Let $I = \int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x)^2}$.
Factor the denominator: $\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x = \sin ^2 x(\sin ^3 x+\cos ^3 x) + \cos ^2 x(\sin ^3 x+\cos ^3 x) = (\sin ^2 x+\cos ^2 x)(\sin ^3 x+\cos ^3 x) = \sin ^3 x+\cos ^3 x$.
Thus,$I = \int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^3 x+\cos ^3 x)^2}$.
Divide the numerator and denominator by $\cos ^6 x$: $I = \int \frac{\tan ^2 x \sec ^2 x \,dx}{(\tan ^3 x+1)^2}$.
Let $u = \tan ^3 x+1$,then $du = 3 \tan ^2 x \sec ^2 x \,dx$,so $\tan ^2 x \sec ^2 x \,dx = \frac{du}{3}$.
Substituting these into the integral: $I = \int \frac{1}{u^2} \cdot \frac{du}{3} = \frac{1}{3} \int u^{-2} \,du = \frac{1}{3} \left(\frac{u^{-1}}{-1}\right) + C = -\frac{1}{3u} + C$.
Substituting back $u = \tan ^3 x+1$: $I = \frac{-1}{3(1+\tan ^3 x)}+C$.

(D) Let $I = \int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} d x$.
Factor the denominator: $\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x = \sin ^2 x(\sin ^3 x + \cos ^3 x) + \cos ^2 x(\sin ^3 x + \cos ^3 x) = (\sin ^2 x + \cos ^2 x)(\sin ^3 x + \cos ^3 x) = (\sin ^3 x + \cos ^3 x)$.
Thus,$I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^3 x + \cos ^3 x)^2} d x$.
Divide the numerator and denominator by $\cos ^6 x$: $I = \int \frac{\tan ^2 x \cdot \sec ^2 x}{(\tan ^3 x + 1)^2} d x$.
Let $t = \tan ^3 x + 1$,then $dt = 3 \tan ^2 x \sec ^2 x d x$,so $\tan ^2 x \sec ^2 x d x = \frac{dt}{3}$.
Substituting these into the integral: $I = \int \frac{1}{t^2} \cdot \frac{dt}{3} = \frac{1}{3} \int t^{-2} dt = \frac{1}{3} \left( \frac{t^{-1}}{-1} \right) + C = -\frac{1}{3t} + C$.
Substituting back $t = \tan ^3 x + 1$,we get $I = \frac{-1}{3(1 + \tan ^3 x)} + C$.

(B) Let $I = \int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \,d x$.
Substitute $t = \cot ^{-1}\left(x^3\right)$.
Then,the derivative is $\frac{dt}{dx} = -\frac{1}{1+(x^3)^2} \cdot 3x^2 = -\frac{3x^2}{1+x^6}$.
This implies $\frac{x^2}{1+x^6} dx = -\frac{1}{3} dt$.
Substituting these into the integral:
$I = 4 \int t \left(-\frac{1}{3} dt\right) = -\frac{4}{3} \int t \,dt$.
Integrating with respect to $t$:
$I = -\frac{4}{3} \cdot \frac{t^2}{2} + C = -\frac{2}{3} t^2 + C$.
Substituting back $t = \cot ^{-1}\left(x^3\right)$:
$I = -\frac{2}{3} \left(\cot ^{-1} x^3\right)^2 + C$.

(C) Let $I = \int \frac{x+1}{\sqrt{2x-1}} \, dx$.
Substitute $2x-1 = t$,which implies $x = \frac{t+1}{2}$ and $dx = \frac{1}{2} dt$.
$I = \int \frac{\frac{t+1}{2} + 1}{\sqrt{t}} \cdot \frac{1}{2} \, dt$
$I = \frac{1}{2} \int \frac{\frac{t+3}{2}}{\sqrt{t}} \, dt = \frac{1}{4} \int \left( \frac{t}{\sqrt{t}} + \frac{3}{\sqrt{t}} \right) \, dt$
$I = \frac{1}{4} \int (t^{1/2} + 3t^{-1/2}) \, dt$
$I = \frac{1}{4} \left( \frac{t^{3/2}}{3/2} + 3 \cdot \frac{t^{1/2}}{1/2} \right) + C$
$I = \frac{1}{4} \left( \frac{2}{3} t^{3/2} + 6 t^{1/2} \right) + C$
$I = \left( \frac{1}{6} t^{3/2} + \frac{3}{2} t^{1/2} \right) + C$
$I = \frac{1}{6} t^{1/2} (t + 9) + C$
Substituting $t = 2x-1$ back:
$I = \frac{1}{6} \sqrt{2x-1} (2x-1+9) + C$
$I = \frac{1}{6} \sqrt{2x-1} (2x+8) + C$
$I = \frac{1}{6} \cdot 2(x+4) \sqrt{2x-1} + C = \frac{1}{3}(x+4) \sqrt{2x-1} + C$
Comparing with $f(x) \sqrt{2x-1} + C$,we get $f(x) = \frac{1}{3}(x+4)$.

(B) We have the integral $I = \int \frac{d x}{3-2 \cos 2 x}$.
Using the identity $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$,we get:
$I = \int \frac{d x}{3-2 \left(\frac{1-\tan^2 x}{1+\tan^2 x}\right)} = \int \frac{1+\tan^2 x}{3(1+\tan^2 x) - 2(1-\tan^2 x)} \,d x$.
Since $1+\tan^2 x = \sec^2 x$,the integral becomes:
$I = \int \frac{\sec^2 x}{3+3\tan^2 x - 2 + 2\tan^2 x} \,d x = \int \frac{\sec^2 x}{5\tan^2 x + 1} \,d x$.
Let $t = \tan x$,then $dt = \sec^2 x \,d x$.
Substituting these into the integral:
$I = \int \frac{dt}{5t^2 + 1} = \int \frac{dt}{(\sqrt{5}t)^2 + 1^2}$.
Using the standard formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = \frac{1}{\sqrt{5}} \tan^{-1}(\sqrt{5}t) + C$.
Substituting $t = \tan x$ back,we get:
$I = \frac{1}{\sqrt{5}} \tan^{-1}(\sqrt{5} \tan x) + C$.

(A) To evaluate the integral $I = \int \cos \sqrt{x} \, dx$,we use the substitution method.
Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$.
Substituting these into the integral,we get:
$I = \int \cos(t) \cdot (2t \, dt) = 2 \int t \cos t \, dt$.
Now,we apply integration by parts,where $\int u \, dv = uv - \int v \, du$.
Let $u = t$ and $dv = \cos t \, dt$. Then $du = dt$ and $v = \sin t$.
$I = 2 [t \sin t - \int \sin t \, dt]$.
$I = 2 [t \sin t - (-\cos t)] + C$.
$I = 2 [t \sin t + \cos t] + C$.
Substituting $t = \sqrt{x}$ back into the expression:
$I = 2[\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + C$.

(A) Let $t = \log \left(x+\sqrt{1+x^2}\right)$.
Then,differentiating both sides with respect to $x$,we get:
$\frac{dt}{dx} = \frac{1}{x+\sqrt{1+x^2}} \cdot \left(1 + \frac{1}{2\sqrt{1+x^2}} \cdot 2x\right) = \frac{1}{x+\sqrt{1+x^2}} \cdot \left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right) = \frac{1}{\sqrt{1+x^2}}$.
Thus,$dt = \frac{dx}{\sqrt{1+x^2}}$.
Substituting these into the integral:
$\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \,dx = \int t \,dt = \frac{t^2}{2} + C$.
Comparing this with the given expression $\frac{1}{2}(g(x))^2 + C$,we have:
$\frac{1}{2}(g(x))^2 = \frac{1}{2} \left(\log \left(x+\sqrt{1+x^2}\right)\right)^2$.
Therefore,$g(x) = \log \left(x+\sqrt{1+x^2}\right)$.

(A) Let $I = \int \frac{x^3}{\sqrt{1+x^2}} dx = \int \frac{x^2 \cdot x}{\sqrt{1+x^2}} dx$.
Substitute $t = \sqrt{1+x^2}$,so $t^2 = 1+x^2$ and $x^2 = t^2 - 1$.
Differentiating,$2t dt = 2x dx$,which implies $x dx = t dt$.
Substituting these into the integral: $I = \int \frac{(t^2 - 1) t dt}{t} = \int (t^2 - 1) dt$.
Integrating,we get $I = \frac{t^3}{3} - t + c$.
Substituting back $t = \sqrt{1+x^2}$,we get $I = \frac{(1+x^2)^{\frac{3}{2}}}{3} - \sqrt{1+x^2} + c$.
Comparing this with the given expression $a(1+x^2)^{\frac{3}{2}} + b \sqrt{1+x^2} + c$,we find $a = \frac{1}{3}$ and $b = -1$.
Therefore,$a+b = \frac{1}{3} - 1 = \frac{-2}{3}$.

(B) Let $I = \int \frac{dx}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}$.
Substitute $x^{\frac{1}{6}} = t$,then $x = t^6$ and $dx = 6t^5 \, dt$.
Also,$x^{\frac{1}{2}} = t^3$ and $x^{\frac{1}{3}} = t^2$.
Substituting these into the integral:
$I = \int \frac{6t^5 \, dt}{t^3 + t^2} = 6 \int \frac{t^5}{t^2(t+1)} \, dt = 6 \int \frac{t^3}{t+1} \, dt$.
Using polynomial division or algebraic manipulation:
$I = 6 \int \frac{(t^3 + 1) - 1}{t+1} \, dt = 6 \int \left( \frac{(t+1)(t^2 - t + 1)}{t+1} - \frac{1}{t+1} \right) \, dt$.
$I = 6 \int (t^2 - t + 1) \, dt - 6 \int \frac{1}{t+1} \, dt$.
$I = 6 \left( \frac{t^3}{3} - \frac{t^2}{2} + t \right) - 6 \log |t+1| + c$.
$I = 2t^3 - 3t^2 + 6t - 6 \log |t+1| + c$.
Substituting back $t = x^{\frac{1}{6}}$:
$I = 2 \sqrt{x} - 3 \sqrt[3]{x} + 6 \sqrt[6]{x} - 6 \log |\sqrt[6]{x} + 1| + c$.

(A) Let $I = \int \frac{\sec^8 x}{\operatorname{cosec} x} \, dx$.
Since $\frac{1}{\operatorname{cosec} x} = \sin x$,we have:
$I = \int \sec^8 x \cdot \sin x \, dx$.
We can rewrite this as:
$I = \int \sec^7 x \cdot (\sec x \tan x) \, dx$.
Let $t = \sec x$,then $dt = \sec x \tan x \, dx$.
Substituting these into the integral:
$I = \int t^7 \, dt$.
Integrating with respect to $t$:
$I = \frac{t^8}{8} + c$.
Wait,let us re-evaluate the substitution:
$I = \int \sec^7 x (\sec x \tan x) \, dx = \frac{\sec^8 x}{8} + c$.
Upon re-checking the original expression $\int \frac{\sec^8 x}{\operatorname{cosec} x} \, dx = \int \sec^7 x \tan x \, dx$,the integral is $\frac{\sec^8 x}{8} + c$.

(C) Let $I = \int \frac{\tan ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x$.
Substitute $\sqrt{x} = t$,then $\frac{1}{2\sqrt{x}} dx = dt$,which implies $\frac{dx}{\sqrt{x}} = 2 dt$.
Substituting these into the integral,we get $I = 2 \int \tan ^4 t \cdot \sec ^2 t dt$.
Now,let $u = \tan t$,then $du = \sec ^2 t dt$.
Substituting $u$ into the integral,we get $I = 2 \int u^4 du$.
Integrating $u^4$ with respect to $u$,we get $I = 2 \cdot \frac{u^5}{5} + c = \frac{2}{5} u^5 + c$.
Substituting back $u = \tan t$ and $t = \sqrt{x}$,we get $I = \frac{2}{5} \tan ^5 \sqrt{x} + c$.

(A) Given the integral $I = \int \cos ^3 x \cdot e^{\log (\sin x)} d x$.
Since $e^{\log (f(x))} = f(x)$,the integral simplifies to:
$I = \int \cos ^3 x \cdot \sin x d x$.
Let $u = \cos x$.
Then $du = -\sin x d x$,which implies $\sin x d x = -du$.
Substituting these into the integral:
$I = \int u^3 (-du) = -\int u^3 du$.
Integrating $u^3$ gives $\frac{u^4}{4}$.
Thus,$I = -\frac{u^4}{4} + c$.
Substituting back $u = \cos x$,we get:
$I = -\frac{\cos ^4 x}{4} + c$.

(C) Let $I = \int \frac{\cos x - \sin x}{8 - \sin 2x} dx$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$. However,it is easier to write $8 - \sin 2x = 9 - (1 + \sin 2x) = 9 - (\sin x + \cos x)^2$.
So,$I = \int \frac{\cos x - \sin x}{3^2 - (\sin x + \cos x)^2} dx$.
Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$.
Substituting these into the integral,we get $I = \int \frac{dt}{3^2 - t^2}$.
Using the standard formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + c$,we have:
$I = \frac{1}{2(3)} \log \left| \frac{3+t}{3-t} \right| + c = \frac{1}{6} \log \left| \frac{3 + \sin x + \cos x}{3 - (\sin x + \cos x)} \right| + c$.
Comparing this with the given expression $\frac{1}{p} \log \left[ \frac{3 + \sin x + \cos x}{3 - \sin x - \cos x} \right] + c$,we find $p = 6$.

(B) Let $I = \int \frac{10^{\frac{x}{2}}}{\sqrt{10^{-x}-10^x}} dx$
We can rewrite the denominator as:
$\sqrt{10^{-x}-10^x} = \sqrt{\frac{1}{10^x} - 10^x} = \sqrt{\frac{1-(10^x)^2}{10^x}} = \frac{\sqrt{1-(10^x)^2}}{10^{\frac{x}{2}}}$
Substituting this into the integral:
$I = \int \frac{10^{\frac{x}{2}}}{\frac{\sqrt{1-(10^x)^2}}{10^{\frac{x}{2}}}} dx = \int \frac{10^x}{\sqrt{1-(10^x)^2}} dx$
Let $t = 10^x$. Then $dt = 10^x (\log 10) dx$,which implies $10^x dx = \frac{dt}{\log 10}$.
Substituting $t$ into the integral:
$I = \frac{1}{\log 10} \int \frac{dt}{\sqrt{1-t^2}} = \frac{1}{\log 10} \sin^{-1}(t) + c$
Replacing $t$ with $10^x$:
$I = \frac{1}{\log 10} \sin^{-1}(10^x) + c$