Integration by Parts Questions in English

(D) Let $I = \int \cos 2\theta \log \left( \frac{\cos \theta + \sin \theta }{\cos \theta - \sin \theta } \right) d\theta$.
First,simplify the logarithmic term: $\log \left( \frac{\cos \theta + \sin \theta }{\cos \theta - \sin \theta } \right) = \log \left( \frac{1 + \tan \theta }{1 - \tan \theta } \right) = \log \tan \left( \frac{\pi }{4} + \theta \right)$.
Let $u = \log \tan \left( \frac{\pi }{4} + \theta \right)$ and $dv = \cos 2\theta d\theta$.
Then $du = \frac{1}{\tan(\frac{\pi}{4} + \theta)} \cdot \sec^2(\frac{\pi}{4} + \theta) d\theta = \frac{\cos(\frac{\pi}{4} + \theta)}{\sin(\frac{\pi}{4} + \theta)} \cdot \frac{1}{\cos^2(\frac{\pi}{4} + \theta)} d\theta = \frac{1}{\sin(\frac{\pi}{4} + \theta)\cos(\frac{\pi}{4} + \theta)} d\theta = \frac{2}{\sin(\frac{\pi}{2} + 2\theta)} d\theta = 2 \sec 2\theta d\theta$.
Also,$v = \int \cos 2\theta d\theta = \frac{1}{2} \sin 2\theta$.
Using integration by parts $\int u dv = uv - \int v du$:
$I = \frac{1}{2} \sin 2\theta \log \tan \left( \frac{\pi }{4} + \theta \right) - \int \left( \frac{1}{2} \sin 2\theta \right) (2 \sec 2\theta) d\theta$.
$I = \frac{1}{2} \sin 2\theta \log \tan \left( \frac{\pi }{4} + \theta \right) - \int \tan 2\theta d\theta$.
$I = \frac{1}{2} \sin 2\theta \log \tan \left( \frac{\pi }{4} + \theta \right) - \frac{1}{2} \log |\sec 2\theta| + C$.

(C) Using integration by parts for $u = \int e^{ax} \cos bx \, dx$:
$u = e^{ax} \frac{\sin bx}{b} - \int a e^{ax} \frac{\sin bx}{b} \, dx = \frac{e^{ax} \sin bx}{b} - \frac{a}{b}v$
Multiplying by $b$,we get $bu + av = e^{ax} \sin bx$ ... $(i)$
Similarly,for $v = \int e^{ax} \sin bx \, dx$:
$v = e^{ax} \left( -\frac{\cos bx}{b} \right) - \int a e^{ax} \left( -\frac{\cos bx}{b} \right) \, dx = -\frac{e^{ax} \cos bx}{b} + \frac{a}{b}u$
Multiplying by $b$,we get $bv - au = -e^{ax} \cos bx$ ... $(ii)$
Squaring and adding equations $(i)$ and $(ii)$:
$(bu + av)^2 + (bv - au)^2 = (e^{ax} \sin bx)^2 + (-e^{ax} \cos bx)^2$
$b^2u^2 + a^2v^2 + 2abuv + b^2v^2 + a^2u^2 - 2abuv = e^{2ax} (\sin^2 bx + \cos^2 bx)$
$(a^2 + b^2)u^2 + (a^2 + b^2)v^2 = e^{2ax}(1)$
$(a^2 + b^2)(u^2 + v^2) = e^{2ax}$

(A) Given ${I_n} = \int {{(\log x)}^n} \, dx$.
Using integration by parts,let $u = {(\log x)^n}$ and $dv = dx$.
Then $du = n{(\log x)}^{n - 1} \cdot \frac{1}{x} \, dx$ and $v = x$.
Applying the formula $\int u \, dv = uv - \int v \, du$:
${I_n} = x{(\log x)^n} - \int x \cdot n{(\log x)}^{n - 1} \cdot \frac{1}{x} \, dx$
${I_n} = x{(\log x)^n} - n \int {{(\log x)}^{n - 1}} \, dx$
Since ${I_{n - 1}} = \int {{(\log x)}^{n - 1}} \, dx$,we have:
${I_n} = x{(\log x)^n} - n{I_{n - 1}}$
Rearranging the terms,we get:
${I_n} + n{I_{n - 1}} = x{(\log x)^n}$.

(B) To evaluate the integral $I = \int_1^e \log x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \log x$ and $dv = dx$.
Then $du = \frac{1}{x} \, dx$ and $v = x$.
Applying the formula:
$I = [x \log x]_1^e - \int_1^e x \cdot \frac{1}{x} \, dx$
$I = [x \log x]_1^e - \int_1^e 1 \, dx$
$I = [x \log x - x]_1^e$
Now,substitute the limits:
$I = (e \log e - e) - (1 \log 1 - 1)$
Since $\log e = 1$ and $\log 1 = 0$:
$I = (e(1) - e) - (0 - 1)$
$I = (e - e) - (-1)$
$I = 0 + 1 = 1$.

(B) Given ${I_m} = \int_1^x {(\log x)^m} dx$.
Using integration by parts,let $u = (\log x)^m$ and $dv = dx$.
Then $du = m(\log x)^{m-1} \cdot \frac{1}{x} dx$ and $v = x$.
${I_m} = [x(\log x)^m]_1^x - \int_1^x x \cdot m(\log x)^{m-1} \cdot \frac{1}{x} dx$.
${I_m} = x(\log x)^m - m \int_1^x (\log x)^{m-1} dx$.
${I_m} = x(\log x)^m - m{I_{m-1}}$.
Comparing this with the given relation ${I_m} = k - l{I_{m-1}}$,we get $k = x(\log x)^m$ and $l = m$.
Thus,the correct option is $l = m$.

(A) We are given $I(m, n) = \int_0^1 t^m (1 + t)^n dt$.
Using integration by parts,let $u = (1 + t)^n$ and $dv = t^m dt$.
Then $du = n(1 + t)^{n-1} dt$ and $v = \frac{t^{m+1}}{m+1}$.
Applying the formula $\int u dv = uv - \int v du$:
$I(m, n) = \left[ \frac{t^{m+1}(1 + t)^n}{m+1} \right]_0^1 - \int_0^1 \frac{t^{m+1}}{m+1} \cdot n(1 + t)^{n-1} dt$.
Evaluating the boundary terms:
At $t=1$,we get $\frac{1^{m+1}(1+1)^n}{m+1} = \frac{2^n}{m+1}$.
At $t=0$,the term is $0$.
So,$I(m, n) = \frac{2^n}{m+1} - \frac{n}{m+1} \int_0^1 t^{m+1}(1 + t)^{n-1} dt$.
Since $I(m+1, n-1) = \int_0^1 t^{m+1}(1 + t)^{n-1} dt$,we have:
$I(m, n) = \frac{2^n}{m+1} - \frac{n}{m+1} I(m+1, n-1)$.

(C) Given $\int f(x) dx = \varphi(x)$.
Let $I = \int x^5 f(x^3) dx$.
Substitute $t = x^3$,then $dt = 3x^2 dx$,which implies $x^2 dx = \frac{1}{3} dt$.
Also,$x^5 dx = x^3 \cdot x^2 dx = t \cdot \frac{1}{3} dt$.
So,$I = \int t f(t) \cdot \frac{1}{3} dt = \frac{1}{3} \int t f(t) dt$.
Using integration by parts,$\int u dv = uv - \int v du$,where $u = t$ and $dv = f(t) dt$ (so $du = dt$ and $v = \varphi(t)$).
$I = \frac{1}{3} [t \varphi(t) - \int \varphi(t) dt] + c$.
Substituting $t = x^3$ back into the expression:
$I = \frac{1}{3} [x^3 \varphi(x^3) - \int \varphi(x^3) \cdot (3x^2 dx)] + c$.
$I = \frac{1}{3} x^3 \varphi(x^3) - \int x^2 \varphi(x^3) dx + c$.

(A) Using integration by parts for both integrals:
$\int u \frac{d^2v}{dx^2} dx = u \frac{dv}{dx} - \int \frac{du}{dx} \frac{dv}{dx} dx$
$\int v \frac{d^2u}{dx^2} dx = v \frac{du}{dx} - \int \frac{dv}{dx} \frac{du}{dx} dx$
Subtracting the two expressions:
$\int u \frac{d^2v}{dx^2} dx - \int v \frac{d^2u}{dx^2} dx = (u \frac{dv}{dx} - \int \frac{du}{dx} \frac{dv}{dx} dx) - (v \frac{du}{dx} - \int \frac{dv}{dx} \frac{du}{dx} dx) + C$
$= u \frac{dv}{dx} - v \frac{du}{dx} - \int \frac{du}{dx} \frac{dv}{dx} dx + \int \frac{du}{dx} \frac{dv}{dx} dx + C$
$= u \frac{dv}{dx} - v \frac{du}{dx} + C$

(B) Let $I = \int x^{-2} \log(x^2 + a^2) \, dx$.
Using integration by parts,let $u = \log(x^2 + a^2)$ and $dv = x^{-2} \, dx$.
Then $du = \frac{2x}{x^2 + a^2} \, dx$ and $v = -\frac{1}{x}$.
Applying the formula $\int u \, dv = uv - \int v \, du$:
$I = -\frac{1}{x} \log(x^2 + a^2) - \int \left(-\frac{1}{x}\right) \left(\frac{2x}{x^2 + a^2}\right) \, dx$
$I = -\frac{1}{x} \log(x^2 + a^2) + \int \frac{2}{x^2 + a^2} \, dx$
$I = -\frac{1}{x} \log(x^2 + a^2) + 2 \left(\frac{1}{a} \tan^{-1} \frac{x}{a}\right) + c$
$I = -\frac{1}{x} \log(x^2 + a^2) + \frac{2}{a} \tan^{-1} \frac{x}{a} + c$.

(C) Let $I = \int \frac{\cot^{-1}(e^x)}{e^x} dx$.
Substitute $e^x = t$,so $e^x dx = dt$,which implies $dx = \frac{dt}{t}$.
Then $I = \int \frac{\cot^{-1}(t)}{t^2} dt$.
Using integration by parts,let $u = \cot^{-1}(t)$ and $dv = t^{-2} dt$. Then $du = -\frac{1}{1+t^2} dt$ and $v = -\frac{1}{t}$.
$I = -\frac{\cot^{-1}(t)}{t} - \int \left(-\frac{1}{t}\right) \left(-\frac{1}{1+t^2}\right) dt = -\frac{\cot^{-1}(t)}{t} - \int \frac{1}{t(1+t^2)} dt$.
Using partial fractions: $\frac{1}{t(1+t^2)} = \frac{1}{t} - \frac{t}{1+t^2}$.
So,$\int \frac{1}{t(1+t^2)} dt = \int \frac{1}{t} dt - \int \frac{t}{1+t^2} dt = \ln|t| - \frac{1}{2} \ln(1+t^2)$.
Substituting back into $I$: $I = -\frac{\cot^{-1}(t)}{t} - (\ln|t| - \frac{1}{2} \ln(1+t^2)) + C = \frac{1}{2} \ln(1+t^2) - \ln|t| - \frac{\cot^{-1}(t)}{t} + C$.
Since $t = e^x$,$\ln|t| = x$ and $t^2 = e^{2x}$.
$I = \frac{1}{2} \ln(1+e^{2x}) - x - \frac{\cot^{-1}(e^x)}{e^x} + C$.

(A) Let $I = \int \frac{x}{\sqrt{1 + x^2}} \ln(x + \sqrt{1 + x^2}) \, dx$.
Using Integration by Parts,let $u = \ln(x + \sqrt{1 + x^2})$ and $dv = \frac{x}{\sqrt{1 + x^2}} \, dx$.
Then $du = \frac{1}{x + \sqrt{1 + x^2}} \cdot (1 + \frac{x}{\sqrt{1 + x^2}}) \, dx = \frac{1}{x + \sqrt{1 + x^2}} \cdot \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}} \, dx = \frac{1}{\sqrt{1 + x^2}} \, dx$.
And $v = \int \frac{x}{\sqrt{1 + x^2}} \, dx = \sqrt{1 + x^2}$.
Applying the formula $\int u \, dv = uv - \int v \, du$:
$I = \sqrt{1 + x^2} \ln(x + \sqrt{1 + x^2}) - \int \sqrt{1 + x^2} \cdot \frac{1}{\sqrt{1 + x^2}} \, dx$.
$I = \sqrt{1 + x^2} \ln(x + \sqrt{1 + x^2}) - \int 1 \, dx$.
$I = \sqrt{1 + x^2} \ln(x + \sqrt{1 + x^2}) - x + c$.

(C) Let $I = \int \frac{\cot ^{-1}(e^x)}{e^x} dx$.
Substitute $e^x = t$,then $e^x dx = dt$,which implies $dx = \frac{dt}{t}$.
Therefore,$I = \int \frac{\cot ^{-1}(t)}{t^2} dt$.
Using integration by parts,let $u = \cot ^{-1}(t)$ and $dv = t^{-2} dt$.
Then $du = -\frac{1}{1+t^2} dt$ and $v = -\frac{1}{t}$.
$I = u v - \int v du = -\frac{\cot ^{-1}(t)}{t} - \int \left(-\frac{1}{t}\right) \left(-\frac{1}{1+t^2}\right) dt$.
$I = -\frac{\cot ^{-1}(t)}{t} - \int \frac{1}{t(1+t^2)} dt$.
Using partial fractions,$\frac{1}{t(1+t^2)} = \frac{1}{t} - \frac{t}{1+t^2}$.
So,$\int \frac{1}{t(1+t^2)} dt = \int \frac{1}{t} dt - \int \frac{t}{1+t^2} dt = \ln|t| - \frac{1}{2} \ln(1+t^2)$.
Substituting back,$I = -\frac{\cot ^{-1}(t)}{t} - (\ln|t| - \frac{1}{2} \ln(1+t^2)) + C$.
$I = \frac{1}{2} \ln(1+t^2) - \ln|t| - \frac{\cot ^{-1}(t)}{t} + C$.
Since $t = e^x$,$\ln|t| = x$.
$I = \frac{1}{2} \ln(1+e^{2x}) - x - \frac{\cot ^{-1}(e^x)}{e^x} + C$.

(B) We need to evaluate the integral $I = \int x \sin x \sec^3 x \, dx$.
First,simplify the integrand:
$I = \int x \tan x \sec^2 x \, dx$.
Using integration by parts,let $u = x$ and $dv = \tan x \sec^2 x \, dx$.
Then $du = dx$ and $v = \int \tan x \sec^2 x \, dx = \frac{\tan^2 x}{2}$.
Applying the formula $\int u \, dv = uv - \int v \, du$:
$I = x \cdot \frac{\tan^2 x}{2} - \int \frac{\tan^2 x}{2} \, dx$.
$I = \frac{x \tan^2 x}{2} - \frac{1}{2} \int (\sec^2 x - 1) \, dx$.
$I = \frac{x \tan^2 x}{2} - \frac{1}{2} (\tan x - x) + C$.
$I = \frac{x \tan^2 x}{2} - \frac{1}{2} \tan x + \frac{x}{2} + C$.
Since $\tan^2 x = \sec^2 x - 1$:
$I = \frac{x(\sec^2 x - 1)}{2} - \frac{1}{2} \tan x + \frac{x}{2} + C$.
$I = \frac{x \sec^2 x}{2} - \frac{x}{2} - \frac{1}{2} \tan x + \frac{x}{2} + C$.
$I = \frac{1}{2} (x \sec^2 x - \tan x) + C$.

(A) We are given the integral $I = \int [(1 - 2x)\cos x + (3 + 2x)\sin x] dx$.
Using integration by parts,let $u = (1 - 2x)$ and $dv = \cos x dx$,then $du = -2 dx$ and $v = \sin x$.
$\int (1 - 2x)\cos x dx = (1 - 2x)\sin x - \int \sin x (-2) dx = (1 - 2x)\sin x - 2\cos x$.
Now,for the second part,let $u = (3 + 2x)$ and $dv = \sin x dx$,then $du = 2 dx$ and $v = -\cos x$.
$\int (3 + 2x)\sin x dx = (3 + 2x)(-\cos x) - \int (-\cos x)(2) dx = -(3 + 2x)\cos x + 2\sin x$.
Adding both parts:
$I = (1 - 2x)\sin x - 2\cos x - (3 + 2x)\cos x + 2\sin x + C$.
$I = \sin x(1 - 2x + 2) + \cos x(-2 - 3 - 2x) + C$.
$I = (3 - 2x)\sin x + (-5 - 2x)\cos x + C$.
Comparing this with $f(x)\sin x + g(x)\cos x + C$,we get $f(x) = 3 - 2x$ and $g(x) = -5 - 2x$.
Thus,option $A$ is correct.

(A) Let $I = \int x \cos^{-1} \left( \frac{1-x^2}{1+x^2} \right) dx$.
Since $x > 0$,we use the substitution $x = \tan \theta$,which implies $\theta = \tan^{-1} x$.
Then $\cos^{-1} \left( \frac{1-x^2}{1+x^2} \right) = \cos^{-1} (\cos 2\theta) = 2\theta = 2\tan^{-1} x$.
Thus,$I = \int x (2\tan^{-1} x) dx = 2 \int x \tan^{-1} x dx$.
Using integration by parts,let $u = \tan^{-1} x$ and $dv = x dx$. Then $du = \frac{1}{1+x^2} dx$ and $v = \frac{x^2}{2}$.
$I = 2 \left[ \frac{x^2}{2} \tan^{-1} x - \int \frac{x^2}{2(1+x^2)} dx \right] + c$.
$I = x^2 \tan^{-1} x - \int \frac{x^2+1-1}{1+x^2} dx + c$.
$I = x^2 \tan^{-1} x - \int \left( 1 - \frac{1}{1+x^2} \right) dx + c$.
$I = x^2 \tan^{-1} x - x + \tan^{-1} x + c$.
$I = (x^2 + 1) \tan^{-1} x - x + c$.

(B) Let $I = \int x^5 e^{-4x^3} dx$.
Substitute $t = -4x^3$,then $dt = -12x^2 dx$,which implies $x^2 dx = -\frac{1}{12} dt$.
Also,$x^3 = -\frac{t}{4}$.
Substituting these into the integral:
$I = \int x^3 \cdot e^{-4x^3} \cdot x^2 dx = \int (-\frac{t}{4}) e^t (-\frac{1}{12} dt) = \frac{1}{48} \int t e^t dt$.
Using integration by parts: $\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t + C_1$.
So,$I = \frac{1}{48} (t e^t - e^t) + C = \frac{1}{48} e^t (t - 1) + C$.
Substituting $t = -4x^3$ back:
$I = \frac{1}{48} e^{-4x^3} (-4x^3 - 1) + C$.
Comparing this with the given form $\frac{1}{48} e^{-4x^3} f(x) + C$,we get $f(x) = -4x^3 - 1$.

(C) Let $I = \int \cos(\log_e x) dx$.
Using integration by parts,let $u = \cos(\log_e x)$ and $dv = dx$.
Then $du = -\sin(\log_e x) \cdot \frac{1}{x} dx$ and $v = x$.
$I = x \cos(\log_e x) - \int x \left( -\sin(\log_e x) \cdot \frac{1}{x} \right) dx$
$I = x \cos(\log_e x) + \int \sin(\log_e x) dx$.
Now,apply integration by parts to $\int \sin(\log_e x) dx$:
Let $u = \sin(\log_e x)$ and $dv = dx$.
Then $du = \cos(\log_e x) \cdot \frac{1}{x} dx$ and $v = x$.
$\int \sin(\log_e x) dx = x \sin(\log_e x) - \int x \left( \cos(\log_e x) \cdot \frac{1}{x} \right) dx = x \sin(\log_e x) - I$.
Substituting this back into the equation for $I$:
$I = x \cos(\log_e x) + x \sin(\log_e x) - I$
$2I = x(\cos(\log_e x) + \sin(\log_e x))$
$I = \frac{x}{2}(\cos(\log_e x) + \sin(\log_e x)) + C$.

(C) Let $t = x^2$,then $dt = 2x dx$,which implies $x dx = \frac{1}{2} dt$.
We can rewrite the integral as $\int x^4 \cdot e^{-x^2} \cdot x dx = \int t^2 \cdot e^{-t} \cdot \frac{1}{2} dt = \frac{1}{2} \int t^2 e^{-t} dt$.
Using integration by parts $\int u dv = uv - \int v du$,let $u = t^2$ and $dv = e^{-t} dt$,so $du = 2t dt$ and $v = -e^{-t}$.
$\frac{1}{2} \int t^2 e^{-t} dt = \frac{1}{2} [ -t^2 e^{-t} - \int (-e^{-t}) 2t dt ] = \frac{1}{2} [ -t^2 e^{-t} + 2 \int t e^{-t} dt ]$.
Applying integration by parts again for $\int t e^{-t} dt$ with $u = t$ and $dv = e^{-t} dt$:
$\int t e^{-t} dt = -t e^{-t} - \int (-e^{-t}) dt = -t e^{-t} - e^{-t}$.
Substituting back: $\frac{1}{2} [ -t^2 e^{-t} + 2(-t e^{-t} - e^{-t}) ] = (-\frac{1}{2} t^2 - t - 1) e^{-t} + c$.
Substituting $t = x^2$: $g(x) = -\frac{x^4}{2} - x^2 - 1$.
Thus,$g(-1) = -\frac{(-1)^4}{2} - (-1)^2 - 1 = -\frac{1}{2} - 1 - 1 = -\frac{5}{2}$.

(N/A) To evaluate the integral $\int_{0}^{1} x e^{x} d x$,we use the method of integration by parts.
The formula for integration by parts is $\int u dv = uv - \int v du$.
Let $u = x$ and $dv = e^{x} dx$.
Then,$du = dx$ and $v = \int e^{x} dx = e^{x}$.
Applying the formula:
$\int x e^{x} dx = x e^{x} - \int e^{x} dx = x e^{x} - e^{x} = e^{x}(x - 1)$.
Now,we apply the limits from $0$ to $1$:
$\int_{0}^{1} x e^{x} dx = [e^{x}(x - 1)]_{0}^{1}$.
Evaluating at the upper limit $(x = 1)$:
$e^{1}(1 - 1) = e(0) = 0$.
Evaluating at the lower limit $(x = 0)$:
$e^{0}(0 - 1) = 1(-1) = -1$.
Subtracting the lower limit value from the upper limit value:
$0 - (-1) = 1$.
Thus,$\int_{0}^{1} x e^{x} dx = 1$.

(C) To evaluate $\int x \cos x \, dx$,we use the integration by parts formula: $\int f(x)g(x) \, dx = f(x) \int g(x) \, dx - \int [f'(x) \int g(x) \, dx] \, dx$.
Using the $LIATE$ rule,we choose $f(x) = x$ (algebraic) and $g(x) = \cos x$ (trigonometric).
Applying the formula:
$\int x \cos x \, dx = x \int \cos x \, dx - \int [\frac{d}{dx}(x) \int \cos x \, dx] \, dx$
$= x \sin x - \int (1) \sin x \, dx$
$= x \sin x - (-\cos x) + C$
$= x \sin x + \cos x + C$
If we had chosen $f(x) = \cos x$ and $g(x) = x$,the integral would become more complex,demonstrating the importance of the $LIATE$ rule.

(B) To evaluate the integral $\int \log x \, dx$,we use the method of integration by parts.
Integration by parts formula is given by $\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$.
Let $u = \log x$ and $v = 1$.
Then,$\frac{du}{dx} = \frac{1}{x}$ and $\int v \, dx = \int 1 \, dx = x$.
Substituting these into the formula:
$\int \log x \cdot 1 \, dx = (\log x) \cdot x - \int \left( \frac{1}{x} \cdot x \right) dx$
$= x \log x - \int 1 \, dx$
$= x \log x - x + C$.

(B) To evaluate the integral $\int x e^{x} d x$,we use the method of integration by parts,which states that $\int u v d x = u \int v d x - \int (\frac{du}{dx} \int v d x) d x$.
Let $u = x$ and $v = e^{x}$.
Then $\frac{du}{dx} = 1$ and $\int v d x = e^{x}$.
Applying the formula:
$\int x e^{x} d x = x e^{x} - \int (1 \cdot e^{x}) d x$
$= x e^{x} - \int e^{x} d x$
$= x e^{x} - e^{x} + C$
$= e^{x}(x - 1) + C$

(N/A) Let the first function be $u = \sin ^{-1} x$ and the second function be $dv = \frac{x}{\sqrt{1-x^{2}}} dx$.
First,we find the integral of the second function:
$\int \frac{x}{\sqrt{1-x^{2}}} dx$
Let $t = 1-x^{2}$,then $dt = -2x dx$,which implies $x dx = -\frac{1}{2} dt$.
$\int \frac{x}{\sqrt{1-x^{2}}} dx = -\frac{1}{2} \int t^{-1/2} dt = -\frac{1}{2} (2t^{1/2}) = -\sqrt{1-x^{2}}$.
Now,apply the integration by parts formula: $\int u dv = uv - \int v du$.
Here,$du = \frac{1}{\sqrt{1-x^{2}}} dx$.
$\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} dx = (\sin ^{-1} x)(-\sqrt{1-x^{2}}) - \int (-\sqrt{1-x^{2}}) \frac{1}{\sqrt{1-x^{2}}} dx$
$= -\sqrt{1-x^{2}} \sin ^{-1} x + \int 1 dx$
$= -\sqrt{1-x^{2}} \sin ^{-1} x + x + C$
$= x - \sqrt{1-x^{2}} \sin ^{-1} x + C$.

Let $I = \int e^{x} \sin x \, dx$. Using the integration by parts formula $\int u \, dv = uv - \int v \, du$,let $u = \sin x$ and $dv = e^{x} \, dx$. Then $du = \cos x \, dx$ and $v = e^{x}$.
$I = e^{x} \sin x - \int e^{x} \cos x \, dx$ ..........$(1)$
Now,evaluate $\int e^{x} \cos x \, dx$ using integration by parts again. Let $u = \cos x$ and $dv = e^{x} \, dx$. Then $du = -\sin x \, dx$ and $v = e^{x}$.
$\int e^{x} \cos x \, dx = e^{x} \cos x - \int e^{x} (-\sin x) \, dx = e^{x} \cos x + \int e^{x} \sin x \, dx$
Substituting this back into equation $(1)$:
$I = e^{x} \sin x - (e^{x} \cos x + I)$
$I = e^{x} \sin x - e^{x} \cos x - I$
$2I = e^{x} (\sin x - \cos x)$
$I = \frac{e^{x}}{2} (\sin x - \cos x) + C$

(A) Let $I = \int x \sin x \, dx$.
Using the integration by parts formula: $\int u v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$.
Taking $u = x$ (first function) and $v = \sin x$ (second function),we get:
$I = x \int \sin x \, dx - \int \left( \frac{d}{dx}(x) \int \sin x \, dx \right) dx$
$I = x(-\cos x) - \int (1 \cdot (-\cos x)) \, dx$
$I = -x \cos x + \int \cos x \, dx$
$I = -x \cos x + \sin x + C$,where $C$ is an arbitrary constant.

(N/A) Let $I = \int x \sin 3x \, dx$.
Using the integration by parts formula: $\int u v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$.
Taking $u = x$ (first function) and $v = \sin 3x$ (second function),we get:
$I = x \int \sin 3x \, dx - \int \left( \frac{d}{dx}(x) \int \sin 3x \, dx \right) dx$
$I = x \left( \frac{-\cos 3x}{3} \right) - \int 1 \cdot \left( \frac{-\cos 3x}{3} \right) dx$
$I = -\frac{x \cos 3x}{3} + \frac{1}{3} \int \cos 3x \, dx$
$I = -\frac{x \cos 3x}{3} + \frac{1}{3} \left( \frac{\sin 3x}{3} \right) + C$
$I = -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} + C$,where $C$ is an arbitrary constant.

Let $I = \int x^{2} e^{x} dx$.
Using the integration by parts formula,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Taking $u = x^{2}$ and $v = e^{x}$,we get:
$I = x^{2} \int e^{x} dx - \int (\frac{d}{dx} x^{2} \cdot \int e^{x} dx) dx$
$I = x^{2} e^{x} - \int 2x e^{x} dx$
$I = x^{2} e^{x} - 2 \int x e^{x} dx$.
Applying integration by parts again for $\int x e^{x} dx$ with $u = x$ and $v = e^{x}$:
$\int x e^{x} dx = x e^{x} - \int (1 \cdot e^{x}) dx = x e^{x} - e^{x}$.
Substituting this back into the expression for $I$:
$I = x^{2} e^{x} - 2(x e^{x} - e^{x}) + C$
$I = x^{2} e^{x} - 2x e^{x} + 2e^{x} + C$
$I = e^{x}(x^{2} - 2x + 2) + C$,where $C$ is an arbitrary constant.

Let $I = \int x \log x \, dx$.
Using the integration by parts formula: $\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$.
Taking $u = \log x$ (first function) and $v = x$ (second function) based on the $LIATE$ rule:
$I = \log x \int x \, dx - \int \left( \frac{d}{dx} \log x \cdot \int x \, dx \right) dx$
$I = \log x \cdot \frac{x^2}{2} - \int \left( \frac{1}{x} \cdot \frac{x^2}{2} \right) dx$
$I = \frac{x^2 \log x}{2} - \int \frac{x}{2} \, dx$
$I = \frac{x^2 \log x}{2} - \frac{1}{2} \cdot \frac{x^2}{2} + C$
$I = \frac{x^2 \log x}{2} - \frac{x^2}{4} + C$,where $C$ is the constant of integration.

Let $I = \int x \log(2x) \, dx$.
Using the integration by parts formula $\int u v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$,where $u = \log(2x)$ and $v = x$.
Then $u' = \frac{1}{2x} \cdot 2 = \frac{1}{x}$ and $\int v \, dx = \frac{x^2}{2}$.
Substituting these into the formula:
$I = \log(2x) \cdot \frac{x^2}{2} - \int \left( \frac{1}{x} \cdot \frac{x^2}{2} \right) \, dx$
$I = \frac{x^2 \log(2x)}{2} - \int \frac{x}{2} \, dx$
$I = \frac{x^2 \log(2x)}{2} - \frac{1}{2} \cdot \frac{x^2}{2} + C$
$I = \frac{x^2 \log(2x)}{2} - \frac{x^2}{4} + C$,where $C$ is an arbitrary constant.

(N/A) Let $I = \int x^{2} \log x \, dx$.
Using the integration by parts formula $\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$,where we choose $u = \log x$ (first function) and $v = x^{2}$ (second function) based on the $LIATE$ rule:
$I = \log x \int x^{2} \, dx - \int \left( \frac{d}{dx}(\log x) \int x^{2} \, dx \right) dx$
$I = \log x \left( \frac{x^{3}}{3} \right) - \int \left( \frac{1}{x} \cdot \frac{x^{3}}{3} \right) dx$
$I = \frac{x^{3} \log x}{3} - \int \frac{x^{2}}{3} \, dx$
$I = \frac{x^{3} \log x}{3} - \frac{1}{3} \left( \frac{x^{3}}{3} \right) + C$
$I = \frac{x^{3} \log x}{3} - \frac{x^{3}}{9} + C$,where $C$ is an arbitrary constant.

Let $I = \int x \sin^{-1} x \, dx$.
Using integration by parts,where $\sin^{-1} x$ is the first function and $x$ is the second function:
$I = \sin^{-1} x \int x \, dx - \int \left( \frac{d}{dx} \sin^{-1} x \int x \, dx \right) dx$
$I = \sin^{-1} x \left( \frac{x^2}{2} \right) - \int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \, dx$
$I = \frac{x^2 \sin^{-1} x}{2} - \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx$
To solve the integral,rewrite the numerator as $x^2 = -(1-x^2) + 1$:
$I = \frac{x^2 \sin^{-1} x}{2} - \frac{1}{2} \int \frac{-(1-x^2) + 1}{\sqrt{1-x^2}} \, dx$
$I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \int \sqrt{1-x^2} \, dx - \frac{1}{2} \int \frac{1}{\sqrt{1-x^2}} \, dx$
Using standard integrals $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$ and $\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1} x$:
$I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \left( \frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin^{-1} x \right) - \frac{1}{2} \sin^{-1} x + C$
$I = \frac{x^2 \sin^{-1} x}{2} + \frac{x}{4} \sqrt{1-x^2} + \frac{1}{4} \sin^{-1} x - \frac{1}{2} \sin^{-1} x + C$
$I = \frac{1}{4} (2x^2 - 1) \sin^{-1} x + \frac{x}{4} \sqrt{1-x^2} + C$,where $C$ is an arbitrary constant.

Let $I = \int x \tan^{-1} x \, dx$.
Using integration by parts,where $\tan^{-1} x$ is the first function and $x$ is the second function:
$I = \tan^{-1} x \int x \, dx - \int \left( \frac{d}{dx} \tan^{-1} x \int x \, dx \right) dx$
$I = \tan^{-1} x \left( \frac{x^2}{2} \right) - \int \left( \frac{1}{1+x^2} \cdot \frac{x^2}{2} \right) dx$
$I = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx$
To integrate $\frac{x^2}{1+x^2}$,we rewrite the numerator as $(x^2+1-1)$:
$I = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \left( \frac{x^2+1}{1+x^2} - \frac{1}{1+x^2} \right) dx$
$I = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \left( 1 - \frac{1}{1+x^2} \right) dx$
$I = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} (x - \tan^{-1} x) + C$
$I = \frac{x^2}{2} \tan^{-1} x - \frac{x}{2} + \frac{1}{2} \tan^{-1} x + C$,where $C$ is the constant of integration.

Let $I = \int x \cos^{-1} x \, dx$.
Using integration by parts,where $\cos^{-1} x$ is the first function and $x$ is the second function:
$I = \cos^{-1} x \int x \, dx - \int \left( \frac{d}{dx} \cos^{-1} x \cdot \int x \, dx \right) dx$
$I = \cos^{-1} x \cdot \frac{x^2}{2} - \int \left( \frac{-1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \right) dx$
$I = \frac{x^2 \cos^{-1} x}{2} + \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx$
To solve $\int \frac{x^2}{\sqrt{1-x^2}} \, dx$,rewrite the numerator as $-(1-x^2) + 1$:
$I = \frac{x^2 \cos^{-1} x}{2} + \frac{1}{2} \int \frac{-(1-x^2) + 1}{\sqrt{1-x^2}} \, dx$
$I = \frac{x^2 \cos^{-1} x}{2} - \frac{1}{2} \int \sqrt{1-x^2} \, dx + \frac{1}{2} \int \frac{1}{\sqrt{1-x^2}} \, dx$
Using the standard integral $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right)$:
$I = \frac{x^2 \cos^{-1} x}{2} - \frac{1}{2} \left( \frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin^{-1} x \right) + \frac{1}{2} \sin^{-1} x + C$
$I = \frac{x^2 \cos^{-1} x}{2} - \frac{x}{4} \sqrt{1-x^2} + \frac{1}{4} \sin^{-1} x + C$

Let $I = \int (\sin^{-1} x)^2 \cdot 1 \, dx$.
Using integration by parts,where $u = (\sin^{-1} x)^2$ and $dv = dx$,we have $du = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}} \, dx$ and $v = x$.
$I = x(\sin^{-1} x)^2 - \int x \cdot \frac{2 \sin^{-1} x}{\sqrt{1-x^2}} \, dx$.
$= x(\sin^{-1} x)^2 + \int \sin^{-1} x \cdot \left( \frac{-2x}{\sqrt{1-x^2}} \right) \, dx$.
Now,apply integration by parts again for the second integral,where $u = \sin^{-1} x$ and $dv = \frac{-2x}{\sqrt{1-x^2}} \, dx$.
Then $du = \frac{1}{\sqrt{1-x^2}} \, dx$ and $v = 2\sqrt{1-x^2}$.
$I = x(\sin^{-1} x)^2 + \left[ \sin^{-1} x \cdot 2\sqrt{1-x^2} - \int \frac{1}{\sqrt{1-x^2}} \cdot 2\sqrt{1-x^2} \, dx \right]$.
$= x(\sin^{-1} x)^2 + 2\sqrt{1-x^2} \sin^{-1} x - \int 2 \, dx$.
$= x(\sin^{-1} x)^2 + 2\sqrt{1-x^2} \sin^{-1} x - 2x + C$,where $C$ is the constant of integration.

Let $I = \int \frac{x \cos^{-1} x}{\sqrt{1-x^{2}}} dx$.
We can rewrite the integral as $I = -\frac{1}{2} \int \frac{-2x}{\sqrt{1-x^{2}}} \cdot \cos^{-1} x dx$.
Using integration by parts,let $u = \cos^{-1} x$ and $dv = \frac{-2x}{\sqrt{1-x^{2}}} dx$.
Then $du = -\frac{1}{\sqrt{1-x^{2}}} dx$ and $v = 2\sqrt{1-x^{2}}$.
Using the formula $\int u dv = uv - \int v du$:
$I = -\frac{1}{2} \left[ \cos^{-1} x \cdot 2\sqrt{1-x^{2}} - \int 2\sqrt{1-x^{2}} \cdot \left( -\frac{1}{\sqrt{1-x^{2}}} \right) dx \right]$.
$I = -\frac{1}{2} \left[ 2\sqrt{1-x^{2}} \cos^{-1} x + \int 2 dx \right]$.
$I = -\frac{1}{2} \left[ 2\sqrt{1-x^{2}} \cos^{-1} x + 2x \right] + C$.
$I = -\sqrt{1-x^{2}} \cos^{-1} x - x + C$,where $C$ is the constant of integration.

(N/A) Let $I = \int x \sec^{2} x \, dx$.
Using the integration by parts formula $\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$,where $u = x$ and $v = \sec^{2} x$:
$I = x \int \sec^{2} x \, dx - \int \left( \frac{d}{dx}(x) \int \sec^{2} x \, dx \right) dx$
$I = x \tan x - \int (1 \cdot \tan x) \, dx$
$I = x \tan x - \int \tan x \, dx$
Since $\int \tan x \, dx = \ln |\sec x| + C$ or $-\ln |\cos x| + C$,we have:
$I = x \tan x + \ln |\cos x| + C$,where $C$ is an arbitrary constant.

Let $I = \int 1 \cdot \tan^{-1} x \, dx$.
Using integration by parts,where $\tan^{-1} x$ is the first function and $1$ is the second function:
$I = \tan^{-1} x \int 1 \, dx - \int \left( \frac{d}{dx} \tan^{-1} x \cdot \int 1 \, dx \right) dx$
$I = x \tan^{-1} x - \int \frac{1}{1+x^2} \cdot x \, dx$
$I = x \tan^{-1} x - \frac{1}{2} \int \frac{2x}{1+x^2} \, dx$
Since the derivative of $1+x^2$ is $2x$,we use the substitution $u = 1+x^2$,$du = 2x \, dx$:
$I = x \tan^{-1} x - \frac{1}{2} \log |1+x^2| + C$
Since $1+x^2 > 0$ for all real $x$,we can write:
$I = x \tan^{-1} x - \frac{1}{2} \log (1+x^2) + C$,where $C$ is an arbitrary constant.

Let $I = \int x(\log x)^{2} \, dx$.
Using integration by parts,where $\int u \, dv = uv - \int v \, du$.
Let $u = (\log x)^{2}$ and $dv = x \, dx$.
Then $du = 2 \log x \cdot \frac{1}{x} \, dx$ and $v = \frac{x^{2}}{2}$.
$I = (\log x)^{2} \cdot \frac{x^{2}}{2} - \int \frac{x^{2}}{2} \cdot 2 \log x \cdot \frac{1}{x} \, dx$
$I = \frac{x^{2}}{2}(\log x)^{2} - \int x \log x \, dx$.
Now,integrate $\int x \log x \, dx$ by parts again.
Let $u = \log x$ and $dv = x \, dx$.
Then $du = \frac{1}{x} \, dx$ and $v = \frac{x^{2}}{2}$.
$\int x \log x \, dx = \log x \cdot \frac{x^{2}}{2} - \int \frac{x^{2}}{2} \cdot \frac{1}{x} \, dx$
$= \frac{x^{2}}{2} \log x - \frac{1}{2} \int x \, dx = \frac{x^{2}}{2} \log x - \frac{x^{2}}{4}$.
Substituting this back into the expression for $I$:
$I = \frac{x^{2}}{2}(\log x)^{2} - \left( \frac{x^{2}}{2} \log x - \frac{x^{2}}{4} \right) + C$
$I = \frac{x^{2}}{2}(\log x)^{2} - \frac{x^{2}}{2} \log x + \frac{x^{2}}{4} + C$.

(N/A) Let $I = \int \left(x^{2}+1\right) \log x \, dx = \int x^{2} \log x \, dx + \int \log x \, dx$.
Let $I = I_{1} + I_{2}$ where $I_{1} = \int x^{2} \log x \, dx$ and $I_{2} = \int \log x \, dx$.
For $I_{1}$,using integration by parts with $\log x$ as the first function and $x^{2}$ as the second function:
$I_{1} = \log x \cdot \frac{x^{3}}{3} - \int \frac{1}{x} \cdot \frac{x^{3}}{3} \, dx = \frac{x^{3}}{3} \log x - \frac{1}{3} \int x^{2} \, dx = \frac{x^{3}}{3} \log x - \frac{x^{3}}{9} + C_{1}$.
For $I_{2}$,using integration by parts with $\log x$ as the first function and $1$ as the second function:
$I_{2} = \log x \cdot x - \int \frac{1}{x} \cdot x \, dx = x \log x - \int 1 \, dx = x \log x - x + C_{2}$.
Combining these results:
$I = \left( \frac{x^{3}}{3} \log x - \frac{x^{3}}{9} \right) + (x \log x - x) + C$
$I = \left( \frac{x^{3}}{3} + x \right) \log x - \frac{x^{3}}{9} - x + C$.

Let $I = \int e^{2x} \sin x \, dx$ ..........$(1)$
Integrating by parts,we use the formula $\int u \, dv = uv - \int v \, du$. Let $u = \sin x$ and $dv = e^{2x} \, dx$. Then $du = \cos x \, dx$ and $v = \frac{e^{2x}}{2}$.
$I = \sin x \cdot \frac{e^{2x}}{2} - \int \cos x \cdot \frac{e^{2x}}{2} \, dx$
$I = \frac{e^{2x} \sin x}{2} - \frac{1}{2} \int e^{2x} \cos x \, dx$
Again integrating by parts for $\int e^{2x} \cos x \, dx$,let $u = \cos x$ and $dv = e^{2x} \, dx$. Then $du = -\sin x \, dx$ and $v = \frac{e^{2x}}{2}$.
$I = \frac{e^{2x} \sin x}{2} - \frac{1}{2} \left[ \cos x \cdot \frac{e^{2x}}{2} - \int (-\sin x) \cdot \frac{e^{2x}}{2} \, dx \right]$
$I = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4} - \frac{1}{4} \int e^{2x} \sin x \, dx$
$I = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4} - \frac{1}{4} I$
$I + \frac{1}{4} I = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4}$
$\frac{5}{4} I = \frac{2e^{2x} \sin x - e^{2x} \cos x}{4}$
$I = \frac{e^{2x}}{5} (2 \sin x - \cos x) + C$,where $C$ is an arbitrary constant.

Let $x = \tan \theta$,then $dx = \sec^2 \theta \, d\theta$.
$\therefore \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) = \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right) = \sin ^{-1}(\sin 2 \theta) = 2 \theta$.
$\int \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) dx = \int 2 \theta \cdot \sec ^{2} \theta \, d\theta = 2 \int \theta \cdot \sec ^{2} \theta \, d\theta$.
Integrating by parts,we obtain:
$2 \left[ \theta \int \sec ^{2} \theta \, d\theta - \int \left( \frac{d}{d \theta} \theta \cdot \int \sec ^{2} \theta \, d\theta \right) d\theta \right]$
$= 2 [ \theta \tan \theta - \int \tan \theta \, d\theta ]$
$= 2 [ \theta \tan \theta + \log |\cos \theta| ] + C$
Since $\tan \theta = x$,we have $\theta = \tan^{-1} x$ and $\cos \theta = \frac{1}{\sqrt{1+x^2}}$.
$= 2 [ x \tan^{-1} x + \log |\frac{1}{\sqrt{1+x^2}}| ] + C$
$= 2 x \tan^{-1} x + 2 [ -\frac{1}{2} \log (1+x^2) ] + C$
$= 2 x \tan^{-1} x - \log (1+x^2) + C$,where $C$ is an arbitrary constant.

Let $I = \int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} \, dx$.
Substitute $x = \cos \theta$,so $dx = -\sin \theta \, d\theta$.
Then $I = \int \tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \cdot (-\sin \theta) \, d\theta$.
Using trigonometric identities $\frac{1-\cos \theta}{1+\cos \theta} = \tan^2 \frac{\theta}{2}$,we get:
$I = -\int \tan ^{-1} \left( \tan \frac{\theta}{2} \right) \sin \theta \, d\theta = -\int \frac{\theta}{2} \sin \theta \, d\theta$.
Using integration by parts:
$I = -\frac{1}{2} \left[ \theta \int \sin \theta \, d\theta - \int \left( \frac{d}{d\theta} \theta \cdot \int \sin \theta \, d\theta \right) d\theta \right]$.
$I = -\frac{1}{2} [-\theta \cos \theta + \int \cos \theta \, d\theta] = -\frac{1}{2} [-\theta \cos \theta + \sin \theta] + C$.
$I = \frac{1}{2} \theta \cos \theta - \frac{1}{2} \sin \theta + C$.
Since $\cos \theta = x$,then $\theta = \cos ^{-1} x$ and $\sin \theta = \sqrt{1-x^2}$.
Thus,$I = \frac{1}{2} x \cos ^{-1} x - \frac{1}{2} \sqrt{1-x^2} + C = \frac{1}{2} (x \cos ^{-1} x - \sqrt{1-x^2}) + C$.

(A) $I(x) = \int \frac{\sec^2 x}{\sin^{2022} x} dx - 2022 \int \frac{1}{\sin^{2022} x} dx$
Using integration by parts on the first integral,let $u = \sin^{-2022} x$ and $dv = \sec^2 x dx$. Then $du = -2022 \sin^{-2023} x \cos x dx$ and $v = \tan x$.
$I(x) = \tan x \sin^{-2022} x - \int \tan x (-2022 \sin^{-2023} x \cos x) dx - 2022 \int \sin^{-2022} x dx$
$I(x) = \tan x \sin^{-2022} x + 2022 \int \frac{\sin x}{\cos x} \cdot \frac{\cos x}{\sin^{2023} x} dx - 2022 \int \sin^{-2022} x dx$
$I(x) = \tan x \sin^{-2022} x + 2022 \int \sin^{-2022} x dx - 2022 \int \sin^{-2022} x dx + C$
$I(x) = \frac{\tan x}{\sin^{2022} x} + C$
Given $I(\pi/4) = 2^{1011}$,we have $\frac{1}{\sin^{2022}(\pi/4)} + C = 2^{1011} \implies (\sqrt{2})^{2022} + C = 2^{1011} \implies 2^{1011} + C = 2^{1011} \implies C = 0$.
Thus,$I(x) = \frac{\tan x}{\sin^{2022} x}$.
$I(\pi/6) = \frac{\tan(\pi/6)}{\sin^{2022}(\pi/6)} = \frac{1/\sqrt{3}}{(1/2)^{2022}} = \frac{2^{2022}}{\sqrt{3}}$.
$I(\pi/3) = \frac{\tan(\pi/3)}{\sin^{2022}(\pi/3)} = \frac{\sqrt{3}}{(\sqrt{3}/2)^{2022}} = \frac{\sqrt{3} \cdot 2^{2022}}{3^{1011}} = \frac{2^{2022}}{3^{1010.5}}$.
Comparing $I(\pi/3)$ and $I(\pi/6)$,we find $3^{1010} I(\pi/3) = I(\pi/6)$.

(A) Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = x^3$ and $dv = \sin x \, dx$. Then $du = 3x^2 \, dx$ and $v = -\cos x$.
$\int x^3 \sin x \, dx = -x^3 \cos x + \int 3x^2 \cos x \, dx$.
Applying integration by parts again for $\int 3x^2 \cos x \, dx$: $u = 3x^2, dv = \cos x \, dx \implies du = 6x \, dx, v = \sin x$.
$= -x^3 \cos x + 3x^2 \sin x - \int 6x \sin x \, dx$.
Applying integration by parts for $\int 6x \sin x \, dx$: $u = 6x, dv = \sin x \, dx \implies du = 6 \, dx, v = -\cos x$.
$= -x^3 \cos x + 3x^2 \sin x - (6x(-\cos x) - \int -6 \cos x \, dx) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C$.
Thus,$g(x) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x$.
$g\left(\frac{\pi}{2}\right) = -\left(\frac{\pi}{2}\right)^3 \cos\left(\frac{\pi}{2}\right) + 3\left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + 6\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) - 6 \sin\left(\frac{\pi}{2}\right) = 0 + \frac{3\pi^2}{4} + 0 - 6 = \frac{3\pi^2}{4} - 6$.
Since $g(x) = \int x^3 \sin x \, dx$,by the Fundamental Theorem of Calculus,$g^{\prime}(x) = x^3 \sin x$.
$g^{\prime}\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^3 \sin\left(\frac{\pi}{2}\right) = \frac{\pi^3}{8}$.
$8\left(g\left(\frac{\pi}{2}\right) + g^{\prime}\left(\frac{\pi}{2}\right)\right) = 8\left(\frac{3\pi^2}{4} - 6 + \frac{\pi^3}{8}\right) = 6\pi^2 - 48 + \pi^3 = 1\pi^3 + 6\pi^2 - 48$.
Comparing with $\alpha \pi^3 + \beta \pi^2 + \gamma$,we get $\alpha = 1, \beta = 6, \gamma = -48$.
$\alpha + \beta - \gamma = 1 + 6 - (-48) = 7 + 48 = 55$.

(C) Let $I = \int \cos (\log _e x) dx$.
Substitute $\log _e x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then,$I = \int e^t \cos t dt$.
Using the integration by parts formula $\int e^t f(t) dt = e^t f(t) - \int e^t f'(t) dt$,we have:
$I = e^t \cos t - \int e^t (-\sin t) dt = e^t \cos t + \int e^t \sin t dt$.
Applying integration by parts again to $\int e^t \sin t dt$:
$I = e^t \cos t + [e^t \sin t - \int e^t \cos t dt]$.
$I = e^t \cos t + e^t \sin t - I$.
$2I = e^t (\cos t + \sin t)$.
$I = \frac{e^t}{2} (\cos t + \sin t) + C$.
Substituting $e^t = x$ and $t = \log _e x$:
$I = \frac{x}{2} [\cos (\log _e x) + \sin (\log _e x)] + C$.

(A) Let $I = \int \sin \sqrt{x} \, dx$.
Substitute $\sqrt{x} = t$,then $x = t^2$,which implies $dx = 2t \, dt$.
Substituting these into the integral:
$I = \int \sin t \cdot (2t) \, dt = 2 \int t \sin t \, dt$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = \sin t \, dt$:
$I = 2 \left[ t(-\cos t) - \int 1 \cdot (-\cos t) \, dt \right]$
$I = 2 [ -t \cos t + \int \cos t \, dt ]$
$I = 2 [ -t \cos t + \sin t ] + c$
$I = 2 \sin t - 2t \cos t + c$.
Substituting back $t = \sqrt{x}$:
$I = 2 \sin \sqrt{x} - 2 \sqrt{x} \cos \sqrt{x} + c$.

(C) Given $\int f(x) dx = \psi(x)$.
Let $I = \int x^5 f(x^3) dx$.
We can rewrite this as $I = \int x^3 \cdot x^2 f(x^3) dx$.
Let $x^3 = t$,then $3x^2 dx = dt$,which implies $x^2 dx = \frac{1}{3} dt$.
Substituting these into the integral:
$I = \int t \cdot f(t) \cdot \frac{1}{3} dt = \frac{1}{3} \int t f(t) dt$.
Using integration by parts,$\int u dv = uv - \int v du$,where $u = t$ and $dv = f(t) dt$:
$I = \frac{1}{3} [t \int f(t) dt - \int (\frac{d}{dt}(t) \cdot \int f(t) dt) dt] + c$.
Since $\int f(t) dt = \psi(t)$:
$I = \frac{1}{3} [t \psi(t) - \int \psi(t) dt] + c$.
Substituting $t = x^3$ and $dt = 3x^2 dx$:
$I = \frac{1}{3} [x^3 \psi(x^3) - \int \psi(x^3) \cdot 3x^2 dx] + c$.
$I = \frac{1}{3} [x^3 \psi(x^3) - 3 \int x^2 \psi(x^3) dx] + c$.
$I = \frac{1}{3} x^3 \psi(x^3) - \int x^2 \psi(x^3) dx + c$.

(A) Let $I = \int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$.
Substitute $x = \tan t$,then $dx = \sec^2 t dt$.
The integral becomes $I = \int \sin^{-1}(\sin 2t) \sec^2 t dt = \int 2t \sec^2 t dt$.
Using integration by parts: $\int u dv = uv - \int v du$,where $u = 2t$ and $dv = \sec^2 t dt$.
$I = 2t \tan t - \int 2 \tan t dt = 2t \tan t - 2 \log |\sec t| + c$.
Since $\sec t = \sqrt{1 + \tan^2 t} = \sqrt{1 + x^2}$,we have $\log |\sec t| = \log (1 + x^2)^{1/2} = \frac{1}{2} \log (1 + x^2)$.
Substituting back: $I = 2x \tan^{-1} x - 2(\frac{1}{2} \log (1 + x^2)) + c = 2x \tan^{-1} x - \log (1 + x^2) + c$.

(C) Let $I = \int \cos (\log x) d x$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then,$I = \int e^t \cos t dt$.
Using the integration by parts formula $\int e^t f(t) dt = e^t \int f(t) dt - \int (e^t \int f(t) dt) dt$ or the standard result $\int e^t (\cos t + \sin t) dt = e^t \cos t + C$,we apply integration by parts:
$I = \cos t \cdot e^t - \int (-\sin t) \cdot e^t dt = e^t \cos t + \int e^t \sin t dt$.
Applying integration by parts again to $\int e^t \sin t dt$:
$I = e^t \cos t + [\sin t \cdot e^t - \int \cos t \cdot e^t dt] = e^t \cos t + e^t \sin t - I$.
Thus,$2I = e^t (\cos t + \sin t) + c_1$.
$I = \frac{e^t}{2} (\cos t + \sin t) + c$.
Substituting back $t = \log x$ and $e^t = x$:
$I = \frac{x}{2} (\cos (\log x) + \sin (\log x)) + c$.

(B) Let $I = \int x^5 e^{-4 x^3} \,d x$.
Substitute $x^3 = t$,then $3x^2 \,d x = dt$,which implies $x^2 \,d x = \frac{1}{3} dt$.
Since $x^5 = x^3 \cdot x^2$,the integral becomes $I = \frac{1}{3} \int t e^{-4 t} dt$.
Using integration by parts,$\int u v \,dt = u \int v \,dt - \int (u' \int v \,dt) dt$,where $u = t$ and $v = e^{-4 t}$.
$I = \frac{1}{3} \left( t \cdot \frac{e^{-4 t}}{-4} - \int 1 \cdot \frac{e^{-4 t}}{-4} dt \right)$.
$I = \frac{1}{3} \left( -\frac{t e^{-4 t}}{4} + \frac{1}{4} \int e^{-4 t} dt \right)$.
$I = \frac{1}{3} \left( -\frac{t e^{-4 t}}{4} - \frac{e^{-4 t}}{16} \right) + c$.
$I = -\frac{t e^{-4 t}}{12} - \frac{e^{-4 t}}{48} + c$.
$I = \frac{e^{-4 t}}{48} (-4 t - 1) + c$.
Substituting $t = x^3$ back,we get $I = \frac{1}{48} e^{-4 x^3} (-4 x^3 - 1) + c$.
Comparing this with the given form,$f(x) = -4 x^3 - 1$.