Definite integration by substitution Questions in English

(B) Let $I = \int_{\pi /3}^{\pi /2} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}} \,dx$.
Multiply the numerator and denominator by $\sqrt{1 - \cos x}$:
$I = \int_{\pi /3}^{\pi /2} \frac{\sqrt{1 - \cos^2 x}}{(1 - \cos x)^{5/2} \sqrt{1 - \cos x}} \,dx = \int_{\pi /3}^{\pi /2} \frac{\sin x}{(1 - \cos x)^3} \,dx$.
Let $1 - \cos x = t$,then $\sin x \,dx = dt$.
When $x = \frac{\pi}{3}$,$t = 1 - \cos(\frac{\pi}{3}) = 1 - \frac{1}{2} = \frac{1}{2}$.
When $x = \frac{\pi}{2}$,$t = 1 - \cos(\frac{\pi}{2}) = 1 - 0 = 1$.
Substituting these into the integral:
$I = \int_{1/2}^{1} t^{-3} \,dt = \left[ \frac{t^{-2}}{-2} \right]_{1/2}^{1} = -\frac{1}{2} \left[ \frac{1}{t^2} \right]_{1/2}^{1}$.
$I = -\frac{1}{2} \left( 1 - 4 \right) = -\frac{1}{2} (-3) = \frac{3}{2}$.

(C) We are given the integral $I = \int_0^{\pi /4} \sec^7 \theta \sin^3 \theta \, d\theta$.
Rewrite the integrand as:
$I = \int_0^{\pi /4} \frac{\sin^3 \theta}{\cos^3 \theta} \cdot \sec^4 \theta \, d\theta = \int_0^{\pi /4} \tan^3 \theta \sec^4 \theta \, d\theta$.
Since $\sec^4 \theta = \sec^2 \theta \cdot \sec^2 \theta = (1 + \tan^2 \theta) \sec^2 \theta$,the integral becomes:
$I = \int_0^{\pi /4} \tan^3 \theta (1 + \tan^2 \theta) \sec^2 \theta \, d\theta$.
Let $t = \tan \theta$,then $dt = \sec^2 \theta \, d\theta$.
When $\theta = 0, t = 0$. When $\theta = \pi /4, t = 1$.
Substituting these into the integral:
$I = \int_0^1 t^3 (1 + t^2) \, dt = \int_0^1 (t^3 + t^5) \, dt$.
Integrating term by term:
$I = \left[ \frac{t^4}{4} + \frac{t^6}{6} \right]_0^1 = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}$.

(A) Let $I = \int_{\pi /4}^{\pi /2} \cos \theta \csc^2 \theta \, d\theta$.
We can rewrite the integrand as $\int_{\pi /4}^{\pi /2} \frac{\cos \theta}{\sin^2 \theta} \, d\theta$.
Let $t = \sin \theta$,then $dt = \cos \theta \, d\theta$.
When $\theta = \pi /4$,$t = \sin(\pi /4) = 1/\sqrt{2}$.
When $\theta = \pi /2$,$t = \sin(\pi /2) = 1$.
Substituting these into the integral,we get:
$I = \int_{1/\sqrt{2}}^{1} \frac{1}{t^2} \, dt = \int_{1/\sqrt{2}}^{1} t^{-2} \, dt$.
Evaluating the integral:
$I = \left[ -\frac{1}{t} \right]_{1/\sqrt{2}}^{1} = \left( -\frac{1}{1} \right) - \left( -\frac{1}{1/\sqrt{2}} \right)$.
$I = -1 + \sqrt{2} = \sqrt{2} - 1$.

(D) Let $I = \int_0^1 \frac{\tan^{-1} x}{1 + x^2} \,dx$.
Substitute $t = \tan^{-1} x$.
Then,$dt = \frac{1}{1 + x^2} \,dx$.
When $x = 0$,$t = \tan^{-1}(0) = 0$.
When $x = 1$,$t = \tan^{-1}(1) = \frac{\pi}{4}$.
Substituting these into the integral,we get:
$I = \int_0^{\pi/4} t \,dt = \left[ \frac{t^2}{2} \right]_0^{\pi/4} = \frac{1}{2} \left( \left( \frac{\pi}{4} \right)^2 - 0^2 \right) = \frac{1}{2} \cdot \frac{\pi^2}{16} = \frac{\pi^2}{32}$.

(B) Let $t = \log x$. Then,$dt = \frac{1}{x} dx$.
When $x = 1$,$t = \log 1 = 0$.
When $x = 2$,$t = \log 2$.
Substituting these into the integral,we get:
$\int_{1}^{2} \frac{\cos(\log x)}{x} \, dx = \int_{0}^{\log 2} \cos(t) \, dt$.
Evaluating the integral:
$\int_{0}^{\log 2} \cos(t) \, dt = [\sin(t)]_{0}^{\log 2} = \sin(\log 2) - \sin(0) = \sin(\log 2) - 0 = \sin(\log 2)$.
Thus,the correct option is $B$.

(D) Let $t = \frac{1}{x}$.
Then,$dt = -\frac{1}{x^2} \,dx$,which implies $-\,dt = \frac{1}{x^2} \,dx$.
When $x = \frac{1}{\pi}$,$t = \pi$.
When $x = \frac{2}{\pi}$,$t = \frac{\pi}{2}$.
Substituting these into the integral:
$\int_{1/\pi}^{2/\pi} \frac{\sin(1/x)}{x^2} \,dx = \int_{\pi}^{\pi/2} \sin(t) \cdot (-dt)$
$= \int_{\pi/2}^{\pi} \sin(t) \,dt$
$= [-\cos(t)]_{\pi/2}^{\pi}$
$= -(\cos(\pi) - \cos(\pi/2))$
$= -(-1 - 0) = 1$.

(D) Let $I = \int_0^{\pi /2} \frac{\sin x \cos x}{1 + \sin^4 x} \, dx$.
Substitute $\sin^2 x = t$.
Then,$2 \sin x \cos x \, dx = dt$,which implies $\sin x \cos x \, dx = \frac{1}{2} \, dt$.
When $x = 0$,$t = \sin^2(0) = 0$.
When $x = \frac{\pi}{2}$,$t = \sin^2(\frac{\pi}{2}) = 1$.
Substituting these into the integral:
$I = \int_0^1 \frac{1}{1 + t^2} \cdot \frac{1}{2} \, dt = \frac{1}{2} \int_0^1 \frac{1}{1 + t^2} \, dt$.
$I = \frac{1}{2} [\tan^{-1} t]_0^1$.
$I = \frac{1}{2} (\tan^{-1}(1) - \tan^{-1}(0)) = \frac{1}{2} (\frac{\pi}{4} - 0) = \frac{\pi}{8}$.

(A) Let $t = \tan x$.
Then,$dt = \sec^2 x \, dx$.
When $x = 0$,$t = \tan(0) = 0$.
When $x = \frac{\pi}{4}$,$t = \tan(\frac{\pi}{4}) = 1$.
Substituting these into the integral:
$\int_0^1 t^6 \, dt = \left[ \frac{t^7}{7} \right]_0^1 = \frac{1}{7}(1^7 - 0^7) = \frac{1}{7}$.
Thus,the correct option is $A$.

(B) Let $I = \int_0^{\pi /6} \frac{\sin x}{\cos^3 x} \, dx = \int_0^{\pi /6} \tan x \sec^2 x \, dx$.
Substitute $t = \tan x$.
Then $dt = \sec^2 x \, dx$.
When $x = 0$,$t = \tan(0) = 0$.
When $x = \pi/6$,$t = \tan(\pi/6) = 1/\sqrt{3}$.
Thus,$I = \int_0^{1/\sqrt{3}} t \, dt = \left[ \frac{t^2}{2} \right]_0^{1/\sqrt{3}} = \frac{1}{2} \left( \frac{1}{\sqrt{3}} \right)^2 - 0 = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$.

(B) Let $I = \int_0^{\log 5} {\frac{{{e^x}\sqrt {{e^x} - 1} }}{{{e^x} + 3}}} \,dx$.
Substitute ${e^x} - 1 = {t^2}$,so ${e^x} = {t^2} + 1$.
Differentiating both sides,${e^x}dx = 2t\,dt$.
When $x = 0$,${t^2} = {e^0} - 1 = 0$,so $t = 0$.
When $x = \log 5$,${t^2} = {e^{\log 5}} - 1 = 5 - 1 = 4$,so $t = 2$.
Substituting these into the integral:
$I = \int_0^2 {\frac{{2t \cdot t}}{{{t^2} + 1 + 3}}} \,dt = \int_0^2 {\frac{{2{t^2}}}{{{t^2} + 4}}} \,dt$.
$I = 2 \int_0^2 {\frac{{{t^2} + 4 - 4}}{{{t^2} + 4}}} \,dt = 2 \left[ {\int_0^2 {1\,dt - 4\int_0^2 {\frac{{dt}}{{{t^2} + 2^2}}} } } \right]$.
$I = 2 \left[ {t|_0^2 - 4 \cdot \frac{1}{2} \tan^{-1}(\frac{t}{2})|_0^2} \right]$.
$I = 2 \left[ {2 - 2 \tan^{-1}(1)} \right] = 2 \left[ {2 - 2(\frac{\pi}{4})} \right] = 4 - \pi$.

(A) Let $I = \int_1^e \frac{1 + \log x}{x} \, dx$.
Substitute $u = \log x$,then $du = \frac{1}{x} \, dx$.
When $x = 1$,$u = \log 1 = 0$.
When $x = e$,$u = \log e = 1$.
Thus,$I = \int_0^1 (1 + u) \, du$.
$I = [u + \frac{u^2}{2}]_0^1$.
$I = (1 + \frac{1}{2}) - (0 + 0) = \frac{3}{2}$.

(A) Let $I = \int_0^2 \frac{3^{\sqrt{x}}}{\sqrt{x}} \, dx$.
Substitute $\sqrt{x} = t$.
Then,$\frac{1}{2\sqrt{x}} \, dx = dt$,which implies $\frac{1}{\sqrt{x}} \, dx = 2 \, dt$.
When $x = 0$,$t = 0$.
When $x = 2$,$t = \sqrt{2}$.
Substituting these into the integral,we get:
$I = \int_0^{\sqrt{2}} 3^t \cdot 2 \, dt$
$I = 2 \int_0^{\sqrt{2}} 3^t \, dt$
Using the formula $\int a^x \, dx = \frac{a^x}{\log a} + C$,we have:
$I = 2 \left[ \frac{3^t}{\log 3} \right]_0^{\sqrt{2}}$
$I = \frac{2}{\log 3} (3^{\sqrt{2}} - 3^0)$
$I = \frac{2}{\log 3} (3^{\sqrt{2}} - 1)$.

(B) Let $I = \int_0^{\pi /2} \frac{\sin x}{1 + \cos^2 x} \, dx$.
Substitute $\cos x = t$. Then,$-\sin x \, dx = dt$,or $\sin x \, dx = -dt$.
Change the limits of integration:
When $x = 0$,$t = \cos(0) = 1$.
When $x = \pi /2$,$t = \cos(\pi /2) = 0$.
Substituting these into the integral:
$I = \int_1^0 \frac{-dt}{1 + t^2} = \int_0^1 \frac{dt}{1 + t^2}$.
Using the standard integral $\int \frac{dt}{1 + t^2} = \tan^{-1} t$,we get:
$I = [\tan^{-1} t]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

(D) Let $1 + \tan x = t$. Then,$\sec^2 x \,dx = dt$.
When $x = 0$,$t = 1 + \tan(0) = 1$.
When $x = \pi/4$,$t = 1 + \tan(\pi/4) = 2$.
The integral becomes $\int_1^2 \frac{dt}{t(t+1)}$.
Using partial fractions,$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$.
Thus,$\int_1^2 \left( \frac{1}{t} - \frac{1}{t+1} \right) dt = [\log_e |t| - \log_e |t+1|]_1^2$.
$= [\log_e |\frac{t}{t+1}|]_1^2 = \log_e \left( \frac{2}{3} \right) - \log_e \left( \frac{1}{2} \right) = \log_e \left( \frac{2/3}{1/2} \right) = \log_e \left( \frac{4}{3} \right)$.

(D) Let $I = \int_0^a {{x^2}\sin {x^3}\,dx}$.
Substitute $t = x^3$,then $dt = 3x^2\,dx$,which implies $x^2\,dx = \frac{dt}{3}$.
When $x = 0$,$t = 0^3 = 0$. When $x = a$,$t = a^3$.
Substituting these into the integral:
$I = \int_0^{a^3} \sin(t) \cdot \frac{dt}{3} = \frac{1}{3} \int_0^{a^3} \sin(t)\,dt$.
Integrating $\sin(t)$ gives $-\cos(t)$:
$I = \frac{1}{3} [-\cos(t)]_0^{a^3} = -\frac{1}{3} [\cos(a^3) - \cos(0)]$.
Since $\cos(0) = 1$,we have:
$I = -\frac{1}{3} [\cos(a^3) - 1] = \frac{1}{3} (1 - \cos(a^3))$.

(B) Let $I = \int_{0}^{1} \frac{\tan^{-1} x}{1 + x^2} dx$.
Substitute $t = \tan^{-1} x$.
Then,$dt = \frac{1}{1 + x^2} dx$.
When $x = 0$,$t = \tan^{-1}(0) = 0$.
When $x = 1$,$t = \tan^{-1}(1) = \pi / 4$.
Therefore,the integral becomes $I = \int_{0}^{\pi / 4} t dt$.
Evaluating the integral,we get $I = \left[ \frac{t^2}{2} \right]_{0}^{\pi / 4}$.
$I = \frac{(\pi / 4)^2}{2} - 0 = \frac{\pi^2 / 16}{2} = \frac{\pi^2}{32}$.

(A) Let $I = \int_{8}^{15} \frac{dx}{(x - 3)\sqrt{x + 1}}$.
Substitute $t = \sqrt{x + 1}$,then $t^2 = x + 1$,so $x = t^2 - 1$ and $dx = 2t \, dt$.
When $x = 8$,$t = \sqrt{8 + 1} = 3$. When $x = 15$,$t = \sqrt{15 + 1} = 4$.
Substituting these into the integral:
$I = \int_{3}^{4} \frac{2t \, dt}{(t^2 - 1 - 3)t} = \int_{3}^{4} \frac{2 \, dt}{t^2 - 4}$.
Using the formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C$:
$I = 2 \times \frac{1}{2(2)} \left[ \log \left| \frac{t - 2}{t + 2} \right| \right]_{3}^{4}$.
$I = \frac{1}{2} \left[ \log \left( \frac{4 - 2}{4 + 2} \right) - \log \left( \frac{3 - 2}{3 + 2} \right) \right]$.
$I = \frac{1}{2} \left[ \log \left( \frac{2}{6} \right) - \log \left( \frac{1}{5} \right) \right] = \frac{1}{2} \left[ \log \left( \frac{1}{3} \right) - \log \left( \frac{1}{5} \right) \right]$.
$I = \frac{1}{2} \log \left( \frac{1/3}{1/5} \right) = \frac{1}{2} \log \frac{5}{3}$.

(A) Let $I = \int_1^{e^2} \frac{dx}{x(1 + \ln x)^2}$.
Substitute $t = 1 + \ln x$.
Then,$dt = \frac{1}{x} dx$.
Change the limits of integration:
When $x = 1$,$t = 1 + \ln(1) = 1 + 0 = 1$.
When $x = e^2$,$t = 1 + \ln(e^2) = 1 + 2 = 3$.
Now,substitute these into the integral:
$I = \int_1^3 \frac{dt}{t^2} = \int_1^3 t^{-2} dt$.
Integrating $t^{-2}$ gives $\left[ \frac{t^{-1}}{-1} \right]_1^3 = \left[ -\frac{1}{t} \right]_1^3$.
Evaluating at the limits:
$I = -\left( \frac{1}{3} - \frac{1}{1} \right) = -\left( \frac{1}{3} - 1 \right) = -\left( -\frac{2}{3} \right) = \frac{2}{3}$.

(C) Let $I = \int_{\log 2}^x \frac{du}{({e^u} - 1)^{1/2}} = \frac{\pi}{6}$.
Substitute $e^u - 1 = t^2$,so $e^u du = 2t dt$,which implies $du = \frac{2t}{e^u} dt = \frac{2t}{t^2 + 1} dt$.
When $u = \log 2$,$t = \sqrt{e^{\log 2} - 1} = \sqrt{2 - 1} = 1$.
When $u = x$,$t = \sqrt{e^x - 1}$.
The integral becomes $\int_{1}^{\sqrt{e^x - 1}} \frac{2t}{t^2 + 1} \cdot \frac{1}{t} dt = \int_{1}^{\sqrt{e^x - 1}} \frac{2}{t^2 + 1} dt = \frac{\pi}{6}$.
$2 [\tan^{-1} t]_{1}^{\sqrt{e^x - 1}} = \frac{\pi}{6}$.
$\tan^{-1}(\sqrt{e^x - 1}) - \tan^{-1}(1) = \frac{\pi}{12}$.
$\tan^{-1}(\sqrt{e^x - 1}) - \frac{\pi}{4} = \frac{\pi}{12}$.
$\tan^{-1}(\sqrt{e^x - 1}) = \frac{\pi}{12} + \frac{\pi}{4} = \frac{\pi + 3\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$.
$\sqrt{e^x - 1} = \tan(\frac{\pi}{3}) = \sqrt{3}$.
$e^x - 1 = 3$.
$e^x = 4$.

(C) Let $I = \frac{1}{c}\int_{ac}^{bc} {f\left( \frac{x}{c} \right)} \,dx$.
Substitute $\frac{x}{c} = t$,which implies $x = ct$.
Then,the differential $dx = c \, dt$.
Now,change the limits of integration:
When $x = ac$,$t = \frac{ac}{c} = a$.
When $x = bc$,$t = \frac{bc}{c} = b$.
Substituting these into the integral,we get:
$I = \frac{1}{c} \int_{a}^{b} f(t) \cdot (c \, dt)$
$I = \int_{a}^{b} f(t) \, dt$.
Since the variable of integration is a dummy variable,we can replace $t$ with $x$:
$I = \int_{a}^{b} f(x) \, dx$.
Thus,the correct option is $C$.

(D) Given that $\frac{d}{dx}F(x) = \frac{e^{\sin x}}{x}$.
We need to evaluate the integral $I = \int_{1}^{4} \frac{3}{x} e^{\sin(x^3)} dx$.
Multiply the numerator and denominator by $x^2$ to facilitate substitution:
$I = \int_{1}^{4} \frac{3x^2}{x^3} e^{\sin(x^3)} dx$.
Let $t = x^3$,then $dt = 3x^2 dx$.
When $x = 1$,$t = 1^3 = 1$.
When $x = 4$,$t = 4^3 = 64$.
Substituting these into the integral:
$I = \int_{1}^{64} \frac{e^{\sin t}}{t} dt$.
Since $\frac{d}{dt}F(t) = \frac{e^{\sin t}}{t}$,the integral becomes:
$I = [F(t)]_{1}^{64} = F(64) - F(1)$.
Comparing this with the given expression $F(k) - F(1)$,we get $k = 64$.

(B) Let $I = \int_0^1 \frac{x^7}{\sqrt{1 - x^4}} dx = \int_0^1 \frac{x^6 \cdot x}{\sqrt{1 - x^4}} dx$.
Substitute $t = x^4$,so $dt = 4x^3 dx$. This approach is slightly complex,so let's use $u = x^4$,$du = 4x^3 dx$. Then $x^4 = u$ and $x^2 = \sqrt{u}$.
$I = \int_0^1 \frac{x^4 \cdot x^3}{\sqrt{1 - x^4}} dx = \int_0^1 \frac{u}{\sqrt{1 - u}} \cdot \frac{1}{4} du$.
Let $1 - u = v^2$,then $du = -2v dv$. When $u=0, v=1$; when $u=1, v=0$.
$I = \frac{1}{4} \int_1^0 \frac{1 - v^2}{v} (-2v) dv = \frac{1}{2} \int_0^1 (1 - v^2) dv$.
$I = \frac{1}{2} [v - \frac{v^3}{3}]_0^1 = \frac{1}{2} (1 - \frac{1}{3}) = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$.

(C) Let $I = \int_a^b f(x)g(x) dx$.
Given that $\frac{d[f(x)]}{dx} = g(x)$,we can write $g(x) dx = d[f(x)]$.
Substituting $f(x) = t$,we get $df(x) = dt$,which implies $g(x) dx = dt$.
When $x = a$,$t = f(a)$,and when $x = b$,$t = f(b)$.
Thus,the integral becomes $I = \int_{f(a)}^{f(b)} t dt$.
Evaluating the integral,we get $I = \left[ \frac{t^2}{2} \right]_{f(a)}^{f(b)} = \frac{[f(b)]^2 - [f(a)]^2}{2}$.

(B) Let $I = \int_0^{\pi /4} \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int_0^{\pi /4} \frac{\sqrt{\tan x} / \cos^2 x}{(\sin x \cos x) / \cos^2 x} \, dx = \int_0^{\pi /4} \frac{\sqrt{\tan x} \sec^2 x}{\tan x} \, dx = \int_0^{\pi /4} \frac{\sec^2 x}{\sqrt{\tan x}} \, dx$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$.
When $x = 0$,$t = 0$. When $x = \pi / 4$,$t = 1$.
Substituting these into the integral:
$I = \int_0^1 \frac{1}{\sqrt{t}} \, dt = \int_0^1 t^{-1/2} \, dt$.
Integrating,we get:
$I = [2t^{1/2}]_0^1 = 2(1) - 2(0) = 2$.

(C) Let $t = e^{x^2}$.
Then,$dt = e^{x^2} \cdot 2x dx$.
When $x = 0$,$t = e^{0^2} = e^0 = 1$.
When $x = \sqrt{\ln(\frac{\pi}{2})}$,$t = e^{(\sqrt{\ln(\frac{\pi}{2})})^2} = e^{\ln(\frac{\pi}{2})} = \frac{\pi}{2}$.
The integral becomes $\int_{1}^{\frac{\pi}{2}} \cos(t) dt$.
Evaluating the integral: $[\sin(t)]_{1}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(1) = 1 - \sin(1)$.

(A) Let $I = \int \frac{e^x \sqrt{e^x - 1}}{e^x + 3} dx$.
Substitute $t = e^x$,then $dt = e^x dx$.
The integral becomes $I = \int \frac{\sqrt{t - 1}}{t + 3} dt$.
Substitute $u = \sqrt{t - 1}$,then $u^2 = t - 1$,so $t = u^2 + 1$ and $2u du = dt$.
$I = \int \frac{u \cdot 2u du}{u^2 + 1 + 3} = 2 \int \frac{u^2}{u^2 + 4} du$.
$I = 2 \int \left( 1 - \frac{4}{u^2 + 4} \right) du = 2u - 8 \int \frac{1}{u^2 + 2^2} du$.
$I = 2u - 8 \cdot \frac{1}{2} \tan^{-1}\left( \frac{u}{2} \right) = 2u - 4 \tan^{-1}\left( \frac{u}{2} \right)$.
Substituting back $u = \sqrt{e^x - 1}$,we get $2 \sqrt{e^x - 1} - 4 \tan^{-1}\left( \frac{\sqrt{e^x - 1}}{2} \right)$.
Evaluating from $0$ to $\ln 5$:
At $x = \ln 5$,$e^x = 5$,so $u = \sqrt{5 - 1} = 2$.
At $x = 0$,$e^x = 1$,so $u = \sqrt{1 - 1} = 0$.
Result $= [2(2) - 4 \tan^{-1}(1)] - [2(0) - 4 \tan^{-1}(0)] = 4 - 4(\frac{\pi}{4}) = 4 - \pi$.

(D) Let $I = \int\limits_0^{{{\left( {\frac{\pi }{2}} \right)}^{\frac{1}{3}}}} {\,{x^5}\cdot\sin {x^3}\,dx} $.
Substitute $t = x^3$,then $dt = 3x^2 dx$,which implies $x^2 dx = \frac{1}{3} dt$.
When $x = 0$,$t = 0$.
When $x = (\frac{\pi}{2})^{1/3}$,$t = \frac{\pi}{2}$.
Substituting these into the integral:
$I = \int\limits_0^{\frac{\pi}{2}} {t \cdot \sin(t) \cdot \frac{1}{3} dt} = \frac{1}{3} \int\limits_0^{\frac{\pi}{2}} {t \sin(t) dt}$.
Using integration by parts $\int u dv = uv - \int v du$,let $u = t$ and $dv = \sin(t) dt$.
Then $du = dt$ and $v = -\cos(t)$.
$I = \frac{1}{3} [ -t \cos(t) |_0^{\frac{\pi}{2}} - \int\limits_0^{\frac{\pi}{2}} {-\cos(t) dt} ]$.
$I = \frac{1}{3} [ (- \frac{\pi}{2} \cos(\frac{\pi}{2}) + 0) + \int\limits_0^{\frac{\pi}{2}} \cos(t) dt ]$.
Since $\cos(\frac{\pi}{2}) = 0$,the first term is $0$.
$I = \frac{1}{3} [ \sin(t) |_0^{\frac{\pi}{2}} ] = \frac{1}{3} [ \sin(\frac{\pi}{2}) - \sin(0) ] = \frac{1}{3} [ 1 - 0 ] = \frac{1}{3}$.

(C) Let $I = \int\limits_0^1 {\frac{{{{\tan }^{ - 1}}x}}{x}\,dx} $.
Substitute $x = \tan \theta$,then $dx = \sec^2 \theta \, d\theta$.
When $x = 0$,$\theta = 0$. When $x = 1$,$\theta = \frac{\pi}{4}$.
$I = \int\limits_0^{\frac{\pi}{4}} {\frac{\theta}{{\tan \theta}} \cdot \sec^2 \theta \, d\theta} = \int\limits_0^{\frac{\pi}{4}} {\frac{\theta}{{\sin \theta \cos \theta}} \, d\theta} = \int\limits_0^{\frac{\pi}{4}} {\frac{{2\theta}}{{\sin 2\theta}} \, d\theta}$.
Let $2\theta = y$,then $2 \, d\theta = dy$ or $d\theta = \frac{1}{2} dy$.
When $\theta = 0$,$y = 0$. When $\theta = \frac{\pi}{4}$,$y = \frac{\pi}{2}$.
$I = \int\limits_0^{\frac{\pi}{2}} {\frac{y}{{\sin y}} \cdot \frac{1}{2} \, dy} = \frac{1}{2} \int\limits_0^{\frac{\pi}{2}} {\frac{y}{{\sin y}} \, dy} = \frac{1}{2} \int\limits_0^{\frac{\pi}{2}} {\frac{x}{{\sin x}} \, dx}$.

(B) Let $I = \int_0^2 \frac{375 x^5}{(1 + x^2)^4} dx$.
Substitute $u = 1 + x^2$,then $du = 2x dx$,so $x^2 = u - 1$ and $x dx = \frac{du}{2}$.
When $x = 0$,$u = 1$. When $x = 2$,$u = 5$.
$I = \int_1^5 \frac{375 (x^2)^2 x}{(1 + x^2)^4} dx = \int_1^5 \frac{375 (u - 1)^2}{u^4} \cdot \frac{du}{2}$.
$I = \frac{375}{2} \int_1^5 \frac{u^2 - 2u + 1}{u^4} du = \frac{375}{2} \int_1^5 (u^{-2} - 2u^{-3} + u^{-4}) du$.
$I = \frac{375}{2} \left[ -u^{-1} + u^{-2} - \frac{1}{3} u^{-3} \right]_1^5$.
$I = \frac{375}{2} \left[ (-\frac{1}{5} + \frac{1}{25} - \frac{1}{375}) - (-1 + 1 - \frac{1}{3}) \right]$.
$I = \frac{375}{2} \left[ \frac{-75 + 15 - 1}{375} + \frac{1}{3} \right] = \frac{375}{2} \left[ \frac{-61}{375} + \frac{125}{375} \right] = \frac{375}{2} \cdot \frac{64}{375} = \frac{64}{2} = 32$.
Since $32 = 2^5$,we have $2^n = 2^5$,which implies $n = 5$.

(A) Let $I = \int\limits_0^{\frac{1}{2}} \frac{1}{1 - x^2} \ln \left( \frac{1 + x}{1 - x} \right) dx$.
Substitute $t = \ln \left( \frac{1 + x}{1 - x} \right) = \ln(1 + x) - \ln(1 - x)$.
Then $dt = \left( \frac{1}{1 + x} - \frac{-1}{1 - x} \right) dx = \left( \frac{1 - x + 1 + x}{1 - x^2} \right) dx = \frac{2}{1 - x^2} dx$.
Thus,$\frac{dx}{1 - x^2} = \frac{1}{2} dt$.
When $x = 0$,$t = \ln(1) = 0$.
When $x = \frac{1}{2}$,$t = \ln \left( \frac{1 + 1/2}{1 - 1/2} \right) = \ln \left( \frac{3/2}{1/2} \right) = \ln 3$.
Therefore,$I = \int\limits_0^{\ln 3} t \cdot \frac{1}{2} dt = \frac{1}{2} \left[ \frac{t^2}{2} \right]_0^{\ln 3} = \frac{1}{4} \ln^2 3$.
Since $\ln^2(1/3) = (\ln 1 - \ln 3)^2 = (-\ln 3)^2 = \ln^2 3$,the result is $\frac{1}{4} \ln^2 3$ or $\frac{1}{4} \ln^2 \left( \frac{1}{3} \right)$.

(D) Let $u = x \tan \theta$. Then $du = \tan \theta \, dx$,which implies $dx = \frac{du}{\tan \theta} = \cot \theta \, du$.
When $x = \sin \theta$,$u = \sin \theta \tan \theta$.
When $x = \cos \theta$,$u = \cos \theta \tan \theta = \sin \theta$.
Substituting these into the integral:
$\int_{\sin \theta}^{\cos \theta} f(x \tan \theta) \, dx = \int_{\sin \theta \tan \theta}^{\sin \theta} f(u) \cot \theta \, du$
$= \cot \theta \int_{\sin \theta \tan \theta}^{\sin \theta} f(u) \, du$
$= -\cot \theta \int_{\sin \theta}^{\sin \theta \tan \theta} f(u) \, du$.
Comparing this with the given options,the correct form is represented by option $D$.

(A) Given $\frac{d}{dx} G(x) = \frac{e^{\tan x}}{x}$ for $x \in (0, \pi/2)$.
Let $I = \int_{1/4}^{1/2} \frac{2}{x} e^{\tan(\pi x^2)} dx$.
Multiply and divide by $\pi$ inside the integral: $I = \int_{1/4}^{1/2} \frac{2\pi x}{\pi x^2} e^{\tan(\pi x^2)} dx$.
Let $t = \pi x^2$. Then $dt = 2\pi x dx$.
When $x = 1/4$,$t = \pi/16$. When $x = 1/2$,$t = \pi/4$.
Substituting these into the integral,we get $I = \int_{\pi/16}^{\pi/4} \frac{e^{\tan t}}{t} dt$.
Since $\frac{d}{dt} G(t) = \frac{e^{\tan t}}{t}$,the integral becomes $[G(t)]_{\pi/16}^{\pi/4} = G(\pi/4) - G(\pi/16)$.

(C) Let $I = \int_{\pi/6}^{\pi/3} \sec^{2/3} x \csc^{4/3} x \, dx$.
We can rewrite the integrand as:
$I = \int_{\pi/6}^{\pi/3} \frac{1}{\cos^{2/3} x \sin^{4/3} x} \, dx = \int_{\pi/6}^{\pi/3} \frac{1}{\cos^{2/3} x \sin^{2/3} x \cdot \sin^{2/3} x} \, dx$
$I = \int_{\pi/6}^{\pi/3} \frac{1}{(\sin x \cos x)^{2/3} \cdot \sin^{2/3} x} \, dx = \int_{\pi/6}^{\pi/3} \frac{1}{\tan^{2/3} x \cdot \sin^2 x} \, dx$
$I = \int_{\pi/6}^{\pi/3} \frac{\sec^2 x}{\tan^{2/3} x} \, dx$.
Let $\tan x = t$,then $\sec^2 x \, dx = dt$.
When $x = \pi/6$,$t = \tan(\pi/6) = 1/\sqrt{3} = 3^{-1/2}$.
When $x = \pi/3$,$t = \tan(\pi/3) = \sqrt{3} = 3^{1/2}$.
$I = \int_{3^{-1/2}}^{3^{1/2}} t^{-2/3} \, dt = \left[ \frac{t^{1/3}}{1/3} \right]_{3^{-1/2}}^{3^{1/2}} = 3 \left[ t^{1/3} \right]_{3^{-1/2}}^{3^{1/2}}$
$I = 3 \left( (3^{1/2})^{1/3} - (3^{-1/2})^{1/3} \right) = 3 \left( 3^{1/6} - 3^{-1/6} \right)$
$I = 3^{1 + 1/6} - 3^{1 - 1/6} = 3^{7/6} - 3^{5/6}$.

(B) To evaluate the integral $I = \int_{3}^{5} \frac{1}{2x + 3} dx$,we use the substitution method.
Let $u = 2x + 3$,then $du = 2 dx$,which implies $dx = \frac{du}{2}$.
When $x = 3$,$u = 2(3) + 3 = 9$.
When $x = 5$,$u = 2(5) + 3 = 13$.
Substituting these into the integral:
$I = \int_{9}^{13} \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} [\ln |u|]_{9}^{13}$.
$I = \frac{1}{2} (\ln 13 - \ln 9) = \frac{1}{2} \ln \left( \frac{13}{9} \right)$.

(A) Let $I = \int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$.
We use the substitution method. Let $t = 30 - x^{\frac{3}{2}}$.
Then,$dt = -\frac{3}{2} x^{\frac{1}{2}} dx$,which implies $\sqrt{x} dx = -\frac{2}{3} dt$.
When $x = 4$,$t = 30 - 4^{\frac{3}{2}} = 30 - 8 = 22$.
When $x = 9$,$t = 30 - 9^{\frac{3}{2}} = 30 - 27 = 3$.
Substituting these into the integral:
$I = \int_{22}^{3} -\frac{2}{3} \frac{dt}{t^2} = \frac{2}{3} \int_{3}^{22} t^{-2} dt$.
$I = \frac{2}{3} \left[ -\frac{1}{t} \right]_{3}^{22} = \frac{2}{3} \left( -\frac{1}{22} - (-\frac{1}{3}) \right) = \frac{2}{3} \left( \frac{1}{3} - \frac{1}{22} \right)$.
$I = \frac{2}{3} \left( \frac{22 - 3}{66} \right) = \frac{2}{3} \left( \frac{19}{66} \right) = \frac{19}{99}$.

(1/8) Let $I = \int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t \,d t$.
Put $\sin 2 t = u$. Then,differentiating both sides with respect to $t$,we get $2 \cos 2 t \,d t = d u$,which implies $\cos 2 t \,d t = \frac{1}{2} d u$.
Now,change the limits of integration:
When $t = 0$,$u = \sin(2 \times 0) = \sin 0 = 0$.
When $t = \frac{\pi}{4}$,$u = \sin(2 \times \frac{\pi}{4}) = \sin \frac{\pi}{2} = 1$.
Substituting these into the integral:
$I = \int_{0}^{1} u^{3} \cdot \frac{1}{2} d u$
$I = \frac{1}{2} \left[ \frac{u^{4}}{4} \right]_{0}^{1}$
$I = \frac{1}{2} \left( \frac{1^{4}}{4} - \frac{0^{4}}{4} \right)$
$I = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

(N/A) Let $I = \int_{2}^{3} \frac{x}{x^{2}+1} dx$.
To evaluate this,we use the substitution method. Let $u = x^{2} + 1$. Then $du = 2x dx$,which implies $x dx = \frac{1}{2} du$.
Changing the limits of integration:
When $x = 2$,$u = 2^{2} + 1 = 5$.
When $x = 3$,$u = 3^{2} + 1 = 10$.
Substituting these into the integral:
$I = \int_{5}^{10} \frac{1}{2u} du = \frac{1}{2} [\ln |u|]_{5}^{10}$.
Applying the fundamental theorem of calculus:
$I = \frac{1}{2} (\ln 10 - \ln 5) = \frac{1}{2} \ln \left(\frac{10}{5}\right) = \frac{1}{2} \ln 2$.

(N/A) Let $I = \int_{0}^{1} x e^{x^{2}} d x$.
Substitute $x^{2} = t$,which implies $2x \, dx = dt$ or $x \, dx = \frac{1}{2} dt$.
Change the limits of integration:
When $x = 0$,$t = 0^{2} = 0$.
When $x = 1$,$t = 1^{2} = 1$.
Substituting these into the integral,we get:
$I = \int_{0}^{1} e^{t} \cdot \frac{1}{2} dt = \frac{1}{2} \int_{0}^{1} e^{t} dt$.
The integral of $e^{t}$ is $e^{t}$.
Applying the limits:
$I = \frac{1}{2} [e^{t}]_{0}^{1} = \frac{1}{2} (e^{1} - e^{0})$.
Since $e^{0} = 1$,we have:
$I = \frac{1}{2} (e - 1)$.

(A) Let $t = x^{5} + 1$. Then,$dt = 5x^{4} dx$.
When $x = -1$,$t = (-1)^{5} + 1 = 0$.
When $x = 1$,$t = (1)^{5} + 1 = 2$.
Therefore,the integral becomes:
$\int_{0}^{2} \sqrt{t} dt = \left[ \frac{2}{3} t^{\frac{3}{2}} \right]_{0}^{2}$
$= \frac{2}{3} (2^{\frac{3}{2}} - 0^{\frac{3}{2}})$
$= \frac{2}{3} (2\sqrt{2}) = \frac{4\sqrt{2}}{3}$.

(C) Let $t = \tan^{-1} x$. Then,$dt = \frac{1}{1+x^2} dx$.
Change the limits of integration:
When $x = 0$,$t = \tan^{-1}(0) = 0$.
When $x = 1$,$t = \tan^{-1}(1) = \frac{\pi}{4}$.
Substitute these into the integral:
$\int_{0}^{1} \frac{\tan^{-1} x}{1+x^2} dx = \int_{0}^{\frac{\pi}{4}} t dt$.
Evaluate the integral:
$\int_{0}^{\frac{\pi}{4}} t dt = \left[ \frac{t^2}{2} \right]_{0}^{\frac{\pi}{4}} = \frac{1}{2} \left( \left( \frac{\pi}{4} \right)^2 - 0^2 \right) = \frac{1}{2} \left( \frac{\pi^2}{16} \right) = \frac{\pi^2}{32}$.

(A) To evaluate the integral $\int_{0}^{1} \frac{x}{x^{2}+1} d x$,we use the method of substitution.
Let $x^{2}+1 = t$. Then,differentiating both sides with respect to $x$,we get $2x dx = dt$,which implies $x dx = \frac{1}{2} dt$.
Now,change the limits of integration:
When $x = 0$,$t = 0^{2} + 1 = 1$.
When $x = 1$,$t = 1^{2} + 1 = 2$.
Substituting these into the integral:
$\int_{0}^{1} \frac{x}{x^{2}+1} d x = \int_{1}^{2} \frac{1}{t} \cdot \frac{1}{2} dt = \frac{1}{2} \int_{1}^{2} \frac{1}{t} dt$.
The integral of $\frac{1}{t}$ is $\log |t|$.
$= \frac{1}{2} [\log |t|]_{1}^{2}$.
$= \frac{1}{2} (\log 2 - \log 1)$.
Since $\log 1 = 0$,we get:
$= \frac{1}{2} \log 2$.

Let $I = \int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$.
Substitute $x = \tan \theta$,then $d x = \sec ^{2} \theta d \theta$.
When $x = 0$,$\theta = 0$. When $x = 1$,$\theta = \frac{\pi}{4}$.
$I = \int_{0}^{\frac{\pi}{4}} \sin ^{-1}(\sin 2 \theta) \sec ^{2} \theta d \theta = \int_{0}^{\frac{\pi}{4}} 2 \theta \sec ^{2} \theta d \theta$.
Using integration by parts,$\int u v d \theta = u \int v d \theta - \int (u' \int v d \theta) d \theta$,where $u = \theta$ and $v = \sec^2 \theta$:
$I = 2 \left[ \theta \tan \theta - \int \tan \theta d \theta \right]_{0}^{\frac{\pi}{4}} = 2 [ \theta \tan \theta + \ln |\cos \theta| ]_{0}^{\frac{\pi}{4}}$.
Evaluating at the limits:
$I = 2 \left[ \frac{\pi}{4} \tan \frac{\pi}{4} + \ln |\cos \frac{\pi}{4}| - (0 + \ln |\cos 0|) \right] = 2 \left[ \frac{\pi}{4} + \ln \frac{1}{\sqrt{2}} - 0 \right]$.
Since $\ln \frac{1}{\sqrt{2}} = -\frac{1}{2} \ln 2$:
$I = 2 \left[ \frac{\pi}{4} - \frac{1}{2} \ln 2 \right] = \frac{\pi}{2} - \ln 2$.

(C) We need to evaluate the integral $I = \int_{0}^{2} x \sqrt{x+2} \, dx$.
Let $x+2 = t^{2}$. Then $x = t^{2}-2$ and $dx = 2t \, dt$.
Change the limits of integration:
When $x=0$,$t^{2} = 0+2 = 2$,so $t = \sqrt{2}$.
When $x=2$,$t^{2} = 2+2 = 4$,so $t = 2$.
Substituting these into the integral:
$I = \int_{\sqrt{2}}^{2} (t^{2}-2) \sqrt{t^{2}} \cdot (2t) \, dt$
$I = \int_{\sqrt{2}}^{2} (t^{2}-2) \cdot t \cdot 2t \, dt$
$I = 2 \int_{\sqrt{2}}^{2} (t^{4}-2t^{2}) \, dt$
Now,integrate with respect to $t$:
$I = 2 \left[ \frac{t^{5}}{5} - \frac{2t^{3}}{3} \right]_{\sqrt{2}}^{2}$
$I = 2 \left[ \left( \frac{32}{5} - \frac{16}{3} \right) - \left( \frac{(\sqrt{2})^{5}}{5} - \frac{2(\sqrt{2})^{3}}{3} \right) \right]$
$I = 2 \left[ \left( \frac{96-80}{15} \right) - \left( \frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3} \right) \right]$
$I = 2 \left[ \frac{16}{15} - \left( \frac{12\sqrt{2}-20\sqrt{2}}{15} \right) \right]$
$I = 2 \left[ \frac{16}{15} - \left( -\frac{8\sqrt{2}}{15} \right) \right] = 2 \left( \frac{16+8\sqrt{2}}{15} \right) = \frac{32+16\sqrt{2}}{15}$

(A) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} \,d x$.
Substitute $\cos x = t$,which implies $-\sin x \,d x = d t$ or $\sin x \,d x = -d t$.
Change the limits of integration:
When $x = 0$,$t = \cos(0) = 1$.
When $x = \frac{\pi}{2}$,$t = \cos(\frac{\pi}{2}) = 0$.
Now,substitute these into the integral:
$I = \int_{1}^{0} \frac{-d t}{1+t^{2}} = \int_{0}^{1} \frac{d t}{1+t^{2}}$.
Using the standard integral formula $\int \frac{1}{1+t^2} \,dt = \tan^{-1} t + C$:
$I = [\tan^{-1} t]_{0}^{1}$.
Evaluate at the limits:
$I = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

(D) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2\sin x \cos x) = 1 - (\sin x - \cos x)^2$.
So,$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{1 - (\sin x - \cos x)^2}} d x$.
Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$.
When $x = \frac{\pi}{6}$,$t = \sin(\frac{\pi}{6}) - \cos(\frac{\pi}{6}) = \frac{1}{2} - \frac{\sqrt{3}}{2} = \frac{1-\sqrt{3}}{2}$.
When $x = \frac{\pi}{3}$,$t = \sin(\frac{\pi}{3}) - \cos(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3}-1}{2}$.
Thus,$I = \int_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{dt}{\sqrt{1-t^2}}$.
Since $f(t) = \frac{1}{\sqrt{1-t^2}}$ is an even function,$I = 2 \int_{0}^{\frac{\sqrt{3}-1}{2}} \frac{dt}{\sqrt{1-t^2}}$.
$I = 2 [\sin^{-1} t]_{0}^{\frac{\sqrt{3}-1}{2}} = 2 \sin^{-1}(\frac{\sqrt{3}-1}{2})$.
Since $\sin(\frac{\pi}{12}) = \sin(15^\circ) = \frac{\sqrt{6}-\sqrt{2}}{4} \neq \frac{\sqrt{3}-1}{2}$,we re-evaluate: $\sin^{-1}(\frac{\sqrt{3}-1}{2}) = \frac{\pi}{12}$.
Therefore,$I = 2 \times \frac{\pi}{12} = \frac{\pi}{6}$.

(D) Let $I = \int_{0}^{\frac{\pi}{2}} \sin 2x \tan^{-1}(\sin x) dx = \int_{0}^{\frac{\pi}{2}} 2 \sin x \cos x \tan^{-1}(\sin x) dx$.
Let $\sin x = t$,then $\cos x dx = dt$.
When $x = 0, t = 0$ and when $x = \frac{\pi}{2}, t = 1$.
Thus,$I = 2 \int_{0}^{1} t \tan^{-1}(t) dt$.....$(1)$.
Using integration by parts for $\int t \tan^{-1} t dt$:
$\int t \tan^{-1} t dt = \tan^{-1} t \cdot \frac{t^2}{2} - \int \frac{1}{1+t^2} \cdot \frac{t^2}{2} dt$
$= \frac{t^2 \tan^{-1} t}{2} - \frac{1}{2} \int \frac{t^2+1-1}{1+t^2} dt$
$= \frac{t^2 \tan^{-1} t}{2} - \frac{1}{2} \int (1 - \frac{1}{1+t^2}) dt$
$= \frac{t^2 \tan^{-1} t}{2} - \frac{1}{2} (t - \tan^{-1} t) = \frac{t^2 \tan^{-1} t}{2} - \frac{t}{2} + \frac{1}{2} \tan^{-1} t$.
Evaluating the definite integral:
$\int_{0}^{1} t \tan^{-1} t dt = [\frac{t^2 \tan^{-1} t}{2} - \frac{t}{2} + \frac{1}{2} \tan^{-1} t]_{0}^{1}$
$= (\frac{1 \cdot \frac{\pi}{4}}{2} - \frac{1}{2} + \frac{1}{2} \cdot \frac{\pi}{4}) - (0) = \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8} = \frac{\pi}{4} - \frac{1}{2}$.
Substituting back into equation $(1)$:
$I = 2 \times (\frac{\pi}{4} - \frac{1}{2}) = \frac{\pi}{2} - 1$.

(C) Let $I = \int \limits_0^1 (x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7} dx$.
We can rewrite the integrand by factoring $x^7$ from the second term:
$I = \int \limits_0^1 (x^{21}+x^{14}+x^7) \cdot x \cdot (2x^7+3+6x^{-7})^{1/7} dx$ is not the most efficient way.
Instead,let $t = 2x^{21} + 3x^{14} + 6x^7$. Then $dt = (42x^{20} + 42x^{13} + 42x^6) dx = 42(x^{20} + x^{13} + x^6) dx$.
However,the original integral is $\int_0^1 (x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7} dx$.
Let $u = 2x^{14} + 3x^7 + 6$. Then $du = (28x^{13} + 21x^6) dx = 7(4x^{13} + 3x^6) dx$. This does not match the first factor.
Re-evaluating: The integral is $\int_0^1 x^7(x^{14}+x^7+1)(2x^{14}+3x^7+6)^{1/7} dx$.
Let $t = 2x^{14} + 3x^7 + 6$. Then $dt = (28x^{13} + 21x^6) dx = 7(4x^{13} + 3x^6) dx$.
Actually,the correct substitution is $t = 2x^{21} + 3x^{14} + 6x^7$.
Then $dt = (42x^{20} + 42x^{13} + 42x^6) dx = 42x^6(x^{14} + x^7 + 1) dx$.
Thus,$\int_0^1 (x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7} dx = \frac{1}{42} \int_0^{11} t^{1/7} dt = \frac{1}{42} [\frac{7}{8} t^{8/7}]_0^{11} = \frac{1}{48} (11)^{8/7}$.
Here $l = 48, m = 8, n = 7$. Since $m, n$ are coprime,$l+m+n = 48+8+7 = 63$.

(B) Given $y=f(x)$,the slope of the tangent is $\frac{dy}{dx} = f'(x)$.
At $x=1$,$f'(1) = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.
At $x=3$,$f'(3) = \tan(\frac{\pi}{4}) = 1$.
Consider the integral $I = \int_1^3 ((f'(t))^2 + 1) f''(t) dt$.
Let $z = f'(t)$,then $dz = f''(t) dt$.
When $t=1$,$z = f'(1) = \frac{1}{\sqrt{3}}$.
When $t=3$,$z = f'(3) = 1$.
Substituting these into the integral:
$I = \int_{1/\sqrt{3}}^1 (z^2 + 1) dz = \left[ \frac{z^3}{3} + z \right]_{1/\sqrt{3}}^1$
$I = (\frac{1}{3} + 1) - (\frac{1}{3} \cdot \frac{1}{3\sqrt{3}} + \frac{1}{\sqrt{3}})$
$I = \frac{4}{3} - (\frac{1}{9\sqrt{3}} + \frac{3}{3\sqrt{3}}) = \frac{4}{3} - \frac{10}{9\sqrt{3}} = \frac{4}{3} - \frac{10\sqrt{3}}{27}$.
Now,$27I = 27(\frac{4}{3} - \frac{10\sqrt{3}}{27}) = 36 - 10\sqrt{3}$.
Comparing with $\alpha + \beta\sqrt{3}$,we get $\alpha = 36$ and $\beta = -10$.
Therefore,$\alpha + \beta = 36 - 10 = 26$.

(D) Given $f(x)=\frac{x}{(1+x^4)^{1/4}}$.
First,calculate $f(f(x)) = \frac{f(x)}{(1+f(x)^4)^{1/4}} = \frac{\frac{x}{(1+x^4)^{1/4}}}{(1+\frac{x^4}{1+x^4})^{1/4}} = \frac{x}{(1+2x^4)^{1/4}}$.
By induction,$g(x) = f(f(f(f(x)))) = \frac{x}{(1+4x^4)^{1/4}}$.
We need to evaluate $I = 18 \int_0^{\sqrt{2\sqrt{5}}} \frac{x^4}{(1+4x^4)^{1/4}} dx$. Note: The integral is $x^3 g(x) dx$.
Let $1+4x^4 = t^4$,then $16x^3 dx = 4t^3 dt$,which implies $x^3 dx = \frac{1}{4} t^3 dt$.
When $x=0$,$t=1$. When $x=\sqrt{2\sqrt{5}}$,$x^4 = 4(5) = 20$,so $t^4 = 1+4(20) = 81$,$t=3$.
$I = 18 \int_1^3 \frac{1}{t} \cdot \frac{1}{4} t^3 dt = \frac{18}{4} \int_1^3 t^2 dt = \frac{9}{2} [\frac{t^3}{3}]_1^3 = \frac{9}{2} \cdot \frac{1}{3} (27-1) = \frac{3}{2} (26) = 39$.

(C) Let $I = \int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x$.
Divide the numerator and denominator by $\cos^6 x$:
$I = \int_0^{\pi / 4} \frac{\frac{\cos ^2 x \sin ^2 x}{\cos^6 x}}{\left(\frac{\cos ^3 x+\sin ^3 x}{\cos^3 x}\right)^2} d x = \int_0^{\pi / 4} \frac{\tan^2 x \sec^2 x}{(1+\tan^3 x)^2} d x$.
Let $t = 1 + \tan^3 x$. Then $dt = 3 \tan^2 x \sec^2 x d x$,which implies $\tan^2 x \sec^2 x d x = \frac{dt}{3}$.
When $x = 0$,$t = 1 + 0 = 1$. When $x = \pi / 4$,$t = 1 + (1)^3 = 2$.
Substituting these into the integral:
$I = \frac{1}{3} \int_1^2 \frac{dt}{t^2} = \frac{1}{3} \left[ -\frac{1}{t} \right]_1^2 = \frac{1}{3} \left( -\frac{1}{2} - (-1) \right) = \frac{1}{3} \left( \frac{1}{2} \right) = \frac{1}{6}$.