Properties of definite integration Questions in English

(A) Let $I = \int_{-2}^{2} |x \cos \pi x| \, dx$. Since $f(x) = |x \cos \pi x|$ is an even function,we have $I = 2 \int_{0}^{2} |x \cos \pi x| \, dx$.
The function $x \cos \pi x$ changes sign at $x = \frac{1}{2}$ and $x = \frac{3}{2}$ in the interval $[0, 2]$.
Thus,$I = 2 \left[ \int_{0}^{1/2} x \cos \pi x \, dx - \int_{1/2}^{3/2} x \cos \pi x \, dx + \int_{3/2}^{2} x \cos \pi x \, dx \right]$.
Using integration by parts,$\int x \cos \pi x \, dx = \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2}$.
Evaluating the integrals:
$1$. $\int_{0}^{1/2} x \cos \pi x \, dx = \left[ \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2} \right]_{0}^{1/2} = (\frac{1}{2\pi} + 0) - (0 + \frac{1}{\pi^2}) = \frac{1}{2\pi} - \frac{1}{\pi^2}$.
$2$. $\int_{1/2}^{3/2} x \cos \pi x \, dx = \left[ \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2} \right]_{1/2}^{3/2} = (-\frac{3}{2\pi} + 0) - (\frac{1}{2\pi} + 0) = -\frac{2}{\pi}$.
$3$. $\int_{3/2}^{2} x \cos \pi x \, dx = \left[ \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2} \right]_{3/2}^{2} = (0 + \frac{1}{\pi^2}) - (-\frac{3}{2\pi} + 0) = \frac{3}{2\pi} + \frac{1}{\pi^2}$.
Substituting these values: $I = 2 [(\frac{1}{2\pi} - \frac{1}{\pi^2}) - (-\frac{2}{\pi}) + (\frac{3}{2\pi} + \frac{1}{\pi^2})] = 2 [\frac{1}{2\pi} + \frac{2}{\pi} + \frac{3}{2\pi}] = 2 [\frac{4}{\pi}] = \frac{8}{\pi}$.

(B) Let $ I = \int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\cot ^{7} x+\tan ^{7} x} d x $.
Using the property $ \int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x $,we get:
$ I = \int_{0}^{\pi / 2} \frac{\tan ^{7}(\frac{\pi}{2}-x)}{\cot ^{7}(\frac{\pi}{2}-x)+\tan ^{7}(\frac{\pi}{2}-x)} d x $.
Since $ \tan(\frac{\pi}{2}-x) = \cot x $ and $ \cot(\frac{\pi}{2}-x) = \tan x $,we have:
$ I = \int_{0}^{\pi / 2} \frac{\cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x $.
Adding the two expressions for $ I $:
$ 2I = \int_{0}^{\pi / 2} \frac{\tan ^{7} x + \cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x = \int_{0}^{\pi / 2} 1 d x $.
$ 2I = [x]_{0}^{\pi / 2} = \frac{\pi}{2} $.
Therefore,$ I = \frac{\pi}{4} $.

(D) Let $ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $.
Using the property $ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx $,we get:
$ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} (\frac{\pi}{2}-x)}{\sin ^{1000} (\frac{\pi}{2}-x)+\cos ^{1000} (\frac{\pi}{2}-x)} \, dx $
Since $ \sin(\frac{\pi}{2}-x) = \cos x $ and $ \cos(\frac{\pi}{2}-x) = \sin x $,we have:
$ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{1000} x}{\cos ^{1000} x+\sin ^{1000} x} \, dx $
Adding the two expressions for $ I $:
$ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x + \cos ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $
$ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} $
Therefore,$ I = \frac{\pi}{4} $.

(B) Let $I = \int_{0}^{\pi} \frac{\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x$.
Since $f(\pi - x) = \frac{\cos^4(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)} = \frac{(-\cos x)^4}{(\sin x)^4 + (-\cos x)^4} = \frac{\cos^4 x}{\sin^4 x + \cos^4 x} = f(x)$,we use the property $\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$.
Thus,$I = 2 \int_{0}^{\pi / 2} \frac{\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x \dots (i)$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$:
$I = 2 \int_{0}^{\pi / 2} \frac{\cos ^{4}(\frac{\pi}{2}-x)}{\sin ^{4}(\frac{\pi}{2}-x)+\cos ^{4}(\frac{\pi}{2}-x)} d x = 2 \int_{0}^{\pi / 2} \frac{\sin ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x \dots (ii)$.
Adding $(i)$ and $(ii)$:
$2I = 2 \int_{0}^{\pi / 2} \frac{\cos^4 x + \sin^4 x}{\sin^4 x + \cos^4 x} dx = 2 \int_{0}^{\pi / 2} 1 dx = 2[x]_{0}^{\pi / 2} = 2(\frac{\pi}{2}) = \pi$.
Therefore,$I = \frac{\pi}{2}$.

(C) Let $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$ $(i)$
Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,we have:
$I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x)}{\sin ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x) + \cos ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x)} d x$
$I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}(\frac{\pi}{2} - x)}{\sin ^{3}(\frac{\pi}{2} - x) + \cos ^{3}(\frac{\pi}{2} - x)} d x$
Since $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$,we get:
$I = \int_{\pi / 6}^{\pi / 3} \frac{\cos ^{3} x}{\cos ^{3} x + \sin ^{3} x} d x$ (ii)
Adding $(i)$ and (ii):
$2I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x + \cos ^{3} x}{\sin ^{3} x + \cos ^{3} x} d x$
$2I = \int_{\pi / 6}^{\pi / 3} 1 d x$
$2I = [x]_{\pi / 6}^{\pi / 3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
$I = \frac{\pi}{12}$

(A) Let $I = \int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} dx \quad \dots(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = 1+3 = 4$.
Substituting $x$ with $(4-x)$:
$I = \int_{1}^{3} \frac{\sqrt{4-(4-x)}}{\sqrt{4-x}+\sqrt{4-(4-x)}} dx$
$I = \int_{1}^{3} \frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}} dx \quad \dots(ii)$
Adding equations $(i)$ and $(ii)$:
$2I = \int_{1}^{3} \left( \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} + \frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}} \right) dx$
$2I = \int_{1}^{3} \frac{\sqrt{4-x}+\sqrt{x}}{\sqrt{x}+\sqrt{4-x}} dx$
$2I = \int_{1}^{3} 1 dx$
$2I = [x]_{1}^{3} = 3 - 1 = 2$
$I = \frac{2}{2} = 1$

(C) Let $I = \int_{-2}^{2}(a x^{3} + b x + c) d x$.
We can split the integral as:
$I = \int_{-2}^{2} a x^{3} d x + \int_{-2}^{2} b x d x + \int_{-2}^{2} c d x$.
We know that for an odd function $f(x)$,$\int_{-k}^{k} f(x) d x = 0$.
Since $a x^{3}$ and $b x$ are odd functions,$\int_{-2}^{2} a x^{3} d x = 0$ and $\int_{-2}^{2} b x d x = 0$.
Thus,$I = 0 + 0 + \int_{-2}^{2} c d x = \int_{-2}^{2} c d x = [c x]_{-2}^{2} = c(2 - (-2)) = 4c$.
Therefore,the value of the integral depends only on the value of $c$.

(D) Let $I = \int_{0}^{\pi / 2} (\sin^{100} x - \cos^{100} x) dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have:
$I = \int_{0}^{\pi / 2} (\sin^{100}(\frac{\pi}{2} - x) - \cos^{100}(\frac{\pi}{2} - x)) dx$
$I = \int_{0}^{\pi / 2} (\cos^{100} x - \sin^{100} x) dx$
$I = -\int_{0}^{\pi / 2} (\sin^{100} x - \cos^{100} x) dx$
$I = -I$
$2I = 0$
$I = 0$.

(C) Let $ I = \int_{-3}^{3} \cot^{-1} x \, dx $.
Using the property $ \int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} f(-x) \, dx $ is not directly applicable here,but we know that $ \cot^{-1}(-x) = \pi - \cot^{-1} x $.
Consider the integral $ I = \int_{-3}^{3} \cot^{-1} x \, dx $.
Using the property $ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx $,we get:
$ I = \int_{-3}^{3} \cot^{-1}(-x) \, dx = \int_{-3}^{3} (\pi - \cot^{-1} x) \, dx $.
$ I = \int_{-3}^{3} \pi \, dx - \int_{-3}^{3} \cot^{-1} x \, dx $.
$ I = \pi [x]_{-3}^{3} - I $.
$ 2I = \pi (3 - (-3)) $.
$ 2I = 6\pi $.
$ I = 3\pi $.

(D) $I_{n} = \int_{0}^{\frac{\pi}{4}} \tan^{n} x \, dx$ ... $(i)$
$I_{n+2} = \int_{0}^{\frac{\pi}{4}} \tan^{n+2} x \, dx$ ... $(ii)$
Adding equations $(i)$ and $(ii)$:
$I_{n} + I_{n+2} = \int_{0}^{\frac{\pi}{4}} \tan^{n} x (1 + \tan^{2} x) \, dx$
$I_{n} + I_{n+2} = \int_{0}^{\frac{\pi}{4}} \tan^{n} x \sec^{2} x \, dx$
Let $u = \tan x$,then $du = \sec^{2} x \, dx$.
When $x = 0, u = 0$ and when $x = \frac{\pi}{4}, u = 1$.
$I_{n} + I_{n+2} = \int_{0}^{1} u^{n} \, du = \left[ \frac{u^{n+1}}{n+1} \right]_{0}^{1} = \frac{1}{n+1}$.
For $n = 8$,we have $I_{8} + I_{10} = \frac{1}{8+1} = \frac{1}{9}$.

(B) We have,$I = \int_{-1 / 2}^{1 / 2} \cos ^{-1} x \, dx$.
Using the property $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$,we get:
$I = \int_{-1 / 2}^{1 / 2} (\frac{\pi}{2} - \sin^{-1} x) \, dx$.
$I = \int_{-1 / 2}^{1 / 2} \frac{\pi}{2} \, dx - \int_{-1 / 2}^{1 / 2} \sin^{-1} x \, dx$.
Since $\sin^{-1} x$ is an odd function,the integral $\int_{-a}^{a} \sin^{-1} x \, dx = 0$.
Therefore,$I = \frac{\pi}{2} [x]_{-1 / 2}^{1 / 2} - 0$.
$I = \frac{\pi}{2} (\frac{1}{2} - (- \frac{1}{2})) = \frac{\pi}{2} (1) = \frac{\pi}{2}$.

(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$.
Using the property $\int_{-a}^{a} f(x) d x = \int_{0}^{a} (f(x) + f(-x)) d x$,we get:
$I = \int_{0}^{\pi / 2} \left( \frac{\cos x}{1+e^{x}} + \frac{\cos(-x)}{1+e^{-x}} \right) d x$.
Since $\cos(-x) = \cos x$ and $\frac{1}{1+e^{-x}} = \frac{e^x}{e^x+1}$,the expression becomes:
$I = \int_{0}^{\pi / 2} \left( \frac{\cos x}{1+e^{x}} + \frac{e^x \cos x}{1+e^{x}} \right) d x$.
$I = \int_{0}^{\pi / 2} \frac{\cos x (1+e^x)}{1+e^x} d x$.
$I = \int_{0}^{\pi / 2} \cos x d x$.
$I = [\sin x]_{0}^{\pi / 2} = \sin(\pi / 2) - \sin(0) = 1 - 0 = 1$.

(D) Let $I = \int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$.
Substitute $x = \tan \theta$,then $d x = \sec^{2} \theta d \theta$.
When $x = 0$,$\theta = 0$,and when $x = 1$,$\theta = \frac{\pi}{4}$.
Thus,$I = \int_{0}^{\pi/4} \frac{\log (1+\tan \theta)}{1+\tan^{2} \theta} (\sec^{2} \theta) d \theta = \int_{0}^{\pi/4} \log (1+\tan \theta) d \theta \quad ...(i)$
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we have:
$I = \int_{0}^{\pi/4} \log \left[1+\tan \left(\frac{\pi}{4}-\theta\right)\right] d \theta$
$I = \int_{0}^{\pi/4} \log \left[1+\frac{1-\tan \theta}{1+\tan \theta}\right] d \theta = \int_{0}^{\pi/4} \log \left(\frac{2}{1+\tan \theta}\right) d \theta$
$I = \int_{0}^{\pi/4} [\log 2 - \log (1+\tan \theta)] d \theta \quad ...(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{0}^{\pi/4} \log 2 d \theta = \log 2 [\theta]_{0}^{\pi/4} = \frac{\pi}{4} \log 2$.
Therefore,$I = \frac{\pi}{8} \log 2$.

(D) Let $ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{dx}{e^{\sin x}+1} $.
Using the property $ \int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx $,we have:
$ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{e^{\sin x}+1} + \frac{1}{e^{\sin(-x)}+1} \right) dx $
$ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{e^{\sin x}+1} + \frac{1}{e^{-\sin x}+1} \right) dx $
$ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{e^{\sin x}+1} + \frac{e^{\sin x}}{1+e^{\sin x}} \right) dx $
$ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1+e^{\sin x}}{1+e^{\sin x}} \right) dx $
$ I = \int_{0}^{\frac{\pi}{2}} 1 dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} $.

(C) Let $I = \int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x$.
Simplifying the integrand,we get $I = \int_{0}^{\pi / 4} \log (1+\tan x) d x$.
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we have:
$I = \int_{0}^{\pi / 4} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x$.
Since $\tan(\frac{\pi}{4}-x) = \frac{1-\tan x}{1+\tan x}$,the integral becomes:
$I = \int_{0}^{\pi / 4} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x = \int_{0}^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x$.
Using the property $\log(\frac{a}{b}) = \log a - \log b$:
$I = \int_{0}^{\pi / 4} \log 2 d x - \int_{0}^{\pi / 4} \log (1+\tan x) d x$.
$I = \log 2 [x]_{0}^{\pi / 4} - I$.
$2I = \frac{\pi}{4} \log 2$.
Therefore,$I = \frac{\pi}{8} \log 2$.

(A) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} \dots (1)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin(\frac{\pi}{2}-x))^{\sqrt{2}} dx}{(\sin(\frac{\pi}{2}-x))^{\sqrt{2}}+(\cos(\frac{\pi}{2}-x))^{\sqrt{2}}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\cos x)^{\sqrt{2}} dx}{(\cos x)^{\sqrt{2}}+(\sin x)^{\sqrt{2}}} \dots (2)$
Adding $(1)$ and $(2)$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
$I = \frac{\pi}{12}$. Thus,Assertion $(A)$ is true.
For Reason $(R)$,using the property $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$:
Here $a = \frac{\pi}{6}$ and $b = \frac{\pi}{3}$,so $\frac{b-a}{2} = \frac{\frac{\pi}{3} - \frac{\pi}{6}}{2} = \frac{\frac{\pi}{6}}{2} = \frac{\pi}{12}$.
Since the general property holds,Reason $(R)$ is true and correctly explains $(A)$.

(C) Let $I = \int \frac{1+x^2}{1+x^4} dx = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx$.
For $A = \int_0^{\infty} \frac{1+x^2}{1+x^4} dx$,we divide numerator and denominator by $x^2$:
$A = \int_0^{\infty} \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 2} dx$.
Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{x^2}) dx$.
As $x \to 0, t \to -\infty$ and as $x \to \infty, t \to \infty$.
$A = \int_{-\infty}^{\infty} \frac{dt}{t^2 + 2} = \left[ \frac{1}{\sqrt{2}} \tan^{-1}(\frac{t}{\sqrt{2}}) \right]_{-\infty}^{\infty} = \frac{1}{\sqrt{2}} (\frac{\pi}{2} - (-\frac{\pi}{2})) = \frac{\pi}{\sqrt{2}}$.
For $B = \int_0^1 \frac{1+x^2}{1+x^4} dx$,let $x = \frac{1}{u}$,then $dx = -\frac{1}{u^2} du$.
$B = \int_{\infty}^1 \frac{1 + \frac{1}{u^2}}{1 + \frac{1}{u^4}} (-\frac{1}{u^2}) du = \int_1^{\infty} \frac{u^2 + 1}{u^4 + 1} du$.
Since $A = \int_0^1 \frac{1+x^2}{1+x^4} dx + \int_1^{\infty} \frac{1+x^2}{1+x^4} dx = B + B = 2B$.
Therefore,$2B = A$.

(B) Given,$I_n = \int_0^{\pi/4} \tan^n x \, dx$.
We know that for $I_n = \int_0^{\pi/4} \tan^n x \, dx$,the reduction formula is $I_n + I_{n-2} = \int_0^{\pi/4} \tan^{n-2} x (\tan^2 x + 1) \, dx = \int_0^{\pi/4} \tan^{n-2} x \sec^2 x \, dx$.
Let $u = \tan x$,then $du = \sec^2 x \, dx$.
When $x = 0, u = 0$ and when $x = \pi/4, u = 1$.
So,$I_n + I_{n-2} = \int_0^1 u^{n-2} \, du = \left[ \frac{u^{n-1}}{n-1} \right]_0^1 = \frac{1}{n-1}$.
For $I_{13} + I_{11}$,we set $n = 13$.
Thus,$I_{13} + I_{11} = \frac{1}{13-1} = \frac{1}{12}$.

(D) Given $I_n = \int_0^{\pi / 4} \tan^n x \, dx$.
Consider $I_n + I_{n+2} = \int_0^{\pi / 4} \tan^n x (1 + \tan^2 x) \, dx = \int_0^{\pi / 4} \tan^n x \sec^2 x \, dx$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$. When $x=0, t=0$ and when $x=\pi/4, t=1$.
Thus,$I_n + I_{n+2} = \int_0^1 t^n \, dt = \left[ \frac{t^{n+1}}{n+1} \right]_0^1 = \frac{1}{n+1}$.
Now,the given expression is $\frac{1}{I_2 + I_4} + \frac{1}{I_3 + I_5} + \frac{1}{I_4 + I_6}$.
Using the result $I_n + I_{n+2} = \frac{1}{n+1}$,we have:
$I_2 + I_4 = \frac{1}{2+1} = \frac{1}{3} \implies \frac{1}{I_2 + I_4} = 3$.
$I_3 + I_5 = \frac{1}{3+1} = \frac{1}{4} \implies \frac{1}{I_3 + I_5} = 4$.
$I_4 + I_6 = \frac{1}{4+1} = \frac{1}{5} \implies \frac{1}{I_4 + I_6} = 5$.
Sum $= 3 + 4 + 5 = 12$.
Checking the options,for $n=11$,$I_{11} + I_{13} = \frac{1}{11+1} = \frac{1}{12}$,so $\frac{1}{I_{11} + I_{13}} = 12$.
Therefore,the correct option is $D$.

(C) To evaluate the integral $I = \int_0^2 x^2(2-x)^5 dx$,we use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.
Substituting $x = 2-t$,we get $dx = -dt$. When $x=0, t=2$ and when $x=2, t=0$.
$I = \int_2^0 (2-t)^2(t)^5 (-dt) = \int_0^2 (4 - 4t + t^2)t^5 dt$.
$I = \int_0^2 (4t^5 - 4t^6 + t^7) dt$.
Integrating term by term:
$I = [\frac{4t^6}{6} - \frac{4t^7}{7} + \frac{t^8}{8}]_0^2$.
$I = [\frac{2(2^6)}{3} - \frac{4(2^7)}{7} + \frac{2^8}{8}] = [\frac{128}{3} - \frac{512}{7} + 32]$.
Finding a common denominator of $21$:
$I = \frac{896 - 1536 + 672}{21} = \frac{32}{21}$.
Thus,the correct option is $C$.

(A) Let $I = \int_0^\pi \frac{x}{\sin x}(3 \cos^2 x + 2 \sin x + \sin^3 x - 3) dx$.
Since $\cos^2 x = 1 - \sin^2 x$,the expression becomes $3(1 - \sin^2 x) + 2 \sin x + \sin^3 x - 3 = -3 \sin^2 x + 2 \sin x + \sin^3 x$.
Dividing by $\sin x$,we get $-3 \sin x + 2 + \sin^2 x$.
Thus,$I = \int_0^\pi x(\sin^2 x - 3 \sin x + 2) dx \dots (1)$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:
$I = \int_0^\pi (\pi - x)(\sin^2 x - 3 \sin x + 2) dx \dots (2)$.
Adding $(1)$ and $(2)$:
$2I = \int_0^\pi \pi(\sin^2 x - 3 \sin x + 2) dx$.
$2I = \pi \int_0^\pi (\frac{1 - \cos 2x}{2} - 3 \sin x + 2) dx = \pi \int_0^\pi (\frac{5}{2} - \frac{1}{2} \cos 2x - 3 \sin x) dx$.
$2I = \pi [\frac{5}{2}x - \frac{1}{4} \sin 2x + 3 \cos x]_0^\pi$.
$2I = \pi [(\frac{5\pi}{2} - 0 + 3(-1)) - (0 - 0 + 3(1))] = \pi [\frac{5\pi}{2} - 3 - 3] = \pi [\frac{5\pi}{2} - 6] = \frac{\pi(5\pi - 12)}{2}$.
Therefore,$I = \frac{\pi(5\pi - 12)}{4}$.

(B) Let $I = \int_0^{2 \pi} \cos m x \cos n x \, dx + \int_{-\pi}^\pi \sin m x \cos n x \, dx$.
Consider the second integral $J = \int_{-\pi}^\pi \sin m x \cos n x \, dx$.
Since $\sin m(-x) \cos n(-x) = -\sin m x \cos n x$,the integrand is an odd function.
Thus,$J = 0$.
Now,consider $I = \int_0^{2 \pi} \cos m x \cos n x \, dx$.
Using the identity $\cos A \cos B = \frac{1}{2} [\cos(A-B) + \cos(A+B)]$,we have:
$I = \frac{1}{2} \int_0^{2 \pi} [\cos((m-n)x) + \cos((m+n)x)] \, dx$.
If $m \neq n$ and $m, n \in \mathbb{Z}$,then $\int_0^{2 \pi} \cos(kx) \, dx = 0$ for $k \neq 0$.
Thus,$I = 0$ if $m \neq n$.
If $m = n \neq 0$,then $I = \int_0^{2 \pi} \cos^2(mx) \, dx = \int_0^{2 \pi} \frac{1 + \cos(2mx)}{2} \, dx = \frac{1}{2} [x + \frac{\sin(2mx)}{2m}]_0^{2 \pi} = \frac{1}{2} (2 \pi) = \pi$.
Therefore,the value is $\pi$ if $m = n \neq 0$.

(A) Let $I = \int_0^{\infty} (x^{12} + x^{-12}) \frac{\log x}{x} dx$.
Substitute $x = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$.
As $x \to 0$,$t \to \infty$,and as $x \to \infty$,$t \to 0$.
Also,$\log x = \log(t^{-1}) = -\log t$.
Substituting these into the integral:
$I = \int_{\infty}^0 (t^{-12} + t^{12}) \frac{-\log t}{1/t} \left(-\frac{1}{t^2}\right) dt$.
$I = \int_{\infty}^0 (t^{-12} + t^{12}) \frac{-\log t}{t} dt$.
$I = -\int_0^{\infty} (t^{-12} + t^{12}) \frac{\log t}{t} dt$.
Since the variable of integration is a dummy variable,we can replace $t$ with $x$:
$I = -\int_0^{\infty} (x^{-12} + x^{12}) \frac{\log x}{x} dx$.
$I = -I$.
$2I = 0 \Rightarrow I = 0$.

(A) Let $I = \int_{0}^{\pi/2} \tan^{n}(x) dx$.
Using the property of definite integration $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we replace $x$ with $\frac{\pi}{2} - x$.
$I = \int_{0}^{\pi/2} \tan^{n}(\frac{\pi}{2} - x) dx$.
Since $\tan(\frac{\pi}{2} - x) = \cot(x)$,we have $I = \int_{0}^{\pi/2} \cot^{n}(x) dx$.
Comparing this with the given equation $\int_{0}^{\pi/2} \tan^{n}(x) dx = k \int_{0}^{\pi/2} \cot^{n}(x) dx$,we see that $I = k \cdot I$.
Therefore,$k = 1$.

(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx$ ...$(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x}{\cos x + \sin x} dx$ ...(ii)
Adding $(i)$ and (ii):
$2I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} dx = \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} dx$
$2I = \frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \frac{1}{\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x} dx = \frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \operatorname{cosec}\left(x + \frac{\pi}{4}\right) dx$
$2I = \frac{1}{\sqrt{2}} \left[ \log \left| \tan \left( \frac{x}{2} + \frac{\pi}{8} \right) \right| \right]_0^{\frac{\pi}{2}}$
Alternatively,using $\int \operatorname{cosec} \theta d\theta = \log |\operatorname{cosec} \theta - \cot \theta |$:
$2I = \frac{1}{\sqrt{2}} \left[ \log \left| \operatorname{cosec} \left( x + \frac{\pi}{4} \right) - \cot \left( x + \frac{\pi}{4} \right) \right| \right]_0^{\frac{\pi}{2}}$
$2I = \frac{1}{\sqrt{2}} [ \log |\operatorname{cosec} \frac{3\pi}{4} - \cot \frac{3\pi}{4}| - \log |\operatorname{cosec} \frac{\pi}{4} - \cot \frac{\pi}{4}| ]$
$2I = \frac{1}{\sqrt{2}} [ \log |\sqrt{2} - (-1)| - \log |\sqrt{2} - 1| ] = \frac{1}{\sqrt{2}} [ \log (\sqrt{2}+1) - \log (\sqrt{2}-1) ]$
Since $\log (\sqrt{2}-1) = \log (\frac{1}{\sqrt{2}+1}) = -\log (\sqrt{2}+1)$:
$2I = \frac{1}{\sqrt{2}} [ 2 \log (\sqrt{2}+1) ] \implies I = \frac{1}{\sqrt{2}} \log (\sqrt{2}+1)$.

(D) Let $I = \int_0^2 x^8 \left(\frac{4}{x^2} - 1\right)^{5/2} dx$.
Substitute $x = 2 \sin \theta$,so $dx = 2 \cos \theta d\theta$.
When $x = 0$,$\theta = 0$. When $x = 2$,$\theta = \frac{\pi}{2}$.
The expression becomes $\frac{4}{x^2} - 1 = \frac{4}{4 \sin^2 \theta} - 1 = \csc^2 \theta - 1 = \cot^2 \theta$.
Substituting these into the integral:
$I = \int_0^{\pi/2} (2 \sin \theta)^8 (\cot^2 \theta)^{5/2} (2 \cos \theta) d\theta$
$I = \int_0^{\pi/2} (2^8 \sin^8 \theta) (\cot^5 \theta) (2 \cos \theta) d\theta$
$I = 2^9 \int_0^{\pi/2} \sin^8 \theta \frac{\cos^5 \theta}{\sin^5 \theta} \cos \theta d\theta$
$I = 2^9 \int_0^{\pi/2} \sin^3 \theta \cos^6 \theta d\theta$
Using the Wallis formula $\int_0^{\pi/2} \sin^m \theta \cos^n \theta d\theta = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$:
$I = 2^9 \times \frac{\Gamma(2) \Gamma(7/2)}{2 \Gamma(11/2)} = 2^9 \times \frac{1! \times \frac{5}{2} \times \frac{3}{2} \times \frac{1}{2} \times \sqrt{\pi}}{2 \times \frac{9}{2} \times \frac{7}{2} \times \frac{5}{2} \times \frac{3}{2} \times \frac{1}{2} \times \sqrt{\pi}}$
$I = 2^9 \times \frac{1}{2 \times \frac{9}{2} \times \frac{7}{2}} = 2^9 \times \frac{4}{63} = \frac{2^{11}}{63} = \frac{2048}{63}$.
Wait,re-evaluating the integral: $I = 2^9 \times \frac{2 \times 1}{9 \times 7 \times 5} = \frac{512 \times 2}{315} = \frac{1024}{315}$.
Actually,calculating $\int_0^{\pi/2} \sin^3 \theta \cos^6 \theta d\theta = \int_0^{\pi/2} (1-\cos^2 \theta) \cos^6 \theta \sin \theta d\theta$.
Let $u = \cos \theta$,$du = -\sin \theta d\theta$.
$I = 2^9 \int_0^1 (1-u^2) u^6 du = 2^9 [\frac{u^7}{7} - \frac{u^9}{9}]_0^1 = 512 (\frac{1}{7} - \frac{1}{9}) = 512 (\frac{2}{63}) = \frac{1024}{63} = \frac{2^{10}}{63}$.

(D) Given $M = \int_0^{\infty} \frac{\log t}{1+t^3} dt$ and $N = \int_{-\infty}^{\infty} \frac{t e^{2 t}}{1+e^{3 t}} dt$.
For $M$,let $t = e^{-x}$,then $dt = -e^{-x} dx$.
When $t = 0, x = \infty$ and when $t = \infty, x = -\infty$.
Substituting these into $M$:
$M = \int_{\infty}^{-\infty} \frac{\log(e^{-x})}{1+(e^{-x})^3} (-e^{-x}) dx = \int_{-\infty}^{\infty} \frac{-x}{1+e^{-3x}} e^{-x} dx$.
Multiply numerator and denominator by $e^{3x}$:
$M = \int_{-\infty}^{\infty} \frac{-x e^{2x}}{e^{3x} + 1} dx = -\int_{-\infty}^{\infty} \frac{x e^{2x}}{1+e^{3x}} dx$.
Comparing this with $N$,we get $M = -N$,which implies $N = -M$.

(D) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot x}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ $(1)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$
$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
$I = \frac{\pi}{12}$

(A) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sum_{n=0}^4 \left(\frac{n \pi}{4}+x\right)}{\cos x+\sin x} dx$.
Since $\sum_{n=0}^4 \left(\frac{n \pi}{4}+x\right) = (0+1+2+3+4) \frac{\pi}{4} + (1+1+1+1+1)x = \frac{10 \pi}{4} + 5x = \frac{5 \pi}{2} + 5x = 5 \left(\frac{\pi}{2} + x\right)$.
Hence,$I = \int_0^{\frac{\pi}{2}} \frac{5 \left(\frac{\pi}{2} + x\right)}{\sqrt{2} \left(\frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x\right)} dx = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2} + x\right) \sec \left(x - \frac{\pi}{4}\right) dx$ ...$(i)$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we replace $x$ by $\left(\frac{\pi}{2} - x\right)$:
$I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left[\frac{\pi}{2} + \left(\frac{\pi}{2} - x\right)\right] \sec \left(\left(\frac{\pi}{2} - x\right) - \frac{\pi}{4}\right) dx = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} (\pi - x) \sec \left(\frac{\pi}{4} - x\right) dx$.
Since $\sec(\alpha) = \sec(-\alpha)$,$\sec \left(\frac{\pi}{4} - x\right) = \sec \left(x - \frac{\pi}{4}\right)$.
So,$I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} (\pi - x) \sec \left(x - \frac{\pi}{4}\right) dx$ ...(ii).
Adding $(i)$ and (ii):
$2I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2} + x + \pi - x\right) \sec \left(x - \frac{\pi}{4}\right) dx = \frac{5}{\sqrt{2}} \cdot \frac{3 \pi}{2} \int_0^{\frac{\pi}{2}} \sec \left(x - \frac{\pi}{4}\right) dx$.
$2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log \left| \sec \left(x - \frac{\pi}{4}\right) + \tan \left(x - \frac{\pi}{4}\right) \right| \right]_0^{\frac{\pi}{2}}$.
$2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log |\sec \frac{\pi}{4} + \tan \frac{\pi}{4}| - \log |\sec \left(-\frac{\pi}{4}\right) + \tan \left(-\frac{\pi}{4}\right)| \right]$.
$2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log |\sqrt{2} + 1| - \log |\sqrt{2} - 1| \right] = \frac{15 \pi}{2 \sqrt{2}} \log \left| \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right| = \frac{15 \pi}{2 \sqrt{2}} \log |(\sqrt{2} + 1)^2| = \frac{15 \pi}{\sqrt{2}} \log |\sqrt{2} + 1|$.
Therefore,$I = \frac{15 \pi}{2 \sqrt{2}} \log |\sqrt{2} + 1|$.

(D) Let $I = \int_{\pi/11}^{9\pi/22} \frac{dx}{1+\sqrt{\tan x}}$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a = \frac{\pi}{11}$ and $b = \frac{9\pi}{22}$,we have $a+b = \frac{2\pi + 9\pi}{22} = \frac{11\pi}{22} = \frac{\pi}{2}$.
Thus,$I = \int_{\pi/11}^{9\pi/22} \frac{dx}{1+\sqrt{\tan(\pi/2 - x)}} = \int_{\pi/11}^{9\pi/22} \frac{dx}{1+\sqrt{\cot x}} = \int_{\pi/11}^{9\pi/22} \frac{\sqrt{\tan x} dx}{\sqrt{\tan x} + 1}$.
Adding the two expressions for $I$:
$2I = \int_{\pi/11}^{9\pi/22} \left( \frac{1}{1+\sqrt{\tan x}} + \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} \right) dx = \int_{\pi/11}^{9\pi/22} 1 dx$.
$2I = [x]_{\pi/11}^{9\pi/22} = \frac{9\pi}{22} - \frac{2\pi}{22} = \frac{7\pi}{22}$.
Therefore,$I = \frac{7\pi}{44}$.

(A) Let $I = \int_0^1 \frac{8 \log (1+x)}{1+x^2} dx$.
Substitute $x = \tan \theta$,so $dx = \sec^2 \theta d\theta$.
When $x = 0$,$\theta = 0$,and when $x = 1$,$\theta = \frac{\pi}{4}$.
Then $I = 8 \int_0^{\pi/4} \frac{\log(1+\tan \theta)}{1+\tan^2 \theta} \sec^2 \theta d\theta = 8 \int_0^{\pi/4} \log(1+\tan \theta) d\theta$.
Using the property $\int_0^a f(\theta) d\theta = \int_0^a f(a-\theta) d\theta$:
$I = 8 \int_0^{\pi/4} \log(1+\tan(\frac{\pi}{4}-\theta)) d\theta$.
Since $\tan(\frac{\pi}{4}-\theta) = \frac{1-\tan \theta}{1+\tan \theta}$,we have:
$I = 8 \int_0^{\pi/4} \log(1 + \frac{1-\tan \theta}{1+\tan \theta}) d\theta = 8 \int_0^{\pi/4} \log(\frac{2}{1+\tan \theta}) d\theta$.
$I = 8 \int_0^{\pi/4} (\log 2 - \log(1+\tan \theta)) d\theta = 8 \int_0^{\pi/4} \log 2 d\theta - I$.
$2I = 8 \log 2 [\theta]_0^{\pi/4} = 8 \log 2 (\frac{\pi}{4}) = 2\pi \log 2$.
Therefore,$I = \pi \log 2$.

(D) Let $I = \int_0^1 \frac{\log _e(1+x)}{1+x^2} d x$.
Substitute $x = \tan \theta$,then $d x = \sec^2 \theta d \theta$.
When $x = 0$,$\theta = 0$,and when $x = 1$,$\theta = \frac{\pi}{4}$.
So,$I = \int_0^{\frac{\pi}{4}} \frac{\log _e(1+\tan \theta)}{1+\tan^2 \theta} \sec^2 \theta d \theta = \int_0^{\frac{\pi}{4}} \log _e(1+\tan \theta) d \theta$ ... $(i)$.
Using the property $\int_0^a f(\theta) d \theta = \int_0^a f(a-\theta) d \theta$,we get:
$I = \int_0^{\frac{\pi}{4}} \log _e(1+\tan(\frac{\pi}{4}-\theta)) d \theta$.
Since $\tan(\frac{\pi}{4}-\theta) = \frac{1-\tan \theta}{1+\tan \theta}$,we have:
$I = \int_0^{\frac{\pi}{4}} \log _e(1 + \frac{1-\tan \theta}{1+\tan \theta}) d \theta = \int_0^{\frac{\pi}{4}} \log _e(\frac{2}{1+\tan \theta}) d \theta$.
$I = \int_0^{\frac{\pi}{4}} \log _e 2 d \theta - \int_0^{\frac{\pi}{4}} \log _e(1+\tan \theta) d \theta$.
$I = \frac{\pi}{4} \log _e 2 - I$.
$2I = \frac{\pi}{4} \log _e 2 \implies I = \frac{\pi}{8} \log _e 2$.
Thus,option $(D)$ is correct.

(C) Let $I = \int_0^{\frac{\pi}{2}} \frac{\cos x \, dx}{\sqrt{1+\cos x \sin x}} \quad \dots (i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^{\frac{\pi}{2}} \frac{\cos(\frac{\pi}{2}-x) \, dx}{\sqrt{1+\cos(\frac{\pi}{2}-x) \sin(\frac{\pi}{2}-x)}} = \int_0^{\frac{\pi}{2}} \frac{\sin x \, dx}{\sqrt{1+\sin x \cos x}} \quad \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\frac{\pi}{2}} \frac{\cos x + \sin x}{\sqrt{1+\cos x \sin x}} \, dx$
Multiply numerator and denominator by $\sqrt{2}$:
$2I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{2}(\cos x + \sin x)}{\sqrt{2+2\sin x \cos x}} \, dx = \int_0^{\frac{\pi}{2}} \frac{\sqrt{2}(\cos x + \sin x)}{\sqrt{3 - (1 - 2\sin x \cos x)}} \, dx$
Since $1 - 2\sin x \cos x = (\sin x - \cos x)^2$:
$2I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{2}(\cos x + \sin x)}{\sqrt{3 - (\sin x - \cos x)^2}} \, dx$
Let $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) \, dx$.
Limits: when $x=0, t=-1$; when $x=\frac{\pi}{2}, t=1$.
$2I = \int_{-1}^1 \frac{\sqrt{2} \, dt}{\sqrt{3-t^2}} = 2 \int_0^1 \frac{\sqrt{2} \, dt}{\sqrt{(\sqrt{3})^2 - t^2}}$
$I = \sqrt{2} \left[ \sin^{-1} \left( \frac{t}{\sqrt{3}} \right) \right]_0^1 = \sqrt{2} \sin^{-1} \left( \frac{1}{\sqrt{3}} \right)$

(D) Let $I = \int_{0}^{\pi} \frac{x \tan x}{\sec x + \tan x} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\pi} \frac{(\pi - x) \tan(\pi - x)}{\sec(\pi - x) + \tan(\pi - x)} dx$.
Since $\tan(\pi - x) = -\tan x$ and $\sec(\pi - x) = -\sec x$,we have:
$I = \int_{0}^{\pi} \frac{(\pi - x)(-\tan x)}{-\sec x - \tan x} dx = \int_{0}^{\pi} \frac{(\pi - x) \tan x}{\sec x + \tan x} dx$.
$I = \pi \int_{0}^{\pi} \frac{\tan x}{\sec x + \tan x} dx - I$.
$2I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \sin x} dx$.
$2I = \pi \int_{0}^{\pi} \frac{\sin x(1 - \sin x)}{\cos^2 x} dx = \pi \int_{0}^{\pi} (\sec x \tan x - \tan^2 x) dx$.
$2I = \pi \int_{0}^{\pi} (\sec x \tan x - (\sec^2 x - 1)) dx$.
$2I = \pi [\sec x - \tan x + x]_{0}^{\pi}$.
$2I = \pi [(\sec \pi - \tan \pi + \pi) - (\sec 0 - \tan 0 + 0)]$.
$2I = \pi [(-1 - 0 + \pi) - (1 - 0 + 0)] = \pi (\pi - 2)$.
$I = \frac{\pi (\pi - 2)}{2}$.

(A) Let $I = \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1+\cos^2(\pi-x)} dx = \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx$.
$I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$.
$I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - I$.
$2I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$.
Let $u = \cos x$,then $du = -\sin x dx$. When $x=0, u=1$; when $x=\pi, u=-1$.
$2I = \pi \int_1^{-1} \frac{-du}{1+u^2} = \pi \int_{-1}^1 \frac{du}{1+u^2}$.
$2I = \pi [\tan^{-1} u]_{-1}^1 = \pi (\tan^{-1}(1) - \tan^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$.
$I = \frac{\pi^2}{4}$.

(D) Let $I = \int_0^{\pi/4} \frac{\sin x + \cos x}{3 + \sin 2x} dx$.
Substitute $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) dx$.
When $x = 0$,$t = \sin 0 - \cos 0 = -1$.
When $x = \pi/4$,$t = \sin(\pi/4) - \cos(\pi/4) = 0$.
Also,$t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - \sin 2x$,so $\sin 2x = 1 - t^2$.
Substituting these into the integral:
$I = \int_{-1}^0 \frac{dt}{3 + (1 - t^2)} = \int_{-1}^0 \frac{dt}{4 - t^2} = \int_{-1}^0 \frac{dt}{2^2 - t^2}$.
Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$:
$I = \left[ \frac{1}{2(2)} \log \left| \frac{2+t}{2-t} \right| \right]_{-1}^0 = \frac{1}{4} \left[ \log \left| \frac{2+0}{2-0} \right| - \log \left| \frac{2-1}{2-(-1)} \right| \right]$.
$I = \frac{1}{4} [ \log(1) - \log(1/3) ] = \frac{1}{4} [ 0 - (-\log 3) ] = \frac{1}{4} \log 3$.

(A) Let $I = \int_0^{\pi / 2} \frac{\sin x}{1+\cos x+\sin x} d x$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\pi / 2} \frac{\sin(\pi/2 - x)}{1+\cos(\pi/2 - x)+\sin(\pi/2 - x)} dx = \int_0^{\pi / 2} \frac{\cos x}{1+\sin x+\cos x} dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi / 2} \frac{\sin x + \cos x}{1+\sin x+\cos x} dx$.
$2I = \int_0^{\pi / 2} \frac{(1+\sin x+\cos x) - 1}{1+\sin x+\cos x} dx = \int_0^{\pi / 2} (1 - \frac{1}{1+\sin x+\cos x}) dx$.
$2I = [x]_0^{\pi / 2} - \int_0^{\pi / 2} \frac{1}{1+2\sin(x/2)\cos(x/2)+2\cos^2(x/2)-1} dx$.
$2I = \frac{\pi}{2} - \int_0^{\pi / 2} \frac{1}{2\cos^2(x/2)(\tan(x/2)+1)} dx = \frac{\pi}{2} - \frac{1}{2} \int_0^{\pi / 2} \frac{\sec^2(x/2)}{\tan(x/2)+1} dx$.
Let $u = \tan(x/2)$,then $du = \frac{1}{2} \sec^2(x/2) dx$.
$2I = \frac{\pi}{2} - \int_0^1 \frac{1}{u+1} du = \frac{\pi}{2} - [\log|u+1|]_0^1 = \frac{\pi}{2} - \log 2$.
Therefore,$I = \frac{\pi}{4} - \frac{1}{2} \log 2$.

(A) Let $I = \int_0^{\pi / 2} \log |\tan x + \cot x| \, dx$.
Since $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$.
Thus,$I = \int_0^{\pi / 2} \log |\frac{2}{\sin 2x}| \, dx = \int_0^{\pi / 2} (\log 2 - \log |\sin 2x|) \, dx$.
$I = \int_0^{\pi / 2} \log 2 \, dx - \int_0^{\pi / 2} \log |\sin 2x| \, dx$.
$I = \frac{\pi}{2} \log 2 - \int_0^{\pi / 2} \log |\sin 2x| \, dx$.
Let $2x = t$,then $2 \, dx = dt$,so $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\pi$.
$\int_0^{\pi / 2} \log |\sin 2x| \, dx = \frac{1}{2} \int_0^{\pi} \log |\sin t| \, dt = \frac{1}{2} \times 2 \int_0^{\pi / 2} \log |\sin t| \, dt = \int_0^{\pi / 2} \log |\sin t| \, dt$.
Using the property $\int_0^{\pi / 2} \log |\sin t| \, dt = -\frac{\pi}{2} \log 2$.
Therefore,$I = \frac{\pi}{2} \log 2 - (-\frac{\pi}{2} \log 2) = \frac{\pi}{2} \log 2 + \frac{\pi}{2} \log 2 = \pi \log 2$.

(C) Let $I = \int_0^{3 \pi / 2} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} d x$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{3 \pi / 2} \frac{\cos ^3 (3 \pi / 2 - x)}{\cos ^3 (3 \pi / 2 - x) + \sin ^3 (3 \pi / 2 - x)} d x$.
Since $\cos(3 \pi / 2 - x) = -\sin x$ and $\sin(3 \pi / 2 - x) = -\cos x$,we have:
$I = \int_0^{3 \pi / 2} \frac{(-\sin x)^3}{(-\sin x)^3 + (-\cos x)^3} d x = \int_0^{3 \pi / 2} \frac{-\sin ^3 x}{-\sin ^3 x - \cos ^3 x} d x = \int_0^{3 \pi / 2} \frac{\sin ^3 x}{\sin ^3 x + \cos ^3 x} d x$.
Adding the two expressions for $I$:
$2I = \int_0^{3 \pi / 2} \frac{\cos ^3 x + \sin ^3 x}{\cos ^3 x + \sin ^3 x} d x = \int_0^{3 \pi / 2} 1 d x$.
$2I = [x]_0^{3 \pi / 2} = \frac{3 \pi}{2}$.
Therefore,$I = \frac{3 \pi}{4}$.

(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x$.
We can split the integral into two parts: $I = \int_{-\pi / 2}^{\pi / 2} \sin ^3 x \cos ^2 x d x + \int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^3 x d x$.
Let $f(x) = \sin ^3 x \cos ^2 x$. Since $f(-x) = \sin ^3(-x) \cos ^2(-x) = -\sin ^3 x \cos ^2 x = -f(x)$,$f(x)$ is an odd function. Thus,$\int_{-\pi / 2}^{\pi / 2} \sin ^3 x \cos ^2 x d x = 0$.
Now consider $g(x) = \sin ^2 x \cos ^3 x$. Since $g(-x) = \sin ^2(-x) \cos ^3(-x) = \sin ^2 x \cos ^3 x = g(x)$,$g(x)$ is an even function. Thus,$\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^3 x d x = 2 \int_{0}^{\pi / 2} \sin ^2 x \cos ^3 x d x$.
Using the reduction formula $\int_{0}^{\pi / 2} \sin^m x \cos^n x dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!}$ for even $m+n$,we have $2 \int_{0}^{\pi / 2} \sin^2 x \cos^3 x dx = 2 \times \frac{(2-1)!!(3-1)!!}{(2+3+1)!!} = 2 \times \frac{1 \times 2}{6 \times 4 \times 2} = 2 \times \frac{2}{48} = \frac{4}{48} = \frac{1}{12}$. Wait,let's re-evaluate: $2 \int_{0}^{\pi / 2} \sin^2 x \cos^2 x \cos x dx$. Let $u = \sin x$,then $du = \cos x dx$. When $x=0, u=0$; when $x=\pi/2, u=1$.
$I = 2 \int_{0}^{1} u^2 (1-u^2) du = 2 \int_{0}^{1} (u^2 - u^4) du = 2 [\frac{u^3}{3} - \frac{u^5}{5}]_{0}^{1} = 2 (\frac{1}{3} - \frac{1}{5}) = 2 (\frac{2}{15}) = \frac{4}{15}$.

(D) We know that $1 - \cos 2x = 2 \sin^2 x$.
Therefore,$\sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x|$.
The integral becomes $I = \int_0^{400 \pi} \sqrt{2} |\sin x| \, dx$.
Since $|\sin x|$ is periodic with period $\pi$,we can write $I = \sqrt{2} \times 400 \int_0^{\pi} |\sin x| \, dx$.
In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$.
Thus,$I = 400 \sqrt{2} \int_0^{\pi} \sin x \, dx$.
Evaluating the integral: $\int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$.
Finally,$I = 400 \sqrt{2} \times 2 = 800 \sqrt{2}$.

(D) Let $I = \int_0^\pi (\sin^5 x \cos^3 x + \sin^4 x \cos^4 x + \sin^3 x \cos^4 x) dx$.
We use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.
For $f_1(x) = \sin^5 x \cos^3 x$,$f_1(\pi-x) = \sin^5 x (-\cos x)^3 = -\sin^5 x \cos^3 x$. Thus,$\int_0^\pi \sin^5 x \cos^3 x dx = 0$.
For $f_3(x) = \sin^3 x \cos^4 x$,$f_3(\pi-x) = \sin^3 x (-\cos x)^4 = \sin^3 x \cos^4 x$.
Using $\int_0^\pi f(x) dx = 2 \int_0^{\pi/2} f(x) dx$ for symmetric functions,$\int_0^\pi \sin^3 x \cos^4 x dx = 2 \int_0^{\pi/2} \sin^3 x \cos^4 x dx = 2 \cdot \frac{\Gamma(2) \Gamma(5/2)}{2 \Gamma(9/2)} = 2 \cdot \frac{1 \cdot (3/4 \cdot 1/2 \cdot \sqrt{\pi})}{7/2 \cdot 5/2 \cdot 3/2 \cdot 1/2 \cdot \sqrt{\pi}} = 2 \cdot \frac{3/8}{105/16} = 2 \cdot \frac{3}{8} \cdot \frac{16}{105} = \frac{4}{35}$.
For $f_2(x) = \sin^4 x \cos^4 x = (\frac{1}{2} \sin 2x)^4 = \frac{1}{16} \sin^4 2x$.
$\int_0^\pi \frac{1}{16} \sin^4 2x dx = \frac{1}{16} \cdot 2 \int_0^{\pi/2} \sin^4 2x dx$. Let $2x = t$,$dx = dt/2$.
$= \frac{1}{8} \cdot \frac{1}{2} \int_0^{\pi} \sin^4 t dt = \frac{1}{16} \cdot 2 \int_0^{\pi/2} \sin^4 t dt = \frac{1}{8} \cdot \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3\pi}{128}$.
Summing these,$I = 0 + \frac{3\pi}{128} + \frac{4}{35} = \frac{3\pi}{128} + \frac{4}{35}$.

(B) Let $I = \int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x$.
Since $f(x) = \frac{x \sin ^3 x}{4-\cos ^2 x}$ is an even function (as $f(-x) = \frac{(-x) \sin ^3(-x)}{4-\cos ^2(-x)} = \frac{(-x)(-\sin^3 x)}{4-\cos^2 x} = f(x)$),we can write:
$I = 2 \int_0^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x$ ....$(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:
$I = 2 \int_0^\pi \frac{(\pi-x) \sin ^3(\pi-x)}{4-\cos ^2(\pi-x)} d x = 2 \int_0^\pi \frac{(\pi-x) \sin ^3 x}{4-\cos ^2 x} d x$ ....(ii)
Adding $(i)$ and (ii):
$2I = 2 \pi \int_0^\pi \frac{\sin ^3 x}{4-\cos ^2 x} d x \Rightarrow I = \pi \int_0^\pi \frac{(1-\cos^2 x) \sin x}{4-\cos ^2 x} d x$.
Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$; when $x=\pi, t=-1$.
$I = -\pi \int_1^{-1} \frac{1-t^2}{4-t^2} dt = \pi \int_{-1}^1 \frac{1-t^2}{4-t^2} dt = 2\pi \int_0^1 \frac{1-t^2}{4-t^2} dt$.
$I = 2\pi \int_0^1 \frac{-(t^2-4)+3}{4-t^2} dt = 2\pi \int_0^1 (1 - \frac{3}{4-t^2}) dt$.
$I = 2\pi [t - \frac{3}{2(2)} \log |\frac{2+t}{2-t}|]_0^1 = 2\pi [1 - \frac{3}{4} \log 3]$.

(D) Let $I = \int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{\tan x}} dx$ $\qquad ....(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:
$I = \int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{\tan(\frac{\pi}{2}-x)}} dx$
$I = \int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{\cot x}} dx = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} dx$ $\qquad ....(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\frac{\pi}{2}} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}} dx$
$2I = \int_0^{\frac{\pi}{2}} 1 dx = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$
$I = \frac{\pi}{4}$

(D) Let $I = \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx \qquad ....(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1+\cos^2(\pi-x)} dx = \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx \qquad ....(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^\pi \frac{\pi \sin x}{1+\cos^2 x} dx$
$I = \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$
Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$ and when $x=\pi, t=-1$.
$I = \frac{\pi}{2} \int_1^{-1} \frac{-dt}{1+t^2} = \frac{\pi}{2} \int_{-1}^1 \frac{dt}{1+t^2}$
$I = \frac{\pi}{2} [\tan^{-1}(t)]_{-1}^1 = \frac{\pi}{2} [\tan^{-1}(1) - \tan^{-1}(-1)]$
$I = \frac{\pi}{2} [\frac{\pi}{4} - (-\frac{\pi}{4})] = \frac{\pi}{2} [\frac{\pi}{2}] = \frac{\pi^2}{4}$

(C) Let $I = \int_{-1}^1 \left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right) dx$.
Define $f(x) = \sqrt{1+x+x^2} - \sqrt{1-x+x^2}$.
We check if $f(x)$ is an odd function:
$f(-x) = \sqrt{1+(-x)+(-x)^2} - \sqrt{1-(-x)+(-x)^2} = \sqrt{1-x+x^2} - \sqrt{1+x+x^2}$.
Since $f(-x) = -(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}) = -f(x)$,the function $f(x)$ is an odd function.
For an odd function,$\int_{-a}^a f(x) dx = 0$.
Therefore,$\int_{-1}^1 \left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right) dx = 0$.

(D) Let $I = \int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx$.
Since $f(x) = \frac{x \sin x}{1+\cos^2 x}$ is an even function because $f(-x) = \frac{(-x) \sin(-x)}{1+\cos^2(-x)} = \frac{(-x)(-\sin x)}{1+\cos^2 x} = f(x)$,we can write:
$I = 2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:
$I = 2 \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1+\cos^2(\pi-x)} dx = 2 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx$.
$I = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - 2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$.
$I = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - I$.
$2I = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx \Rightarrow I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$.
Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$; when $x=\pi, t=-1$.
$I = \pi \int_1^{-1} \frac{-dt}{1+t^2} = \pi \int_{-1}^1 \frac{dt}{1+t^2} = \pi [\tan^{-1} t]_{-1}^1$.
$I = \pi (\tan^{-1}(1) - \tan^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$.

(A) $I = \int_0^\pi \frac{x \sin x}{4 \cos^2 x + 3 \sin^2 x} dx$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:
$I = \int_0^\pi \frac{(\pi - x) \sin x}{4 \cos^2(\pi - x) + 3 \sin^2(\pi - x)} dx = \int_0^\pi \frac{(\pi - x) \sin x}{4 \cos^2 x + 3 \sin^2 x} dx$
Adding the two expressions for $I$:
$2I = \int_0^\pi \frac{\pi \sin x}{4 \cos^2 x + 3 \sin^2 x} dx = \pi \int_0^\pi \frac{\sin x}{4 \cos^2 x + 3(1 - \cos^2 x)} dx$
$2I = \pi \int_0^\pi \frac{\sin x}{\cos^2 x + 3} dx$
Let $t = \cos x$,then $dt = -\sin x dx$. When $x=0, t=1$; when $x=\pi, t=-1$.
$2I = -\pi \int_1^{-1} \frac{dt}{t^2 + 3} = \pi \int_{-1}^1 \frac{dt}{t^2 + 3}$
$2I = \pi \left[ \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{t}{\sqrt{3}} \right) \right]_{-1}^1 = \frac{\pi}{\sqrt{3}} \left( \tan^{-1} \frac{1}{\sqrt{3}} - \tan^{-1} \left( -\frac{1}{\sqrt{3}} \right) \right)$
$2I = \frac{\pi}{\sqrt{3}} \left( \frac{\pi}{6} - (-\frac{\pi}{6}) \right) = \frac{\pi}{\sqrt{3}} \left( \frac{\pi}{3} \right) = \frac{\pi^2}{3 \sqrt{3}}$
$I = \frac{\pi^2}{6 \sqrt{3}}$

(B) We know that $1 - \cos 2x = 2 \sin^2 x$.
Substituting this into the integral,we get:
$I = \int_0^{50 \pi} \sqrt{2 \sin^2 x} \, dx = \sqrt{2} \int_0^{50 \pi} |\sin x| \, dx$.
Since $|\sin x|$ is a periodic function with period $\pi$,we can write:
$I = \sqrt{2} \times 50 \int_0^{\pi} |\sin x| \, dx$.
In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$.
$I = 50 \sqrt{2} \int_0^{\pi} \sin x \, dx$.
Evaluating the integral:
$I = 50 \sqrt{2} [-\cos x]_0^{\pi} = 50 \sqrt{2} (-(\cos \pi - \cos 0)) = 50 \sqrt{2} (-(-1 - 1)) = 50 \sqrt{2} (2) = 100 \sqrt{2}$.

(A) Let $I = \int_{-5}^5 \frac{x^3+5}{\sqrt{12+x}} dx$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have:
$I = \int_{-5}^5 \frac{(-5+5-x)^3+5}{\sqrt{12+(-5+5-x)}} dx = \int_{-5}^5 \frac{-x^3+5}{\sqrt{12-x}} dx$.
Adding the two expressions for $I$:
$2I = \int_{-5}^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$.
Since the integrand is even,we can write:
$2I = 2 \int_0^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$.
Thus,$I = \int_0^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$.
Given $\int_{-5}^5 f(x) dx = \int_0^5 (f(x) + g(x)) dx$,we compare the integrands:
$\int_0^5 (f(x) + g(x)) dx = \int_0^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$.
Therefore,$g(x) = \frac{5-x^3}{\sqrt{12-x}}$.