Properties of definite integration Questions in English

(D) Let $I = \int_0^{\frac{\pi}{4}} \log (1+\tan x) \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) \, dx$.
Since $\tan(\frac{\pi}{4}-x) = \frac{\tan(\frac{\pi}{4})-\tan x}{1+\tan(\frac{\pi}{4})\tan x} = \frac{1-\tan x}{1+\tan x}$,we have:
$I = \int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) \, dx = \int_0^{\frac{\pi}{4}} \log \left(\frac{1+\tan x+1-\tan x}{1+\tan x}\right) \, dx$.
$I = \int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan x}\right) \, dx = \int_0^{\frac{\pi}{4}} (\log 2 - \log(1+\tan x)) \, dx$.
$I = \int_0^{\frac{\pi}{4}} \log 2 \, dx - \int_0^{\frac{\pi}{4}} \log(1+\tan x) \, dx$.
$I = [x \log 2]_0^{\frac{\pi}{4}} - I$.
$2I = \frac{\pi}{4} \log 2$.
$I = \frac{\pi}{8} \log 2$.

(D) Consider the integral $I = \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} [2 \sin x] dx$.
In the interval $\left[\frac{\pi}{2}, \pi\right]$,$0 \le \sin x \le 1$,so $0 \le 2 \sin x \le 2$.
Specifically,for $x \in \left[\frac{\pi}{2}, \pi\right)$,$0 \le 2 \sin x < 2$,so $[2 \sin x] = 0$ or $1$.
More precisely,$2 \sin x = 1 \implies \sin x = \frac{1}{2} \implies x = \frac{5\pi}{6}$.
Thus,$[2 \sin x] = 1$ for $x \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$ is not applicable here.
For $x \in \left[\frac{\pi}{2}, \frac{5\pi}{6}\right)$,$1 \le 2 \sin x < 2 \implies [2 \sin x] = 1$.
For $x \in \left[\frac{5\pi}{6}, \pi\right]$,$0 \le 2 \sin x < 1 \implies [2 \sin x] = 0$.
For $x \in \left[\pi, \frac{7\pi}{6}\right]$,$-1 \le 2 \sin x < 0 \implies [2 \sin x] = -1$.
For $x \in \left[\frac{7\pi}{6}, \frac{3\pi}{2}\right]$,$-2 \le 2 \sin x < -1 \implies [2 \sin x] = -2$.
Calculating the integral: $\int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} 1 dx + \int_{\frac{5\pi}{6}}^{\pi} 0 dx + \int_{\pi}^{\frac{7\pi}{6}} (-1) dx + \int_{\frac{7\pi}{6}}^{\frac{3\pi}{2}} (-2) dx = \left(\frac{5\pi}{6} - \frac{\pi}{2}\right) + 0 - \left(\frac{7\pi}{6} - \pi\right) - 2\left(\frac{3\pi}{2} - \frac{7\pi}{6}\right) = \frac{\pi}{3} - \frac{\pi}{6} - 2\left(\frac{\pi}{3}\right) = \frac{\pi}{6} - \frac{2\pi}{3} = -\frac{\pi}{2} \neq 0$.
Thus,$A$ is false.
The function $f(x) = 2 \sin x$ has derivative $f'(x) = 2 \cos x$. In $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$,$\cos x \le 0$,so $f'(x) \le 0$,meaning $f(x)$ is decreasing. Thus,$R$ is true.

(C) Let $I = \int_{- 1 / 2}^{1 / 2} \{ [x] + \log (\frac{1 + x}{1 - x}) \} dx$.
We can split the integral as $I = \int_{- 1 / 2}^{1 / 2} [x] dx + \int_{- 1 / 2}^{1 / 2} \log (\frac{1 + x}{1 - x}) dx$.
Let $f(x) = \log (\frac{1 + x}{1 - x})$. Then $f(- x) = \log (\frac{1 - x}{1 + x}) = \log (\frac{1 + x}{1 - x})^{- 1} = - \log (\frac{1 + x}{1 - x}) = - f(x)$.
Since $f(x)$ is an odd function,$\int_{- 1 / 2}^{1 / 2} \log (\frac{1 + x}{1 - x}) dx = 0$.
Now,consider $\int_{- 1 / 2}^{1 / 2} [x] dx$.
For $x \in [- 1 / 2, 0)$,$[x] = - 1$.
For $x \in [0, 1 / 2]$,$[x] = 0$.
Thus,$\int_{- 1 / 2}^{1 / 2} [x] dx = \int_{- 1 / 2}^{0} (- 1) dx + \int_{0}^{1 / 2} 0 dx = [- x]_{- 1 / 2}^{0} = 0 - (1 / 2) = - 1 / 2$.
Therefore,$I = - 1 / 2 + 0 = - 1 / 2$.

(B) We are given $I_n = \int_0^{\pi / 2} \sin^n(x) dx$.
Using integration by parts,let $u = \sin^{n-1}(x)$ and $dv = \sin(x) dx$.
Then $du = (n-1) \sin^{n-2}(x) \cos(x) dx$ and $v = -\cos(x)$.
$I_n = [-\sin^{n-1}(x) \cos(x)]_0^{\pi / 2} + \int_0^{\pi / 2} (n-1) \sin^{n-2}(x) \cos^2(x) dx$.
Since $\cos(\pi/2) = 0$ and $\sin(0) = 0$,the boundary term is $0$.
$I_n = (n-1) \int_0^{\pi / 2} \sin^{n-2}(x) (1 - \sin^2(x)) dx$.
$I_n = (n-1) \int_0^{\pi / 2} \sin^{n-2}(x) dx - (n-1) \int_0^{\pi / 2} \sin^n(x) dx$.
$I_n = (n-1) I_{n-2} - (n-1) I_n$.
$I_n + (n-1) I_n = (n-1) I_{n-2}$.
$n I_n = (n-1) I_{n-2}$.
$I_n = \frac{n-1}{n} I_{n-2}$.
Comparing this with $I_n = k I_{n-2}$,we get $k = \frac{n-1}{n}$.
Hence,option $B$ is correct.

(B) Let $I = \int_0^{\pi / 2} |\sin t - \cos t| \, dt$.
Since $\sin t - \cos t \le 0$ for $t \in [0, \pi / 4]$ and $\sin t - \cos t \ge 0$ for $t \in [\pi / 4, \pi / 2]$,we split the integral:
$I = \int_0^{\pi / 4} (\cos t - \sin t) \, dt + \int_{\pi / 4}^{\pi / 2} (\sin t - \cos t) \, dt$.
Evaluating the first part: $[\sin t + \cos t]_0^{\pi / 4} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.
Evaluating the second part: $[-\cos t - \sin t]_{\pi / 4}^{\pi / 2} = (0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 + \sqrt{2} = \sqrt{2} - 1$.
Adding both parts: $I = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)$.
Thus,option $B$ is correct.

(D) Let $I = \int_0^{\pi / 2} \frac{\sin ^3 x \cos x}{\sin ^4 x+\cos ^4 x} \, dx \quad \dots (i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:
$I = \int_0^{\pi / 2} \frac{\sin ^3(\frac{\pi}{2}-x) \cos(\frac{\pi}{2}-x)}{\sin ^4(\frac{\pi}{2}-x)+\cos ^4(\frac{\pi}{2}-x)} \, dx$
$I = \int_0^{\pi / 2} \frac{\cos ^3 x \sin x}{\cos ^4 x+\sin ^4 x} \, dx \quad \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\pi / 2} \frac{\sin x \cos x (\sin ^2 x + \cos ^2 x)}{\sin ^4 x+\cos ^4 x} \, dx$
$2I = \int_0^{\pi / 2} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x} \, dx$
Divide numerator and denominator by $\cos^4 x$:
$2I = \int_0^{\pi / 2} \frac{\tan x \sec^2 x}{\tan ^4 x+1} \, dx$
Let $\tan^2 x = t$,then $2 \tan x \sec^2 x \, dx = dt$,so $\tan x \sec^2 x \, dx = \frac{dt}{2}$.
When $x=0, t=0$ and when $x=\frac{\pi}{2}, t \to \infty$.
$2I = \int_0^{\infty} \frac{1}{t^2+1} \cdot \frac{dt}{2} = \frac{1}{2} [\tan^{-1} t]_0^{\infty} = \frac{1}{2} (\frac{\pi}{2} - 0) = \frac{\pi}{4}$.
$2I = \frac{\pi}{4} \implies I = \frac{\pi}{8}$.

(C) Given $I_n = \int_0^{\pi / 4} \tan^n x \, dx$.
Consider $I_r + I_{r+2} = \int_0^{\pi / 4} \tan^r x \, dx + \int_0^{\pi / 4} \tan^{r+2} x \, dx$.
$I_r + I_{r+2} = \int_0^{\pi / 4} \tan^r x (1 + \tan^2 x) \, dx$.
Since $1 + \tan^2 x = \sec^2 x$,we have $I_r + I_{r+2} = \int_0^{\pi / 4} \tan^r x \sec^2 x \, dx$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$. When $x=0, t=0$ and when $x=\pi/4, t=1$.
$I_r + I_{r+2} = \int_0^1 t^r \, dt = \left[ \frac{t^{r+1}}{r+1} \right]_0^1 = \frac{1}{r+1}$.
Thus,$I_2+I_4 = \frac{1}{3}$,$I_3+I_5 = \frac{1}{4}$,$I_4+I_6 = \frac{1}{5}$,and so on.
The terms are $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots$,which are in harmonic progression $(HP)$ because their reciprocals $3, 4, 5, \ldots$ are in arithmetic progression.

(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:
$I = \int_0^\pi \frac{(\pi - x) \sin(\pi - x)}{1+\cos^2(\pi - x)} dx = \int_0^\pi \frac{(\pi - x) \sin x}{1+\cos^2 x} dx$.
Adding the two expressions for $I$:
$2I = \int_0^\pi \frac{x \sin x + (\pi - x) \sin x}{1+\cos^2 x} dx = \int_0^\pi \frac{\pi \sin x}{1+\cos^2 x} dx$.
$2I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$.
Let $u = \cos x$,then $du = -\sin x dx$.
When $x=0, u=1$; when $x=\pi, u=-1$.
$2I = \pi \int_1^{-1} \frac{-du}{1+u^2} = \pi \int_{-1}^1 \frac{du}{1+u^2}$.
$2I = \pi [\tan^{-1}(u)]_{-1}^1 = \pi (\tan^{-1}(1) - \tan^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$.
Therefore,$I = \frac{\pi^2}{4}$.

(C) Let $I = \int_0^\pi \frac{x \sin x}{\sin^2 x + 2 \cos^2 x} dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^\pi \frac{(\pi - x) \sin(\pi - x)}{\sin^2(\pi - x) + 2 \cos^2(\pi - x)} dx = \int_0^\pi \frac{(\pi - x) \sin x}{\sin^2 x + 2 \cos^2 x} dx$.
Adding the two expressions for $I$:
$2I = \int_0^\pi \frac{x \sin x + (\pi - x) \sin x}{\sin^2 x + 2 \cos^2 x} dx = \pi \int_0^\pi \frac{\sin x}{\sin^2 x + 2 \cos^2 x} dx$.
Since $\sin^2 x = 1 - \cos^2 x$,the denominator becomes $1 - \cos^2 x + 2 \cos^2 x = 1 + \cos^2 x$.
$2I = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} dx$.
Let $u = \cos x$,then $du = -\sin x dx$. When $x=0, u=1$; when $x=\pi, u=-1$.
$2I = \pi \int_1^{-1} \frac{-du}{1 + u^2} = \pi \int_{-1}^1 \frac{du}{1 + u^2} = \pi [\tan^{-1} u]_{-1}^1$.
$2I = \pi (\tan^{-1}(1) - \tan^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$.
Therefore,$I = \frac{\pi^2}{4}$.

(C) Let $I = \int_{5 \pi}^{25 \pi} |\sin 2x + \cos 2x| dx$.
We know that $\sin 2x + \cos 2x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin 2x + \frac{1}{\sqrt{2}} \cos 2x \right) = \sqrt{2} \sin(2x + \frac{\pi}{4})$.
So,$I = \int_{5 \pi}^{25 \pi} |\sqrt{2} \sin(2x + \frac{\pi}{4})| dx = \sqrt{2} \int_{5 \pi}^{25 \pi} |\sin(2x + \frac{\pi}{4})| dx$.
Let $2x + \frac{\pi}{4} = t$,then $2 dx = dt$,so $dx = \frac{dt}{2}$.
When $x = 5\pi$,$t = 10\pi + \frac{\pi}{4} = \frac{41\pi}{4}$.
When $x = 25\pi$,$t = 50\pi + \frac{\pi}{4} = \frac{201\pi}{4}$.
$I = \frac{\sqrt{2}}{2} \int_{41\pi/4}^{201\pi/4} |\sin t| dt$.
The period of $|\sin t|$ is $\pi$. The length of the interval is $\frac{201\pi}{4} - \frac{41\pi}{4} = \frac{160\pi}{4} = 40\pi$.
Since the integral of $|\sin t|$ over one period $[0, \pi]$ is $\int_0^{\pi} \sin t dt = 2$,the integral over $40$ periods is $40 \times 2 = 80$.
Thus,$I = \frac{\sqrt{2}}{2} \times 80 = 40\sqrt{2}$.

(C) Given $f(t) = \int_0^t \tan^{(2n-1)} x \, dx$.
We need to evaluate $f(t+\pi) = \int_0^{t+\pi} \tan^{(2n-1)} x \, dx$.
Using the property of definite integrals,$\int_0^{t+\pi} = \int_0^t + \int_t^{t+\pi}$.
So,$f(t+\pi) = \int_0^t \tan^{(2n-1)} x \, dx + \int_t^{t+\pi} \tan^{(2n-1)} x \, dx$.
Since $\tan x$ has a period of $\pi$,$\int_t^{t+\pi} \tan^{(2n-1)} x \, dx = \int_0^{\pi} \tan^{(2n-1)} x \, dx$.
Thus,$f(t+\pi) = f(t) + \int_0^{\pi} \tan^{(2n-1)} x \, dx$.
Since $f(\pi) = \int_0^{\pi} \tan^{(2n-1)} x \, dx$,we have $f(t+\pi) = f(t) + f(\pi)$.

(A) Let $I = \int_0^{\frac{\pi}{2}} \sqrt{\tan x} \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^{\frac{\pi}{2}} \sqrt{\tan(\frac{\pi}{2} - x)} \, dx = \int_0^{\frac{\pi}{2}} \sqrt{\cot x} \, dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\frac{\pi}{2}} (\sqrt{\tan x} + \sqrt{\cot x}) \, dx = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$.
Multiply numerator and denominator by $\sqrt{2}$:
$2I = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$.
Let $u = \sin x - \cos x$,then $du = (\cos x + \sin x) \, dx$.
When $x=0, u=-1$; when $x=\frac{\pi}{2}, u=1$.
$2I = \sqrt{2} \int_{-1}^1 \frac{du}{\sqrt{1 - u^2}} = \sqrt{2} [\sin^{-1} u]_{-1}^1 = \sqrt{2} (\frac{\pi}{2} - (-\frac{\pi}{2})) = \sqrt{2} \pi$.
Thus,$I = \frac{\sqrt{2} \pi}{2} = \frac{\pi}{\sqrt{2}}$.

(D) Let $I = \int_0^2 x^3(2-x)^4 \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^2 (2-x)^3(2-(2-x))^4 \, dx = \int_0^2 (2-x)^3 x^4 \, dx$.
Expanding the terms:
$I = \int_0^2 (8 - 12x + 6x^2 - x^3) x^4 \, dx = \int_0^2 (8x^4 - 12x^5 + 6x^6 - x^7) \, dx$.
Integrating term by term:
$I = \left[ \frac{8x^5}{5} - \frac{12x^6}{6} + \frac{6x^7}{7} - \frac{x^8}{8} \right]_0^2$.
$I = \left[ \frac{8(32)}{5} - 2(64) + \frac{6(128)}{7} - \frac{256}{8} \right]$.
$I = \frac{256}{5} - 128 + \frac{768}{7} - 32 = \frac{256}{5} + \frac{768}{7} - 160$.
$I = \frac{1792 + 3840 - 5600}{35} = \frac{32}{35}$.

(A) Let $I = \int_0^{\frac{\pi}{2}} \frac{x \tan x \sec^2 x}{\tan^4 x + 1} dx$ ...$(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - x) \tan(\frac{\pi}{2} - x) \sec^2(\frac{\pi}{2} - x)}{\tan^4(\frac{\pi}{2} - x) + 1} dx$
$I = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - x) \cot x \csc^2 x}{\cot^4 x + 1} dx$
Since $\cot x = \frac{1}{\tan x}$ and $\csc^2 x = \frac{\sec^2 x}{\tan^2 x}$,the expression becomes:
$I = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - x) \frac{1}{\tan x} \cdot \frac{\sec^2 x}{\tan^2 x}}{\frac{1}{\tan^4 x} + 1} dx = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - x) \tan x \sec^2 x}{1 + \tan^4 x} dx$ ...(ii)
Adding $(i)$ and (ii):
$2I = \int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2} \tan x \sec^2 x}{\tan^4 x + 1} dx$
$I = \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\tan x \sec^2 x}{\tan^4 x + 1} dx$
Let $u = \tan^2 x$,then $du = 2 \tan x \sec^2 x dx$,so $\tan x \sec^2 x dx = \frac{du}{2}$.
$I = \frac{\pi}{4} \int_0^{\infty} \frac{1}{u^2 + 1} \cdot \frac{du}{2} = \frac{\pi}{8} [\tan^{-1} u]_0^{\infty} = \frac{\pi}{8} \cdot \frac{\pi}{2} = \frac{\pi^2}{16}$

(A) Let $I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$.
Split the integral into two parts:
$I=\frac{3}{\pi} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x + \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} \log \left(\frac{4+\sin x}{4-\sin x}\right) d x$.
The second integral is zero because the integrand is an odd function.
Thus,$I=\frac{3}{\pi} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x = \frac{6}{\pi} \int_0^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x$.
Using $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$,we get:
$I=\frac{6}{\pi} \int_0^{\frac{\pi}{4}} \frac{\sec^2 x}{2(1+\tan^2 x) - (1-\tan^2 x)} d x = \frac{6}{\pi} \int_0^{\frac{\pi}{4}} \frac{\sec^2 x}{1+3\tan^2 x} d x$.
Let $\tan x = t$,then $\sec^2 x d x = d t$. As $x: 0 \to \frac{\pi}{4}$,$t: 0 \to 1$.
$I=\frac{6}{\pi} \int_0^1 \frac{d t}{1+3t^2} = \frac{6}{\pi} \left[ \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}t) \right]_0^1 = \frac{6}{\pi \sqrt{3}} \tan^{-1}(\sqrt{3}) = \frac{2\sqrt{3}}{\pi} \cdot \frac{\pi}{3} = \frac{2}{\sqrt{3}}$.
Therefore,$3I^2 = 3 \left( \frac{2}{\sqrt{3}} \right)^2 = 3 \cdot \frac{4}{3} = 4$.

(A) Let $I = \int_5^9 \frac{\log_3 x^2}{\log_3 x^2 + \log_3(588 - 84x + 3x^2)} dx$.
Using the property $\log_3(588 - 84x + 3x^2) = \log_3(3(196 - 28x + x^2)) = \log_3(3(14 - x)^2)$,we have:
$I = \int_5^9 \frac{\log_3 x^2}{\log_3 x^2 + \log_3(3(14 - x)^2)} dx \dots (1)$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a + b - x) dx$,where $a=5$ and $b=9$,$a+b-x = 14-x$:
$I = \int_5^9 \frac{\log_3(14 - x)^2}{\log_3(14 - x)^2 + \log_3(3(14 - (14 - x))^2)} dx$.
$I = \int_5^9 \frac{\log_3(14 - x)^2}{\log_3(14 - x)^2 + \log_3(3x^2)} dx \dots (2)$.
Adding $(1)$ and $(2)$:
$2I = \int_5^9 \frac{\log_3 x^2 + \log_3(3(14 - x)^2)}{\log_3 x^2 + \log_3(3(14 - x)^2)} dx$.
$2I = \int_5^9 1 dx = [x]_5^9 = 9 - 5 = 4$.
$I = \frac{4}{2} = 2$.

(A) Let $I = \int_3^5 (x-3)^3 (5-x)^5 dx$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $I = \int_3^5 (5-x)^3 (x-3)^5 dx$.
Alternatively,use the substitution $x-3 = 2t$,so $x=3+2t$ and $dx=2dt$.
When $x=3, t=0$ and when $x=5, t=1$.
$I = \int_0^1 (2t)^3 (5-(3+2t))^5 (2dt) = \int_0^1 8t^3 (2-2t)^5 (2dt) = 16 \times 2^5 \int_0^1 t^3 (1-t)^5 dt$.
$I = 512 \int_0^1 t^3 (1-t)^5 dt$.
Using the Beta function $\int_0^1 x^{m-1} (1-x)^{n-1} dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} = \frac{(m-1)!(n-1)!}{(m+n-1)!}$.
Here $m-1=3 \implies m=4$ and $n-1=5 \implies n=6$.
$I = 512 \times \frac{3! \times 5!}{9!} = 512 \times \frac{6 \times 120}{362880} = 512 \times \frac{720}{362880} = 512 \times \frac{1}{504} = \frac{512}{504} = \frac{64}{63}$.

(A) Let $I = \int_{\frac{1}{3}}^3 \frac{1}{x} \sin \left(\frac{1}{x}-x\right) d x$.
Substitute $x = \frac{1}{t}$,which implies $dx = -\frac{1}{t^2} dt$.
When $x = \frac{1}{3}$,$t = 3$,and when $x = 3$,$t = \frac{1}{3}$.
Substituting these into the integral:
$I = \int_3^{\frac{1}{3}} t \sin \left(t - \frac{1}{t}\right) \left(-\frac{1}{t^2}\right) dt$
$I = \int_3^{\frac{1}{3}} -\frac{1}{t} \sin \left(t - \frac{1}{t}\right) dt$
Using the property $\sin(- \theta) = -\sin(\theta)$,we have $\sin(t - \frac{1}{t}) = -\sin(\frac{1}{t} - t)$.
$I = \int_3^{\frac{1}{3}} \frac{1}{t} \sin \left(\frac{1}{t} - t\right) dt$
Reversing the limits of integration changes the sign:
$I = -\int_{\frac{1}{3}}^3 \frac{1}{t} \sin \left(\frac{1}{t} - t\right) dt$
Since the variable of integration is a dummy variable,we can replace $t$ with $x$:
$I = -\int_{\frac{1}{3}}^3 \frac{1}{x} \sin \left(\frac{1}{x} - x\right) dx$
$I = -I$
$2I = 0 \implies I = 0$.

(C) Let $I = \int_0^{\pi / 4} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x$.
Dividing numerator and denominator by $\cos^2 x$,we get:
$I = \int_0^{\pi / 4} \frac{1}{1+4 \tan ^2 x} d x$.
Let $\tan x = t$,then $\sec^2 x dx = dt$,which implies $dx = \frac{dt}{1+t^2}$.
When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=1$.
So,$I = \int_0^1 \frac{1}{(1+t^2)(1+4t^2)} dt$.
Using partial fractions: $\frac{1}{(1+t^2)(1+4t^2)} = \frac{1}{3} \left( \frac{4}{1+4t^2} - \frac{1}{1+t^2} \right)$.
$I = \frac{1}{3} \int_0^1 \left( \frac{4}{1+4t^2} - \frac{1}{1+t^2} \right) dt$.
$I = \frac{1}{3} \left[ 2 \tan^{-1}(2t) - \tan^{-1}(t) \right]_0^1$.
$I = \frac{1}{3} \left[ (2 \tan^{-1}(2) - \tan^{-1}(1)) - (0 - 0) \right]$.
$I = \frac{1}{3} \left[ 2 \tan^{-1}(2) - \frac{\pi}{4} \right] = \frac{2}{3} \tan^{-1}(2) - \frac{\pi}{12}$.

(D) Let $I = \int_0^{\pi / 2} \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x$.
We know that $1-\sin 2x = (\cos x - \sin x)^2$ and $1+\sin 2x = (\cos x + \sin x)^2$.
So,$\sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} = \left| \frac{\cos x - \sin x}{\cos x + \sin x} \right| = |\tan(\frac{\pi}{4} - x)|$.
Since $\tan(\frac{\pi}{4} - x) \ge 0$ for $x \in [0, \frac{\pi}{4}]$ and $\tan(\frac{\pi}{4} - x) < 0$ for $x \in (\frac{\pi}{4}, \frac{\pi}{2}]$,we split the integral:
$I = \int_0^{\pi / 4} \tan(\frac{\pi}{4} - x) dx + \int_{\pi / 4}^{\pi / 2} -\tan(\frac{\pi}{4} - x) dx$.
Using $\int \tan(ax+b) dx = -\frac{1}{a} \log|\cos(ax+b)| + C$:
$I = [\log|\cos(\frac{\pi}{4} - x)|]_0^{\pi / 4} - [\log|\cos(\frac{\pi}{4} - x)|]_{\pi / 4}^{\pi / 2}$.
$I = (\log 1 - \log \frac{1}{\sqrt{2}}) - (\log \frac{1}{\sqrt{2}} - \log 1) = \log \sqrt{2} + \log \sqrt{2} = 2 \log \sqrt{2} = \log 2$.
Therefore,the original expression is $e^I = e^{\log 2} = 2$.

(A) Let $I = \int_{-\pi}^{\pi} \frac{\sin^2 x}{1+a^x} dx$ $(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we replace $x$ with $-\pi + \pi - x = -x$:
$I = \int_{-\pi}^{\pi} \frac{\sin^2(-x)}{1+a^{-x}} dx = \int_{-\pi}^{\pi} \frac{\sin^2 x}{1+\frac{1}{a^x}} dx = \int_{-\pi}^{\pi} \frac{a^x \sin^2 x}{a^x+1} dx$ $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{-\pi}^{\pi} \frac{\sin^2 x + a^x \sin^2 x}{1+a^x} dx = \int_{-\pi}^{\pi} \frac{(1+a^x) \sin^2 x}{1+a^x} dx = \int_{-\pi}^{\pi} \sin^2 x dx$
Since $\sin^2 x$ is an even function:
$2I = 2 \int_{0}^{\pi} \sin^2 x dx = \int_{0}^{\pi} (1 - \cos 2x) dx$
$2I = [x - \frac{\sin 2x}{2}]_{0}^{\pi} = (\pi - 0) - (0 - 0) = \pi$
$I = \frac{\pi}{2}$

(D) Given $\int_0^{2 \pi} \sin^m x \cos^n x \, dx = 4 \int_0^{\pi/2} \sin^m x \cos^n x \, dx$. Since $\int_0^{2 \pi} f(x) \, dx = 4 \int_0^{\pi/2} f(x) \, dx$ holds for even functions of $\sin$ and $\cos$ over quadrants,this implies $m$ and $n$ must be even.
Given $\int_0^{2 \pi} \sin^p x \cos^n x \, dx = 0$. Since $n$ is even,$\cos^n x \ge 0$. For the integral to be $0$,$\sin^p x$ must be odd,so $p$ is odd.
Given $\int_0^{\pi} \sin^p x \cos^q x \, dx = 0$. Since $p$ is odd,$\sin^p x$ is symmetric about $\pi/2$ but changes sign. For the integral to be $0$,$\cos^q x$ must be odd,so $q$ is odd.
Now,$a = m + n + p = \text{even} + \text{even} + \text{odd} = \text{odd}$.
And $b = m + n + q = \text{even} + \text{even} + \text{odd} = \text{odd}$.
Thus,both $a$ and $b$ are odd numbers.

(D) Let $I = \int_{-1}^1 \frac{\log 2 - \log(1+x)}{\sqrt{1-x^2}} dx$.
Substitute $x = \cos \theta$,then $dx = -\sin \theta d\theta$.
When $x = -1$,$\theta = \pi$. When $x = 1$,$\theta = 0$.
$I = \int_{\pi}^0 \frac{\log 2 - \log(1+\cos \theta)}{\sqrt{1-\cos^2 \theta}} (-\sin \theta) d\theta = \int_0^{\pi} \frac{\log 2 - \log(2 \cos^2(\theta/2))}{\sin \theta} \sin \theta d\theta$.
$I = \int_0^{\pi} (\log 2 - \log 2 - 2 \log(\cos(\theta/2))) d\theta = -2 \int_0^{\pi} \log(\cos(\theta/2)) d\theta$.
Let $u = \theta/2$,then $d\theta = 2du$.
$I = -2 \int_0^{\pi/2} \log(\cos u) (2 du) = -4 \int_0^{\pi/2} \log(\cos u) du$.
Using the standard integral $\int_0^{\pi/2} \log(\cos u) du = -\frac{\pi}{2} \log 2$.
$I = -4 \times (-\frac{\pi}{2} \log 2) = 2 \pi \log 2$.

(C) Let $I = \int_0^{\pi / 2} \tan^{14}(\frac{x}{2}) dx$. Let $t = \frac{x}{2}$,then $dx = 2dt$. When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\frac{\pi}{4}$.
$I = 2 \int_0^{\pi/4} \tan^{14}(t) dt$.
Using the reduction formula $I_n = \int \tan^n(t) dt = \frac{\tan^{n-1}(t)}{n-1} - I_{n-2}$,we have $\int_0^{\pi/4} \tan^n(t) dt = \frac{1}{n-1} - \int_0^{\pi/4} \tan^{n-2}(t) dt$.
Applying this repeatedly for $n=14, 12, \dots, 2$:
$I_{14} = \frac{1}{13} - I_{12} = \frac{1}{13} - (\frac{1}{11} - I_{10}) = \frac{1}{13} - \frac{1}{11} + \frac{1}{9} - \frac{1}{7} + \frac{1}{5} - \frac{1}{3} + \frac{1}{1} - \int_0^{\pi/4} 1 dt$.
Since $\int_0^{\pi/4} 1 dt = \frac{\pi}{4}$,we get $I = 2 [ \sum_{k=1}^7 \frac{(-1)^{k+1}}{2k-1} - \frac{\pi}{4} ]$.
Comparing this with the given expression $2 [ \sum_{n=1}^7 f(n) - \frac{\pi}{4} ]$,we find $f(n) = \frac{(-1)^{n+1}}{2n-1}$.

(C) Let $I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin ^4(4 x)}{1+e^{4 x}} d x$ ....$(i)$
Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,where $a = -\frac{\pi}{8}$ and $b = \frac{\pi}{8}$,we have $a+b = 0$.
$I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin ^4(-4 x)}{1+e^{-4 x}} d x = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin ^4(4 x)}{1+\frac{1}{e^{4 x}}} d x = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{e^{4 x} \sin ^4(4 x)}{e^{4 x}+1} d x$ ....(ii)
Adding $(i)$ and (ii):
$2I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{(1+e^{4 x}) \sin ^4(4 x)}{1+e^{4 x}} d x = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \sin ^4(4 x) d x$
Since $\sin^4(4x)$ is an even function,$2I = 2 \int_0^{\frac{\pi}{8}} \sin ^4(4 x) d x$,so $I = \int_0^{\frac{\pi}{8}} \sin ^4(4 x) d x$.
Let $4x = t$,then $4 dx = dt$,so $dx = \frac{1}{4} dt$. When $x=0, t=0$; when $x=\frac{\pi}{8}, t=\frac{\pi}{2}$.
$I = \frac{1}{4} \int_0^{\frac{\pi}{2}} \sin ^4 t d t$.
Using Wallis' formula $\int_0^{\frac{\pi}{2}} \sin^n t dt = \frac{(n-1)!!}{n!!} \times \frac{\pi}{2}$ for even $n$:
$I = \frac{1}{4} \times \left( \frac{3 \times 1}{4 \times 2} \times \frac{\pi}{2} \right) = \frac{1}{4} \times \frac{3 \pi}{16} = \frac{3 \pi}{64}$.

(D) Let $I = \int_{-23}^{23} \frac{dx}{3+f(x)} \dots (i)$
Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,where $a = -23$ and $b = 23$,we have $a+b = 0$.
So,$I = \int_{-23}^{23} \frac{dx}{3+f(-x)}$.
Given $f(x)f(-x) = 9$,we have $f(-x) = \frac{9}{f(x)}$.
Substituting this into the integral:
$I = \int_{-23}^{23} \frac{dx}{3 + \frac{9}{f(x)}} = \int_{-23}^{23} \frac{f(x) dx}{3f(x) + 9} = \int_{-23}^{23} \frac{f(x) dx}{3(f(x) + 3)} \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{-23}^{23} \left( \frac{1}{3+f(x)} + \frac{f(x)}{3(f(x)+3)} \right) dx$
$2I = \int_{-23}^{23} \frac{3 + f(x)}{3(3+f(x))} dx = \int_{-23}^{23} \frac{1}{3} dx$
$2I = \frac{1}{3} [x]_{-23}^{23} = \frac{1}{3} (23 - (-23)) = \frac{46}{3}$
$I = \frac{46}{6}$

(B) Let $I = \int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec (2023 x)}{1+(2023)^{(2023 x)}} d x$.
Using the property $\int_{-a}^{a} f(x) d x = \int_{0}^{a} [f(x) + f(-x)] d x$,we have:
$f(x) = \frac{\sec (2023 x)}{1+(2023)^{(2023 x)}}$
$f(-x) = \frac{\sec (-2023 x)}{1+(2023)^{(-2023 x)}} = \frac{\sec (2023 x)}{1+\frac{1}{(2023)^{(2023 x)}}} = \frac{(2023)^{(2023 x)} \sec (2023 x)}{(2023)^{(2023 x)}+1}$
Adding $f(x)$ and $f(-x)$:
$f(x) + f(-x) = \frac{\sec (2023 x) [1 + (2023)^{(2023 x)}]}{1+(2023)^{(2023 x)}} = \sec (2023 x)$
Therefore,$I = \int_{0}^{\frac{\pi}{8092}} \sec (2023 x) d x$
$= \left[ \frac{1}{2023} \log |\sec (2023 x) + \tan (2023 x)| \right]_{0}^{\frac{\pi}{8092}}$
$= \frac{1}{2023} [\log |\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| - \log |\sec(0) + \tan(0)|]$
$= \frac{1}{2023} [\log (\sqrt{2} + 1) - \log (1 + 0)]$
$= \frac{\log (\sqrt{2} + 1)}{2023}$

(A) Let $I = \int_0^\pi \frac{x \cos^2 x}{1+\sin x} dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^\pi \frac{(\pi-x) \cos^2(\pi-x)}{1+\sin(\pi-x)} dx = \int_0^\pi \frac{(\pi-x) \cos^2 x}{1+\sin x} dx$.
Adding the two expressions for $I$:
$2I = \int_0^\pi \frac{\pi \cos^2 x}{1+\sin x} dx = \pi \int_0^\pi \frac{1-\sin^2 x}{1+\sin x} dx$.
Since $1-\sin^2 x = (1-\sin x)(1+\sin x)$,we have:
$2I = \pi \int_0^\pi (1-\sin x) dx$.
$2I = \pi [x + \cos x]_0^\pi$.
$2I = \pi [(\pi + \cos \pi) - (0 + \cos 0)] = \pi [(\pi - 1) - (1)] = \pi(\pi - 2)$.
Therefore,$I = \frac{\pi(\pi-2)}{2}$.

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(\sin x + \cos x) dx$ ...$(i)$
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a = -\frac{\pi}{2}$ and $b = \frac{\pi}{2}$,we have $a+b = 0$.
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2(-x) \cos^2(-x)(\sin(-x) + \cos(-x)) dx$
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(-\sin x + \cos x) dx$ ...(ii)
Adding $(i)$ and (ii):
$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(\sin x + \cos x - \sin x + \cos x) dx$
$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(2 \cos x) dx$
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^3 x dx$
Since $\sin^2 x \cos^3 x$ is an even function,$I = 2 \int_0^{\frac{\pi}{2}} \sin^2 x \cos^3 x dx$.
Let $\sin x = t$,then $\cos x dx = dt$. When $x=0, t=0$; when $x=\frac{\pi}{2}, t=1$.
$I = 2 \int_0^1 t^2 (1-t^2) dt = 2 \int_0^1 (t^2 - t^4) dt$
$I = 2 \left[ \frac{t^3}{3} - \frac{t^5}{5} \right]_0^1 = 2 \left( \frac{1}{3} - \frac{1}{5} \right) = 2 \left( \frac{2}{15} \right) = \frac{4}{15}$.

(C) The function $f(x) = x|x|$ is an odd function because $f(-x) = (-x)|-x| = -x|x| = -f(x)$.
For an odd function,the definite integral over a symmetric interval $[-a, a]$ is always zero.
Alternatively,we can evaluate it as follows:
$\int_{-1}^1 x|x| \, dx = \int_{-1}^0 x(-x) \, dx + \int_0^1 x(x) \, dx$
$= -\int_{-1}^0 x^2 \, dx + \int_0^1 x^2 \, dx$
$= -\left[\frac{x^3}{3}\right]_{-1}^0 + \left[\frac{x^3}{3}\right]_0^1$
$= -\left(0 - \left(-\frac{1}{3}\right)\right) + \left(\frac{1}{3} - 0\right)$
$= -\frac{1}{3} + \frac{1}{3} = 0$

(B) Let $I = \int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ ...$(i)$
Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,where $a=3$ and $b=6$,we have $a+b-x = 9-x$.
Substituting this into the integral:
$I = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{9-(9-x)}+\sqrt{9-x}} d x$
$I = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$ ...(ii)
Adding equations $(i)$ and (ii):
$2I = \int_3^6 \frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$
$2I = \int_3^6 1 d x$
$2I = [x]_3^6 = 6 - 3 = 3$
$I = \frac{3}{2}$

(B) We are given the integral $I = \int_0^{2a} f(x) dx$.
Using the property of definite integrals,we can split the interval:
$I = \int_0^a f(x) dx + \int_a^{2a} f(x) dx$.
In the second integral,let $x = 2a - t$. Then $dx = -dt$.
When $x = a$,$t = a$,and when $x = 2a$,$t = 0$.
Substituting these into the second integral:
$\int_a^{2a} f(x) dx = \int_a^0 f(2a - t) (-dt) = \int_0^a f(2a - t) dt = \int_0^a f(2a - x) dx$.
Thus,$I = \int_0^a f(x) dx + \int_0^a f(2a - x) dx = \int_0^a (f(x) + f(2a - x)) dx$.
Comparing this with the given options,option $(b)$ is the correct representation.

(C) We know that the property of definite integrals states that $\int_{-a}^a f(x) dx = \int_0^a f(x) dx + \int_0^a f(-x) dx$.
Comparing this with the given equation $\int_{-a}^a f(x) dx = \int_0^a f(x) dx + \int_0^a g(x) dx$,we can clearly see that $g(x) = f(-x)$.

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

(D) Let $I = \int_0^{\pi} x f(\sin x) \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^{\pi} (\pi - x) f(\sin(\pi - x)) \, dx$.
Since $\sin(\pi - x) = \sin x$,this becomes:
$I = \int_0^{\pi} (\pi - x) f(\sin x) \, dx = \pi \int_0^{\pi} f(\sin x) \, dx - \int_0^{\pi} x f(\sin x) \, dx$.
$I = \pi \int_0^{\pi} f(\sin x) \, dx - I$.
$2I = \pi \int_0^{\pi} f(\sin x) \, dx$.
Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$,here $2a = \pi$,so $a = \frac{\pi}{2}$.
Since $f(\sin(\pi - x)) = f(\sin x)$,we have:
$2I = \pi \cdot 2 \int_0^{\frac{\pi}{2}} f(\sin x) \, dx$.
$I = \pi \int_0^{\frac{\pi}{2}} f(\sin x) \, dx$.

(D) Let $I = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{dx}{1+\cos x} \quad \dots (i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi$.
So,$I = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{dx}{1+\cos(\pi-x)} = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{dx}{1-\cos x} \quad \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \left( \frac{1}{1+\cos x} + \frac{1}{1-\cos x} \right) dx$
$2I = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{1-\cos x + 1+\cos x}{1-\cos^2 x} dx = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{2}{\sin^2 x} dx$
$2I = 2 \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \operatorname{cosec}^2 x dx$
$I = [-\cot x]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}$
$I = -(\cot \frac{3\pi}{4} - \cot \frac{\pi}{4}) = -(-1 - 1) = 2$
Since $2 \sin \frac{\pi}{2} = 2(1) = 2$,the correct option is $D$.

(D) Let $I = \int_{\pi / 4}^{\pi / 2} \frac{3 \, dx}{1+e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}} \quad \dots (i)$
Using the property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$,where $a = \frac{\pi}{4}$ and $b = \frac{\pi}{2}$,we have $a+b-x = \frac{3\pi}{4} - x$.
$I = \int_{\pi / 4}^{\pi / 2} \frac{3 \, dx}{1+e^{\sqrt{8} \sin \left(\frac{3 \pi}{4} - x - \frac{3 \pi}{8}\right)}}$
$I = \int_{\pi / 4}^{\pi / 2} \frac{3 \, dx}{1+e^{\sqrt{8} \sin \left(\frac{3 \pi}{8} - x\right)}}$
Since $\sin(\theta) = -\sin(-\theta)$,we have $\sin(\frac{3\pi}{8} - x) = -\sin(x - \frac{3\pi}{8})$.
$I = \int_{\pi / 4}^{\pi / 2} \frac{3 \, dx}{1+e^{-\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}} = \int_{\pi / 4}^{\pi / 2} \frac{3 e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)} \, dx}{e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)} + 1} \quad \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{\pi / 4}^{\pi / 2} \frac{3(1 + e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)})}{1 + e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}} \, dx$
$2I = \int_{\pi / 4}^{\pi / 2} 3 \, dx = 3[x]_{\pi / 4}^{\pi / 2} = 3(\frac{\pi}{2} - \frac{\pi}{4}) = 3(\frac{\pi}{4}) = \frac{3\pi}{4}$
$I = \frac{3\pi}{8}$

(C) For Assertion $(A)$: We know that $\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$.
In the first integral,let $x = -t$,then $dx = -dt$. When $x = -a, t = a$ and when $x = 0, t = 0$.
So,$\int_{-a}^0 f(x) dx = \int_a^0 f(-t) (-dt) = \int_0^a f(-t) dt = \int_0^a f(-x) dx$.
Thus,$\int_{-a}^a f(x) dx = \int_0^a f(-x) dx + \int_0^a f(x) dx = \int_0^a (f(x) + f(-x)) dx$. Hence,$(A)$ is true.
For Reason $(R)$: The substitution rule for definite integrals states that if $x = g(u)$,then $dx = g'(u) du$. The limits change from $a$ to $b$ to $g^{-1}(a)$ to $g^{-1}(b)$.
The formula provided in $(R)$ is $\int_a^b f(x) dx = \int_{g(a)}^{g(b)} f(g(u)) g'(u) du$,which is incorrect because the limits of integration on the right side should be $g^{-1}(a)$ and $g^{-1}(b)$,not $g(a)$ and $g(b)$. Thus,$(R)$ is false.

(B) Given that $f(x)$ is an even function with period $2$.
Since $f(x)$ is periodic with period $2$,we have $\int_0^2 f(t) dt = \int_a^{a+2} f(t) dt$ for any $a$.
Also,for an even function,$\int_{-a}^a f(t) dt = 2 \int_0^a f(t) dt$.
Specifically,$\int_{-1}^1 f(t) dt = 2 \int_0^1 f(t) dt$.
Since the period is $2$,$\int_0^2 f(t) dt = \int_{-1}^1 f(t) dt = 2 \int_0^1 f(t) dt$.
Now,$g(x+2) = \int_0^{x+2} f(t) dt = \int_0^2 f(t) dt + \int_2^{x+2} f(t) dt$.
Using the periodicity property $\int_2^{x+2} f(t) dt = \int_0^x f(t) dt = g(x)$.
Thus,$g(x+2) = g(2) + g(x)$.

(B) $I = \int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\sin x+\cos x} d x \quad \dots(1)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\frac{\pi}{2}} \frac{\cos ^3(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x)+\cos(\frac{\pi}{2}-x)} dx = \int_0^{\frac{\pi}{2}} \frac{\sin ^3 x}{\cos x+\sin x} dx \quad \dots(2)$
Adding $(1)$ and $(2)$:
$2I = \int_0^{\frac{\pi}{2}} \frac{\sin ^3 x + \cos ^3 x}{\sin x+\cos x} dx$
Using the identity $a^3+b^3 = (a+b)(a^2+b^2-ab)$:
$2I = \int_0^{\frac{\pi}{2}} \frac{(\sin x+\cos x)(\sin^2 x+\cos^2 x - \sin x \cos x)}{\sin x+\cos x} dx$
$2I = \int_0^{\frac{\pi}{2}} (1 - \sin x \cos x) dx$
$2I = \int_0^{\frac{\pi}{2}} 1 dx - \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(2x) dx$
$2I = [x]_0^{\frac{\pi}{2}} - \frac{1}{2} [-\frac{\cos(2x)}{2}]_0^{\frac{\pi}{2}}$
$2I = \frac{\pi}{2} + \frac{1}{4} [\cos(\pi) - \cos(0)]$
$2I = \frac{\pi}{2} + \frac{1}{4} [-1 - 1] = \frac{\pi}{2} - \frac{2}{4} = \frac{\pi-1}{2}$
$I = \frac{\pi-1}{4}$

(D) Let $I = \int_0^{\pi / 2} \frac{1}{1+(\tan x)^{\sqrt{2018}}} dx$ $(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^{\pi / 2} \frac{1}{1+(\tan(\pi / 2 - x))^{\sqrt{2018}}} dx$
$I = \int_0^{\pi / 2} \frac{1}{1+(\cot x)^{\sqrt{2018}}} dx$
$I = \int_0^{\pi / 2} \frac{1}{1+\frac{1}{(\tan x)^{\sqrt{2018}}}} dx$
$I = \int_0^{\pi / 2} \frac{(\tan x)^{\sqrt{2018}}}{1+(\tan x)^{\sqrt{2018}}} dx$ $(ii)$
Adding equations $(i)$ and $(ii)$:
$2I = \int_0^{\pi / 2} \left( \frac{1}{1+(\tan x)^{\sqrt{2018}}} + \frac{(\tan x)^{\sqrt{2018}}}{1+(\tan x)^{\sqrt{2018}}} \right) dx$
$2I = \int_0^{\pi / 2} \frac{1+(\tan x)^{\sqrt{2018}}}{1+(\tan x)^{\sqrt{2018}}} dx$
$2I = \int_0^{\pi / 2} 1 dx$
$2I = [x]_0^{\pi / 2} = \frac{\pi}{2}$
$I = \frac{\pi}{4}$

(A) Let $I = \int_0^\pi \frac{x \, dx}{4 \cos^2 x + 9 \sin^2 x}$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^\pi \frac{(\pi - x) \, dx}{4 \cos^2(\pi - x) + 9 \sin^2(\pi - x)} = \int_0^\pi \frac{(\pi - x) \, dx}{4 \cos^2 x + 9 \sin^2 x}$.
Adding the two expressions for $I$:
$2I = \int_0^\pi \frac{\pi \, dx}{4 \cos^2 x + 9 \sin^2 x} = \pi \int_0^\pi \frac{\sec^2 x \, dx}{4 + 9 \tan^2 x}$.
Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$:
$2I = 2\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{4 + 9 \tan^2 x} \Rightarrow I = \pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{4 + 9 \tan^2 x}$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$. Limits change from $[0, \pi/2]$ to $[0, \infty]$.
$I = \pi \int_0^\infty \frac{dt}{4 + 9t^2} = \frac{\pi}{9} \int_0^\infty \frac{dt}{(2/3)^2 + t^2}$.
Using $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a})$:
$I = \frac{\pi}{9} \cdot \frac{3}{2} \left[ \tan^{-1}(\frac{3t}{2}) \right]_0^\infty = \frac{\pi}{6} \cdot \frac{\pi}{2} = \frac{\pi^2}{12}$.

(C) Let $I = \int_0^{2 \pi} \sin ^6 x \cos ^5 x \, dx$.
Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$,we check $f(x) = \sin^6 x \cos^5 x$.
Since $f(2\pi - x) = \sin^6(2\pi - x) \cos^5(2\pi - x) = (-\sin x)^6 (\cos x)^5 = \sin^6 x \cos^5 x = f(x)$,we have $I = 2 \int_0^{\pi} \sin^6 x \cos^5 x \, dx$.
Now,using the property $\int_0^a f(x) \, dx = 0$ if $f(a-x) = -f(x)$,we check $f(\pi - x)$.
$f(\pi - x) = \sin^6(\pi - x) \cos^5(\pi - x) = (\sin x)^6 (-\cos x)^5 = -\sin^6 x \cos^5 x = -f(x)$.
Therefore,$\int_0^{\pi} \sin^6 x \cos^5 x \, dx = 0$.
Hence,$I = 2 \times 0 = 0$.

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