Properties of definite integration Questions in English

(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$\sin \left(\frac{\pi}{4}+x\right) = \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}} (\cos x + \sin x)$.
$\sin \left(\frac{3 \pi}{4}+x\right) = \sin \left(\pi - (\frac{\pi}{4} - x)\right) = \sin (\frac{\pi}{4} - x) = \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}} (\cos x - \sin x)$.
Substituting these into the integral:
$I = \int_0^{\frac{\pi}{2}} \frac{\frac{1}{\sqrt{2}} (\cos x + \sin x) + \frac{1}{\sqrt{2}} (\cos x - \sin x)}{\cos x + \sin x} d x$.
$I = \int_0^{\frac{\pi}{2}} \frac{\frac{2}{\sqrt{2}} \cos x}{\cos x + \sin x} d x = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} d x$.
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$:
$I = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\cos(\frac{\pi}{2}-x)}{\cos(\frac{\pi}{2}-x) + \sin(\frac{\pi}{2}-x)} d x = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} d x$.
Adding the two expressions for $I$:
$2I = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\cos x + \sin x}{\cos x + \sin x} d x = \sqrt{2} \int_0^{\frac{\pi}{2}} 1 d x = \sqrt{2} [x]_0^{\frac{\pi}{2}} = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$.
Therefore,$I = \frac{\pi}{2 \sqrt{2}}$.

(B) Let $I = \int_0^{2024 \pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$.
Since the period of the integrand is $\pi$,we can write $I = 2024 \int_0^{\pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$.
Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$,we have $I = 2024 \times 2 \int_0^{\pi/2} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$.
Let $J = \int_0^{\pi/2} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get $J = \int_0^{\pi/2} \frac{2023^{\cos ^2 x}}{2023^{\cos ^2 x}+2023^{\sin ^2 x}} d x$.
Adding the two expressions for $J$,we get $2J = \int_0^{\pi/2} 1 dx = \frac{\pi}{2}$,so $J = \frac{\pi}{4}$.
Thus,$I = 2024 \times 2 \times \frac{\pi}{4} = 1012 \pi$.
Given $k = I = 1012 \pi$,we calculate $\frac{2k}{\pi} + 1 = \frac{2(1012 \pi)}{\pi} + 1 = 2024 + 1 = 2025$.

(A) Let $I = \int_0^\pi x \sin^3 x \cos^2 x \, dx$ ...$(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$:
$I = \int_0^\pi (\pi - x) \sin^3(\pi - x) \cos^2(\pi - x) \, dx$
Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we have:
$I = \int_0^\pi (\pi - x) \sin^3 x \cos^2 x \, dx$ ...(ii)
Adding $(i)$ and (ii):
$2I = \int_0^\pi \pi \sin^3 x \cos^2 x \, dx = \pi \int_0^\pi (1 - \cos^2 x) \cos^2 x \sin x \, dx$
Let $t = \cos x$,then $dt = -\sin x \, dx$. Limits change from $0 \to \pi$ to $1 \to -1$:
$2I = \pi \int_1^{-1} (1 - t^2) t^2 (-dt) = \pi \int_{-1}^1 (t^2 - t^4) \, dt$
$2I = \pi \left[ \frac{t^3}{3} - \frac{t^5}{5} \right]_{-1}^1 = \pi \left( (\frac{1}{3} - \frac{1}{5}) - (-\frac{1}{3} + \frac{1}{5}) \right) = \pi \left( \frac{2}{15} + \frac{2}{15} \right) = \frac{4\pi}{15}$
$I = \frac{2\pi}{15}$

(D) Let $I = \int_0^{2 \pi} |x \sin x| \, dx$. Since $x \ge 0$ in the interval $[0, 2 \pi]$,we have $I = \int_0^{2 \pi} x |\sin x| \, dx$.
In the interval $[0, \pi]$,$\sin x \ge 0$,and in the interval $[\pi, 2 \pi]$,$\sin x \le 0$.
Thus,$I = \int_0^{\pi} x \sin x \, dx - \int_{\pi}^{2 \pi} x \sin x \, dx$.
Using integration by parts,$\int x \sin x \, dx = -x \cos x + \sin x$.
Evaluating the first integral: $\int_0^{\pi} x \sin x \, dx = [-x \cos x + \sin x]_0^{\pi} = (-\pi(-1) + 0) - (0 + 0) = \pi$.
Evaluating the second integral: $\int_{\pi}^{2 \pi} x \sin x \, dx = [-x \cos x + \sin x]_{\pi}^{2 \pi} = (-2 \pi(1) + 0) - (-\pi(-1) + 0) = -2 \pi - \pi = -3 \pi$.
Therefore,$I = \pi - (-3 \pi) = 4 \pi$.
Given $\int_0^{2 \pi} |x \sin x| \, dx = k \pi$,we have $4 \pi = k \pi$,which implies $k = 4$.

(D) Let $f(x) = \tan ^9 x \sin ^6 x \cos ^3 x$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = [\tan(-x)]^9 [\sin(-x)]^6 [\cos(-x)]^3$
Since $\tan(-x) = -\tan x$,$\sin(-x) = -\sin x$,and $\cos(-x) = \cos x$,we have:
$f(-x) = (-\tan x)^9 (-\sin x)^6 (\cos x)^3$
$f(-x) = -\tan^9 x \cdot \sin^6 x \cdot \cos^3 x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-4 \pi}^{4 \pi} \tan ^9 x \sin ^6 x \cos ^3 x \, dx = 0$.

(C) Given $S_n = \int_0^{\pi/2} \frac{\sin((2n-1)x)}{\sin x} dx$.
Then $S_{n+1} = \int_0^{\pi/2} \frac{\sin((2n+1)x)}{\sin x} dx$.
Now,consider the difference:
$S_{n+1} - S_n = \int_0^{\pi/2} \frac{\sin((2n+1)x) - \sin((2n-1)x)}{\sin x} dx$.
Using the trigonometric identity $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$,we have:
$\sin((2n+1)x) - \sin((2n-1)x) = 2 \cos(2nx) \sin(x)$.
Substituting this into the integral:
$S_{n+1} - S_n = \int_0^{\pi/2} \frac{2 \cos(2nx) \sin x}{\sin x} dx = \int_0^{\pi/2} 2 \cos(2nx) dx$.
Evaluating the integral:
$S_{n+1} - S_n = [\frac{2 \sin(2nx)}{2n}]_0^{\pi/2} = [\frac{\sin(2nx)}{n}]_0^{\pi/2}$.
Substituting the limits:
$S_{n+1} - S_n = \frac{\sin(n\pi) - \sin(0)}{n} = \frac{0 - 0}{n} = 0$,since $n$ is an integer.

(A) Given $a=2n$ and $b=2m+1$ where $m, n \in N$.
Let $I = \int_{-\pi}^{\pi} f(x) dx$,where $f(x) = e^{\sin^a x} \cdot \cot^b((2n+1)x)$.
We check the parity of the function $f(x)$:
$f(-x) = e^{\sin^a(-x)} \cdot \cot^b((2n+1)(-x))$.
Since $a=2n$ is an even number,$\sin^a(-x) = (\sin(-x))^a = (-\sin x)^a = \sin^a x$.
Since $b=2m+1$ is an odd number,$\cot^b((2n+1)(-x)) = (\cot(-(2n+1)x))^b = (-\cot((2n+1)x))^b = -\cot^b((2n+1)x)$.
Therefore,$f(-x) = e^{\sin^a x} \cdot (-\cot^b((2n+1)x)) = -f(x)$.
Since $f(x)$ is an odd function,the integral over the symmetric interval $[-\pi, \pi]$ is zero.
Thus,$I = 0$.

(A) Let $I = \int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} (|\cos t| \sin t + |\sin t| \cos t) dt$.
We know that $\frac{d}{dt} (\sin t \cos t) = \cos^2 t - \sin^2 t = \cos(2t)$.
However,it is easier to use the property $\int f(t) dt = \int f(t) dt$.
Note that the integrand is $f(t) = |\cos t| \sin t + |\sin t| \cos t$.
For $t \in [\frac{\pi}{4}, \frac{\pi}{2}]$,$\cos t > 0$ and $\sin t > 0$,so $f(t) = \cos t \sin t + \sin t \cos t = 2 \sin t \cos t = \sin(2t)$.
For $t \in [\frac{\pi}{2}, \pi]$,$\cos t < 0$ and $\sin t > 0$,so $f(t) = -\cos t \sin t + \sin t \cos t = 0$.
For $t \in [\pi, \frac{5\pi}{4}]$,$\cos t < 0$ and $\sin t < 0$,so $f(t) = -\cos t \sin t - \sin t \cos t = -2 \sin t \cos t = -\sin(2t)$.
Thus,$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(2t) dt + \int_{\frac{\pi}{2}}^{\pi} 0 dt + \int_{\pi}^{\frac{5\pi}{4}} -\sin(2t) dt$.
$I = \left[ -\frac{\cos(2t)}{2} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} + 0 + \left[ \frac{\cos(2t)}{2} \right]_{\pi}^{\frac{5\pi}{4}}$.
$I = (-\frac{\cos(\pi)}{2} - (-\frac{\cos(\frac{\pi}{2})}{2})) + (\frac{\cos(\frac{5\pi}{2})}{2} - \frac{\cos(2\pi)}{2})$.
$I = (\frac{1}{2} - 0) + (0 - \frac{1}{2}) = 0$.

(C) Let $I = \int_{-1}^1 (x[1+\sin(\pi x)]+1) dx$.
Using the property $\int_{-a}^a f(x) dx = \int_{-a}^a f(-x) dx$,we can split the integral.
However,let us evaluate the function $f(x) = x[1+\sin(\pi x)]+1$.
For $x \in [-1, 0]$,$\sin(\pi x) \in [-1, 0]$,so $1+\sin(\pi x) \in [0, 1]$. Thus,$[1+\sin(\pi x)] = 0$ for $x \in (-1, 0)$.
For $x \in [0, 1]$,$\sin(\pi x) \in [0, 1]$,so $1+\sin(\pi x) \in [1, 2]$. Thus,$[1+\sin(\pi x)] = 1$ for $x \in (0, 1)$.
Therefore,$I = \int_{-1}^0 (x \cdot 0 + 1) dx + \int_0^1 (x \cdot 1 + 1) dx$.
$I = \int_{-1}^0 1 dx + \int_0^1 (x+1) dx$.
$I = [x]_{-1}^0 + [\frac{x^2}{2} + x]_0^1$.
$I = (0 - (-1)) + ((\frac{1}{2} + 1) - 0)$.
$I = 1 + \frac{3}{2} = \frac{5}{2}$.

(B) Let $I = \int_{-\pi}^\pi \frac{2x(1+\sin x)}{1+\cos^2 x} dx$.
We can split the integral into two parts:
$I = \int_{-\pi}^\pi \frac{2x}{1+\cos^2 x} dx + \int_{-\pi}^\pi \frac{2x\sin x}{1+\cos^2 x} dx = I_1 + I_2$.
For $I_1 = \int_{-\pi}^\pi \frac{2x}{1+\cos^2 x} dx$,let $f(x) = \frac{2x}{1+\cos^2 x}$.
Since $f(-x) = \frac{2(-x)}{1+\cos^2(-x)} = -\frac{2x}{1+\cos^2 x} = -f(x)$,$f(x)$ is an odd function.
Therefore,$I_1 = 0$.
For $I_2 = \int_{-\pi}^\pi \frac{2x\sin x}{1+\cos^2 x} dx$,let $g(x) = \frac{2x\sin x}{1+\cos^2 x}$.
Since $g(-x) = \frac{2(-x)\sin(-x)}{1+\cos^2(-x)} = \frac{2x\sin x}{1+\cos^2 x} = g(x)$,$g(x)$ is an even function.
Therefore,$I_2 = 2 \int_0^\pi \frac{2x\sin x}{1+\cos^2 x} dx = 4 \int_0^\pi \frac{x\sin x}{1+\cos^2 x} dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I_2 = 4 \int_0^\pi \frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)} dx = 4 \int_0^\pi \frac{(\pi-x)\sin x}{1+\cos^2 x} dx$.
Adding the two expressions for $I_2$:
$2I_2 = 4 \int_0^\pi \frac{\pi\sin x}{1+\cos^2 x} dx = 4\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$.
Let $u = \cos x$,then $du = -\sin x dx$.
When $x=0, u=1$; when $x=\pi, u=-1$.
$2I_2 = 4\pi \int_1^{-1} \frac{-du}{1+u^2} = 4\pi \int_{-1}^1 \frac{du}{1+u^2} = 4\pi [\tan^{-1} u]_{-1}^1$.
$2I_2 = 4\pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = 4\pi (\frac{\pi}{2}) = 2\pi^2$.
Thus,$I_2 = \pi^2$.
Finally,$I = I_1 + I_2 = 0 + \pi^2 = \pi^2$.

(D) Let $I = \int_{-1}^1 \frac{\sin x-x^2}{3-|x|} d x$.
We can split the integral into two parts:
$I = \int_{-1}^1 \frac{\sin x}{3-|x|} d x - \int_{-1}^1 \frac{x^2}{3-|x|} d x$.
Since $f(x) = \frac{\sin x}{3-|x|}$ is an odd function (because $\sin(-x) = -\sin x$ and $|-x| = |x|$),the integral $\int_{-1}^1 \frac{\sin x}{3-|x|} d x = 0$.
Since $g(x) = \frac{x^2}{3-|x|}$ is an even function,$\int_{-1}^1 \frac{x^2}{3-|x|} d x = 2 \int_0^1 \frac{x^2}{3-x} d x$.
Thus,$I = -2 \int_0^1 \frac{x^2}{3-x} d x = 2 \int_0^1 \frac{x^2}{x-3} d x$.
Performing polynomial division: $\frac{x^2}{x-3} = x + 3 + \frac{9}{x-3}$.
So,$I = 2 \int_0^1 (x + 3 + \frac{9}{x-3}) d x = 2 [\frac{x^2}{2} + 3x + 9 \ln|x-3|]_0^1$.
Evaluating at limits: $I = 2 [(\frac{1}{2} + 3 + 9 \ln 2) - (0 + 0 + 9 \ln 3)] = 2 [\frac{7}{2} + 9 \ln(\frac{2}{3})] = 7 + 18 \ln(\frac{2}{3}) = 7 - 18 \ln(\frac{3}{2})$.

(C) We know that the property of definite integrals states: $\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$.
By substituting $x = -t$ in the first integral $\int_{-a}^0 f(x) dx$,we get $\int_a^0 f(-t) (-dt) = \int_0^a f(-t) dt = \int_0^a f(-x) dx$.
Therefore,$\int_{-a}^a f(x) dx = \int_0^a f(-x) dx + \int_0^a f(x) dx$.
Substituting this into the given expression: $(\int_0^a f(-x) dx + \int_0^a f(x) dx) - \int_0^a f(-x) dx = \int_0^a f(x) dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,the result is $\int_0^a f(a-x) dx$.

(C) Let $I = \int_0^\pi \frac{d x}{1+2 \sin ^2 x}$.
Since the integrand $f(x) = \frac{1}{1+2 \sin ^2 x}$ satisfies $f(\pi - x) = f(x)$,we can write:
$I = 2 \int_0^{\pi / 2} \frac{d x}{1+2 \sin ^2 x}$.
Dividing the numerator and denominator by $\cos ^2 x$:
$I = 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{\sec ^2 x + 2 \tan ^2 x} d x = 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{1 + \tan ^2 x + 2 \tan ^2 x} d x = 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{1 + 3 \tan ^2 x} d x$.
Let $\tan x = t$,then $\sec ^2 x d x = d t$. When $x = 0, t = 0$ and when $x \to \pi / 2, t \to \infty$.
$I = 2 \int_0^{\infty} \frac{d t}{1 + 3 t^2} = 2 \int_0^{\infty} \frac{d t}{1 + (\sqrt{3} t)^2}$.
Using the formula $\int \frac{dx}{1+a^2x^2} = \frac{1}{a} \tan^{-1}(ax)$:
$I = 2 \left[ \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3} t) \right]_0^{\infty} = \frac{2}{\sqrt{3}} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{\sqrt{3}}$.
Given $k = \frac{\pi}{\sqrt{3}} \approx \frac{3.14159}{1.732} \approx 1.81$.
The greatest integer less than or equal to $k$ is $\lfloor 1.81 \rfloor = 1$.

(B) Let $I_n = \int_0^\pi \frac{\sin nx}{\sin x} dx$.
Using the given recurrence relation: $I_n - I_{n-2} = \int_0^\pi \frac{\sin nx - \sin(n-2)x}{\sin x} dx = \int_0^\pi \frac{2 \cos(n-1)x \sin x}{\sin x} dx = \int_0^\pi 2 \cos(n-1)x dx$.
For $n > 1$,$I_n - I_{n-2} = \left[ \frac{2 \sin(n-1)x}{n-1} \right]_0^\pi = 0$.
Thus,$I_n = I_{n-2}$ for all $n > 1$.
For $n=1$,$I_1 = \int_0^\pi \frac{\sin x}{\sin x} dx = \int_0^\pi 1 dx = \pi$.
For $n=2$,$I_2 = \int_0^\pi \frac{\sin 2x}{\sin x} dx = \int_0^\pi 2 \cos x dx = [2 \sin x]_0^\pi = 0$.
If $n$ is odd,$I_n = I_1 = \pi = 2 \cdot \frac{\pi}{2} = (1 - (-1)^n) \frac{\pi}{2}$.
If $n$ is even,$I_n = I_2 = 0 = (1 - (-1)^n) \frac{\pi}{2}$.
Therefore,$k = 1 - (-1)^n$.

(B) Let $I = \int_{0}^{\pi / 4} \log (1 + \tan x) dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have:
$I = \int_{0}^{\pi / 4} \log (1 + \tan (\frac{\pi}{4} - x)) dx$.
Since $\tan (\frac{\pi}{4} - x) = \frac{1 - \tan x}{1 + \tan x}$,we get:
$I = \int_{0}^{\pi / 4} \log (1 + \frac{1 - \tan x}{1 + \tan x}) dx = \int_{0}^{\pi / 4} \log (\frac{2}{1 + \tan x}) dx$.
$I = \int_{0}^{\pi / 4} (\log 2 - \log (1 + \tan x)) dx = \frac{\pi}{4} \log 2 - I$.
$2I = \frac{\pi}{4} \log 2 \implies I = \frac{\pi}{8} \log 2$.
Given $\int_{0}^{\pi} \log (\sin x) dx = 8k$. We know $\int_{0}^{\pi} \log (\sin x) dx = -\pi \log 2$.
So,$8k = -\pi \log 2 \implies \pi \log 2 = -8k$.
Substituting this into $I = \frac{\pi}{8} \log 2$,we get $I = \frac{1}{8} (-8k) = -k$.

(D) We are given the integral $I = \int_{0}^{1} x^{m} (1 - x)^{n} dx$.
Using the property of definite integrals,$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we substitute $x$ with $(1 - x)$:
$I = \int_{0}^{1} (1 - x)^{m} (1 - (1 - x))^{n} dx$
$I = \int_{0}^{1} (1 - x)^{m} (x)^{n} dx$
$I = \int_{0}^{1} x^{n} (1 - x)^{m} dx$.
Comparing this with the given equation $\int_{0}^{1} x^{m} (1 - x)^{n} dx = k \int_{0}^{1} x^{n} (1 - x)^{m} dx$,we see that $I = k \cdot I$.
Since the integral is non-zero,we conclude that $k = 1$.

(A) Let $I = \int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x \quad ...(i)$
Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,we get:
$I = \int_0^\pi \frac{(\pi-x) \tan(\pi-x)}{\sec(\pi-x)+\tan(\pi-x)} d x$
Since $\tan(\pi-x) = -\tan x$ and $\sec(\pi-x) = -\sec x$,we have:
$I = \int_0^\pi \frac{(\pi-x)(-\tan x)}{-\sec x-\tan x} d x = \int_0^\pi \frac{(\pi-x) \tan x}{\sec x+\tan x} d x \quad ...(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^\pi \frac{x \tan x + (\pi-x) \tan x}{\sec x+\tan x} d x = \int_0^\pi \frac{\pi \tan x}{\sec x+\tan x} d x$
$I = \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\sin x} d x$
Multiply numerator and denominator by $(1-\sin x)$:
$I = \frac{\pi}{2} \int_0^\pi \frac{\sin x(1-\sin x)}{1-\sin^2 x} d x = \frac{\pi}{2} \int_0^\pi \frac{\sin x - \sin^2 x}{\cos^2 x} d x$
$I = \frac{\pi}{2} \int_0^\pi (\sec x \tan x - \tan^2 x) d x = \frac{\pi}{2} \int_0^\pi (\sec x \tan x - (\sec^2 x - 1)) d x$
$I = \frac{\pi}{2} [\sec x - \tan x + x]_0^\pi$
$I = \frac{\pi}{2} [(\sec \pi - \tan \pi + \pi) - (\sec 0 - \tan 0 + 0)]$
$I = \frac{\pi}{2} [(-1 - 0 + \pi) - (1 - 0 + 0)] = \frac{\pi}{2} (\pi - 2)$

(A) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x \tan \left(1+x^2\right) d x$.
Define the integrand as $f(x) = x \tan \left(1+x^2\right)$.
To check if the function is odd or even,we evaluate $f(-x)$:
$f(-x) = (-x) \tan \left(1+(-x)^2\right) = -x \tan \left(1+x^2\right) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) d x = 0$.
Therefore,$I = 0$.

(B) We have $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin(\pi - x)) + \cos^2(\cos(\pi - x)) \right\} dx$
$I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin x) + \cos^2(-\cos x) \right\} dx$
$I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$
Adding the two expressions for $I$:
$2I = \int_0^\pi \pi \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$
$I = \frac{\pi}{2} \int_0^\pi \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$
Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$:
$I = \frac{\pi}{2} \cdot 2 \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$
$I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$
Now,using $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ again:
$I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin(\pi/2 - x)) + \cos^2(\cos(\pi/2 - x)) \right\} dx$
$I = \pi \int_0^{\pi/2} \left\{ \sin^2(\cos x) + \cos^2(\sin x) \right\} dx$
Adding these two forms of $I$:
$2I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) + \sin^2(\cos x) + \cos^2(\sin x) \right\} dx$
$2I = \pi \int_0^{\pi/2} \left\{ (\sin^2(\sin x) + \cos^2(\sin x)) + (\cos^2(\cos x) + \sin^2(\cos x)) \right\} dx$
$2I = \pi \int_0^{\pi/2} (1 + 1) dx = 2\pi \int_0^{\pi/2} dx = 2\pi \cdot \frac{\pi}{2} = \pi^2$
Thus,$I = \frac{\pi^2}{2}$.
Since $\pi^2 \approx 9.869$,then $I \approx \frac{9.869}{2} = 4.9345$.
Therefore,$[I] = [4.9345] = 4$.

(C) Let $f(x) = x^2 \sin x$.
Check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = (-x)^2 \sin(-x) = x^2 (-\sin x) = -x^2 \sin x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x) = x^2 \sin x$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-\pi}^{\pi} x^2 \sin x \, dx = 0$.

(A) Let $I = \int_{-\pi / 4}^{\pi / 4} x^3 \sin^4(x) dx$.
We know that for a definite integral,$\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function,i.e.,$f(-x) = -f(x)$.
Here,$f(x) = x^3 \sin^4(x)$.
Calculating $f(-x)$:
$f(-x) = (-x)^3 \sin^4(-x) = -x^3 (\sin(x))^4 = -x^3 \sin^4(x) = -f(x)$.
Since $f(x)$ is an odd function,the integral over the symmetric interval $[-\pi/4, \pi/4]$ is $0$.
Therefore,$I = 0$.

(A) Let $I = \int_0^{b-c} f(x+c) dx$.
Substitute $x+c = t$,then $dx = dt$.
When $x = 0$,$t = c$.
When $x = b-c$,$t = b$.
Substituting these into the integral,we get:
$I = \int_c^b f(t) dt$.
Since the variable of integration is a dummy variable,we can write $\int_c^b f(t) dt = \int_c^b f(x) dx$.
Thus,$\int_0^{b-c} f(x+c) dx = 1 \cdot \int_c^b f(x) dx$.
Comparing this with the given expression $\int_0^{b-c} f(x+c) dx = k \int_c^b f(x) dx$,we find that $k = 1$.

(C) Let $I = \int_0^{\pi / 2} \frac{1}{1+\tan ^{2020} x} d x$.
Since $\tan x = \frac{\sin x}{\cos x}$,we can write $I = \int_0^{\pi / 2} \frac{\cos ^{2020} x}{\cos ^{2020} x+\sin ^{2020} x} d x$ $(i)$.
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get $I = \int_0^{\pi / 2} \frac{\sin ^{2020} x}{\sin ^{2020} x+\cos ^{2020} x} d x$ (ii).
Adding equations $(i)$ and (ii),we get $2I = \int_0^{\pi / 2} \frac{\cos ^{2020} x + \sin ^{2020} x}{\cos ^{2020} x + \sin ^{2020} x} d x = \int_0^{\pi / 2} 1 d x$.
Thus,$2I = [x]_0^{\pi / 2} = \frac{\pi}{2}$,which implies $I = \frac{\pi}{4}$.
Therefore,option $C$ is correct.

(B) Let $I = \int_{-\pi / 2}^{\pi / 2} (2 \sin |x| + \cos |x|) dx$.
Since $f(x) = 2 \sin |x| + \cos |x|$ is an even function because $f(-x) = 2 \sin |-x| + \cos |-x| = 2 \sin |x| + \cos |x| = f(x)$,we can use the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$.
$I = 2 \int_{0}^{\pi / 2} (2 \sin x + \cos x) dx$.
Integrating the terms,we get $I = 2 [-2 \cos x + \sin x]_{0}^{\pi / 2}$.
Evaluating at the limits,$I = 2 [(-2 \cos(\pi / 2) + \sin(\pi / 2)) - (-2 \cos(0) + \sin(0))]$.
$I = 2 [(-2 \times 0 + 1) - (-2 \times 1 + 0)]$.
$I = 2 [1 - (-2)] = 2 [1 + 2] = 2 \times 3 = 6$.

(D) Let $I = \int_0^{2 \pi} \frac{x \cos x}{1+\cos x} d x$ ... $(i)$
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get:
$I = \int_0^{2 \pi} \frac{(2 \pi - x) \cos(2 \pi - x)}{1 + \cos(2 \pi - x)} d x$
Since $\cos(2 \pi - x) = \cos x$,we have:
$I = \int_0^{2 \pi} \frac{(2 \pi - x) \cos x}{1 + \cos x} d x$ ... (ii)
Adding $(i)$ and (ii):
$2I = \int_0^{2 \pi} \frac{2 \pi \cos x}{1 + \cos x} d x = 2 \pi \int_0^{2 \pi} \frac{\cos x}{1 + \cos x} d x$
$I = \pi \int_0^{2 \pi} \frac{\cos x}{1 + \cos x} d x = 2 \pi \int_0^{\pi} \frac{\cos x}{1 + \cos x} d x$
$I = 2 \pi \int_0^{\pi} (1 - \frac{1}{1 + \cos x}) d x = 2 \pi \int_0^{\pi} (1 - \frac{1}{2 \cos^2(x/2)}) d x$
$I = 2 \pi \int_0^{\pi} (1 - \frac{1}{2} \sec^2(x/2)) d x$
$I = 2 \pi [x - \tan(x/2)]_0^{\pi}$
Evaluating at limits: $I = 2 \pi [(\pi - \tan(\pi/2)) - (0 - \tan(0))]$. Since $\tan(\pi/2)$ is undefined,the integral diverges. Thus,the correct option is $D$.

(A) Let $I = \int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x$.
We can rewrite the integrand as $I = \int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$ $(i)$.
Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$,we get:
$I = \int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)}+\sqrt{\cos(\frac{\pi}{2}-x)}} d x = \int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$ (ii).
Adding $(i)$ and (ii):
$2I = \int_{\pi / 6}^{\pi / 3} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) d x = \int_{\pi / 6}^{\pi / 3} 1 d x$.
$2I = [x]_{\pi / 6}^{\pi / 3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$I = \frac{\pi}{12}$.

(A) We are given the property of definite integrals: $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$.
Applying this property to the integral $I = \int_{0}^{\pi/2} \cos^{4}(x) \cdot \sin^{2}(x) dx$,we replace $x$ with $(\frac{\pi}{2} - x)$.
Since $\cos(\frac{\pi}{2} - x) = \sin(x)$ and $\sin(\frac{\pi}{2} - x) = \cos(x)$,the integral becomes:
$I = \int_{0}^{\pi/2} \sin^{4}(x) \cdot \cos^{2}(x) dx$.
Given that $\int_{0}^{\pi/2} \sin^{4}(x) \cdot \cos^{2}(x) dx = \frac{\pi}{32}$,it follows that $I = \frac{\pi}{32}$.

(D) Let $I = \int_{-a}^{a} \sqrt{\frac{a - x}{a + x}} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,where $a = -a$ and $b = a$,we have $f(x) \rightarrow f(-a + a - x) = f(-x)$.
$I = \int_{-a}^{a} \sqrt{\frac{a - (-x)}{a + (-x)}} dx = \int_{-a}^{a} \sqrt{\frac{a + x}{a - x}} dx$.
Adding the two expressions for $I$:
$2I = \int_{-a}^{a} \left( \sqrt{\frac{a - x}{a + x}} + \sqrt{\frac{a + x}{a - x}} \right) dx$.
$2I = \int_{-a}^{a} \frac{(a - x) + (a + x)}{\sqrt{(a + x)(a - x)}} dx = \int_{-a}^{a} \frac{2a}{\sqrt{a^2 - x^2}} dx$.
Since the integrand is an even function,$2I = 2 \int_{0}^{a} \frac{2a}{\sqrt{a^2 - x^2}} dx$.
$I = 2a \int_{0}^{a} \frac{1}{\sqrt{a^2 - x^2}} dx$.
$I = 2a [\sin^{-1}(\frac{x}{a})]_{0}^{a}$.
$I = 2a (\sin^{-1}(1) - \sin^{-1}(0)) = 2a (\frac{\pi}{2} - 0) = a \pi$.

(C) Let $I = \int_0^{\pi / 2} \frac{2 \sin (x)+3 \cos (x)}{\sin (x)+\cos (x)} d x$.
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get:
$I = \int_0^{\pi / 2} \frac{2 \sin (\pi/2-x)+3 \cos (\pi/2-x)}{\sin (\pi/2-x)+\cos (\pi/2-x)} d x = \int_0^{\pi / 2} \frac{2 \cos (x)+3 \sin (x)}{\cos (x)+\sin (x)} d x$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi / 2} \frac{(2 \sin x + 3 \cos x) + (2 \cos x + 3 \sin x)}{\sin x + \cos x} d x$.
$2I = \int_0^{\pi / 2} \frac{5 \sin x + 5 \cos x}{\sin x + \cos x} d x = \int_0^{\pi / 2} 5 d x$.
$2I = [5x]_0^{\pi / 2} = 5(\frac{\pi}{2} - 0) = \frac{5\pi}{2}$.
Therefore,$I = \frac{5\pi}{4}$.

(A) We are given the expression: $I = \int_0^{2a} f(x) dx - \int_a^{2a} f(x) dx$.
Using the property of definite integrals,$\int_0^{2a} f(x) dx = \int_0^a f(x) dx + \int_a^{2a} f(x) dx$.
Substituting this into the expression,we get: $I = (\int_0^a f(x) dx + \int_a^{2a} f(x) dx) - \int_a^{2a} f(x) dx$.
Simplifying the expression,the term $\int_a^{2a} f(x) dx$ cancels out.
Therefore,$I = \int_0^a f(x) dx$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} dx$ ... $(i)$
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 $a+b = 0$.
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos(-x)}{1+e^{-x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+\frac{1}{e^x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^x \cos x}{e^x+1} dx$ ... (ii)
Adding $(i)$ and (ii):
$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x + e^x \cos x}{1+e^x} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x(1+e^x)}{1+e^x} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x dx$
$2I = [\sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2}) = 1 - (-1) = 2$
$I = 1$
Therefore,$\tan^{-1}(I) = \tan^{-1}(1) = \frac{\pi}{4}$.

(D) Let $I = \int_0^{\alpha / 3} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x$ $(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we replace $x$ with $(\frac{\alpha}{3} - x)$:
$I = \int_0^{\alpha / 3} \frac{f(\frac{\alpha}{3} - x)}{f(\frac{\alpha}{3} - x) + f(\frac{\alpha - 3(\frac{\alpha}{3} - x)}{3})} dx$
$I = \int_0^{\alpha / 3} \frac{f(\frac{\alpha}{3} - x)}{f(\frac{\alpha}{3} - x) + f(x)} dx$ $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\alpha / 3} \frac{f(x) + f(\frac{\alpha}{3} - x)}{f(x) + f(\frac{\alpha}{3} - x)} dx$
$2I = \int_0^{\alpha / 3} 1 dx$
$2I = [x]_0^{\alpha / 3} = \frac{\alpha}{3}$
$I = \frac{\alpha}{6}$

(D) Let $I = \int_0^{\pi / 2} \log _e(\sin 2 x) d x$.
Substitute $2x = t$,so $dx = \frac{1}{2} dt$.
When $x = 0$,$t = 0$. When $x = \frac{\pi}{2}$,$t = \pi$.
Thus,$I = \frac{1}{2} \int_0^{\pi} \log _e(\sin t) dt$.
Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$,we have $\int_0^{\pi} \log _e(\sin t) dt = 2 \int_0^{\pi/2} \log _e(\sin t) dt$.
So,$I = \frac{1}{2} \times 2 \int_0^{\pi/2} \log _e(\sin t) dt = \int_0^{\pi/2} \log _e(\sin t) dt$.
It is a standard result that $\int_0^{\pi/2} \log _e(\sin t) dt = -\frac{\pi}{2} \log _e 2$.
Therefore,$I = -\frac{\pi}{2} \log _e 2$.

(C) Let $I = \int_0^1 f(k-1+x) d x$.
Substitute $k-1+x = t$,then $d x = d t$.
When $x=0$,$t=k-1$.
When $x=1$,$t=k$.
Therefore,$I = \int_{k-1}^k f(t) d t = \int_{k-1}^k f(x) d x$.
Now,we need to evaluate the summation:
$\sum_{k=1}^{10} \int_0^1 f(k-1+x) d x = \sum_{k=1}^{10} \int_{k-1}^k f(x) d x$.
Expanding the summation:
$= \int_0^1 f(x) d x + \int_1^2 f(x) d x + \dots + \int_9^{10} f(x) d x$.
Using the property of definite integrals $\int_a^b f(x) d x + \int_b^c f(x) d x = \int_a^c f(x) d x$,we get:
$= \int_0^{10} f(x) d x$.
Given that $\int_0^{10} f(x) d x = 5$,the final value is $5$.

(A) We are given that $\int_{-1}^4 f(x) dx = 4$.
Using the property of definite integrals,we can split the interval:
$\int_{-1}^4 f(x) dx = \int_{-1}^2 f(x) dx + \int_2^4 f(x) dx = 4$.
We are also given $\int_2^4 (3 - f(x)) dx = 7$.
This can be written as:
$\int_2^4 3 dx - \int_2^4 f(x) dx = 7$.
Evaluating the first part:
$[3x]_2^4 - \int_2^4 f(x) dx = 7
(3 \times 4) - (3 \times 2) - \int_2^4 f(x) dx = 7
12 - 6 - \int_2^4 f(x) dx = 7
6 - \int_2^4 f(x) dx = 7
\int_2^4 f(x) dx = 6 - 7 = -1$.
Now,substitute $\int_2^4 f(x) dx = -1$ into the first equation:
$\int_{-1}^2 f(x) dx + (-1) = 4
\int_{-1}^2 f(x) dx = 4 + 1 = 5$.

(D) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2x} dx$.
We can split this into two integrals: $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x}{2-\cos 2x} dx + \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{\pi}{4}}{2-\cos 2x} dx$.
Let $f(x) = \frac{x}{2-\cos 2x}$. Since $f(-x) = \frac{-x}{2-\cos(-2x)} = -f(x)$,$f(x)$ is an odd function. Thus,$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(x) dx = 0$.
Now,$I = \frac{\pi}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2x} dx$. Since the integrand is an even function,$I = 2 \times \frac{\pi}{4} \int_{0}^{\frac{\pi}{4}} \frac{1}{2-\cos 2x} dx = \frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \frac{1}{2-\cos 2x} dx$.
Using $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$,we get $I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \frac{1+\tan^2 x}{2(1+\tan^2 x) - (1-\tan^2 x)} dx = \frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \frac{\sec^2 x}{1+3\tan^2 x} dx$.
Let $t = \tan x$,then $dt = \sec^2 x dx$. Limits change from $[0, \frac{\pi}{4}]$ to $[0, 1]$.
$I = \frac{\pi}{2} \int_{0}^{1} \frac{dt}{1+3t^2} = \frac{\pi}{2} \left[ \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}t) \right]_{0}^{1} = \frac{\pi}{2\sqrt{3}} \tan^{-1}(\sqrt{3}) = \frac{\pi}{2\sqrt{3}} \times \frac{\pi}{3} = \frac{\pi^2}{6\sqrt{3}}$.

(C) Let $I = \int_{-1}^1 f(x) dx$,where $f(x) = \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \frac{\sqrt{1+(-x)+(-x)^2}-\sqrt{1-(-x)+(-x)^2}}{\sqrt{1+(-x)+(-x)^2}+\sqrt{1-(-x)+(-x)^2}}$
$f(-x) = \frac{\sqrt{1-x+x^2}-\sqrt{1+x+x^2}}{\sqrt{1-x+x^2}+\sqrt{1+x+x^2}}$
$f(-x) = -\left( \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} \right) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
By the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^a f(x) dx = 0$.
Therefore,$\int_{-1}^1 f(x) dx = 0$.

(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$.
Since $f(x) = \sin |x|$ is an even function because $f(-x) = \sin |-x| = \sin |x| = f(x)$,we can use the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$.
Thus,$I = 2 \int_{0}^{\pi / 2} \sin |x| \, dx$.
For $x \in [0, \pi / 2]$,$|x| = x$,so $I = 2 \int_{0}^{\pi / 2} \sin x \, dx$.
Evaluating the integral,$I = 2 [-\cos x]_{0}^{\pi / 2}$.
$I = 2 [-\cos(\pi / 2) - (-\cos(0))]$.
$I = 2 [0 - (-1)] = 2(1) = 2$.

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

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$.
Define $f(\theta) = \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right)$.
Now,check for parity by calculating $f(-\theta)$:
$f(-\theta) = \log \left(\frac{2-\sin(-\theta)}{2+\sin(-\theta)}\right) = \log \left(\frac{2+\sin \theta}{2-\sin \theta}\right)$.
Using the property $\log(x^{-1}) = -\log(x)$,we get:
$f(-\theta) = \log \left(\left(\frac{2-\sin \theta}{2+\sin \theta}\right)^{-1}\right) = -\log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) = -f(\theta)$.
Since $f(-\theta) = -f(\theta)$,the function $f(\theta)$ is an odd function.
According to the property of definite integrals,$\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function.
Therefore,$I = 0$.

(B) The function $f(x) = e^{x-[x]}$ is a periodic function with period $T = 1$.
We can split the integral over the interval $[0, 1000]$ into $1000$ integrals of length $1$:
$\int_0^{1000} e^{x-[x]} dx = \sum_{k=0}^{999} \int_k^{k+1} e^{x-[x]} dx$.
For $x \in [k, k+1)$,$[x] = k$,so the integrand becomes $e^{x-k}$.
Thus,$\int_k^{k+1} e^{x-k} dx = [e^{x-k}]_k^{k+1} = e^{(k+1)-k} - e^{k-k} = e^1 - e^0 = e-1$.
Since there are $1000$ such intervals,the total sum is $1000 \times (e-1)$.

(B) Let $I = \int_0^\pi x (\sin^2(\sin x) + \cos^2(\cos x)) dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^\pi (\pi - x) (\sin^2(\sin(\pi - x)) + \cos^2(\cos(\pi - x))) dx$
Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we get:
$I = \int_0^\pi (\pi - x) (\sin^2(\sin x) + \cos^2(-\cos x)) dx$
$I = \int_0^\pi (\pi - x) (\sin^2(\sin x) + \cos^2(\cos x)) dx$
Adding the two expressions for $I$:
$2I = \int_0^\pi \pi (\sin^2(\sin x) + \cos^2(\cos x)) dx$
$2I = \pi \int_0^\pi (\sin^2(\sin x) + \cos^2(\cos x)) dx$
Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$:
$2I = 2\pi \int_0^{\frac{\pi}{2}} (\sin^2(\sin x) + \cos^2(\cos x)) dx$
$I = \pi \int_0^{\frac{\pi}{2}} (\sin^2(\sin x) + \cos^2(\cos x)) dx . . . (i)$
Applying the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ again:
$I = \pi \int_0^{\frac{\pi}{2}} (\sin^2(\sin(\frac{\pi}{2}-x)) + \cos^2(\cos(\frac{\pi}{2}-x))) dx$
$I = \pi \int_0^{\frac{\pi}{2}} (\sin^2(\cos x) + \cos^2(\sin x)) dx . . . (ii)$
Adding $(i)$ and $(ii)$:
$2I = \pi \int_0^{\frac{\pi}{2}} (\sin^2(\sin x) + \cos^2(\cos x) + \sin^2(\cos x) + \cos^2(\sin x)) dx$
$2I = \pi \int_0^{\frac{\pi}{2}} ((\sin^2(\sin x) + \cos^2(\sin x)) + (\sin^2(\cos x) + \cos^2(\cos x))) dx$
Since $\sin^2 \theta + \cos^2 \theta = 1$:
$2I = \pi \int_0^{\frac{\pi}{2}} (1 + 1) dx = 2\pi \int_0^{\frac{\pi}{2}} dx$
$2I = 2\pi [x]_0^{\frac{\pi}{2}} = 2\pi (\frac{\pi}{2}) = \pi^2$
$I = \frac{\pi^2}{2}$

(C) Given,$I = \int_0^T f(x) dx$.
Since $f(x+T) = f(x)$,$f$ is a periodic function with period $T$.
We need to evaluate $J = \int_0^{5T} f(2x) dx$.
Let $2x = y$,then $dx = \frac{1}{2} dy$.
When $x = 0$,$y = 0$. When $x = 5T$,$y = 10T$.
Substituting these into the integral:
$J = \int_0^{10T} f(y) \cdot \frac{1}{2} dy = \frac{1}{2} \int_0^{10T} f(y) dy$.
Using the property of periodic functions $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$:
$J = \frac{1}{2} \times 10 \int_0^T f(y) dy = 5 \int_0^T f(x) dx = 5I$.

(B) Given the integral $I = \int_0^{50\pi} \sqrt{1-\cos 2x} dx$.
Using the identity $1-\cos 2x = 2\sin^2 x$,we have $\sqrt{1-\cos 2x} = \sqrt{2\sin^2 x} = \sqrt{2}|\sin x|$.
Since the function $f(x) = \sqrt{2}|\sin x|$ has a period $T = \pi$,we can use the property $\int_0^{NT} f(x) dx = N \int_0^T f(x) dx$.
Here $N = 50$ and $T = \pi$,so $I = 50 \int_0^{\pi} \sqrt{2}|\sin x| dx$.
In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$.
Thus,$I = 50\sqrt{2} \int_0^{\pi} \sin x dx$.
$I = 50\sqrt{2} [-\cos x]_0^{\pi}$.
$I = 50\sqrt{2} (-(\cos \pi - \cos 0)) = 50\sqrt{2} (-(-1 - 1)) = 50\sqrt{2} (2) = 100\sqrt{2}$.

(B) Let $I = \int_{-\pi / 2}^{2 \pi} \sin ^{-1}(\sin x) d x$.
We split the integral into intervals where $\sin^{-1}(\sin x)$ has a linear form:
$I = \int_{-\pi / 2}^{\pi / 2} x d x + \int_{\pi / 2}^{3 \pi / 2} (\pi - x) d x + \int_{3 \pi / 2}^{2 \pi} (x - 2 \pi) d x$
Evaluating each integral:
$\int_{-\pi / 2}^{\pi / 2} x d x = 0$ (since it is an odd function over a symmetric interval).
$\int_{\pi / 2}^{3 \pi / 2} (\pi - x) d x = [\pi x - \frac{x^2}{2}]_{\pi / 2}^{3 \pi / 2} = (\frac{3 \pi^2}{2} - \frac{9 \pi^2}{8}) - (\frac{\pi^2}{2} - \frac{\pi^2}{8}) = \frac{3 \pi^2}{8} - \frac{3 \pi^2}{8} = 0$.
$\int_{3 \pi / 2}^{2 \pi} (x - 2 \pi) d x = [\frac{x^2}{2} - 2 \pi x]_{3 \pi / 2}^{2 \pi} = (2 \pi^2 - 4 \pi^2) - (\frac{9 \pi^2}{8} - 3 \pi^2) = -2 \pi^2 - (-\frac{15 \pi^2}{8}) = -2 \pi^2 + \frac{15 \pi^2}{8} = -\frac{\pi^2}{8}$.
Thus,$I = 0 + 0 - \frac{\pi^2}{8} = -\frac{\pi^2}{8}$.

(D) Let $I = \int_{-1/24}^{1/24} \sec x \log \left(\frac{1-x}{1+x}\right) dx$.
Define $f(x) = \sec x \log \left(\frac{1-x}{1+x}\right)$.
We check if the function is odd or even by evaluating $f(-x)$:
$f(-x) = \sec(-x) \log \left(\frac{1-(-x)}{1+(-x)}\right) = \sec x \log \left(\frac{1+x}{1-x}\right)$.
Using the property $\log(a^{-1}) = -\log a$,we have:
$f(-x) = \sec x \log \left(\frac{1-x}{1+x}\right)^{-1} = -\sec x \log \left(\frac{1-x}{1+x}\right) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
For an odd function,the integral over a symmetric interval $[-a, a]$ is always $0$.
Therefore,$\int_{-1/24}^{1/24} f(x) dx = 0$.