Properties of definite integration Questions in English

(A) Let $I = \int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^x-1}{e^x+1}\right) d x$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we note that the limits are $a = \log \frac{1}{2} = -\log 2$ and $b = \log 2$.
Thus,$a+b = 0$.
So,$I = \int_{-\log 2}^{\log 2} \sin \left(\frac{e^{-x}-1}{e^{-x}+1}\right) d x$.
Simplifying the argument of the sine function: $\frac{e^{-x}-1}{e^{-x}+1} = \frac{\frac{1}{e^x}-1}{\frac{1}{e^x}+1} = \frac{1-e^x}{1+e^x} = -\left(\frac{e^x-1}{e^x+1}\right)$.
Since $\sin(- \theta) = -\sin(\theta)$,the integrand is an odd function.
Therefore,$I = \int_{-\log 2}^{\log 2} -\sin \left(\frac{e^x-1}{e^x+1}\right) d x = -I$.
This implies $2I = 0$,so $I = 0$.

(A) Let $I = \int_{\frac{1}{2}}^2 \frac{1}{x} \operatorname{cosec}^{101}\left(x-\frac{1}{x}\right) d x$.
Using the property $\int_a^b f(x) dx = \int_a^b f\left(\frac{ab}{x}\right) \frac{ab}{x^2} dx$,we set $a = \frac{1}{2}$ and $b = 2$,so $ab = 1$.
Then $I = \int_{\frac{1}{2}}^2 \frac{1}{1/x} \operatorname{cosec}^{101}\left(\frac{1}{x}-x\right) \frac{1}{x^2} dx$.
$I = \int_{\frac{1}{2}}^2 x \operatorname{cosec}^{101}\left(-(x-\frac{1}{x})\right) \frac{1}{x^2} dx$.
Since $\operatorname{cosec}(- \theta) = -\operatorname{cosec}(\theta)$ and the power $101$ is odd,$\operatorname{cosec}^{101}(- \theta) = -\operatorname{cosec}^{101}(\theta)$.
Thus,$I = - \int_{\frac{1}{2}}^2 \frac{1}{x} \operatorname{cosec}^{101}\left(x-\frac{1}{x}\right) dx = -I$.
$2I = 0 \implies I = 0$.

(A) Let $I = \int_0^{\frac{\pi}{2}} \frac{300 \sin x + 100 \cos x}{\sin x + \cos x} dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we get:
$I = \int_0^{\frac{\pi}{2}} \frac{300 \sin(\frac{\pi}{2}-x) + 100 \cos(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x) + \cos(\frac{\pi}{2}-x)} dx = \int_0^{\frac{\pi}{2}} \frac{300 \cos x + 100 \sin x}{\cos x + \sin x} dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\frac{\pi}{2}} \frac{(300 \sin x + 100 \cos x) + (300 \cos x + 100 \sin x)}{\sin x + \cos x} dx$.
$2I = \int_0^{\frac{\pi}{2}} \frac{400 \sin x + 400 \cos x}{\sin x + \cos x} dx$.
$2I = \int_0^{\frac{\pi}{2}} 400 dx$.
$2I = 400 [x]_0^{\frac{\pi}{2}} = 400 \times \frac{\pi}{2} = 200 \pi$.
Therefore, $I = 100 \pi$.

(D) Let $I = \int_2^3 x f(x) dx$.
Using the property $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$,we have:
$I = \int_2^3 (2+3-x) f(2+3-x) dx = \int_2^3 (5-x) f(5-x) dx$.
Given $f(5-x) = f(x)$,we substitute this into the integral:
$I = \int_2^3 (5-x) f(x) dx = 5 \int_2^3 f(x) dx - \int_2^3 x f(x) dx$.
Since $\int_2^3 f(x) dx = 2$,we get:
$I = 5(2) - I$.
$2I = 10$.
$I = 5$.

(C) Let $I = \int_0^a f(x) g(x) dx$.
Using the property $\int_0^a h(x) dx = \int_0^a h(a-x) dx$,we have:
$I = \int_0^a f(a-x) g(a-x) dx$.
Since $f(x) = f(a-x)$ and $g(a-x) = 4 - g(x)$,we substitute these into the integral:
$I = \int_0^a f(x) (4 - g(x)) dx$.
$I = 4 \int_0^a f(x) dx - \int_0^a f(x) g(x) dx$.
$I = 4 \int_0^a f(x) dx - I$.
$2I = 4 \int_0^a f(x) dx$.
$I = 2 \int_0^a f(x) dx$.

(A) Let $I=\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^2}{\sin x^2+\sin (\log 6-x^2)} d x$.
Put $x^2=t$,then $2x \, dx = dt$,so $x \, dx = \frac{1}{2} dt$.
The limits change as follows: when $x = \sqrt{\log 2}$,$t = \log 2$; when $x = \sqrt{\log 3}$,$t = \log 3$.
Thus,$I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t}{\sin t + \sin (\log 6 - t)} dt \quad \dots (i)$
Using the property $\int_a^b f(t) dt = \int_a^b f(a+b-t) dt$,we get:
$I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 2 + \log 3 - t)}{\sin (\log 2 + \log 3 - t) + \sin (\log 6 - (\log 2 + \log 3 - t))} dt$
$I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 6 - t)}{\sin (\log 6 - t) + \sin t} dt \quad \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t + \sin (\log 6 - t)}{\sin t + \sin (\log 6 - t)} dt$
$2I = \frac{1}{2} \int_{\log 2}^{\log 3} 1 \, dt = \frac{1}{2} [t]_{\log 2}^{\log 3} = \frac{1}{2} (\log 3 - \log 2) = \frac{1}{2} \log \left(\frac{3}{2}\right)$
$I = \frac{1}{4} \log \left(\frac{3}{2}\right)$.

(A) Let $I = \int_{\frac{-1}{2}}^{\frac{1}{2}} \left([x] + \log_{e}\left(\frac{1+x}{1-x}\right)\right) dx$.
We can split the integral as:
$I = \int_{\frac{-1}{2}}^{0} [x] dx + \int_{0}^{\frac{1}{2}} [x] dx + \int_{\frac{-1}{2}}^{\frac{1}{2}} \log_{e}\left(\frac{1+x}{1-x}\right) dx$.
Let $g(x) = \log_{e}\left(\frac{1+x}{1-x}\right)$.
Then $g(-x) = \log_{e}\left(\frac{1-x}{1+x}\right) = \log_{e}\left(\left(\frac{1+x}{1-x}\right)^{-1}\right) = -\log_{e}\left(\frac{1+x}{1-x}\right) = -g(x)$.
Since $g(-x) = -g(x)$,$g(x)$ is an odd function.
Therefore,$\int_{\frac{-1}{2}}^{\frac{1}{2}} g(x) dx = 0$.
Now,for the greatest integer function $[x]$:
In the interval $[-\frac{1}{2}, 0)$,$[x] = -1$.
In the interval $[0, \frac{1}{2}]$,$[x] = 0$.
Thus,$I = \int_{\frac{-1}{2}}^{0} (-1) dx + \int_{0}^{\frac{1}{2}} (0) dx + 0$.
$I = [-x]_{\frac{-1}{2}}^{0} = -(0 - (-\frac{1}{2})) = -\frac{1}{2}$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \,dx \quad ...(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$, we get:
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(-x)^2 \cos(-x)}{1+e^{-(-x)}} \,dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} \,dx \quad ...(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \left( \frac{1}{1+e^{-x}} + \frac{1}{1+e^x} \right) dx$
Since $\frac{1}{1+e^{-x}} + \frac{1}{1+e^x} = \frac{e^x}{e^x+1} + \frac{1}{1+e^x} = 1$, we have:
$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \,dx$
Since $f(x) = x^2 \cos x$ is an even function, $2I = 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x \,dx$, so $I = \int_{0}^{\frac{\pi}{2}} x^2 \cos x \,dx$
Using integration by parts:
$I = [x^2 \sin x]_0^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2x \sin x \,dx$
$I = [x^2 \sin x]_0^{\frac{\pi}{2}} - 2([-x \cos x]_0^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} \cos x \,dx)$
$I = [x^2 \sin x + 2x \cos x - 2 \sin x]_0^{\frac{\pi}{2}}$
$I = ((\frac{\pi}{2})^2 \sin(\frac{\pi}{2}) + 2(\frac{\pi}{2}) \cos(\frac{\pi}{2}) - 2 \sin(\frac{\pi}{2})) - (0 + 0 - 0)$
$I = \frac{\pi^2}{4}(1) + 0 - 2(1) = \frac{\pi^2}{4} - 2$

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

(C) 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 $\tan \left( \frac{\pi}{4} - x \right) = \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x} = \frac{1 - \tan x}{1 + \tan x}$,we have:
$I = \int_0^{\frac{\pi}{4}} \log \left( 1 + \frac{1 - \tan x}{1 + \tan x} \right) \, dx$
$I = \int_0^{\frac{\pi}{4}} \log \left( \frac{1 + \tan x + 1 - \tan x}{1 + \tan x} \right) \, dx$
$I = \int_0^{\frac{\pi}{4}} \log \left( \frac{2}{1 + \tan x} \right) \, dx$
$I = \int_0^{\frac{\pi}{4}} (\log 2 - \log (1 + \tan x)) \, dx$
$I = \int_0^{\frac{\pi}{4}} \log 2 \, dx - \int_0^{\frac{\pi}{4}} \log (1 + \tan x) \, dx$
$I = \log 2 [x]_0^{\frac{\pi}{4}} - I$
$2I = \frac{\pi}{4} \log 2$
$I = \frac{\pi}{8} \log 2$

(D) Let $h(x) = [f(x)+f(-x)][g(x)-g(-x)]$.
We evaluate $h(-x)$:
$h(-x) = [f(-x)+f(x)][g(-x)-g(x)]$.
This can be rewritten as:
$h(-x) = [f(x)+f(-x)] \cdot -[g(x)-g(-x)]$.
$h(-x) = -[f(x)+f(-x)][g(x)-g(-x)] = -h(x)$.
Since $h(-x) = -h(x)$,the function $h(x)$ is an odd function.
According to the property of definite integrals,if $h(x)$ is an odd function,then $\int_{-a}^{a} h(x) \, dx = 0$.
Therefore,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x)+f(-x)][g(x)-g(-x)] \, dx = 0$.

(D) Let $I = \int_{-1}^3 \left(\cot^{-1}\left(\frac{x}{x^2+1}\right) + \cot^{-1}\left(\frac{x^2+1}{x}\right)\right) dx$.
Since $\cot^{-1}(u) = \tan^{-1}(\frac{1}{u})$ for $u > 0$,we observe the term $\cot^{-1}(\frac{x}{x^2+1})$.
For $x > 0$,$\cot^{-1}(\frac{x}{x^2+1}) = \tan^{-1}(\frac{x^2+1}{x})$.
However,the identity $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ holds for all $u \in \mathbb{R}$.
Thus,the integrand simplifies to $\frac{\pi}{2}$ for all $x \neq 0$.
$I = \int_{-1}^3 \frac{\pi}{2} dx = \frac{\pi}{2} [x]_{-1}^3$.
$I = \frac{\pi}{2} (3 - (-1)) = \frac{\pi}{2} (4) = 2\pi$.

(C) We are given $I_n = \int_0^{\frac{\pi}{4}} \tan^n \theta \, d\theta$.
Consider the sum $I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} (\tan^n \theta + \tan^{n-2} \theta) \, d\theta$.
Factor out $\tan^{n-2} \theta$:
$I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} \tan^{n-2} \theta (\tan^2 \theta + 1) \, d\theta$.
Since $1 + \tan^2 \theta = \sec^2 \theta$,we have:
$I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} \tan^{n-2} \theta \sec^2 \theta \, d\theta$.
Let $u = \tan \theta$,then $du = \sec^2 \theta \, d\theta$.
When $\theta = 0$,$u = 0$. When $\theta = \frac{\pi}{4}$,$u = 1$.
Thus,$I_n + I_{n-2} = \int_0^1 u^{n-2} \, du = \left[ \frac{u^{n-1}}{n-1} \right]_0^1 = \frac{1}{n-1}$.
For $n = 12$,we have $I_{12} + I_{10} = \frac{1}{12-1} = \frac{1}{11}$.

(A) Given $R_1 = \int_{-1}^2 x f(x) dx$ $(i)$
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a=-1$ and $b=2$,we have $a+b = 1$.
So,$R_1 = \int_{-1}^2 (1-x) f(1-x) dx$.
Since $f(1-x) = f(x)$,this becomes $R_1 = \int_{-1}^2 (1-x) f(x) dx = \int_{-1}^2 f(x) dx - \int_{-1}^2 x f(x) dx$.
$R_1 = \int_{-1}^2 f(x) dx - R_1$.
$2 R_1 = \int_{-1}^2 f(x) dx$.
Since $R_2$ is the area bounded by $y=f(x)$,$x=-1$,$x=2$ and the $X$-axis,$R_2 = \int_{-1}^2 f(x) dx$.
Therefore,$2 R_1 = R_2$.

(B) Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$\int_0^1 x(1-x)^n dx = \int_0^1 (1-x)(1-(1-x))^n dx$
$= \int_0^1 (1-x)x^n dx$
$= \int_0^1 (x^n - x^{n+1}) dx$
$= \left[ \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_0^1$
$= \left( \frac{1}{n+1} - \frac{1}{n+2} \right) - (0 - 0)$
$= \frac{(n+2) - (n+1)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)}$

(D) Let $I = \int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^{x}-1}{e^{x}+1}\right) dx$.
Note that $\log \frac{1}{2} = \log 1 - \log 2 = -\log 2$.
So,$I = \int_{-\log 2}^{\log 2} f(x) dx$,where $f(x) = \sin \left(\frac{e^{x}-1}{e^{x}+1}\right)$.
Check if $f(x)$ is an odd function: $f(-x) = \sin \left(\frac{e^{-x}-1}{e^{-x}+1}\right) = \sin \left(\frac{\frac{1}{e^{x}}-1}{\frac{1}{e^{x}}+1}\right) = \sin \left(\frac{1-e^{x}}{1+e^{x}}\right) = \sin \left(-\frac{e^{x}-1}{e^{x}+1}\right) = -\sin \left(\frac{e^{x}-1}{e^{x}+1}\right) = -f(x)$.
Since $f(x)$ is an odd function and the interval is symmetric $[-\log 2, \log 2]$,the integral evaluates to $0$.

(C) Given the integral $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin |x| + \cos |x|) dx$.
Since $f(x) = \sin |x| + \cos |x|$ is an even function because $f(-x) = \sin |-x| + \cos |-x| = \sin |x| + \cos |x| = f(x)$,we can use the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$.
Thus,$I = 2 \int_{0}^{\frac{\pi}{2}} (\sin |x| + \cos |x|) dx$.
For $x \in [0, \frac{\pi}{2}]$,$|x| = x$,so the integral becomes $I = 2 \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) dx$.
Evaluating the integral: $I = 2 [-\cos x + \sin x]_{0}^{\frac{\pi}{2}}$.
Substituting the limits: $I = 2 [(-\cos \frac{\pi}{2} + \sin \frac{\pi}{2}) - (-\cos 0 + \sin 0)]$.
$I = 2 [(0 + 1) - (-1 + 0)] = 2 [1 + 1] = 2(2) = 4$.

(C) Let $f(x) = \log \left(\frac{2-x}{2+x}\right)$.
Check if the function is odd or even by evaluating $f(-x)$:
$f(-x) = \log \left(\frac{2-(-x)}{2+(-x)}\right) = \log \left(\frac{2+x}{2-x}\right)$.
Using the property $\log(a/b) = -\log(b/a)$,we get:
$f(-x) = -\log \left(\frac{2-x}{2+x}\right) = -f(x)$.
Since $f(-x) = -f(x)$,the function is an odd function.
For an odd function,the property of definite integration states that $\int_{-a}^a f(x) dx = 0$.
Therefore,$\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right) d x = 0$.

(D) Let $I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1+\alpha^x} \, dx$ --- $(i)$
Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we get:
$I = \int_{-\pi}^{\pi} \frac{\cos^2(-x)}{1+\alpha^{-x}} \, dx = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1+\frac{1}{\alpha^x}} \, dx = \int_{-\pi}^{\pi} \frac{\alpha^x \cos^2 x}{\alpha^x + 1} \, dx$ --- $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{-\pi}^{\pi} \frac{\cos^2 x + \alpha^x \cos^2 x}{1+\alpha^x} \, dx = \int_{-\pi}^{\pi} \frac{\cos^2 x(1+\alpha^x)}{1+\alpha^x} \, dx = \int_{-\pi}^{\pi} \cos^2 x \, dx$
Since $\cos^2 x$ is an even function,$2I = 2 \int_{0}^{\pi} \cos^2 x \, dx$
$I = \int_{0}^{\pi} \cos^2 x \, dx = \int_{0}^{\pi} \frac{1+\cos 2x}{2} \, dx = \left[ \frac{x}{2} + \frac{\sin 2x}{4} \right]_{0}^{\pi} = \frac{\pi}{2} + 0 - (0 + 0) = \frac{\pi}{2}$

(A) Let $f(x) = \log \left(\frac{1+x}{1-x}\right)$.
Check if the function is odd or even by evaluating $f(-x)$:
$f(-x) = \log \left(\frac{1+(-x)}{1-(-x)}\right) = \log \left(\frac{1-x}{1+x}\right)$.
Using the property $\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)$,we get:
$f(-x) = -\log \left(\frac{1+x}{1-x}\right) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ 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,$\int_{-1/2}^{1/2} \log \left(\frac{1+x}{1-x}\right) dx = 0$.

(A) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{e^x(x \sin x)}{e^{2x}-1} dx$.
Consider the integrand $f(x) = \frac{e^x(x \sin x)}{e^{2x}-1}$.
We can rewrite $f(x)$ as:
$f(x) = \frac{e^x(x \sin x)}{e^x(e^x - e^{-x})} = \frac{x \sin x}{e^x - e^{-x}} = \frac{x \sin x}{2 \sinh x}$.
Now,check if $f(x)$ is an even or odd function by evaluating $f(-x)$:
$f(-x) = \frac{(-x) \sin(-x)}{e^{-x} - e^x} = \frac{(-x)(-\sin x)}{-(e^x - e^{-x})} = \frac{x \sin x}{-(e^x - e^{-x})} = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ 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) Let $I = \int_0^{\frac{\pi}{2}} \frac{dx}{1+\tan^3 x}$.
We can rewrite the integrand as: $I = \int_0^{\frac{\pi}{2}} \frac{\cos^3 x}{\cos^3 x + \sin^3 x} dx$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have:
$I = \int_0^{\frac{\pi}{2}} \frac{\cos^3(\frac{\pi}{2}-x)}{\cos^3(\frac{\pi}{2}-x) + \sin^3(\frac{\pi}{2}-x)} dx = \int_0^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin^3 x + \cos^3 x} dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\frac{\pi}{2}} \frac{\cos^3 x + \sin^3 x}{\cos^3 x + \sin^3 x} dx = \int_0^{\frac{\pi}{2}} 1 dx = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$.
Therefore,$I = \frac{\pi}{4}$.

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

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

(A) Let $f(x) = \frac{x \sin x}{1+\cos^2 x}$.
Since $f(-x) = \frac{(-x) \sin(-x)}{1+\cos^2(-x)} = \frac{(-x)(-\sin x)}{1+\cos^2 x} = f(x)$,the function is even.
Using the property $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$,we have $I = 2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$.
Applying $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get $I = 2 \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1+\cos^2(\pi-x)} dx = 2 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx$.
Adding the two expressions for $I$:
$2I = 2 \int_0^\pi \frac{\pi \sin x}{1+\cos^2 x} dx \Rightarrow I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$.
Let $t = \cos x$,then $dt = -\sin x dx$. Limits change from $0$ to $\pi$ to $1$ to $-1$.
$I = \pi \int_1^{-1} \frac{-dt}{1+t^2} = \pi \int_{-1}^1 \frac{dt}{1+t^2} = 2\pi \int_0^1 \frac{dt}{1+t^2}$.
$I = 2\pi [\tan^{-1} t]_0^1 = 2\pi (\frac{\pi}{4} - 0) = \frac{\pi^2}{2}$.

(A) Let $I = \int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ ... $(1)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\pi / 2} \log \left(\frac{4+3 \sin(\pi/2 - x)}{4+3 \cos(\pi/2 - x)}\right) d x$
$I = \int_0^{\pi / 2} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x$ ... $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_0^{\pi / 2} \left[ \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) + \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) \right] d x$
$2I = \int_0^{\pi / 2} \log \left( \frac{4+3 \sin x}{4+3 \cos x} \times \frac{4+3 \cos x}{4+3 \sin x} \right) d x$
$2I = \int_0^{\pi / 2} \log(1) d x$
Since $\log(1) = 0$,we have $2I = 0$,which implies $I = 0$.

(C) Let $I = \int_0^\pi x \sin x \cos^4 x \, dx \quad \dots(1)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^\pi (\pi - x) \sin(\pi - x) \cos^4(\pi - x) \, dx$
$I = \int_0^\pi (\pi - x) \sin x (-\cos x)^4 \, dx$
$I = \int_0^\pi (\pi - x) \sin x \cos^4 x \, dx \quad \dots(2)$
Adding $(1)$ and $(2)$:
$2I = \int_0^\pi (x + \pi - x) \sin x \cos^4 x \, dx$
$2I = \pi \int_0^\pi \sin x \cos^4 x \, dx$
Let $t = \cos x$,then $dt = -\sin x \, dx$. When $x=0, t=1$ and when $x=\pi, t=-1$.
$2I = \pi \int_1^{-1} t^4 (-dt) = \pi \int_{-1}^1 t^4 \, dt$
Since $t^4$ is an even function,$\int_{-1}^1 t^4 \, dt = 2 \int_0^1 t^4 \, dt$.
$2I = 2\pi \left[ \frac{t^5}{5} \right]_0^1 = 2\pi \left( \frac{1}{5} \right) = \frac{2\pi}{5}$
$I = \frac{\pi}{5}$

(A) Let $f(x) = \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}}$.
Then,$f(-x) = \frac{e^{-x} + e^{x}}{e^{-x} - e^{x}} = \frac{e^{-x} + e^{x}}{-(e^{x} - e^{-x})} = -\frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
For an odd function,the property of definite integrals states that $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is continuous on $[-a, a]$.
However,note that $f(x)$ has a discontinuity at $x = 0$ because the denominator $e^{x} - e^{-x} = 0$ at $x = 0$.
Therefore,the integral $\int_{-5}^{5} \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} dx$ is an improper integral.
Evaluating the principal value: $\int_{-5}^{5} \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} dx = \left[ \ln|e^{x} - e^{-x}| \right]_{-5}^{5} = \ln|e^{5} - e^{-5}| - \ln|e^{-5} - e^{5}| = 0$.

(C) Let $f(x) = \sin^{2} x$.
Since $f(-x) = [\sin(-x)]^{2} = (-\sin x)^{2} = \sin^{2} x$,the function $f(x)$ is an even function.
Using the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ for an even function:
$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin^{2} x \, dx = 2 \int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx$.
Using the identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$:
$= 2 \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx = \int_{0}^{\frac{\pi}{2}} (1 - \cos 2x) \, dx$.
Integrating term by term:
$= [x - \frac{\sin 2x}{2}]_{0}^{\frac{\pi}{2}}$.
Evaluating at the limits:
$= (\frac{\pi}{2} - \frac{\sin \pi}{2}) - (0 - \frac{\sin 0}{2}) = (\frac{\pi}{2} - 0) - (0 - 0) = \frac{\pi}{2}$.

(B) Let $I = \int_{0}^{a} (a-x)^{\frac{3}{2}} x^{2} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{a} (a-(a-x))^{\frac{3}{2}} (a-x)^{2} dx = \int_{0}^{a} x^{\frac{3}{2}} (a^{2} - 2ax + x^{2}) dx$.
$I = \int_{0}^{a} (a^{2} x^{\frac{3}{2}} - 2a x^{\frac{5}{2}} + x^{\frac{7}{2}}) dx$.
Integrating term by term:
$I = a^{2} \left[ \frac{x^{\frac{5}{2}}}{\frac{5}{2}} \right]_{0}^{a} - 2a \left[ \frac{x^{\frac{7}{2}}}{\frac{7}{2}} \right]_{0}^{a} + \left[ \frac{x^{\frac{9}{2}}}{\frac{9}{2}} \right]_{0}^{a}$.
$I = \frac{2}{5} a^{2} (a^{\frac{5}{2}}) - 2a \left( \frac{2}{7} a^{\frac{7}{2}} \right) + \frac{2}{9} a^{\frac{9}{2}}$.
$I = \frac{2}{5} a^{\frac{9}{2}} - \frac{4}{7} a^{\frac{9}{2}} + \frac{2}{9} a^{\frac{9}{2}}$.
$I = a^{\frac{9}{2}} \left( \frac{2}{5} - \frac{4}{7} + \frac{2}{9} \right) = a^{\frac{9}{2}} \left( \frac{126 - 180 + 70}{315} \right) = \frac{16}{315} a^{\frac{9}{2}}$.

(A) Let $I = \int_{0}^{1} \tan ^{-1}\left[\frac{2 x-1}{1+x-x^{2}}\right] d x$.
We can rewrite the argument of the $\tan ^{-1}$ function as:
$\frac{2x-1}{1+x-x^2} = \frac{x - (1-x)}{1 + x(1-x)}$.
Using the property $\tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a-b}{1+ab} \right)$,we get:
$I = \int_{0}^{1} [\tan^{-1} x - \tan^{-1}(1-x)] dx \quad \dots(1)$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have:
$I = \int_{0}^{1} [\tan^{-1}(1-x) - \tan^{-1}(1-(1-x))] dx = \int_{0}^{1} [\tan^{-1}(1-x) - \tan^{-1} x] dx \quad \dots(2)$.
Adding equations $(1)$ and $(2)$:
$2I = \int_{0}^{1} [\tan^{-1} x - \tan^{-1}(1-x) + \tan^{-1}(1-x) - \tan^{-1} x] dx = \int_{0}^{1} 0 dx = 0$.
Therefore,$I = 0$.

(A) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{2}{3}} x}{\sin^{\frac{2}{3}} x + \cos^{\frac{2}{3}} x} dx \quad ...(1)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{2}{3}}(\frac{\pi}{2}-x)}{\sin^{\frac{2}{3}}(\frac{\pi}{2}-x) + \cos^{\frac{2}{3}}(\frac{\pi}{2}-x)} dx$
Since $\sin(\frac{\pi}{2}-x) = \cos x$ and $\cos(\frac{\pi}{2}-x) = \sin x$,we have:
$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{\frac{2}{3}} x}{\cos^{\frac{2}{3}} x + \sin^{\frac{2}{3}} x} dx \quad ...(2)$
Adding equations $(1)$ and $(2)$:
$2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{2}{3}} x + \cos^{\frac{2}{3}} x}{\sin^{\frac{2}{3}} x + \cos^{\frac{2}{3}} x} dx$
$2I = \int_{0}^{\frac{\pi}{2}} 1 dx$
$2I = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$
$I = \frac{\pi}{4}$

(B) Let $f(x) = x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right)$.
Now,check for parity by calculating $f(-x)$:
$f(-x) = (-x)^{2}\left(\frac{e^{(-x)^{3}}-e^{-(-x)^{3}}}{e^{(-x)^{3}}+e^{-(-x)^{3}}}\right)$
$f(-x) = x^{2}\left(\frac{e^{-x^{3}}-e^{x^{3}}}{e^{-x^{3}}+e^{x^{3}}}\right)$
$f(-x) = x^{2}\left(\frac{-(e^{x^{3}}-e^{-x^{3}})}{e^{x^{3}}+e^{-x^{3}}}\right)$
$f(-x) = -x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) = -f(x)$.
Since $f(-x) = -f(x)$,the function 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,$\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x = 0$.

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{(\sec x)^{1/3}}{(\sec x)^{1/3} + (\operatorname{cosec} x)^{1/3}} dx \quad ...(1)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have:
$I = \int_{0}^{\frac{\pi}{2}} \frac{(\sec(\frac{\pi}{2}-x))^{1/3}}{(\sec(\frac{\pi}{2}-x))^{1/3} + (\operatorname{cosec}(\frac{\pi}{2}-x))^{1/3}} dx$
Since $\sec(\frac{\pi}{2}-x) = \operatorname{cosec} x$ and $\operatorname{cosec}(\frac{\pi}{2}-x) = \sec x$,we get:
$I = \int_{0}^{\frac{\pi}{2}} \frac{(\operatorname{cosec} x)^{1/3}}{(\operatorname{cosec} x)^{1/3} + (\sec x)^{1/3}} dx \quad ...(2)$
Adding equation $(1)$ and $(2)$:
$2I = \int_{0}^{\frac{\pi}{2}} \frac{(\sec x)^{1/3} + (\operatorname{cosec} x)^{1/3}}{(\sec x)^{1/3} + (\operatorname{cosec} x)^{1/3}} dx$
$2I = \int_{0}^{\frac{\pi}{2}} 1 dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$
Therefore,$I = \frac{\pi}{4}$.

(A) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x$.
Converting to sine and cosine:
$I = \int_{0}^{\frac{\pi}{2}} \frac{1-\frac{\cos x}{\sin x}}{\frac{1}{\sin x}+\cos x} d x = \int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x \quad ...(1)$
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$:
$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)-\cos(\frac{\pi}{2}-x)}{1+\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2}-x)} d x$
$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\cos x \sin x} d x \quad ...(2)$
Adding equations $(1)$ and $(2)$:
$2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x + \cos x-\sin x}{1+\sin x \cos x} d x$
$2I = \int_{0}^{\frac{\pi}{2}} 0 d x = 0$
Therefore,$I = 0$.

(B) Let $f(x) = \frac{2x}{1+\cos^2 x}$.
Check if the function is odd or even by evaluating $f(-x)$:
$f(-x) = \frac{2(-x)}{1+\cos^2(-x)} = \frac{-2x}{1+\cos^2 x} = -f(x)$.
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,$\int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x} dx = 0$.

(D) Let $I = \int_{-1}^{1} \left( \sqrt{1+x+x^{2}} - \sqrt{1-x+x^{2}} \right) dx$.
Define $f(x) = \sqrt{1+x+x^{2}} - \sqrt{1-x+x^{2}}$.
Now,calculate $f(-x)$:
$f(-x) = \sqrt{1+(-x)+(-x)^{2}} - \sqrt{1-(-x)+(-x)^{2}} = \sqrt{1-x+x^{2}} - \sqrt{1+x+x^{2}}$.
This can be written as $f(-x) = - \left( \sqrt{1+x+x^{2}} - \sqrt{1-x+x^{2}} \right) = -f(x)$.
Since $f(x)$ is an odd function,by the property of definite integrals $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is odd,we have $I = 0$.

(B) Let $I = \int_{0}^{\pi} \frac{x \cos x \sin x}{\cos^{3} x + \cos x} dx$.
Simplifying the integrand: $I = \int_{0}^{\pi} \frac{x \sin x}{\cos^{2} x + 1} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$:
$I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{\cos^{2}(\pi - x) + 1} dx = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{\cos^{2} x + 1} dx$.
$I = \pi \int_{0}^{\pi} \frac{\sin x}{\cos^{2} x + 1} dx - \int_{0}^{\pi} \frac{x \sin x}{\cos^{2} x + 1} dx$.
$I = \pi \int_{0}^{\pi} \frac{\sin x}{\cos^{2} x + 1} dx - I$.
$2I = \pi \int_{0}^{\pi} \frac{\sin x}{\cos^{2} x + 1} dx$.
Let $t = \cos x$,then $dt = -\sin x dx$.
When $x = 0, t = 1$; when $x = \pi, t = -1$.
$2I = -\pi \int_{1}^{-1} \frac{dt}{t^{2} + 1} = \pi \int_{-1}^{1} \frac{dt}{t^{2} + 1} = 2\pi \int_{0}^{1} \frac{dt}{t^{2} + 1}$.
$2I = 2\pi [\tan^{-1} t]_{0}^{1} = 2\pi (\frac{\pi}{4} - 0) = \frac{\pi^{2}}{2}$.
$I = \frac{\pi^{2}}{4}$.

(B) Let $I = \int_{0}^{\frac{\pi}{2}} (e^{\sin x} - e^{\cos x}) dx$ ....$(1)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\frac{\pi}{2}} (e^{\sin(\frac{\pi}{2}-x)} - e^{\cos(\frac{\pi}{2}-x)}) dx$
$I = \int_{0}^{\frac{\pi}{2}} (e^{\cos x} - e^{\sin x}) dx$ ....$(2)$
Adding equation $(1)$ and $(2)$:
$2I = \int_{0}^{\frac{\pi}{2}} (e^{\sin x} - e^{\cos x} + e^{\cos x} - e^{\sin x}) dx$
$2I = \int_{0}^{\frac{\pi}{2}} (0) dx$
$2I = 0 \Rightarrow I = 0$

(D) Let $f(x) = \log \left(\frac{8-x}{8+x}\right)$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \log \left(\frac{8-(-x)}{8+(-x)}\right) = \log \left(\frac{8+x}{8-x}\right)$.
Using the property $\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)$,we get:
$f(-x) = -\log \left(\frac{8-x}{8+x}\right) = -f(x)$.
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,$\int_{-4}^{4} \log \left(\frac{8-x}{8+x}\right) d x = 0$.

(B) Let $I = \int_{-5}^{5} \log \left(\frac{7-x}{7+x}\right) dx$.
Define $f(x) = \log \left(\frac{7-x}{7+x}\right)$.
Check if the function is even or odd by calculating $f(-x)$:
$f(-x) = \log \left(\frac{7-(-x)}{7+(-x)}\right) = \log \left(\frac{7+x}{7-x}\right)$.
Using the property $\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)$,we get:
$f(-x) = -\log \left(\frac{7-x}{7+x}\right) = -f(x)$.
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 $\int_{-a}^{a} f(x) dx = 0$.
Therefore,$I = 0$.

(D) Let $I = \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{\tan x}{\tan x + \cot x} \, dx \quad \dots(1)$
Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,where $a = \frac{\pi}{5}$ and $b = \frac{3 \pi}{10}$,we have $a+b = \frac{\pi}{5} + \frac{3 \pi}{10} = \frac{2\pi + 3\pi}{10} = \frac{5\pi}{10} = \frac{\pi}{2}$.
So,$I = \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{\tan(\frac{\pi}{2}-x)}{\tan(\frac{\pi}{2}-x) + \cot(\frac{\pi}{2}-x)} \, dx$.
Since $\tan(\frac{\pi}{2}-x) = \cot x$ and $\cot(\frac{\pi}{2}-x) = \tan x$,we get:
$I = \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{\cot x}{\cot x + \tan x} \, dx \quad \dots(2)$
Adding equations $(1)$ and $(2)$:
$2I = \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{\tan x + \cot x}{\tan x + \cot x} \, dx = \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} 1 \, dx$
$2I = [x]_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} = \frac{3 \pi}{10} - \frac{\pi}{5} = \frac{3 \pi - 2 \pi}{10} = \frac{\pi}{10}$
$I = \frac{\pi}{20}$

(A) Let $I = \int_{0}^{1} \tan ^{-1}\left(\frac{2x-1}{1+x-x^{2}}\right) dx$.
We can rewrite the argument of the inverse tangent function as:
$\frac{2x-1}{1+x-x^{2}} = \frac{x - (1-x)}{1 + x(1-x)}$.
Using the identity $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$,we have:
$I = \int_{0}^{1} (\tan^{-1} x - \tan^{-1}(1-x)) dx$ ... $(1)$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{1} (\tan^{-1}(1-x) - \tan^{-1}(1-(1-x))) dx = \int_{0}^{1} (\tan^{-1}(1-x) - \tan^{-1} x) dx$ ... $(2)$.
Adding equations $(1)$ and $(2)$:
$2I = \int_{0}^{1} (\tan^{-1} x - \tan^{-1}(1-x) + \tan^{-1}(1-x) - \tan^{-1} x) dx = \int_{0}^{1} 0 dx = 0$.
Therefore,$I = 0$.

(C) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}} dx$ ... $(1)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin(\frac{\pi}{2}-x)}}{\sqrt[7]{\sin(\frac{\pi}{2}-x)}+\sqrt[7]{\cos(\frac{\pi}{2}-x)}} dx$
$I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\cos x}}{\sqrt[7]{\cos x}+\sqrt[7]{\sin x}} dx$ ... $(2)$
Adding equations $(1)$ and $(2)$:
$2I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x} + \sqrt[7]{\cos x}}{\sqrt[7]{\sin x} + \sqrt[7]{\cos x}} dx$
$2I = \int_{0}^{\frac{\pi}{2}} 1 dx$
$2I = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$
$I = \frac{\pi}{4}$

(D) Let $I = \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x \quad ...(1)$
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:
$I = \int_{0}^{\pi} \frac{e^{\cos(\pi-x)}}{e^{\cos(\pi-x)}+e^{-\cos(\pi-x)}} d x$
Since $\cos(\pi-x) = -\cos x$,we have:
$I = \int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}} d x \quad ...(2)$
Adding equations $(1)$ and $(2)$:
$2I = \int_{0}^{\pi} \left( \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} + \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}} \right) d x$
$2I = \int_{0}^{\pi} \frac{e^{\cos x} + e^{-\cos x}}{e^{\cos x} + e^{-\cos x}} d x$
$2I = \int_{0}^{\pi} 1 d x$
$2I = [x]_{0}^{\pi} = \pi - 0 = \pi$
$I = \frac{\pi}{2}$

(B) Let $I = \int_{-8}^{8} \frac{x^{5}+x^{3}}{4-x^{2}} \, dx$.
Define $f(x) = \frac{x^{5}+x^{3}}{4-x^{2}}$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \frac{(-x)^{5}+(-x)^{3}}{4-(-x)^{2}} = \frac{-(x^{5}+x^{3})}{4-x^{2}} = -f(x)$.
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,$I = 0$.

(B) Let $I = \int_{0}^{1} x(1 - x)^{5} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have:
$I = \int_{0}^{1} (1 - x)(1 - (1 - x))^{5} dx$
$I = \int_{0}^{1} (1 - x)(x)^{5} dx$
$I = \int_{0}^{1} (x^{5} - x^{6}) dx$
Integrating term by term:
$I = \left[ \frac{x^{6}}{6} - \frac{x^{7}}{7} \right]_{0}^{1}$
$I = \left( \frac{1}{6} - \frac{1}{7} \right) - (0 - 0)$
$I = \frac{7 - 6}{42} = \frac{1}{42}$.

(B) Let $I = \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} dx$ . . . .$(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we get:
$I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{a + b - (a + b - x)}} dx$
$I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{x}} dx$ . . . .$(ii)$
Adding equations $(i)$ and $(ii)$:
$2I = \int_{a}^{b} \frac{\sqrt{x} + \sqrt{a + b - x}}{\sqrt{x} + \sqrt{a + b - x}} dx$
$2I = \int_{a}^{b} 1 dx = [x]_{a}^{b} = b - a$
$I = \frac{b - a}{2}$

(B) Let $I = \int_{-3}^{3} (ax^5 + bx^3 + cx + k) dx$.
We can split the integral as:
$I = \int_{-3}^{3} (ax^5 + bx^3 + cx) dx + \int_{-3}^{3} k dx$.
Since $f(x) = ax^5 + bx^3 + cx$ is an odd function (i.e.,$f(-x) = -f(x)$),the integral of this function over the symmetric interval $[-3, 3]$ is $0$.
Therefore,$I = 0 + \int_{-3}^{3} k dx = \int_{-3}^{3} k dx$.
Evaluating this,we get $I = [kx]_{-3}^{3} = k(3) - k(-3) = 3k + 3k = 6k$.
Thus,the value of the integral depends only on the constant $k$.