Properties of definite integration Questions in English

(B) Let $I = \int_{0}^{\pi} \log (1+\cos x) d x \dots (i)$
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:
$I = \int_{0}^{\pi} \log (1+\cos(\pi-x)) d x = \int_{0}^{\pi} \log (1-\cos x) d x \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{0}^{\pi} [\log(1+\cos x) + \log(1-\cos x)] d x$
$2I = \int_{0}^{\pi} \log(1-\cos^2 x) d x = \int_{0}^{\pi} \log(\sin^2 x) d x$
$2I = 2 \int_{0}^{\pi} \log(\sin x) d x$
$I = \int_{0}^{\pi} \log(\sin x) d x$
Using the property $\int_{0}^{2a} f(x) d x = 2 \int_{0}^{a} f(x) d x$ if $f(2a-x) = f(x)$:
$I = 2 \int_{0}^{\pi/2} \log(\sin x) d x$
Since $\int_{0}^{\pi/2} \log(\sin x) d x = -\frac{\pi}{2} \log 2$:
$I = 2 \times (-\frac{\pi}{2} \log 2) = -\pi \log 2 = \pi \log(\frac{1}{2})$

(D) Let $I = \int_{0}^{\pi / 2} \frac{d x}{1+\tan x}$.
We can write $\tan x = \frac{\sin x}{\cos x}$,so $I = \int_{0}^{\pi / 2} \frac{\cos x}{\sin x + \cos x} d x$ ... $(i)$.
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:
$I = \int_{0}^{\pi / 2} \frac{\cos(\pi / 2 - x)}{\sin(\pi / 2 - x) + \cos(\pi / 2 - x)} d x = \int_{0}^{\pi / 2} \frac{\sin x}{\cos x + \sin x} d x$ ... $(ii)$.
Adding $(i)$ and $(ii)$:
$2I = \int_{0}^{\pi / 2} \frac{\cos x + \sin x}{\sin x + \cos x} d x = \int_{0}^{\pi / 2} 1 d x$.
$2I = [x]_{0}^{\pi / 2} = \pi / 2$.
Therefore,$I = \pi / 4$.

(A) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$ $(i)$
Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,where $a = -\pi / 2$ and $b = \pi / 2$,we have $a+b = 0$.
So,$I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos(-x)}{1+e^{-x}} d x = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{-x}} d x$
Multiplying the numerator and denominator by $e^x$,we get:
$I = \int_{-\pi / 2}^{\pi / 2} \frac{e^x \cos x}{e^x + 1} d x$ $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x + e^x \cos x}{1+e^x} d x = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x(1+e^x)}{1+e^x} d x$
$2I = \int_{-\pi / 2}^{\pi / 2} \cos x d x$
Since $\cos x$ is an even function,$2I = 2 \int_{0}^{\pi / 2} \cos x d x$
$2I = 2[\sin x]_{0}^{\pi / 2} = 2(1 - 0) = 2$
Therefore,$I = 1$.

(C) Let $I = \int_{4}^{7} \frac{(11-x)^{2}}{x^{2}+(11-x)^{2}} d x$ $(i)$
Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,we get:
$I = \int_{4}^{7} \frac{(11-(11-x))^{2}}{(11-x)^{2}+(11-(11-x))^{2}} d x$
$I = \int_{4}^{7} \frac{x^{2}}{(11-x)^{2}+x^{2}} d x$ $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{4}^{7} \frac{(11-x)^{2} + x^{2}}{x^{2}+(11-x)^{2}} d x$
$2I = \int_{4}^{7} 1 d x$
$2I = [x]_{4}^{7} = 7 - 4 = 3$
$I = \frac{3}{2}$

(A) Let $I = \int_{0}^{\pi / 2} \log (\operatorname{cosec} x) d x$.
Since $\operatorname{cosec} x = \frac{1}{\sin x}$,we have $\log (\operatorname{cosec} x) = \log (\sin x)^{-1} = -\log \sin x$.
Therefore,$I = -\int_{0}^{\pi / 2} \log \sin x d x$.
Using the standard definite integral result $\int_{0}^{\pi / 2} \log \sin x d x = -\frac{\pi}{2} \log 2$,we get:
$I = -(-\frac{\pi}{2} \log 2) = \frac{\pi}{2} \log 2$.

(A) Let $I = \int_{0}^{\pi / 2} \frac{\sin x - \cos x}{1 - \sin x \cos x} d x$ $(i)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we replace $x$ with $(\frac{\pi}{2} - x)$:
$I = \int_{0}^{\pi / 2} \frac{\sin(\frac{\pi}{2} - x) - \cos(\frac{\pi}{2} - x)}{1 - \sin(\frac{\pi}{2} - x) \cos(\frac{\pi}{2} - x)} dx$
Since $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$,we get:
$I = \int_{0}^{\pi / 2} \frac{\cos x - \sin x}{1 - \cos x \sin x} dx$
$I = - \int_{0}^{\pi / 2} \frac{\sin x - \cos x}{1 - \sin x \cos x} dx$
$I = -I$ (ii)
Adding equation $(i)$ and (ii):
$I + I = 0$
$2I = 0$
$I = 0$

(B) Let $I = \int_{0}^{\pi} x \sin^{3} x \, dx$ ...$(i)$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get:
$I = \int_{0}^{\pi} (\pi - x) \sin^{3}(\pi - x) \, dx = \int_{0}^{\pi} (\pi - x) \sin^{3} x \, dx$ ...(ii)
Adding $(i)$ and (ii):
$2I = \int_{0}^{\pi} (x + \pi - x) \sin^{3} x \, dx = \pi \int_{0}^{\pi} \sin^{3} x \, dx$
Using $\sin 3x = 3 \sin x - 4 \sin^{3} x$,we have $\sin^{3} x = \frac{3 \sin x - \sin 3x}{4}$.
$2I = \frac{\pi}{4} \int_{0}^{\pi} (3 \sin x - \sin 3x) \, dx$
$2I = \frac{\pi}{4} \left[ -3 \cos x + \frac{\cos 3x}{3} \right]_{0}^{\pi}$
$2I = \frac{\pi}{4} \left[ (-3(-1) + \frac{-1}{3}) - (-3(1) + \frac{1}{3}) \right]$
$2I = \frac{\pi}{4} \left[ (3 - \frac{1}{3}) - (-3 + \frac{1}{3}) \right] = \frac{\pi}{4} \left[ \frac{8}{3} + \frac{8}{3} \right] = \frac{\pi}{4} \times \frac{16}{3} = \frac{4 \pi}{3}$
$I = \frac{2 \pi}{3}$

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

(C) We need to evaluate the integral $I = \int_0^{2 \pi} (\sin x + |\sin x|) \, dx$.
Since the function $|\sin x|$ changes behavior at $x = \pi$,we split the integral:
$I = \int_0^{\pi} (\sin x + |\sin x|) \, dx + \int_{\pi}^{2 \pi} (\sin x + |\sin x|) \, dx$.
In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$.
In the interval $[\pi, 2 \pi]$,$\sin x \le 0$,so $|\sin x| = -\sin x$.
Thus,$I = \int_0^{\pi} (\sin x + \sin x) \, dx + \int_{\pi}^{2 \pi} (\sin x - \sin x) \, dx$.
$I = \int_0^{\pi} 2 \sin x \, dx + \int_{\pi}^{2 \pi} 0 \, dx$.
$I = 2 [-\cos x]_0^{\pi} + 0$.
$I = 2 [-\cos(\pi) - (-\cos(0))] = 2 [-(-1) - (-1)] = 2 [1 + 1] = 4$.

(A) Given $g(x) = \int_0^x \cos^4 t \,dt$.
We need to find $g(x+\pi) = \int_0^{x+\pi} \cos^4 t \,dt$.
Using the property of definite integrals, $\int_0^{x+\pi} f(t) \,dt = \int_0^x f(t) \,dt + \int_x^{x+\pi} f(t) \,dt$.
Thus, $g(x+\pi) = g(x) + \int_x^{x+\pi} \cos^4 t \,dt$.
Since $\cos^4 t$ is a periodic function with period $\pi$, the integral over any interval of length $\pi$ is equal to the integral over $[0, \pi]$.
Therefore, $\int_x^{x+\pi} \cos^4 t \,dt = \int_0^\pi \cos^4 t \,dt = g(\pi)$.
Hence, $g(x+\pi) = g(x) + g(\pi)$.

(C) Let $ I = \int_0^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 - \sin x \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{\sin(\frac{\pi}{2} - x) - \cos(\frac{\pi}{2} - x)}{1 - \sin(\frac{\pi}{2} - x) \cos(\frac{\pi}{2} - x)} dx $
$ I = \int_0^{\frac{\pi}{2}} \frac{\cos x - \sin x}{1 - \cos x \sin x} dx $ $(2)$
Adding equation $(1)$ and $(2)$:
$ 2I = \int_0^{\frac{\pi}{2}} \left( \frac{\sin x - \cos x}{1 - \sin x \cos x} + \frac{\cos x - \sin x}{1 - \sin x \cos x} \right) dx $
$ 2I = \int_0^{\frac{\pi}{2}} \frac{\sin x - \cos x + \cos x - \sin x}{1 - \sin x \cos x} dx $
$ 2I = \int_0^{\frac{\pi}{2}} 0 dx = 0 $
Therefore,$ I = 0 $.

(C) 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 get:
$I = \int_{0}^{1} (1-x)(1-(1-x))^{5} dx$
$I = \int_{0}^{1} (1-x)x^{5} dx$
$I = \int_{0}^{1} (x^{5} - x^{6}) dx$
$I = \left[ \frac{x^{6}}{6} - \frac{x^{7}}{7} \right]_{0}^{1}$
$I = \left( \frac{1}{6} - \frac{1}{7} \right) - (0 - 0)$
$I = \frac{7-6}{42} = \frac{1}{42}$.

(C) Given $I_1 = \int_{1-h}^{h} x f(x(1-x)) dx$ and $I_2 = \int_{1-h}^{h} f(x(1-x)) dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = (1-h) + h = 1$.
Applying this to $I_1$:
$I_1 = \int_{1-h}^{h} (1-x) f((1-x)(1-(1-x))) dx$
$I_1 = \int_{1-h}^{h} (1-x) f((1-x)x) dx$
$I_1 = \int_{1-h}^{h} f(x(1-x)) dx - \int_{1-h}^{h} x f(x(1-x)) dx$
$I_1 = I_2 - I_1$
$2I_1 = I_2$
Therefore,$\frac{I_1}{I_2} = \frac{1}{2}$.

(A) We need to evaluate the integral $I = \int_1^2 |2x - [3x]| dx$.
The function $[3x]$ changes its value at $x = \frac{n}{3}$ for integers $n$. In the interval $[1, 2]$,the points of discontinuity are $x = \frac{4}{3}, \frac{5}{3}$.
We split the integral as follows:
$I = \int_1^{4/3} |2x - 3| dx + \int_{4/3}^{5/3} |2x - 4| dx + \int_{5/3}^2 |2x - 5| dx$
Since $2x < 3$ for $x \in [1, 4/3]$,$2x < 4$ for $x \in [4/3, 5/3]$,and $2x < 5$ for $x \in [5/3, 2]$,we have:
$I = \int_1^{4/3} (3 - 2x) dx + \int_{4/3}^{5/3} (4 - 2x) dx + \int_{5/3}^2 (5 - 2x) dx$
$I = [3x - x^2]_1^{4/3} + [4x - x^2]_{4/3}^{5/3} + [5x - x^2]_{5/3}^2$
$I = (4 - 16/9) - (3 - 1) + (20/3 - 25/9) - (16/3 - 16/9) + (10 - 4) - (25/3 - 25/9)$
$I = (20/9 - 2) + (35/9 - 32/9) + (6 - 50/9) = 20/9 - 18/9 + 3/9 + 54/9 - 50/9 = 9/9 = 1$.

(B) Let $I = \int_{0}^{1} \tan ^{-1}\left(\frac{1-x}{1+x}\right) \, dx$.
Using the trigonometric identity $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1}\left(\frac{A-B}{1+AB}\right)$,we can write $\tan ^{-1}\left(\frac{1-x}{1+x}\right) = \tan ^{-1}(1) - \tan ^{-1}(x)$.
Since $\tan ^{-1}(1) = \frac{\pi}{4}$,the integral becomes:
$I = \int_{0}^{1} \left( \frac{\pi}{4} - \tan ^{-1}(x) \right) \, dx$.
$I = \int_{0}^{1} \frac{\pi}{4} \, dx - \int_{0}^{1} \tan ^{-1}(x) \, dx$.
$I = \left[ \frac{\pi}{4} x \right]_{0}^{1} - p$.
$I = \frac{\pi}{4}(1 - 0) - p$.
$I = \frac{\pi}{4} - p$.

(B) 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:
$\frac{2x-1}{1+x-x^2} = \frac{x + (x-1)}{1 - x(x-1)}$.
Using the identity $\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$,we have:
$I = \int_{0}^{1} (\tan^{-1}(x) + \tan^{-1}(x-1)) d x$.
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,let $f(x) = \tan^{-1}(x-1)$.
Then $\int_{0}^{1} \tan^{-1}(x-1) d x = \int_{0}^{1} \tan^{-1}((1-x)-1) d x = \int_{0}^{1} \tan^{-1}(-x) d x = -\int_{0}^{1} \tan^{-1}(x) d x$.
Substituting this back into the integral:
$I = \int_{0}^{1} \tan^{-1}(x) d x - \int_{0}^{1} \tan^{-1}(x) d x = 0$.

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

(D) Let $I = \int_0^{\pi / 2} \frac{\cos x}{3 \cos x + \sin x} dx$.
We express the numerator as: $\cos x = A(3 \cos x + \sin x) + B \frac{d}{dx}(3 \cos x + \sin x)$.
$\cos x = A(3 \cos x + \sin x) + B(-3 \sin x + \cos x)$.
Comparing coefficients of $\cos x$ and $\sin x$:
$3A + B = 1$ and $A - 3B = 0$.
From the second equation,$A = 3B$. Substituting into the first: $3(3B) + B = 1 \implies 10B = 1 \implies B = \frac{1}{10}$.
Then $A = 3(\frac{1}{10}) = \frac{3}{10}$.
Now,$I = \int_0^{\pi / 2} \left( \frac{3}{10} \frac{3 \cos x + \sin x}{3 \cos x + \sin x} + \frac{1}{10} \frac{-3 \sin x + \cos x}{3 \cos x + \sin x} \right) dx$.
$I = \frac{3}{10} \int_0^{\pi / 2} dx + \frac{1}{10} \int_0^{\pi / 2} \frac{d(3 \cos x + \sin x)}{3 \cos x + \sin x}$.
$I = \frac{3}{10} [x]_0^{\pi / 2} + \frac{1}{10} [\log |3 \cos x + \sin x|]_0^{\pi / 2}$.
$I = \frac{3}{10} (\frac{\pi}{2} - 0) + \frac{1}{10} (\log |3(0) + 1| - \log |3(1) + 0|)$.
$I = \frac{3 \pi}{20} + \frac{1}{10} (\log 1 - \log 3) = \frac{3 \pi}{20} - \frac{\log 3}{10}$.

(C) Let $I = \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{2\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ . . . $(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a = \frac{\pi}{18}$ and $b = \frac{4\pi}{9}$,we have $a+b = \frac{\pi}{18} + \frac{8\pi}{18} = \frac{9\pi}{18} = \frac{\pi}{2}$.
So,$I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} dx$
$I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ . . . $(ii)$
Adding $(i)$ and $(ii)$:
$2I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) dx$
$2I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} 1 dx = 2 [x]_{\frac{\pi}{18}}^{\frac{4\pi}{9}}$
$2I = 2 \left( \frac{4\pi}{9} - \frac{\pi}{18} \right) = 2 \left( \frac{8\pi - \pi}{18} \right) = 2 \left( \frac{7\pi}{18} \right) = \frac{7\pi}{9}$
$I = \frac{7\pi}{18}$

(A) Let $I = \int_{0}^{\pi} \frac{x \, dx}{1+\cos \alpha \sin x} \quad \dots (i)$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get
$I = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{1+\cos \alpha \sin(\pi-x)} = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{1+\cos \alpha \sin x} \quad \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \pi \int_{0}^{\pi} \frac{dx}{1+\cos \alpha \sin x}$
Using $\sin x = \frac{2 \tan(x/2)}{1+\tan^2(x/2)}$ and $dx = \frac{2 \sec^2(x/2) \, d(x/2)}{1}$,let $t = \tan(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) \, dx$:
$2I = \pi \int_{0}^{\infty} \frac{2 \, dt}{1+t^2 + 2t \cos \alpha} = 2\pi \int_{0}^{\infty} \frac{dt}{(t+\cos \alpha)^2 + \sin^2 \alpha}$
$I = \pi \left[ \frac{1}{\sin \alpha} \tan^{-1} \left( \frac{t+\cos \alpha}{\sin \alpha} \right) \right]_{0}^{\infty}$
$I = \frac{\pi}{\sin \alpha} \left( \frac{\pi}{2} - \tan^{-1} \left( \cot \alpha \right) \right) = \frac{\pi}{\sin \alpha} \left( \frac{\pi}{2} - (\frac{\pi}{2} - \alpha) \right) = \frac{\pi \alpha}{\sin \alpha}$

(A) We know that $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for all $u \in \mathbb{R}$.
Given the integral $I = \int_1^3 \left[ \tan^{-1} \left( \frac{x}{x^2-1} \right) + \tan^{-1} \left( \frac{x^2-1}{x} \right) \right] dx$.
Since $\tan^{-1} \left( \frac{x^2-1}{x} \right) = \cot^{-1} \left( \frac{x}{x^2-1} \right)$,the expression becomes $\tan^{-1} \left( \frac{x}{x^2-1} \right) + \cot^{-1} \left( \frac{x}{x^2-1} \right)$.
Thus,the integrand simplifies to $\frac{\pi}{2}$.
Therefore,$I = \int_1^3 \frac{\pi}{2} dx = \frac{\pi}{2} [x]_1^3 = \frac{\pi}{2} (3-1) = \frac{\pi}{2} \times 2 = \pi$.

(B) We know that $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for any $u > 0$.
Given the integral $I = \int_{-1}^{3} \left[ \tan^{-1} \left( \frac{x}{x^{2}+1} \right) + \tan^{-1} \left( \frac{x^{2}+1}{x} \right) \right] dx$.
Since $\tan^{-1} \left( \frac{x^{2}+1}{x} \right) = \cot^{-1} \left( \frac{x}{x^{2}+1} \right)$ for $x > 0$,the expression inside the integral becomes $\tan^{-1} \left( \frac{x}{x^{2}+1} \right) + \cot^{-1} \left( \frac{x}{x^{2}+1} \right) = \frac{\pi}{2}$.
Thus,$I = \int_{-1}^{3} \frac{\pi}{2} dx = \frac{\pi}{2} [x]_{-1}^{3} = \frac{\pi}{2} (3 - (-1)) = \frac{\pi}{2} (4) = 2 \pi$.

(B) Let $I = \int_0^1 \frac{8 \log (1+x)}{1+x^2} \,dx$.
Substitute $x = \tan \theta$, so $dx = \sec^2 \theta \,d\theta$.
When $x = 0$, $\theta = 0$; when $x = 1$, $\theta = \frac{\pi}{4}$.
$I = \int_0^{\frac{\pi}{4}} \frac{8 \log (1+\tan \theta)}{1+\tan^2 \theta} \cdot \sec^2 \theta \,d\theta = \int_0^{\frac{\pi}{4}} 8 \log (1+\tan \theta) \,d\theta \quad \dots(i)$
Using the property $\int_0^a f(\theta) \,d\theta = \int_0^a f(a-\theta) \,d\theta$:
$I = \int_0^{\frac{\pi}{4}} 8 \log \left(1 + \tan\left(\frac{\pi}{4} - \theta\right)\right) \,d\theta$
Since $\tan\left(\frac{\pi}{4} - \theta\right) = \frac{1-\tan \theta}{1+\tan \theta}$, we have:
$I = \int_0^{\frac{\pi}{4}} 8 \log \left(1 + \frac{1-\tan \theta}{1+\tan \theta}\right) \,d\theta = \int_0^{\frac{\pi}{4}} 8 \log \left(\frac{2}{1+\tan \theta}\right) \,d\theta \quad \dots(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\frac{\pi}{4}} 8 \left[ \log(1+\tan \theta) + \log\left(\frac{2}{1+\tan \theta}\right) \right] \,d\theta$
$2I = 8 \int_0^{\frac{\pi}{4}} \log \left( (1+\tan \theta) \cdot \frac{2}{1+\tan \theta} \right) \,d\theta$
$2I = 8 \int_0^{\frac{\pi}{4}} \log 2 \,d\theta = 8 \log 2 [\theta]_0^{\frac{\pi}{4}} = 8 \log 2 \cdot \frac{\pi}{4} = 2\pi \log 2$.
Therefore, $I = \pi \log 2$.

(C) Let $f(x) = x^5 - x^3 \cos x + \sin^3 x - 3$.
We observe that $f(-x) = (-x)^5 - (-x)^3 \cos(-x) + \sin^3(-x) - 3 = -x^5 + x^3 \cos x - \sin^3 x - 3$.
This does not immediately show $f(x)$ is even or odd,so we split the integral:
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^5 \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 \cos x \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^3 x \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 \, dx$.
Since $x^5$,$x^3 \cos x$,and $\sin^3 x$ are odd functions,their integrals over the symmetric interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ are $0$.
Thus,$I = 0 - 0 + 0 - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 \, dx$.
$I = -3 \times [x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = -3 \times (\frac{\pi}{2} - (-\frac{\pi}{2})) = -3 \times \pi = -3\pi$.

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\tan^4 x} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$,we have:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\tan^4(\frac{\pi}{2}-x)} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\cot^4 x} dx$.
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\tan^4 x}{1+\tan^4 x} dx$.
Adding the two expressions for $I$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1+\tan^4 x}{1+\tan^4 x} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$.
$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$I = \frac{\pi}{12}$.

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{13} + x \cos x + \tan^{15} x + 1) \, dx$.
Using the property $\int_{-a}^{a} f(x) \, dx = 0$ if $f(x)$ is an odd function,and $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(x)$ is an even function.
Let $f(x) = x^{13} + x \cos x + \tan^{15} x + 1$.
We can split the integral as $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{13} + x \cos x + \tan^{15} x) \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx$.
Since $x^{13}$,$x \cos x$,and $\tan^{15} x$ are all odd functions,their integral over the symmetric interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is $0$.
Thus,$I = 0 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = [x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2019-x}{2019+x}\right) d x$.
Consider the function $f(x) = \log \left(\frac{2019-x}{2019+x}\right)$.
Check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \log \left(\frac{2019-(-x)}{2019+(-x)}\right) = \log \left(\frac{2019+x}{2019-x}\right)$.
Using the property $\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)$,we get:
$f(-x) = -\log \left(\frac{2019-x}{2019+x}\right) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
For an odd function,the definite integral over a symmetric interval $[-a, a]$ is always $0$,i.e.,$\int_{-a}^{a} f(x) dx = 0$.
Therefore,$I = 0$.

(C) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\tan x}}$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a = \frac{\pi}{6}$ and $b = \frac{\pi}{3}$,we have $a+b = \frac{\pi}{2}$.
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\tan(\frac{\pi}{2}-x)}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\cot x}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\frac{1}{\sqrt{\tan x}}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}}{\sqrt{\tan x}+1} dx$.
Adding the two expressions for $I$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \frac{1}{1+\sqrt{\tan x}} + \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} \right) dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$.
$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$I = \frac{\pi}{12}$.

(D) Let $I = \int_0^{2 \pi} \sin ^3 x \cos ^2 x \, dx$.
Using the property $\int_0^{2a} f(x) \, dx = 0$ if $f(2a - x) = -f(x)$.
Here,$f(x) = \sin ^3 x \cos ^2 x$.
$f(2 \pi - x) = \sin ^3(2 \pi - x) \cos ^2(2 \pi - x) = (-\sin x)^3 (\cos x)^2 = -\sin ^3 x \cos ^2 x = -f(x)$.
Since $f(2 \pi - x) = -f(x)$,the integral evaluates to $0$.

(C) Let $f(x) = \sin^7 x \cdot \cos^6 x$.
Check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \sin^7(-x) \cdot \cos^6(-x)$
Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$,we have:
$f(-x) = (-\sin x)^7 \cdot (\cos x)^6 = -\sin^7 x \cdot \cos^6 x = -f(x)$.
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,$\int_{-1}^1 \sin^7 x \cdot \cos^6 x \, dx = 0$.

(C) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx$.
Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we have $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$.
So,$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}}(\frac{\pi}{2}-x)}{\cos^{\frac{3}{2}}(\frac{\pi}{2}-x) + \sin^{\frac{3}{2}}(\frac{\pi}{2}-x)} \, dx$.
Since $\cos(\frac{\pi}{2}-x) = \sin x$ and $\sin(\frac{\pi}{2}-x) = \cos x$,we get $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \, dx$.
Adding the two expressions for $I$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, dx$.
$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$I = \frac{\pi}{12}$.

(A) Let $I = \int_a^b x f(x) d x$.
Using the property $\int_a^b g(x) d x = \int_a^b g(a+b-x) d x$,we get:
$I = \int_a^b (a+b-x) f(a+b-x) d x$.
Since $f(a+b-x) = f(x)$,this becomes:
$I = \int_a^b (a+b-x) f(x) d x$.
$I = (a+b) \int_a^b f(x) d x - \int_a^b x f(x) d x$.
$I = (a+b) \int_a^b f(x) d x - I$.
$2I = (a+b) \int_a^b f(x) d x$.
$I = \frac{a+b}{2} \int_a^b f(x) d x$.
Thus,the correct option is $A$.

(D) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$.
Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,we have $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$.
So,$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)}+\sqrt{\cos(\frac{\pi}{2}-x)}} d x$.
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$.
Adding the two expressions for $I$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 d x$.
$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$I = \frac{\pi}{12}$.

(B) Let $f(x) = \sin^5 x \cos^4 x$.
We check if the function is even or odd.
$f(-x) = \sin^5(-x) \cos^4(-x) = (-\sin x)^5 (\cos x)^4 = -\sin^5 x \cos^4 x = -f(x)$.
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,$\int_{-1}^1 \sin^5 x \cos^4 x \, dx = 0$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+\cos x+\tan^5 x) dx$.
We can split this into three integrals: $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan^5 x dx$.
Recall the property of definite integrals: $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function,and $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(x)$ is an even function.
For $f_1(x) = x^3$,$f_1(-x) = (-x)^3 = -x^3 = -f_1(x)$,so it is an odd function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 dx = 0$.
For $f_2(x) = \cos x$,$f_2(-x) = \cos(-x) = \cos x = f_2(x)$,so it is an even function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x dx = 2 \int_{0}^{\frac{\pi}{2}} \cos x dx = 2[\sin x]_{0}^{\frac{\pi}{2}} = 2(1-0) = 2$.
For $f_3(x) = \tan^5 x$,$f_3(-x) = \tan^5(-x) = -\tan^5 x = -f_3(x)$,so it is an odd function. Thus,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan^5 x dx = 0$.
Therefore,$I = 0 + 2 + 0 = 2$.

(C) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\cot x}}$.
We can write $\sqrt{\cot x} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}$,so $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$,we get:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$.
Adding the two expressions for $I$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$.
$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$I = \frac{\pi}{12}$.

(A) Let $I = \int_0^\pi \sin^2 x \cos^3 x \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:
$I = \int_0^\pi \sin^2(\pi - x) \cos^3(\pi - x) \, dx$.
Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we get:
$I = \int_0^\pi \sin^2 x (-\cos x)^3 \, dx = -\int_0^\pi \sin^2 x \cos^3 x \, dx$.
Thus,$I = -I$,which implies $2I = 0$,so $I = 0$.

(D) Let $I = \int_{e}^{\Pi} x f(x) dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we have:
$I = \int_{e}^{\Pi} (e + \Pi - x) f(e + \Pi - x) dx$.
Since $f(e + \Pi - x) = f(x)$,this becomes:
$I = \int_{e}^{\Pi} (e + \Pi - x) f(x) dx = (e + \Pi) \int_{e}^{\Pi} f(x) dx - \int_{e}^{\Pi} x f(x) dx$.
$I = (e + \Pi) \left( \frac{2}{e + \Pi} \right) - I$.
$2I = 2$.
$I = 1$.

(A) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log (\sec \theta - \tan \theta) \, d\theta$.
Consider the function $f(\theta) = \log (\sec \theta - \tan \theta)$.
We check if $f(\theta)$ is an odd function by evaluating $f(-\theta)$:
$f(-\theta) = \log (\sec(-\theta) - \tan(-\theta)) = \log (\sec \theta + \tan \theta)$.
Since $\sec \theta + \tan \theta = \frac{1}{\sec \theta - \tan \theta}$,we have:
$f(-\theta) = \log \left( \frac{1}{\sec \theta - \tan \theta} \right) = -\log (\sec \theta - \tan \theta) = -f(\theta)$.
Since $f(\theta)$ is an odd function and the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$ is symmetric about the origin,the integral of an odd function over this interval is $0$.
Therefore,$I = 0$.

(B) Let $I = \int_{-2}^0 (x^3+3x^2+3x+3+(x+1) \cos(x+1)) \, dx$.
We can rewrite the integrand as:
$I = \int_{-2}^0 ((x+1)^3 + 2 + (x+1) \cos(x+1)) \, dx$.
Substitute $t = x+1$,so $dt = dx$.
When $x = -2$,$t = -1$. When $x = 0$,$t = 1$.
Thus,$I = \int_{-1}^1 (t^3 + 2 + t \cos t) \, dt$.
We can split the integral:
$I = \int_{-1}^1 t^3 \, dt + \int_{-1}^1 2 \, dt + \int_{-1}^1 t \cos t \, dt$.
Since $f(t) = t^3$ and $g(t) = t \cos t$ are odd functions,their integrals over the symmetric interval $[-1, 1]$ are $0$.
Therefore,$I = 0 + [2t]_{-1}^1 + 0 = 2(1 - (-1)) = 2(2) = 4$.

(D) Let $f(x) = \sin^{103} x \cdot \cos^{101} x$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \sin^{103}(-x) \cdot \cos^{101}(-x)$
Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$,we have:
$f(-x) = (-\sin x)^{103} \cdot (\cos x)^{101} = -\sin^{103} x \cdot \cos^{101} x = -f(x)$.
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,$\int_{-\pi/4}^{\pi/4} \sin^{103} x \cdot \cos^{101} x \, dx = 0$.

(D) Let $ I = \int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x \quad (1) $
Using the property $ \int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x $,where $ a=2 $ and $ b=8 $,we have $ a+b-x = 2+8-x = 10-x $.
Substituting this into the integral,we get:
$ I = \int_{2}^{8} \frac{\sqrt{10-(10-x)}}{\sqrt{10-x}+\sqrt{10-(10-x)}} d x $
$ I = \int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x \quad (2) $
Adding equations $(1)$ and $(2)$:
$ 2I = \int_{2}^{8} \frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{x}+\sqrt{10-x}} d x $
$ 2I = \int_{2}^{8} 1 d x $
$ 2I = [x]_{2}^{8} = 8 - 2 = 6 $
$ I = \frac{6}{2} = 3 $

(D) Let $f(x) = \sin^5 x \cos^4 x$.
We check if the function is even or odd by evaluating $f(-x)$.
$f(-x) = \sin^5(-x) \cos^4(-x) = (-\sin x)^5 (\cos x)^4 = -\sin^5 x \cos^4 x = -f(x)$.
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,$\int_{-1}^1 \sin^5 x \cos^4 x \, dx = 0$.

(B) Let $I = \int_0^1 \log \left(\frac{1-x}{x}\right) d x$.
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get:
$I = \int_0^1 \log \left(\frac{1-(1-x)}{1-x}\right) d x$
$I = \int_0^1 \log \left(\frac{x}{1-x}\right) d x$
Adding the two expressions for $I$:
$2I = \int_0^1 \left[ \log \left(\frac{1-x}{x}\right) + \log \left(\frac{x}{1-x}\right) \right] d x$
$2I = \int_0^1 \log \left( \frac{1-x}{x} \times \frac{x}{1-x} \right) d x$
$2I = \int_0^1 \log(1) d x$
Since $\log(1) = 0$,we have $2I = 0$,which implies $I = 0$.

(D) Let $f(x) = (1-x^2) \sin x \cdot \cos^2 x$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = (1-(-x)^2) \sin(-x) \cdot \cos^2(-x)$
Since $(-x)^2 = x^2$,$\sin(-x) = -\sin x$,and $\cos(-x) = \cos x$:
$f(-x) = (1-x^2) (-\sin x) \cdot (\cos x)^2$
$f(-x) = -(1-x^2) \sin x \cdot \cos^2 x = -f(x)$.
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,$\int_{-\pi}^{\pi} (1-x^2) \sin x \cdot \cos^2 x \, dx = 0$.

(C) Let $I = \int_2^8 \frac{5^{\sqrt{10-x}}}{5^{\sqrt{x}} + 5^{\sqrt{10-x}}} dx$ $(i)$
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$:
$I = \int_2^8 \frac{5^{\sqrt{10-(2+8-x)}}}{5^{\sqrt{2+8-x}} + 5^{\sqrt{10-(2+8-x)}}} dx$
$I = \int_2^8 \frac{5^{\sqrt{10-10+x}}}{5^{\sqrt{10-x}} + 5^{\sqrt{10-10+x}}} dx$
$I = \int_2^8 \frac{5^{\sqrt{x}}}{5^{\sqrt{10-x}} + 5^{\sqrt{x}}} dx$ (ii)
Adding equations $(i)$ and (ii):
$2I = \int_2^8 \frac{5^{\sqrt{10-x}} + 5^{\sqrt{x}}}{5^{\sqrt{x}} + 5^{\sqrt{10-x}}} dx$
$2I = \int_2^8 1 dx$
$2I = [x]_2^8 = 8 - 2 = 6$
$I = \frac{6}{2} = 3$

(A) Let $I = \int_0^\pi \frac{x \tan x}{\sec x \cdot \operatorname{cosec} x} d x$ $(i)$
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we have:
$I = \int_0^\pi \frac{(\pi-x) \tan(\pi-x)}{\sec(\pi-x) \operatorname{cosec}(\pi-x)} d x$
Since $\tan(\pi-x) = -\tan x$,$\sec(\pi-x) = -\sec x$,and $\operatorname{cosec}(\pi-x) = \operatorname{cosec} x$,we get:
$I = \int_0^\pi \frac{(\pi-x)(-\tan x)}{(-\sec x)(\operatorname{cosec} x)} d x = \int_0^\pi \frac{(\pi-x) \tan x}{\sec x \operatorname{cosec} x} d x$ (ii)
Adding $(i)$ and (ii):
$2I = \int_0^\pi \frac{x \tan x + (\pi-x) \tan x}{\sec x \operatorname{cosec} x} d x = \int_0^\pi \frac{\pi \tan x}{\sec x \operatorname{cosec} x} d x$
Since $\frac{\tan x}{\sec x \operatorname{cosec} x} = \frac{\sin x / \cos x}{(1 / \cos x)(1 / \sin x)} = \sin^2 x$:
$2I = \pi \int_0^\pi \sin^2 x d x = \pi \int_0^\pi \frac{1 - \cos 2x}{2} d x$
$2I = \frac{\pi}{2} \left[ x - \frac{\sin 2x}{2} \right]_0^\pi = \frac{\pi}{2} [(\pi - 0) - (0 - 0)] = \frac{\pi^2}{2}$
$I = \frac{\pi^2}{4}$

(B) Let $I = \int_{0}^{4042} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4042-x}} \, dx \quad \dots (i)$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get:
$I = \int_{0}^{4042} \frac{\sqrt{4042-x}}{\sqrt{4042-x}+\sqrt{x}} \, dx \quad \dots (ii)$
Adding equations $(i)$ and $(ii)$:
$2I = \int_{0}^{4042} \frac{\sqrt{x}+\sqrt{4042-x}}{\sqrt{x}+\sqrt{4042-x}} \, dx$
$2I = \int_{0}^{4042} 1 \, dx$
$2I = [x]_{0}^{4042} = 4042$
$I = \frac{4042}{2} = 2021$