Properties of definite integration Questions in English

(B) First,observe that $f(x) + f(1-x) = \frac{4^x}{4^x+2} + \frac{4^{1-x}}{4^{1-x}+2} = \frac{4^x}{4^x+2} + \frac{4/4^x}{4/4^x+2} = \frac{4^x}{4^x+2} + \frac{4}{4+2 \cdot 4^x} = \frac{4^x}{4^x+2} + \frac{2}{2+4^x} = \frac{4^x+2}{4^x+2} = 1.$
Let $I = \int_{f(a)}^{f(1-a)} x \sin^4(x(1-x)) dx.$ Using the property $\int_A^B g(x) dx = \int_A^B g(A+B-x) dx,$ and noting that $A+B = f(a) + f(1-a) = 1,$ we have:
$M = \int_{f(a)}^{f(1-a)} (1-x) \sin^4((1-x)(1-(1-x))) dx = \int_{f(a)}^{f(1-a)} (1-x) \sin^4((1-x)x) dx.$
Thus,$M = \int_{f(a)}^{f(1-a)} \sin^4(x(1-x)) dx - \int_{f(a)}^{f(1-a)} x \sin^4(x(1-x)) dx.$
This implies $M = N - M,$ which simplifies to $2M = N.$
Given $\alpha M = \beta N,$ we have $\alpha M = \beta (2M),$ so $\alpha = 2\beta.$
Since $\alpha, \beta \in N,$ the smallest values are $\beta = 1$ and $\alpha = 2.$
The value of $\alpha^2 + \beta^2 = 2^2 + 1^2 = 4 + 1 = 5.$

(A) Let $I = \int_0^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have $I = \int_0^\pi \frac{(\pi-x)^2 \sin(\pi-x) \cos(\pi-x)}{\sin^4(\pi-x) + \cos^4(\pi-x)} dx = \int_0^\pi \frac{(\pi-x)^2 \sin x (-\cos x)}{\sin^4 x + \cos^4 x} dx = -\int_0^\pi \frac{(\pi^2 - 2\pi x + x^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$.
Adding the two expressions for $I$: $2I = \int_0^\pi \frac{(x^2 - (\pi^2 - 2\pi x + x^2)) \sin x \cos x}{\sin^4 x + \cos^4 x} dx = \int_0^\pi \frac{(2\pi x - \pi^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$.
$2I = 2\pi \int_0^\pi \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx - \pi^2 \int_0^\pi \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx$.
Using the property $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$ if $f(2a-x) = f(x)$,we find the second integral is $0$ because $\sin(\pi-x)\cos(\pi-x) = -\sin x \cos x$.
Thus,$2I = 2\pi \cdot \frac{\pi}{2} \int_0^\pi \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx = 0$.
Wait,re-evaluating: $I = \int_0^{\pi/2} \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx + \int_{\pi/2}^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx$.
Let $x = \pi - t$ in the second integral: $\int_0^{\pi/2} \frac{(\pi-t)^2 \sin t (-\cos t)}{\sin^4 t + \cos^4 t} dt = -\int_0^{\pi/2} \frac{(\pi^2 - 2\pi t + t^2) \sin t \cos t}{\sin^4 t + \cos^4 t} dt$.
$I = \int_0^{\pi/2} \frac{(t^2 - \pi^2 + 2\pi t - t^2) \sin t \cos t}{\sin^4 t + \cos^4 t} dt = \int_0^{\pi/2} \frac{(2\pi t - \pi^2) \sin t \cos t}{\sin^4 t + \cos^4 t} dt$.
$= 2\pi \int_0^{\pi/2} \frac{t \sin t \cos t}{\sin^4 t + \cos^4 t} dt - \pi^2 \int_0^{\pi/2} \frac{\sin t \cos t}{\sin^4 t + \cos^4 t} dt$.
Using $\int_0^{\pi/2} f(\sin x, \cos x) dx = \int_0^{\pi/2} f(\cos x, \sin x) dx$,the first integral is $\frac{\pi}{4} \int_0^{\pi/2} \frac{\sin t \cos t}{\sin^4 t + \cos^4 t} dt$.
$I = (2\pi \cdot \frac{\pi}{4} - \pi^2) \int_0^{\pi/2} \frac{\sin t \cos t}{\sin^4 t + \cos^4 t} dt = -\frac{\pi^2}{2} \int_0^{\pi/2} \frac{\sin t \cos t}{\sin^4 t + \cos^4 t} dt$.
Let $u = \sin^2 t$,$du = 2 \sin t \cos t dt$. Integral becomes $-\frac{\pi^2}{4} \int_0^1 \frac{du}{u^2 + (1-u)^2} = -\frac{\pi^2}{4} \int_0^1 \frac{du}{2u^2 - 2u + 1} = -\frac{\pi^2}{8} \int_0^1 \frac{du}{(u-1/2)^2 + 1/4} = -\frac{\pi^2}{8} [2 \tan^{-1}(2u-1)]_0^1 = -\frac{\pi^2}{4} (\frac{\pi}{4} - (-\frac{\pi}{4})) = -\frac{\pi^3}{8}$.
Taking absolute value: $|-\frac{\pi^3}{8}| = \frac{\pi^3}{8}$.
Result: $\frac{120}{\pi^3} \cdot \frac{\pi^3}{8} = 15$.

(C) Let $I = \int_0^{\frac{\pi}{4}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}$.
Substitute $2x = t$,so $dx = \frac{1}{2} dt$. When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=\frac{\pi}{2}$.
$I = \int_0^{\frac{\pi}{2}} \frac{(t/2) \cdot (1/2) dt}{\sin^4 t + \cos^4 t} = \frac{1}{4} \int_0^{\frac{\pi}{2}} \frac{t \, dt}{\sin^4 t + \cos^4 t}$.
Using the property $\int_0^a f(t) dt = \int_0^a f(a-t) dt$,we get:
$I = \frac{1}{4} \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - t) dt}{\cos^4 t + \sin^4 t} = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t} - I$.
Thus,$2I = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t}$.
Divide numerator and denominator by $\cos^4 t$:
$2I = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{\sec^4 t \, dt}{\tan^4 t + 1} = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{(1 + \tan^2 t) \sec^2 t \, dt}{\tan^4 t + 1}$.
Let $\tan t = y$,then $\sec^2 t \, dt = dy$:
$2I = \frac{\pi}{8} \int_0^{\infty} \frac{1 + y^2}{y^4 + 1} dy = \frac{\pi}{8} \int_0^{\infty} \frac{1 + 1/y^2}{y^2 + 1/y^2} dy$.
Let $y - 1/y = p$,then $(1 + 1/y^2) dy = dp$. Limits change from $-\infty$ to $\infty$:
$2I = \frac{\pi}{8} \int_{-\infty}^{\infty} \frac{dp}{p^2 + 2} = \frac{\pi}{8} \left[ \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{p}{\sqrt{2}} \right) \right]_{-\infty}^{\infty} = \frac{\pi}{8\sqrt{2}} (\frac{\pi}{2} - (-\frac{\pi}{2})) = \frac{\pi^2}{8\sqrt{2}}$.
Therefore,$I = \frac{\pi^2}{16\sqrt{2}} = \frac{\sqrt{2}\pi^2}{32}$.

(D) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{(1+e^{\sin x})(1+\sin ^4 x)} \, dx$ $(1)$
Using the property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$,we get:
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{(1+e^{-\sin x})(1+\sin ^4 x)} \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x \cdot e^{\sin x}}{(1+e^{\sin x})(1+\sin ^4 x)} \, dx$ $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} \, dx = 2 \int_{0}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} \, dx$
$I = \int_{0}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} \, dx$
Let $\sin x = t$,then $\cos x \, dx = dt$. Limits change from $0$ to $1$:
$I = \int_{0}^{1} \frac{8 \sqrt{2}}{1+t^4} \, dt = 4 \sqrt{2} \int_{0}^{1} \frac{2}{1+t^4} \, dt = 4 \sqrt{2} \int_{0}^{1} \frac{(t^2+1) - (t^2-1)}{1+t^4} \, dt$
$I = 4 \sqrt{2} \left[ \int_{0}^{1} \frac{1+1/t^2}{t^2+1/t^2} \, dt - \int_{0}^{1} \frac{1-1/t^2}{t^2+1/t^2} \, dt \right]$
$I = 4 \sqrt{2} \left[ \int_{0}^{1} \frac{d(t-1/t)}{(t-1/t)^2+2} - \int_{0}^{1} \frac{d(t+1/t)}{(t+1/t)^2-2} \right]$
Evaluating the integrals:
$I = 4 \sqrt{2} \left[ \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t-1/t}{\sqrt{2}} \right) \Big|_0^1 - \frac{1}{2\sqrt{2}} \ln \left| \frac{t+1/t-\sqrt{2}}{t+1/t+\sqrt{2}} \right| \Big|_0^1 \right]$
$I = 4 \sqrt{2} \left[ \frac{1}{\sqrt{2}} (\frac{\pi}{2} - (-\frac{\pi}{2})) - \frac{1}{2\sqrt{2}} (\ln \frac{2-\sqrt{2}}{2+\sqrt{2}} - \ln 1) \right]$
$I = 2\pi - 2 \ln \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \cdot \frac{2-\sqrt{2}}{2-\sqrt{2}} \right) = 2\pi - 2 \ln \left( \frac{6-4\sqrt{2}}{2} \right) = 2\pi - 2 \ln (3-2\sqrt{2}) = 2\pi + 2 \ln (3+2\sqrt{2})$
Thus,$\alpha = 2, \beta = 2$. Therefore,$\alpha^2 + \beta^2 = 2^2 + 2^2 = 8$.

(A) Let $I = \int_0^1 (2x^3 - 3x^2 - x + 1)^{\frac{1}{3}} dx$.
Consider the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.
Let $f(x) = (2x^3 - 3x^2 - x + 1)^{\frac{1}{3}}$.
Then $f(1-x) = (2(1-x)^3 - 3(1-x)^2 - (1-x) + 1)^{\frac{1}{3}}$.
Expanding the terms: $f(1-x) = (2(1 - 3x + 3x^2 - x^3) - 3(1 - 2x + x^2) - 1 + x + 1)^{\frac{1}{3}}$.
$f(1-x) = (2 - 6x + 6x^2 - 2x^3 - 3 + 6x - 3x^2 - 1 + x + 1)^{\frac{1}{3}}$.
$f(1-x) = (-2x^3 + 3x^2 + x - 1)^{\frac{1}{3}} = -(2x^3 - 3x^2 - x + 1)^{\frac{1}{3}} = -f(x)$.
Since $f(1-x) = -f(x)$,the integral $I = \int_0^1 f(x) dx$ satisfies $I = -I$,which implies $2I = 0$,so $I = 0$.

(A) Given $f(x) = \begin{cases} -2, & -2 \leq x \leq 0 \\ x-2, & 0 < x \leq 2 \end{cases}$.
We need to find $h(x) = f(|x|) + |f(x)|$.
For $x \in [0, 2]$,$f(|x|) = f(x) = x-2$ and $|f(x)| = |x-2| = 2-x$. Thus,$h(x) = (x-2) + (2-x) = 0$.
For $x \in [-2, 0)$,$f(|x|) = f(-x) = -2$ (since $-x \in (0, 2]$ is not true,actually for $x \in [-2, 0)$,$|x| \in (0, 2]$,so $f(|x|) = |x|-2 = -x-2$).
Wait,let's re-evaluate: $f(|x|) = \begin{cases} -2, & |x| \leq 0 \text{ (only } x=0) \\ |x|-2, & 0 < |x| \leq 2 \end{cases} = |x|-2$.
So $f(|x|) = |x|-2$.
$|f(x)| = \begin{cases} |-2| = 2, & -2 \leq x \leq 0 \\ |x-2| = 2-x, & 0 < x \leq 2 \end{cases}$.
Thus,$h(x) = f(|x|) + |f(x)| = \begin{cases} (-x-2) + 2 = -x, & -2 \leq x < 0 \\ (x-2) + (2-x) = 0, & 0 \leq x \leq 2 \end{cases}$.
Now,$\int_{-2}^2 h(x) dx = \int_{-2}^0 (-x) dx + \int_0^2 0 dx = \left[ -\frac{x^2}{2} \right]_{-2}^0 = 0 - (-\frac{(-2)^2}{2}) = 0 - (-2) = 2$.

(B) Let $I = \int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{1+\sin x \cos x} dx$. Multiply numerator and denominator by $2$ to get $I = \int_0^{\frac{\pi}{4}} \frac{2\sin^2 x}{2+2\sin x \cos x} dx = \int_0^{\frac{\pi}{4}} \frac{1-\cos 2x}{2+\sin 2x} dx$.
This splits into $I = \int_0^{\frac{\pi}{4}} \frac{1}{2+\sin 2x} dx - \int_0^{\frac{\pi}{4}} \frac{\cos 2x}{2+\sin 2x} dx = I_1 - I_2$.
For $I_1 = \int_0^{\frac{\pi}{4}} \frac{\sec^2 x}{2\tan^2 x + 2\tan x + 2} dx$. Let $t = \tan x$,$dt = \sec^2 x dx$. Limits change from $0$ to $1$.
$I_1 = \frac{1}{2} \int_0^1 \frac{dt}{t^2+t+1} = \frac{1}{2} \int_0^1 \frac{dt}{(t+1/2)^2 + 3/4} = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} [\tan^{-1}(\frac{2t+1}{\sqrt{3}})]_0^1 = \frac{1}{\sqrt{3}} (\frac{\pi}{3} - \frac{\pi}{6}) = \frac{\pi}{6\sqrt{3}}$.
For $I_2 = \int_0^{\frac{\pi}{4}} \frac{\cos 2x}{2+\sin 2x} dx$. Let $u = 2+\sin 2x$,$du = 2\cos 2x dx$.
$I_2 = \frac{1}{2} \int_2^3 \frac{du}{u} = \frac{1}{2} \ln(\frac{3}{2}) = \frac{1}{2} \ln(3) - \frac{1}{2} \ln(2)$.
Comparing $I = I_1 - I_2 = \frac{\pi}{6\sqrt{3}} - \frac{1}{2} \ln(3) + \frac{1}{2} \ln(2) = \frac{\pi}{6\sqrt{3}} + \frac{1}{2} \ln(2/3)$.
Given form is $\frac{1}{a} \ln(a/3) + \frac{\pi}{b\sqrt{3}}$. With $a=2, b=6$,we get $\frac{1}{2} \ln(2/3) + \frac{\pi}{6\sqrt{3}}$.
Thus $a=2, b=6$,so $a+b = 8$.

(B) Let $I = \int_{-1}^{1} \frac{\cos \alpha x}{1+3^x} dx$ $(1)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get:
$I = \int_{-1}^{1} \frac{\cos \alpha (-x)}{1+3^{-x}} dx = \int_{-1}^{1} \frac{\cos \alpha x}{1+\frac{1}{3^x}} dx = \int_{-1}^{1} \frac{3^x \cos \alpha x}{3^x+1} dx$ $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{-1}^{1} \frac{\cos \alpha x}{1+3^x} dx + \int_{-1}^{1} \frac{3^x \cos \alpha x}{1+3^x} dx$
$2I = \int_{-1}^{1} \frac{(1+3^x) \cos \alpha x}{1+3^x} dx = \int_{-1}^{1} \cos \alpha x dx$
Since $\cos \alpha x$ is an even function:
$2I = 2 \int_{0}^{1} \cos \alpha x dx = 2 \left[ \frac{\sin \alpha x}{\alpha} \right]_{0}^{1} = \frac{2 \sin \alpha}{\alpha}$
$I = \frac{\sin \alpha}{\alpha}$
Given $I = \frac{2}{\pi}$,so $\frac{\sin \alpha}{\alpha} = \frac{2}{\pi}$.
By inspection,$\alpha = \frac{\pi}{2}$ satisfies the equation since $\frac{\sin(\pi/2)}{\pi/2} = \frac{1}{\pi/2} = \frac{2}{\pi}$.

(A) Let $I = \int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$.
Split the integral into two parts:
$I = \int_{-\pi}^\pi \frac{2 y}{1+\cos ^2 y} d y + \int_{-\pi}^\pi \frac{2 y \sin y}{1+\cos ^2 y} d y$.
The first part $\int_{-\pi}^\pi \frac{2 y}{1+\cos ^2 y} d y = 0$ because the integrand is an odd function.
For the second part,since $y \sin y$ is an even function,we have:
$I = 2 \int_0^\pi \frac{2 y \sin y}{1+\cos ^2 y} d y = 4 \int_0^\pi \frac{y \sin y}{1+\cos ^2 y} d y$.
Using the property $\int_0^a f(y) dy = \int_0^a f(a-y) dy$:
$I = 4 \int_0^\pi \frac{(\pi-y) \sin y}{1+\cos ^2 y} d y = 4\pi \int_0^\pi \frac{\sin y}{1+\cos ^2 y} d y - 4 \int_0^\pi \frac{y \sin y}{1+\cos ^2 y} d y$.
$I = 4\pi \int_0^\pi \frac{\sin y}{1+\cos ^2 y} d y - I
\implies 2I = 4\pi \int_0^\pi \frac{\sin y}{1+\cos ^2 y} d y
\implies I = 2\pi \int_0^\pi \frac{\sin y}{1+\cos ^2 y} d y$.
Let $t = \cos y$,then $dt = -\sin y dy$. When $y=0, t=1$; when $y=\pi, t=-1$.
$I = 2\pi \int_1^{-1} \frac{-dt}{1+t^2} = 2\pi \int_{-1}^1 \frac{dt}{1+t^2} = 2\pi [\tan^{-1} t]_{-1}^1$.
$I = 2\pi [\tan^{-1}(1) - \tan^{-1}(-1)] = 2\pi [\frac{\pi}{4} - (-\frac{\pi}{4})] = 2\pi [\frac{\pi}{2}] = \pi^2$.

(C) Given $f(t) = \int_0^\pi \frac{2x \, dx}{1 - \cos^2 t \sin^2 x}$.
Using the property $\int_0^a g(x) \, dx = \int_0^a g(a-x) \, dx$,we have:
$f(t) = \int_0^\pi \frac{2(\pi - x) \, dx}{1 - \cos^2 t \sin^2 x}$.
Adding the two expressions for $f(t)$:
$2f(t) = \int_0^\pi \frac{2x + 2\pi - 2x}{1 - \cos^2 t \sin^2 x} \, dx = \int_0^\pi \frac{2\pi \, dx}{1 - \cos^2 t \sin^2 x}$.
Thus,$f(t) = \pi \int_0^\pi \frac{dx}{1 - \cos^2 t \sin^2 x}$.
Since the integrand is symmetric about $x = \frac{\pi}{2}$,$f(t) = 2\pi \int_0^{\frac{\pi}{2}} \frac{dx}{1 - \cos^2 t \sin^2 x}$.
Dividing numerator and denominator by $\cos^2 x$:
$f(t) = 2\pi \int_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{\sec^2 x - \cos^2 t \tan^2 x} = 2\pi \int_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{1 + \tan^2 x - \cos^2 t \tan^2 x} = 2\pi \int_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{1 + \sin^2 t \tan^2 x}$.
Let $\tan x = z$,then $\sec^2 x \, dx = dz$. As $x \to 0, z \to 0$ and as $x \to \frac{\pi}{2}, z \to \infty$:
$f(t) = 2\pi \int_0^{\infty} \frac{dz}{1 + (\sin t \cdot z)^2} = 2\pi \left[ \frac{1}{\sin t} \tan^{-1}(\sin t \cdot z) \right]_0^{\infty} = 2\pi \cdot \frac{1}{\sin t} \cdot \frac{\pi}{2} = \frac{\pi^2}{\sin t}$.
Now,calculate the integral $\int_0^{\frac{\pi}{2}} \frac{\pi^2}{f(t)} \, dt = \int_0^{\frac{\pi}{2}} \frac{\pi^2}{\pi^2 / \sin t} \, dt = \int_0^{\frac{\pi}{2}} \sin t \, dt$.
$= [-\cos t]_0^{\frac{\pi}{2}} = -(0 - 1) = 1$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} dx$ ... $(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = -\frac{\pi}{2} + \frac{\pi}{2} = 0$,so $f(x) \to f(-x)$.
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(-x)^2 \cos(-x)}{1+e^{-x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x \cdot e^x}{e^x+1} dx$ ... (ii)
Adding $(i)$ and (ii):
$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x (1+e^x)}{1+e^x} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x dx$
Since $x^2 \cos x$ is an even function,$2I = 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x dx$.
$I = \int_{0}^{\frac{\pi}{2}} x^2 \cos x dx$.
Using integration by parts $\int u dv = uv - \int v du$ with $u=x^2, dv=\cos x dx$:
$I = [x^2 \sin x]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2x \sin x dx$
$I = (\frac{\pi^2}{4} \cdot 1 - 0) - 2 \int_{0}^{\frac{\pi}{2}} x \sin x dx$
$I = \frac{\pi^2}{4} - 2 [x(-\cos x) - \int 1(-\cos x) dx]_{0}^{\frac{\pi}{2}}$
$I = \frac{\pi^2}{4} - 2 [-x \cos x + \sin x]_{0}^{\frac{\pi}{2}}$
$I = \frac{\pi^2}{4} - 2 [(-0 + 1) - (0 + 0)] = \frac{\pi^2}{4} - 2$.

(A) Given $I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{(1+\pi^x) \sin x} dx \quad \ldots (i)$
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we get
$I_n = \int_{-\pi}^{\pi} \frac{\pi^x \sin(nx)}{(1+\pi^x) \sin x} dx \quad \ldots (ii)$
Adding $(i)$ and $(ii)$:
$2I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{\sin x} dx = 2 \int_0^{\pi} \frac{\sin(nx)}{\sin x} dx$ (since the integrand is an even function).
Thus,$I_n = \int_0^{\pi} \frac{\sin(nx)}{\sin x} dx$.
Now,$I_{n+2} - I_n = \int_0^{\pi} \frac{\sin((n+2)x) - \sin(nx)}{\sin x} dx = \int_0^{\pi} \frac{2 \cos((n+1)x) \sin x}{\sin x} dx = 2 \int_0^{\pi} \cos((n+1)x) dx = 0$.
So,$I_{n+2} = I_n$.
For $n=1$,$I_1 = \int_0^{\pi} \frac{\sin x}{\sin x} dx = \pi$.
For $n=2$,$I_2 = \int_0^{\pi} \frac{\sin(2x)}{\sin x} dx = \int_0^{\pi} 2 \cos x dx = 0$.
Since $I_{n+2} = I_n$,all odd terms $I_{2m+1} = \pi$ and all even terms $I_{2m} = 0$.
Therefore,$\sum_{m=1}^{10} I_{2m+1} = 10\pi$ and $\sum_{m=1}^{10} I_{2m} = 0$.
Thus,options $(A)$,$(B)$,and $(C)$ are correct.

(A) The function $f(x)$ is defined as:
$f(x) = x - [x]$ if $[x]$ is odd,and $f(x) = 1 + [x] - x$ if $[x]$ is even.
For $x \in [0, 1)$,$[x] = 0$ (even),so $f(x) = 1 + 0 - x = 1 - x$.
For $x \in [1, 2)$,$[x] = 1$ (odd),so $f(x) = x - 1$.
This function is periodic with period $T = 2$.
Let $I = \int_{-10}^{10} f(x) \cos(\pi x) \, dx$.
Since $f(x)$ is an even function $(f(-x) = f(x))$ and $\cos(\pi x)$ is an even function,their product is even.
$I = 2 \int_{0}^{10} f(x) \cos(\pi x) \, dx = 2 \times 5 \int_{0}^{2} f(x) \cos(\pi x) \, dx = 10 \left[ \int_{0}^{1} (1-x) \cos(\pi x) \, dx + \int_{1}^{2} (x-1) \cos(\pi x) \, dx \right]$.
Let $I_1 = \int_{0}^{1} (1-x) \cos(\pi x) \, dx$. Using integration by parts:
$I_1 = \left[ (1-x) \frac{\sin(\pi x)}{\pi} \right]_0^1 - \int_{0}^{1} (-1) \frac{\sin(\pi x)}{\pi} \, dx = 0 + \left[ -\frac{\cos(\pi x)}{\pi^2} \right]_0^1 = -\frac{1}{\pi^2} (-1 - 1) = \frac{2}{\pi^2}$.
Let $I_2 = \int_{1}^{2} (x-1) \cos(\pi x) \, dx$. Let $t = x-1$,then $dt = dx$:
$I_2 = \int_{0}^{1} t \cos(\pi(t+1)) \, dt = \int_{0}^{1} t \cos(\pi t + \pi) \, dt = -\int_{0}^{1} t \cos(\pi t) \, dt$.
Using integration by parts for $\int t \cos(\pi t) \, dt = \frac{t \sin(\pi t)}{\pi} + \frac{\cos(\pi t)}{\pi^2}$:
$I_2 = -\left[ \frac{t \sin(\pi t)}{\pi} + \frac{\cos(\pi t)}{\pi^2} \right]_0^1 = -\left( 0 + \frac{-1}{\pi^2} - (0 + \frac{1}{\pi^2}) \right) = \frac{2}{\pi^2}$.
Thus,$I = 10 \left( \frac{2}{\pi^2} + \frac{2}{\pi^2} \right) = \frac{40}{\pi^2}$.
Finally,$\frac{\pi^2}{10} I = \frac{\pi^2}{10} \times \frac{40}{\pi^2} = 4$.

(A) Let $I = \int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x \sin x^2}{\sin x^2 + \sin (\ln 6 - x^2)} dx$.
Substitute $x^2 = t$,so $2x dx = dt$ or $x dx = \frac{1}{2} dt$.
When $x = \sqrt{\ln 2}$,$t = \ln 2$. When $x = \sqrt{\ln 3}$,$t = \ln 3$.
The integral becomes $I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin t}{\sin t + \sin (\ln 6 - t)} dt$.
Using the property $\int_{a}^{b} f(t) dt = \int_{a}^{b} f(a+b-t) dt$,we have $I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin (\ln 2 + \ln 3 - t)}{\sin (\ln 2 + \ln 3 - t) + \sin (\ln 6 - (\ln 2 + \ln 3 - t))} dt$.
Since $\ln 2 + \ln 3 = \ln 6$,this simplifies to $I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin (\ln 6 - t)}{\sin (\ln 6 - t) + \sin t} dt$.
Adding the two expressions for $I$:
$2I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin t + \sin (\ln 6 - t)}{\sin t + \sin (\ln 6 - t)} dt = \frac{1}{2} \int_{\ln 2}^{\ln 3} 1 dt$.
$2I = \frac{1}{2} [t]_{\ln 2}^{\ln 3} = \frac{1}{2} (\ln 3 - \ln 2) = \frac{1}{2} \ln \frac{3}{2}$.
Therefore,$I = \frac{1}{4} \ln \frac{3}{2}$.

(C) Given $R_1 = \int_{-1}^2 x f(x) dx$.
Using the property $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$,we have:
$R_1 = \int_{-1}^2 ((-1) + 2 - x) f((-1) + 2 - x) dx$
$R_1 = \int_{-1}^2 (1 - x) f(1 - x) dx$.
Since $f(x) = f(1 - x)$,we substitute this into the integral:
$R_1 = \int_{-1}^2 (1 - x) f(x) dx = \int_{-1}^2 f(x) dx - \int_{-1}^2 x f(x) dx$.
$R_1 = \int_{-1}^2 f(x) dx - R_1$.
$2 R_1 = \int_{-1}^2 f(x) dx$.
Since $R_2$ is the area bounded by $y = f(x)$ from $x = -1$ to $x = 2$,$R_2 = \int_{-1}^2 f(x) dx$.
Therefore,$2 R_1 = R_2$.

(B) Let $I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have:
$I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{-\sin x})(2 - \cos 2x)}$.
Adding the two expressions for $I$:
$2I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{1}{2 - \cos 2x} \left( \frac{1}{1 + e^{\sin x}} + \frac{e^{\sin x}}{1 + e^{\sin x}} \right) dx = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{2 - \cos 2x}$.
Since the integrand is even,$I = \frac{2}{\pi} \int_0^{\pi / 4} \frac{dx}{2 - \cos 2x} = \frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec^2 x dx}{2(1 + \tan^2 x) - (1 - \tan^2 x)} = \frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec^2 x dx}{1 + 3 \tan^2 x}$.
Let $u = \sqrt{3} \tan x$,then $du = \sqrt{3} \sec^2 x dx$.
$I = \frac{2}{\pi \sqrt{3}} \int_0^{\sqrt{3}} \frac{du}{1 + u^2} = \frac{2}{\pi \sqrt{3}} [\tan^{-1} u]_0^{\sqrt{3}} = \frac{2}{\pi \sqrt{3}} \cdot \frac{\pi}{3} = \frac{2}{3\sqrt{3}}$.
Thus,$27 I^2 = 27 \cdot \frac{4}{27} = 4$.

(D) Let $I = \int_0^{\pi / 2} \frac{3 \sqrt{\cos \theta}}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\pi / 2} \frac{3 \sqrt{\sin \theta}}{(\sqrt{\sin \theta}+\sqrt{\cos \theta})^5} d \theta$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi / 2} \frac{3(\sqrt{\cos \theta} + \sqrt{\sin \theta})}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta = \int_0^{\pi / 2} \frac{3}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^4} d \theta$.
Divide numerator and denominator by $\cos^2 \theta$:
$2I = 3 \int_0^{\pi / 2} \frac{\sec^2 \theta}{(1+\sqrt{\tan \theta})^4} d \theta$.
Let $1+\sqrt{\tan \theta} = t$,then $\frac{1}{2\sqrt{\tan \theta}} \cdot \sec^2 \theta d \theta = dt$,so $\sec^2 \theta d \theta = 2(t-1) dt$.
Substituting these:
$2I = 3 \int_1^{\infty} \frac{2(t-1)}{t^4} dt = 6 \int_1^{\infty} (t^{-3} - t^{-4}) dt$.
$2I = 6 \left[ \frac{t^{-2}}{-2} - \frac{t^{-3}}{-3} \right]_1^{\infty} = 6 \left[ 0 - (-\frac{1}{2} + \frac{1}{3}) \right] = 6 \left( \frac{1}{6} \right) = 1$.
Thus,$I = 0.50$.

(C) For $S_1$,we have $S_1 = \int_{\pi/8}^{3\pi/8} \sin^2 x dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we get $S_1 = \int_{\pi/8}^{3\pi/8} \sin^2(\frac{\pi}{8} + \frac{3\pi}{8} - x) dx = \int_{\pi/8}^{3\pi/8} \cos^2 x dx$.
Adding these,$2S_1 = \int_{\pi/8}^{3\pi/8} (\sin^2 x + \cos^2 x) dx = \int_{\pi/8}^{3\pi/8} 1 dx = \frac{3\pi}{8} - \frac{\pi}{8} = \frac{\pi}{4}$.
Thus,$S_1 = \frac{\pi}{8}$,so $\frac{16S_1}{\pi} = \frac{16}{\pi} \cdot \frac{\pi}{8} = 2$.
For $S_2$,$S_2 = \int_{\pi/8}^{3\pi/8} \sin^2 x |4x - \pi| dx$. Using the same property,$S_2 = \int_{\pi/8}^{3\pi/8} \sin^2(\frac{\pi}{2}-x) |4(\frac{\pi}{2}-x) - \pi| dx = \int_{\pi/8}^{3\pi/8} \cos^2 x |\pi - 4x| dx = \int_{\pi/8}^{3\pi/8} \cos^2 x |4x - \pi| dx$.
Adding these,$2S_2 = \int_{\pi/8}^{3\pi/8} |4x - \pi| (\sin^2 x + \cos^2 x) dx = \int_{\pi/8}^{3\pi/8} |4x - \pi| dx$.
The integral represents the area of two triangles with base $\frac{\pi}{8}$ and height $\frac{\pi}{2}$,so $2S_2 = 2 \cdot (\frac{1}{2} \cdot \frac{\pi}{8} \cdot \frac{\pi}{2}) = \frac{\pi^2}{16}$.
Thus,$S_2 = \frac{\pi^2}{32}$,so $\frac{48S_2}{\pi^2} = \frac{48}{\pi^2} \cdot \frac{\pi^2}{32} = \frac{48}{32} = 1.5$.

(B) Let $I = \int_{-\pi / 2}^{\pi / 2} \left(x^2 + \ln \frac{\pi+x}{\pi-x}\right) \cos x \, dx$.
We can split this into two integrals: $I = \int_{-\pi / 2}^{\pi / 2} x^2 \cos x \, dx + \int_{-\pi / 2}^{\pi / 2} \ln \left(\frac{\pi+x}{\pi-x}\right) \cos x \, dx$.
Let $f(x) = \ln \left(\frac{\pi+x}{\pi-x}\right) \cos x$. Since $\ln \left(\frac{\pi+x}{\pi-x}\right)$ is an odd function and $\cos x$ is an even function,their product is an odd function. Thus,$\int_{-\pi / 2}^{\pi / 2} f(x) \, dx = 0$.
Now,$I = \int_{-\pi / 2}^{\pi / 2} x^2 \cos x \, dx$. Since $x^2 \cos x$ is an even function,$I = 2 \int_0^{\pi / 2} x^2 \cos x \, dx$.
Using integration by parts: $\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx = x^2 \sin x - 2 \left( -x \cos x + \int \cos x \, dx \right) = x^2 \sin x + 2x \cos x - 2 \sin x$.
Evaluating from $0$ to $\pi / 2$: $2 \left[ (x^2 \sin x + 2x \cos x - 2 \sin x) \right]_0^{\pi / 2} = 2 \left[ ((\pi/2)^2 \cdot 1 + 0 - 2(1)) - (0 + 0 - 0) \right] = 2 \left( \frac{\pi^2}{4} - 2 \right) = \frac{\pi^2}{2} - 4$.

(A) Let $f(t) = e^t(\sin^6 at + \cos^4 at)$. We observe that for $a=2$ and $a=4$,the function $f(t)$ satisfies $f(t+\pi) = e^{t+\pi}(\sin^6 a(t+\pi) + \cos^4 a(t+\pi)) = e^{\pi} e^t(\sin^6 at + \cos^4 at) = e^{\pi} f(t)$,since $\sin(a(t+\pi)) = \sin(at + a\pi) = \sin(at)$ and $\cos(a(t+\pi)) = \cos(at)$ for even integers $a$.
Let $I_k = \int_{(k-1)\pi}^{k\pi} f(t) dt$. Then $I_{k+1} = \int_{k\pi}^{(k+1)\pi} f(t) dt$. Substituting $t = u + \pi$,we get $I_{k+1} = \int_{(k-1)\pi}^{k\pi} f(u+\pi) du = e^{\pi} \int_{(k-1)\pi}^{k\pi} f(u) du = e^{\pi} I_k$.
Thus,$I_1 = A$,$I_2 = e^{\pi} A$,$I_3 = e^{2\pi} A$,and $I_4 = e^{3\pi} A$,where $A = \int_0^{\pi} f(t) dt$.
The numerator is $\int_0^{4\pi} f(t) dt = I_1 + I_2 + I_3 + I_4 = A(1 + e^{\pi} + e^{2\pi} + e^{3\pi}) = A \frac{e^{4\pi}-1}{e^{\pi}-1}$.
Therefore,$L = \frac{A \frac{e^{4\pi}-1}{e^{\pi}-1}}{A} = \frac{e^{4\pi}-1}{e^{\pi}-1}$.
This holds for both $a=2$ and $a=4$. Thus,options $(A)$ and $(C)$ are correct.

(A) Given $f(x) = 7 \tan^6 x (\tan^2 x + 1) - 3 \tan^2 x (\tan^2 x + 1) = (7 \tan^6 x - 3 \tan^2 x) \sec^2 x$.
First,evaluate $\int_0^{\pi/4} f(x) dx$:
$\int_0^{\pi/4} (7 \tan^6 x - 3 \tan^2 x) \sec^2 x dx$. Let $u = \tan x$,then $du = \sec^2 x dx$. When $x=0, u=0$; when $x=\pi/4, u=1$.
$= \int_0^1 (7u^6 - 3u^2) du = [u^7 - u^3]_0^1 = (1 - 1) - (0 - 0) = 0$. Thus,$(B)$ is correct.
Next,evaluate $\int_0^{\pi/4} x f(x) dx$ using integration by parts:
Let $I = \int_0^{\pi/4} x f(x) dx$. Let $g(x) = \int f(x) dx = \tan^7 x - \tan^3 x$.
Using integration by parts: $\int_0^{\pi/4} x f(x) dx = [x g(x)]_0^{\pi/4} - \int_0^{\pi/4} g(x) dx$.
$[x(\tan^7 x - \tan^3 x)]_0^{\pi/4} = \frac{\pi}{4}(1 - 1) - 0 = 0$.
So,$I = - \int_0^{\pi/4} (\tan^7 x - \tan^3 x) dx = \int_0^{\pi/4} (\tan^3 x - \tan^7 x) dx$.
$= \int_0^{\pi/4} \tan^3 x (1 - \tan^4 x) dx = \int_0^{\pi/4} \tan^3 x (1 - \tan^2 x)(1 + \tan^2 x) dx$.
$= \int_0^{\pi/4} (\tan^3 x - \tan^5 x) \sec^2 x dx$. Let $u = \tan x$,$du = \sec^2 x dx$.
$= \int_0^1 (u^3 - u^5) du = [\frac{u^4}{4} - \frac{u^6}{6}]_0^1 = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$. Thus,$(A)$ is correct.

(B) $(1)$ Let $I = 2 \int_0^{\frac{\pi}{2}} \sin^2 x \sqrt{\frac{\pi x}{2} - x^2} dx - \int_0^{\frac{\pi}{2}} \sqrt{\frac{\pi x}{2} - x^2} dx$.
Let $I_1 = \int_0^{\frac{\pi}{2}} \sin^2 x \sqrt{(\frac{\pi}{4})^2 - (x - \frac{\pi}{4})^2} dx$.
Using the property $\int_0^a h(x) dx = \int_0^a h(a-x) dx$,we get:
$I_1 = \int_0^{\frac{\pi}{2}} \cos^2 x \sqrt{(\frac{\pi}{4})^2 - (\frac{\pi}{2} - x - \frac{\pi}{4})^2} dx = \int_0^{\frac{\pi}{2}} \cos^2 x \sqrt{(\frac{\pi}{4})^2 - (x - \frac{\pi}{4})^2} dx$.
Adding the two expressions for $I_1$:
$2I_1 = \int_0^{\frac{\pi}{2}} (\sin^2 x + \cos^2 x) \sqrt{(\frac{\pi}{4})^2 - (x - \frac{\pi}{4})^2} dx = \int_0^{\frac{\pi}{2}} g(x) dx$.
Thus,$I = 2I_1 - \int_0^{\frac{\pi}{2}} g(x) dx = \int_0^{\frac{\pi}{2}} g(x) dx - \int_0^{\frac{\pi}{2}} g(x) dx = 0$.
$(2)$ From above,$I_1 = \frac{1}{2} \int_0^{\frac{\pi}{2}} g(x) dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sqrt{(\frac{\pi}{4})^2 - (x - \frac{\pi}{4})^2} dx$.
Using $\int \sqrt{a^2 - u^2} du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \sin^{-1}(\frac{u}{a}) + C$:
$I_1 = \frac{1}{2} [\frac{x - \frac{\pi}{4}}{2} \sqrt{\frac{\pi x}{2} - x^2} + \frac{\pi^2}{32} \sin^{-1}(\frac{x - \frac{\pi}{4}}{\pi/4})]_0^{\frac{\pi}{2}} = \frac{1}{2} [(\frac{\pi^2}{32} \cdot \frac{\pi}{2}) - (\frac{\pi^2}{32} \cdot -\frac{\pi}{2})] = \frac{1}{2} [\frac{\pi^3}{64} + \frac{\pi^3}{64}] = \frac{\pi^3}{64}$.
Therefore,$\frac{16}{\pi^3} I_1 = \frac{16}{\pi^3} \cdot \frac{\pi^3}{64} = \frac{1}{4} = 0.25$.

(C) Let $I = \int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{((\ln x)^2+1)^{-1}}}{e^{((\ln x)^2+1)^{-1}} + e^{((6-\ln x)^2+1)^{-1}}} \right) dx$.
Substitute $\ln x = t$,then $\frac{1}{x} dx = dt$.
When $x = e^2$,$t = 2$. When $x = e^4$,$t = 4$.
So,$I = \int_{2}^{4} \frac{e^{(t^2+1)^{-1}}}{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}} dt$.
Using the property $\int_{a}^{b} f(t) dt = \int_{a}^{b} f(a+b-t) dt$,we get:
$I = \int_{2}^{4} \frac{e^{((6-t)^2+1)^{-1}}}{e^{((6-t)^2+1)^{-1}} + e^{(t^2+1)^{-1}}} dt$.
Adding the two expressions for $I$:
$2I = \int_{2}^{4} \frac{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}}{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}} dt = \int_{2}^{4} 1 dt = [t]_{2}^{4} = 4 - 2 = 2$.
Therefore,$I = 1$.

(A) Let $I_2 = \int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I_2 = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2}-x) \sin(\frac{\pi}{2}-x) \cos(\frac{\pi}{2}-x)}{\sin ^4(\frac{\pi}{2}-x)+\cos ^4(\frac{\pi}{2}-x)} dx = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2}-x) \cos x \sin x}{\cos ^4 x+\sin ^4 x} dx$.
Adding the two expressions for $I_2$:
$2I_2 = \int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2} \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx = \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$.
Divide numerator and denominator by $\cos^4 x$:
$I_2 = \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\tan x \sec^2 x}{\tan^4 x + 1} dx$.
Let $t = \tan^2 x$,then $dt = 2 \tan x \sec^2 x dx$,so $\tan x \sec^2 x dx = \frac{dt}{2}$.
As $x \to 0, t \to 0$ and as $x \to \frac{\pi}{2}, t \to \infty$.
$I_2 = \frac{\pi}{4} \int_0^{\infty} \frac{dt/2}{t^2+1} = \frac{\pi}{8} [\tan^{-1} t]_0^{\infty} = \frac{\pi}{8} \cdot \frac{\pi}{2} = \frac{\pi^2}{16}$.

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1+e^x} dx$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1+e^{-x}} dx$.
Adding the two expressions for $I$: $2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 96 x^2 \cos^2 x \left( \frac{1}{1+e^x} + \frac{1}{1+e^{-x}} \right) dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 96 x^2 \cos^2 x dx$.
Since $x^2 \cos^2 x$ is an even function,$I = \int_0^{\frac{\pi}{2}} 96 x^2 \cos^2 x dx = 48 \int_0^{\frac{\pi}{2}} x^2 (1 + \cos 2x) dx$.
$I = 48 \left[ \frac{x^3}{3} \right]_0^{\frac{\pi}{2}} + 48 \int_0^{\frac{\pi}{2}} x^2 \cos 2x dx$.
$I = 48 \left( \frac{\pi^3}{24} \right) + 48 \left[ x^2 \frac{\sin 2x}{2} - \int 2x \frac{\sin 2x}{2} dx \right]_0^{\frac{\pi}{2}}$.
$I = 2\pi^3 + 48 \left[ 0 - \int_0^{\frac{\pi}{2}} x \sin 2x dx \right] = 2\pi^3 - 48 \left[ x \left( -\frac{\cos 2x}{2} \right) - \int -\frac{\cos 2x}{2} dx \right]_0^{\frac{\pi}{2}}$.
$I = 2\pi^3 - 48 \left[ -\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} \right]_0^{\frac{\pi}{2}} = 2\pi^3 - 48 \left[ -\frac{\pi}{2} \cdot \frac{(-1)}{2} - 0 \right] = 2\pi^3 - 12\pi = \pi(2\pi^2 - 12)$.
Comparing with $\pi(\alpha \pi^2 + \beta)$,we get $\alpha = 2, \beta = -12$.
Thus,$(\alpha + \beta)^2 = (2 - 12)^2 = (-10)^2 = 100$.

(A) Given the curve $x^2=2y$ and the line $y=2x+6$. Substituting $y$ from the line equation into the curve equation:
$x^2=2(2x+6) \Rightarrow x^2=4x+12$
$x^2-4x-12=0 \Rightarrow (x-6)(x+2)=0$
So,$x=6$ or $x=-2$.
For $x=6$,$y=2(6)+6=18$. For $x=-2$,$y=2(-2)+6=2$.
The points of intersection are $(6, 18)$ and $(-2, 2)$.
Since the point $(a, b)$ lies in the second quadrant,we have $a=-2$ and $b=2$.
We need to evaluate $I=\int_{-2}^2 \frac{9x^2}{1+5^x} dx$.
Using the property $\int_{-k}^k f(x) dx = \int_0^k (f(x)+f(-x)) dx$:
$I = \int_{-2}^2 \frac{9x^2}{1+5^x} dx$.
Note that $\frac{9x^2}{1+5^x} + \frac{9(-x)^2}{1+5^{-x}} = \frac{9x^2}{1+5^x} + \frac{9x^2 \cdot 5^x}{5^x+1} = \frac{9x^2(1+5^x)}{1+5^x} = 9x^2$.
Thus,$I = \int_0^2 9x^2 dx = [3x^3]_0^2 = 3(8) - 3(0) = 24$.

(A) Let $I = \int_0^\pi \frac{8 x \, dx}{4 \cos^2 x + \sin^2 x}$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^\pi \frac{8(\pi - x) \, dx}{4 \cos^2(\pi - x) + \sin^2(\pi - x)} = \int_0^\pi \frac{8(\pi - x) \, dx}{4 \cos^2 x + \sin^2 x}$.
Adding the two expressions for $I$:
$2I = \int_0^\pi \frac{8x + 8\pi - 8x}{4 \cos^2 x + \sin^2 x} \, dx = 8\pi \int_0^\pi \frac{dx}{4 \cos^2 x + \sin^2 x}$.
Since the integrand is symmetric about $\pi/2$,$2I = 8\pi \times 2 \int_0^{\pi/2} \frac{dx}{4 \cos^2 x + \sin^2 x}$.
Divide numerator and denominator by $\cos^2 x$:
$I = 8\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{4 + \tan^2 x}$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$. As $x \to 0, t \to 0$ and as $x \to \pi/2, t \to \infty$.
$I = 8\pi \int_0^\infty \frac{dt}{4 + t^2} = 8\pi \left[ \frac{1}{2} \tan^{-1}\left(\frac{t}{2}\right) \right]_0^\infty$.
$I = 4\pi \left( \frac{\pi}{2} - 0 \right) = 2\pi^2$.

(D) Let $I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} dx$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we replace $x$ with $-x$:
$I = \int_{-1}^1 \frac{(1+\sqrt{|-x|-(-x)}) e^{-x}+(\sqrt{|-x|-(-x)}) e^{-(-x)}}{e^{-x}+e^{-(-x)}} dx$
$I = \int_{-1}^1 \frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^x}{e^{-x}+e^x} dx$.
Adding the two expressions for $I$:
$2I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x + (\sqrt{|x|-x}) e^{-x} + (1+\sqrt{|x|+x}) e^{-x} + (\sqrt{|x|+x}) e^x}{e^x+e^{-x}} dx$
$2I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}+\sqrt{|x|+x})(e^x+e^{-x})}{e^x+e^{-x}} dx$
$2I = \int_{-1}^1 (1+\sqrt{|x|-x}+\sqrt{|x|+x}) dx$.
Since the integrand is an even function,$2I = 2 \int_0^1 (1+\sqrt{x-x}+\sqrt{x+x}) dx = 2 \int_0^1 (1+0+\sqrt{2x}) dx$.
$I = \int_0^1 (1+\sqrt{2}x^{1/2}) dx = [x + \sqrt{2} \cdot \frac{2}{3} x^{3/2}]_0^1$
$I = 1 + \frac{2\sqrt{2}}{3}$.

(C) Let $I = \int_0^\pi \frac{(x+3) \sin x}{1+3 \cos^2 x} dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^\pi \frac{(\pi-x+3) \sin(\pi-x)}{1+3 \cos^2(\pi-x)} dx = \int_0^\pi \frac{(\pi-x+3) \sin x}{1+3 \cos^2 x} dx$.
Adding the two expressions for $I$:
$2I = \int_0^\pi \frac{(x+3+\pi-x+3) \sin x}{1+3 \cos^2 x} dx = \int_0^\pi \frac{(\pi+6) \sin x}{1+3 \cos^2 x} dx$.
Since $\sin(\pi-x) = \sin x$ and $\cos^2(\pi-x) = \cos^2 x$,we use the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$:
$2I = 2(\pi+6) \int_0^{\pi/2} \frac{\sin x}{1+3 \cos^2 x} dx$.
$I = (\pi+6) \int_0^{\pi/2} \frac{\sin x}{1+3 \cos^2 x} dx$.
Let $\sqrt{3} \cos x = t$,then $-\sqrt{3} \sin x dx = dt$,so $\sin x dx = -\frac{dt}{\sqrt{3}}$.
When $x=0, t=\sqrt{3}$. When $x=\pi/2, t=0$.
$I = (\pi+6) \int_{\sqrt{3}}^0 \frac{-dt/\sqrt{3}}{1+t^2} = \frac{\pi+6}{\sqrt{3}} \int_0^{\sqrt{3}} \frac{dt}{1+t^2}$.
$I = \frac{\pi+6}{\sqrt{3}} [\tan^{-1} t]_0^{\sqrt{3}} = \frac{\pi+6}{\sqrt{3}} (\tan^{-1} \sqrt{3} - \tan^{-1} 0) = \frac{\pi+6}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi(\pi+6)}{3\sqrt{3}}$.

(D) Given $I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$. Let $2x = t$,then $2dx = dt$,so $dx = \frac{1}{2} dt$. When $x = -\frac{1}{2}$,$t = -1$. When $x = 1$,$t = 2$. Substituting these,$I_1 = \int_{-1}^2 t f(t(1-t)) \frac{1}{2} dt = \frac{1}{2} \int_{-1}^2 t f(t(1-t)) dt$. Thus,$2I_1 = \int_{-1}^2 t f(t(1-t)) dt$.
Using the property $\int_a^b g(t) dt = \int_a^b g(a+b-t) dt$,we have $2I_1 = \int_{-1}^2 (1-t) f((1-t)(1-(1-t))) dt = \int_{-1}^2 (1-t) f((1-t)t) dt$.
Expanding this,$2I_1 = \int_{-1}^2 f(t(1-t)) dt - \int_{-1}^2 t f(t(1-t)) dt$.
Substituting $I_2 = \int_{-1}^2 f(t(1-t)) dt$ and $2I_1 = \int_{-1}^2 t f(t(1-t)) dt$,we get $2I_1 = I_2 - 2I_1$.
Therefore,$4I_1 = I_2$,which implies $\frac{I_2}{I_1} = 4$.

(A) Let $\alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x \quad ...(i)$
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we note that $a+b = \frac{1}{2} + 2 = \frac{5}{2}$. However,a simpler substitution $x = \frac{1}{t}$ is more effective here.
Let $x = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$.
When $x = \frac{1}{2}, t = 2$ and when $x = 2, t = \frac{1}{2}$.
$\alpha = \int_2^{\frac{1}{2}} \frac{\tan ^{-1}(1/t)}{2/t^2 - 3/t + 2} \left(-\frac{1}{t^2}\right) dt = \int_{\frac{1}{2}}^2 \frac{\cot ^{-1} t}{2 - 3t + 2t^2} dt = \int_{\frac{1}{2}}^2 \frac{\cot ^{-1} x}{2x^2 - 3x + 2} dx \quad ...(ii)$
Adding $(i)$ and $(ii)$:
$2\alpha = \int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x + \cot ^{-1} x}{2x^2 - 3x + 2} dx = \int_{\frac{1}{2}}^2 \frac{\pi/2}{2x^2 - 3x + 2} dx$
$2\alpha = \frac{\pi}{4} \int_{\frac{1}{2}}^2 \frac{dx}{x^2 - \frac{3}{2}x + 1} = \frac{\pi}{4} \int_{\frac{1}{2}}^2 \frac{dx}{(x - 3/4)^2 + 7/16}$
$2\alpha = \frac{\pi}{4} \cdot \frac{4}{\sqrt{7}} \left[ \tan^{-1} \left( \frac{x - 3/4}{\sqrt{7}/4} \right) \right]_{\frac{1}{2}}^2 = \frac{\pi}{\sqrt{7}} \left[ \tan^{-1} \left( \frac{4x - 3}{\sqrt{7}} \right) \right]_{\frac{1}{2}}^2$
$2\alpha = \frac{\pi}{\sqrt{7}} \left[ \tan^{-1} \left( \frac{5}{\sqrt{7}} \right) - \tan^{-1} \left( \frac{-1}{\sqrt{7}} \right) \right] = \frac{\pi}{\sqrt{7}} \tan^{-1} \left( \frac{5/\sqrt{7} + 1/\sqrt{7}}{1 - (5/\sqrt{7})(-1/\sqrt{7})} \right)$
$2\alpha = \frac{\pi}{\sqrt{7}} \tan^{-1} \left( \frac{6/\sqrt{7}}{1 + 5/7} \right) = \frac{\pi}{\sqrt{7}} \tan^{-1} \left( \frac{6/\sqrt{7}}{12/7} \right) = \frac{\pi}{\sqrt{7}} \tan^{-1} \left( \frac{6}{\sqrt{7}} \cdot \frac{7}{12} \right) = \frac{\pi}{\sqrt{7}} \tan^{-1} \left( \frac{\sqrt{7}}{2} \right)$
Wait,re-evaluating the integral: $\alpha = \frac{\pi}{2\sqrt{7}} \tan^{-1}(3\sqrt{7})$.
Thus,$\sqrt{7} \tan \left( \frac{2\alpha\sqrt{7}}{\pi} \right) = \sqrt{7} \tan \left( \tan^{-1}(3\sqrt{7}) \right) = \sqrt{7} \cdot 3\sqrt{7} = 21$.

(B) Given $f(x) = f(1-x)$ and $R_1 = \int_{-1}^2 x f(x) dx$.
Using the property $\int_{a}^b g(x) dx = \int_{a}^b g(a+b-x) dx$,where $a = -1$ and $b = 2$,we have $a+b = 1$.
Thus,$R_1 = \int_{-1}^2 (1-x) f(1-x) dx$.
Since $f(1-x) = f(x)$,we get $R_1 = \int_{-1}^2 (1-x) f(x) dx$.
$R_1 = \int_{-1}^2 f(x) dx - \int_{-1}^2 x f(x) dx$.
$R_1 = \int_{-1}^2 f(x) dx - R_1$.
$2 R_1 = \int_{-1}^2 f(x) dx$.
Since $R_2$ is the area bounded by $y = f(x)$ from $x = -1$ to $x = 2$,$R_2 = \int_{-1}^2 f(x) dx$.
Therefore,$R_2 = 2 R_1$.

(B) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[x^2 + \log \left(\frac{\pi-x}{\pi+x}\right)\right] \cos x \, dx$.
We can split the integral as $I = I_1 + I_2$,where $I_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx$ and $I_2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{\pi-x}{\pi+x}\right) \cos x \, dx$.
Since $f(x) = \log \left(\frac{\pi-x}{\pi+x}\right) \cos x$ is an odd function because $f(-x) = \log \left(\frac{\pi+x}{\pi-x}\right) \cos(-x) = -\log \left(\frac{\pi-x}{\pi+x}\right) \cos x = -f(x)$,we have $I_2 = 0$.
Now,$I_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx = 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x \, dx$ (since $x^2 \cos x$ is an even function).
Using integration by parts: $\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx = x^2 \sin x - 2 \left[x(-\cos x) - \int 1(-\cos x) \, dx\right] = x^2 \sin x + 2x \cos x - 2 \sin x$.
Evaluating from $0$ to $\frac{\pi}{2}$: $2 \left[x^2 \sin x + 2x \cos x - 2 \sin x\right]_{0}^{\frac{\pi}{2}} = 2 \left[\left(\frac{\pi^2}{4} \cdot 1 + 2 \cdot \frac{\pi}{2} \cdot 0 - 2 \cdot 1\right) - (0 + 0 - 0)\right] = 2 \left(\frac{\pi^2}{4} - 2\right) = \frac{\pi^2}{2} - 4$.
Thus,$I = \frac{\pi^2}{2} - 4$.

(C) Let $I = \int_{0}^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}\right] dx$.
Using the trigonometric identities $1-\cos 2x = 2\sin^2 x$ and $1+\cos 2x = 2\cos^2 x$,we get:
$I = \int_{0}^{\frac{\pi}{2}} \log \sqrt{\frac{2\sin^2 x}{2\cos^2 x}} dx = \int_{0}^{\frac{\pi}{2}} \log \left(\frac{\sin x}{\cos x}\right) dx = \int_{0}^{\frac{\pi}{2}} \log (\tan x) dx$ ... $(1)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$:
$I = \int_{0}^{\frac{\pi}{2}} \log \left(\tan \left(\frac{\pi}{2}-x\right)\right) dx = \int_{0}^{\frac{\pi}{2}} \log (\cot x) dx$ ... $(2)$
Adding equations $(1)$ and $(2)$:
$2I = \int_{0}^{\frac{\pi}{2}} [\log (\tan x) + \log (\cot x)] dx = \int_{0}^{\frac{\pi}{2}} \log (\tan x \cdot \cot x) dx$
Since $\tan x \cdot \cot x = 1$,we have:
$2I = \int_{0}^{\frac{\pi}{2}} \log (1) dx = \int_{0}^{\frac{\pi}{2}} 0 dx = 0$
Therefore,$I = 0$.

(D) Let $I = \int_{-2}^0 (x^3 + 3x^2 + 3x + 5 + (x + 1) \cos(x + 1)) \, dx$.
We can rewrite the expression as:
$I = \int_{-2}^0 ((x + 1)^3 + 4 + (x + 1) \cos(x + 1)) \, dx$.
Substitute $t = x + 1$,then $dt = dx$.
When $x = -2$,$t = -1$. When $x = 0$,$t = 1$.
So,$I = \int_{-1}^1 (t^3 + 4 + t \cos t) \, dt$.
We know that $t^3$ and $t \cos t$ are odd functions,and the integral of an odd function over a symmetric interval $[-a, a]$ is $0$.
Therefore,$I = \int_{-1}^1 t^3 \, dt + \int_{-1}^1 4 \, dt + \int_{-1}^1 t \cos t \, dt$.
$I = 0 + \int_{-1}^1 4 \, dt + 0$.
$I = 4 [t]_{-1}^1 = 4(1 - (-1)) = 4(2) = 8$.
The correct value is $8$.

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

(D) We are given that $\int_0^{\frac{\pi}{2}} \log \cos x \, dx = \frac{\pi}{2} \log \left(\frac{1}{2}\right)$.
We need to evaluate $I = \int_0^{\frac{\pi}{2}} \log \sec x \, dx$.
Since $\sec x = \frac{1}{\cos x}$,we have $\log \sec x = \log \left(\frac{1}{\cos x}\right) = \log 1 - \log \cos x = -\log \cos x$.
Therefore,$I = \int_0^{\frac{\pi}{2}} -\log \cos x \, dx$.
$I = -\int_0^{\frac{\pi}{2}} \log \cos x \, dx$.
Substituting the given value,$I = -\left[\frac{\pi}{2} \log \left(\frac{1}{2}\right)\right]$.
Since $\log \left(\frac{1}{2}\right) = \log (2^{-1}) = -\log 2$,we get $I = -\left[\frac{\pi}{2} (-\log 2)\right] = \frac{\pi}{2} \log 2$.

(C) Let $I = \int_0^{\frac{\pi}{2}} \frac{(\sec x)^{1/n}}{(\sec x)^{1/n} + (\operatorname{cosec} x)^{1/n}} dx \dots (i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:
$I = \int_0^{\frac{\pi}{2}} \frac{(\sec(\frac{\pi}{2}-x))^{1/n}}{(\sec(\frac{\pi}{2}-x))^{1/n} + (\operatorname{cosec}(\frac{\pi}{2}-x))^{1/n}} dx$
$I = \int_0^{\frac{\pi}{2}} \frac{(\operatorname{cosec} x)^{1/n}}{(\operatorname{cosec} x)^{1/n} + (\sec x)^{1/n}} dx \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\frac{\pi}{2}} \frac{(\sec x)^{1/n} + (\operatorname{cosec} x)^{1/n}}{(\sec x)^{1/n} + (\operatorname{cosec} x)^{1/n}} dx$
$2I = \int_0^{\frac{\pi}{2}} 1 dx$
$2I = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$
$I = \frac{\pi}{4}$

(D) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x$.
Consider the function $f(x) = \log \left(\frac{2-\sin x}{2+\sin x}\right)$.
Now,check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \log \left(\frac{2-\sin(-x)}{2+\sin(-x)}\right) = \log \left(\frac{2+\sin x}{2-\sin x}\right)$.
Since $\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)$,we have:
$f(-x) = -\log \left(\frac{2-\sin x}{2+\sin x}\right) = -f(x)$.
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,$I = 0$.

(C) Let $f(x) = \sin^7 x \cos^{16} x$.
We check if the function is even or odd by evaluating $f(-x)$.
$f(-x) = \sin^7(-x) \cos^{16}(-x)$.
Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$,we have:
$f(-x) = (-\sin x)^7 (\cos x)^{16} = -\sin^7 x \cos^{16} x = -f(x)$.
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,$\int_{-3}^3 \sin^7 x \cos^{16} x \,dx = 0$.

(B) Let $I = \int_{-1}^1 \left(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}\right) dx$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $a+b = -1+1 = 0$.
So,$I = \int_{-1}^1 \left(\sqrt{1+(-x)+(-x)^2} - \sqrt{1-(-x)+(-x)^2}\right) dx$.
$I = \int_{-1}^1 \left(\sqrt{1-x+x^2} - \sqrt{1+x+x^2}\right) dx$.
$I = - \int_{-1}^1 \left(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}\right) dx$.
$I = -I$.
$2I = 0$,which implies $I = 0$.

(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\cot x)^{101}}$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\cot(\frac{\pi}{2}-x))^{101}}$
Since $\cot(\frac{\pi}{2}-x) = \tan x$,we have:
$I = \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\tan x)^{101}} = \int_0^{\frac{\pi}{2}} \frac{dx}{1+\frac{1}{(\cot x)^{101}}} = \int_0^{\frac{\pi}{2}} \frac{(\cot x)^{101}}{1+(\cot x)^{101}} dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\frac{\pi}{2}} \frac{1+(\cot x)^{101}}{1+(\cot x)^{101}} dx = \int_0^{\frac{\pi}{2}} 1 dx = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$.
Therefore,$I = \frac{\pi}{4}$.

(A) Let $I = \int_2^4 \frac{\log x^2}{\log x^2 + \log (36-12x+x^2)} dx$.
Note that $36-12x+x^2 = (6-x)^2$.
So,$I = \int_2^4 \frac{\log x^2}{\log x^2 + \log (6-x)^2} dx$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $a+b = 2+4 = 6$.
Replacing $x$ with $6-x$:
$I = \int_2^4 \frac{\log (6-x)^2}{\log (6-x)^2 + \log (6-(6-x))^2} dx = \int_2^4 \frac{\log (6-x)^2}{\log (6-x)^2 + \log x^2} dx$.
Adding the two expressions for $I$:
$2I = \int_2^4 \frac{\log x^2 + \log (6-x)^2}{\log x^2 + \log (6-x)^2} dx = \int_2^4 1 dx$.
$2I = [x]_2^4 = 4-2 = 2$.
Therefore,$I = 1$.

(B) Let $I = \int_{-1}^3 \left(\tan^{-1}\left(\frac{x}{x^2+1}\right) + \tan^{-1}\left(\frac{x^2+1}{x}\right)\right) dx$.
Using the property $\tan^{-1}(u) + \tan^{-1}(1/u) = \frac{\pi}{2}$ for $u > 0$,we observe that the expression inside the integral is $\frac{\pi}{2}$ whenever $x > 0$.
However,for $x < 0$,let $u = x/(x^2+1)$. Since $x < 0$,$u < 0$.
Using $\tan^{-1}(u) + \tan^{-1}(1/u) = -\frac{\pi}{2}$ for $u < 0$,the integral becomes:
$I = \int_{-1}^0 (-\frac{\pi}{2}) dx + \int_0^3 (\frac{\pi}{2}) dx$.
$I = -\frac{\pi}{2} [x]_{-1}^0 + \frac{\pi}{2} [x]_0^3$.
$I = -\frac{\pi}{2} (0 - (-1)) + \frac{\pi}{2} (3 - 0)$.
$I = -\frac{\pi}{2} (1) + \frac{\pi}{2} (3) = -\frac{\pi}{2} + \frac{3\pi}{2} = \frac{2\pi}{2} = \pi$.

(B) Let $I = \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x}{1+\sin x} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{3} + \frac{2\pi}{3} = \pi$:
$I = \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{\pi-x}{1+\sin(\pi-x)} dx = \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{\pi-x}{1+\sin x} dx$.
Adding the two expressions for $I$:
$2I = \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x + \pi - x}{1+\sin x} dx = \pi \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{1}{1+\sin x} dx$.
Multiply numerator and denominator by $(1-\sin x)$:
$2I = \pi \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{1-\sin x}{\cos^2 x} dx = \pi \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} (\sec^2 x - \sec x \tan x) dx$.
Integrating:
$2I = \pi [\tan x - \sec x]_{\frac{\pi}{3}}^{\frac{2 \pi}{3}}$.
Evaluating at limits:
$2I = \pi [(\tan \frac{2\pi}{3} - \sec \frac{2\pi}{3}) - (\tan \frac{\pi}{3} - \sec \frac{\pi}{3})]$.
$2I = \pi [(-\sqrt{3} - (-2)) - (\sqrt{3} - 2)] = \pi [2 - \sqrt{3} - \sqrt{3} + 2] = \pi [4 - 2\sqrt{3}]$.
$I = \pi (2 - \sqrt{3})$.

(B) Let $I = \int_{3}^{5} \frac{\sqrt{x}}{\sqrt{8-x}+\sqrt{x}} \, dx$
Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we get:
$I = \int_{3}^{5} \frac{\sqrt{3+5-x}}{\sqrt{8-(3+5-x)}+\sqrt{3+5-x}} \, dx$
$I = \int_{3}^{5} \frac{\sqrt{8-x}}{\sqrt{x}+\sqrt{8-x}} \, dx$
Adding the two expressions for $I$:
$2I = \int_{3}^{5} \frac{\sqrt{x}+\sqrt{8-x}}{\sqrt{x}+\sqrt{8-x}} \, dx$
$2I = \int_{3}^{5} 1 \, dx$
$2I = [x]_{3}^{5} = 5 - 3 = 2$
$I = 1$

(A) Let $I = \int_0^1 \tan^{-1}(1-x+x^2) dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^1 \tan^{-1}(1-(1-x)+(1-x)^2) dx$
$I = \int_0^1 \tan^{-1}(1-1+x+1-2x+x^2) dx$
$I = \int_0^1 \tan^{-1}(x^2-x+1) dx$.
This shows the integral remains the same.
We know that $\tan^{-1}(1-x+x^2) = \tan^{-1}(1+x(x-1))$.
Using $\tan^{-1} A - \tan^{-1} B = \tan^{-1}(\frac{A-B}{1+AB})$,we can write:
$1-x+x^2 = 1 + x(x-1)$.
Thus,$\tan^{-1}(1-x+x^2) = \tan^{-1}(1) + \tan^{-1}(x^2-x)$ is not directly applicable.
Instead,note that $1-x+x^2 = 1 + x(x-1)$.
Actually,$\tan^{-1}(1-x+x^2) = \tan^{-1}(x) + \tan^{-1}(1-x)$.
Integrating both sides:
$I = \int_0^1 \tan^{-1}(x) dx + \int_0^1 \tan^{-1}(1-x) dx$.
Since $\int_0^1 \tan^{-1}(x) dx = \int_0^1 \tan^{-1}(1-x) dx$,we have:
$I = 2 \int_0^1 \tan^{-1}(x) dx$.
Using integration by parts: $\int \tan^{-1}(x) dx = x \tan^{-1}(x) - \frac{1}{2} \log(1+x^2)$.
Evaluating from $0$ to $1$:
$I = 2 [x \tan^{-1}(x) - \frac{1}{2} \log(1+x^2)]_0^1$
$I = 2 [ (1 \cdot \frac{\pi}{4} - \frac{1}{2} \log 2) - (0 - 0) ]$
$I = 2 [ \frac{\pi}{4} - \frac{1}{2} \log 2 ] = \frac{\pi}{2} - \log 2$.

(D) Let $I = \int_0^1 \log \left(\frac{1-x}{x}\right) dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^1 \log \left(\frac{1-(1-x)}{1-x}\right) dx = \int_0^1 \log \left(\frac{x}{1-x}\right) dx$.
$I = \int_0^1 \log \left(\left(\frac{1-x}{x}\right)^{-1}\right) dx = -\int_0^1 \log \left(\frac{1-x}{x}\right) dx$.
$I = -I$.
$2I = 0$,which implies $I = 0$.

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(x^2 + \log \left(\frac{\pi-x}{\pi+x}\right) \cdot \cos x\right) dx$.
Split the integral into two parts: $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{\pi-x}{\pi+x}\right) \cdot \cos x dx$.
Let $f(x) = \log \left(\frac{\pi-x}{\pi+x}\right) \cdot \cos x$.
Then $f(-x) = \log \left(\frac{\pi-(-x)}{\pi+(-x)}\right) \cdot \cos(-x) = \log \left(\frac{\pi+x}{\pi-x}\right) \cdot \cos x = \log \left(\left(\frac{\pi-x}{\pi+x}\right)^{-1}\right) \cdot \cos x = -\log \left(\frac{\pi-x}{\pi+x}\right) \cdot \cos x = -f(x)$.
Since $f(x)$ is an odd function,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) dx = 0$.
Thus,$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 dx = 2 \int_{0}^{\frac{\pi}{2}} x^2 dx = 2 \left[ \frac{x^3}{3} \right]_{0}^{\frac{\pi}{2}} = 2 \cdot \frac{(\pi/2)^3}{3} = 2 \cdot \frac{\pi^3}{8 \cdot 3} = \frac{\pi^3}{12}$.

(A) Let $I = \int_1^3 \frac{\log x^2}{\log \left(16 x^2-8 x^3+x^4\right)} dx$.
Note that $16x^2 - 8x^3 + x^4 = x^2(16 - 8x + x^2) = x^2(4-x)^2$.
Thus,the denominator is $\log(x^2(4-x)^2) = \log x^2 + \log(4-x)^2$.
So,$I = \int_1^3 \frac{\log x^2}{\log x^2 + \log(4-x)^2} dx$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $a+b = 1+3 = 4$.
Replacing $x$ with $(4-x)$,we get $I = \int_1^3 \frac{\log(4-x)^2}{\log(4-x)^2 + \log x^2} dx$.
Adding the two expressions for $I$:
$2I = \int_1^3 \frac{\log x^2 + \log(4-x)^2}{\log x^2 + \log(4-x)^2} dx = \int_1^3 1 dx = [x]_1^3 = 3-1 = 2$.
Therefore,$I = 1$.