Properties of definite integration Questions in English

(B) Given integral is $I = \int_0^{2n\pi } {\left( {|\sin x| - \frac{1}{2}|\sin x|} \right)} \;dx$.
Simplifying the integrand,we get $I = \int_0^{2n\pi } {\frac{1}{2}|\sin x|} \;dx$.
Since $|\sin x|$ is a periodic function with period $\pi$,we use the property $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$.
Here $T = \pi$,so $I = \frac{1}{2} \times n \int_0^{\pi} |\sin x| \;dx$.
Since $\sin x \ge 0$ for $x \in [0, \pi]$,$|\sin x| = \sin x$.
$I = \frac{n}{2} \int_0^{\pi} \sin x \;dx = \frac{n}{2} [-\cos x]_0^{\pi}$.
$I = \frac{n}{2} (-(-1) - (-1)) = \frac{n}{2} (1 + 1) = \frac{n}{2} (2) = 2n$ is incorrect,let us re-evaluate:
$I = \frac{n}{2} [-\cos x]_0^{\pi} = \frac{n}{2} (-(-1) - (-1)) = \frac{n}{2} (2) = n$.
Wait,the integral $\int_0^{\pi} \sin x dx = 2$.
So $I = \frac{n}{2} \times 2 = n$.
Re-checking the provided solution: The step $2n \int_0^{\pi/2} \sin x dx = 2n(1) = 2n$ is used.
Actually,$\int_0^{2n\pi} |\sin x| dx = 2n \int_0^{\pi} \sin x dx = 2n(2) = 4n$.
Then $I = \frac{1}{2} (4n) = 2n$.
The correct answer is $2n$.

(A) Let $f(x) = \frac{1}{x + x^3}$.
Check if the function is even or odd.
$f(-x) = \frac{1}{(-x) + (-x)^3} = \frac{1}{-(x + x^3)} = -\frac{1}{x + x^3} = -f(x)$.
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,$\int_{-a}^a \frac{1}{x + x^3} dx = 0$.

(A) Let $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\tan x}}$.
$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ ...$(i)$
Using the property $\int_a^b f(x) dx = \int_a^b f(a + b - x) dx$,we have:
$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos(\pi/2 - x)}}{\sqrt{\cos(\pi/2 - x)} + \sqrt{\sin(\pi/2 - x)}} dx$
$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ ...(ii)
Adding $(i)$ and $(ii)$:
$2I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$
$2I = \int_{\pi /6}^{\pi /3} 1 dx$
$2I = [x]_{\pi /6}^{\pi /3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
$I = \frac{\pi}{12}$.

(D) Let $I = \int_{-\pi}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$.
Since the integrand $f(x) = \frac{\sin^4 x}{\sin^4 x + \cos^4 x}$ is an even function,we have $I = 2 \int_{0}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$.
Using the property $\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(2a-x) = f(x)$,we note that $f(\pi - x) = \frac{\sin^4(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)} = \frac{\sin^4 x}{\sin^4 x + \cos^4 x} = f(x)$.
Thus,$I = 2 \times 2 \int_{0}^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx = 4 \int_{0}^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$ ..... $(i)$.
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get $I = 4 \int_{0}^{\pi/2} \frac{\cos^4 x}{\cos^4 x + \sin^4 x} \, dx$ ..... $(ii)$.
Adding $(i)$ and $(ii)$,we get $2I = 4 \int_{0}^{\pi/2} \frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x} \, dx = 4 \int_{0}^{\pi/2} 1 \, dx = 4 \times \frac{\pi}{2} = 2\pi$.
Therefore,$I = \pi$.

(D) Given that $f$ is a continuous function.
Consider the integral $I = \int_{-3}^{5} f(x) dx$.
Let $x = t - 1$,then $dx = dt$.
When $x = -3$,$t = -2$.
When $x = 5$,$t = 6$.
Substituting these into the integral,we get:
$I = \int_{-2}^{6} f(t - 1) dt$.
Since the variable of integration is a dummy variable,we can replace $t$ with $x$:
$I = \int_{-2}^{6} f(x - 1) dx$.
Thus,option $(d)$ is correct.

(B) Let $I = \int_0^\infty \frac{\log(1 + x^2)}{1 + x^2} \,dx$.
Substitute $x = \tan \theta$,then $dx = \sec^2 \theta \,d\theta$.
When $x = 0$,$\theta = 0$,and when $x \to \infty$,$\theta = \frac{\pi}{2}$.
Thus,$I = \int_0^{\pi/2} \frac{\log(1 + \tan^2 \theta)}{1 + \tan^2 \theta} \cdot \sec^2 \theta \,d\theta$.
Since $1 + \tan^2 \theta = \sec^2 \theta$,we have $I = \int_0^{\pi/2} \log(\sec^2 \theta) \,d\theta$.
Using the property $\log(a^b) = b \log a$,we get $I = 2 \int_0^{\pi/2} \log(\sec \theta) \,d\theta$.
Since $\sec \theta = \frac{1}{\cos \theta}$,we have $\log(\sec \theta) = -\log(\cos \theta)$.
So,$I = -2 \int_0^{\pi/2} \log(\cos \theta) \,d\theta$.
Using the standard integral result $\int_0^{\pi/2} \log(\cos \theta) \,d\theta = -\frac{\pi}{2} \log 2$,we get:
$I = -2 \left( -\frac{\pi}{2} \log 2 \right) = \pi \log 2$.

(C) Let $f(x) = \tan^{-1} \frac{1}{x}$. The integral is $\int_{-1}^{1} f'(x) dx = [f(x)]_{-1}^{1} = f(1) - f(-1)$.
Since $f(1) = \tan^{-1}(1) = \frac{\pi}{4}$ and $f(-1) = \tan^{-1}(-1) = -\frac{\pi}{4}$.
Therefore,$\int_{-1}^{1} f'(x) dx = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$.
However,note that the function $\tan^{-1} \frac{1}{x}$ is discontinuous at $x = 0$. The integral is an improper integral: $\int_{-1}^{1} f'(x) dx = \lim_{\epsilon \to 0^+} \int_{-1}^{-\epsilon} f'(x) dx + \lim_{\delta \to 0^+} \int_{\delta}^{1} f'(x) dx$.
$= [f(-\epsilon) - f(-1)] + [f(1) - f(\delta)] = (\tan^{-1}(-1/\epsilon) - \tan^{-1}(-1)) + (\tan^{-1}(1) - \tan^{-1}(1/\delta))$.
As $\epsilon, \delta \to 0$,$\tan^{-1}(-1/\epsilon) \to -\frac{\pi}{2}$ and $\tan^{-1}(1/\delta) \to \frac{\pi}{2}$.
Result $= (-\frac{\pi}{2} - (-\frac{\pi}{4})) + (\frac{\pi}{4} - \frac{\pi}{2}) = (-\frac{\pi}{2} + \frac{\pi}{4}) + (\frac{\pi}{4} - \frac{\pi}{2}) = -\frac{\pi}{4} - \frac{\pi}{4} = -\frac{\pi}{2}$.

(C) Let $I_k = \int_0^1 {f(k - 1 + x)\,dx} $.
Substitute $t = k - 1 + x$,then $dt = dx$.
When $x = 0$,$t = k - 1$. When $x = 1$,$t = k$.
So,$I_k = \int_{k - 1}^k {f(t)\,dt} = \int_{k - 1}^k {f(x)\,dx} $.
Now,the given sum is $\sum\limits_{k = 1}^n {I_k} = \sum\limits_{k = 1}^n {\int_{k - 1}^k {f(x)\,dx} } $.
Expanding the summation,we get:
$\int_0^1 {f(x)\,dx} + \int_1^2 {f(x)\,dx} + \dots + \int_{n - 1}^n {f(x)\,dx} $.
By the property of definite integrals,$\int_a^b f(x)dx + \int_b^c f(x)dx = \int_a^c f(x)dx$.
Therefore,the sum equals $\int_0^n {f(x)\,dx} $.

(C) Let $I = \int_a^{a + (\pi /2)} (\sin^4 x + \cos^4 x) \, dx$.
Since the integrand $f(x) = \sin^4 x + \cos^4 x$ is a periodic function with period $\frac{\pi}{2}$,the integral over any interval of length $\frac{\pi}{2}$ is constant and equal to the integral over $[0, \pi/2]$.
Thus,$I = \int_0^{\pi/2} (\sin^4 x + \cos^4 x) \, dx$.
Using the identity $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2(2x)$.
Therefore,$I = \int_0^{\pi/2} (1 - \frac{1}{2}\sin^2(2x)) \, dx = \int_0^{\pi/2} 1 \, dx - \frac{1}{2} \int_0^{\pi/2} \sin^2(2x) \, dx$.
$I = [x]_0^{\pi/2} - \frac{1}{2} \int_0^{\pi/2} \frac{1 - \cos(4x)}{2} \, dx = \frac{\pi}{2} - \frac{1}{4} [x - \frac{\sin(4x)}{4}]_0^{\pi/2}$.
$I = \frac{\pi}{2} - \frac{1}{4} (\frac{\pi}{2} - 0) = \frac{\pi}{2} - \frac{\pi}{8} = \frac{3\pi}{8}$.

(A) Let $I = \int_0^\infty \log \left( x + \frac{1}{x} \right) \frac{dx}{1 + x^2}$.
Substitute $x = \tan \theta$,then $dx = \sec^2 \theta \, d\theta$.
As $x \to 0, \theta \to 0$ and as $x \to \infty, \theta \to \pi / 2$.
$I = \int_0^{\pi / 2} \log (\tan \theta + \cot \theta) \frac{\sec^2 \theta}{1 + \tan^2 \theta} \, d\theta = \int_0^{\pi / 2} \log \left( \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \right) d\theta$.
$I = \int_0^{\pi / 2} \log \left( \frac{1}{\sin \theta \cos \theta} \right) d\theta = - \int_0^{\pi / 2} \log (\sin \theta \cos \theta) \, d\theta$.
$I = - \int_0^{\pi / 2} (\log \sin \theta + \log \cos \theta) \, d\theta$.
Using the property $\int_0^{\pi / 2} \log \sin \theta \, d\theta = \int_0^{\pi / 2} \log \cos \theta \, d\theta = -\frac{\pi}{2} \log 2$.
$I = - \left( -\frac{\pi}{2} \log 2 - \frac{\pi}{2} \log 2 \right) = - (-\pi \log 2) = \pi \log 2$.

(A) Let $I = \int_{0}^{\infty} \frac{x \ln x}{(1 + x^2)^2} \, dx$.
Substitute $x = \tan \theta$,then $dx = \sec^2 \theta \, d\theta$.
As $x \to 0$,$\theta \to 0$,and as $x \to \infty$,$\theta \to \frac{\pi}{2}$.
Substituting these into the integral:
$I = \int_{0}^{\pi/2} \frac{\tan \theta \ln(\tan \theta)}{(1 + \tan^2 \theta)^2} \sec^2 \theta \, d\theta$
Since $1 + \tan^2 \theta = \sec^2 \theta$,we have:
$I = \int_{0}^{\pi/2} \frac{\tan \theta \ln(\tan \theta)}{\sec^4 \theta} \sec^2 \theta \, d\theta$
$I = \int_{0}^{\pi/2} \frac{\tan \theta}{\sec^2 \theta} \ln(\tan \theta) \, d\theta$
$I = \int_{0}^{\pi/2} \sin \theta \cos \theta \ln(\tan \theta) \, d\theta$
Using $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$:
$I = \frac{1}{2} \int_{0}^{\pi/2} \sin 2\theta \ln(\tan \theta) \, d\theta$
Let $J = \int_{0}^{\pi/2} \sin 2\theta \ln(\tan \theta) \, d\theta$. Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$:
$J = \int_{0}^{\pi/2} \sin(2(\frac{\pi}{2} - \theta)) \ln(\tan(\frac{\pi}{2} - \theta)) \, d\theta$
$J = \int_{0}^{\pi/2} \sin(\pi - 2\theta) \ln(\cot \theta) \, d\theta$
$J = \int_{0}^{\pi/2} \sin 2\theta \ln(\cot \theta) \, d\theta$
$J = \int_{0}^{\pi/2} \sin 2\theta (-\ln(\tan \theta)) \, d\theta = -J$
Thus,$2J = 0$,which implies $J = 0$. Therefore,$I = 0$.

(D) For $0 < x < 1$,we have $\frac{1}{2}{x^2} < {x^2} < x$.
This implies $- \frac{1}{2}{x^2} > - {x^2} > - x$.
Consequently,${e^{ - \frac{1}{2}{x^2}}} > {e^{ - {x^2}}} > {e^{ - x}}$.
Since ${\cos ^2}x \le 1$,we have ${e^{ - {x^2}}}{\cos ^2}x \le {e^{ - {x^2}}} < {e^{ - \frac{1}{2}{x^2}}}$.
Integrating from $0$ to $1$,we get $\int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} < \int_0^1 {{e^{ - {x^2}}}{{\cos }^2}x\,dx} < \int_0^1 {{e^{ - {x^2}}}} dx < \int_0^1 {{e^{ - {x^2}/2}}dx}$.
Thus,${I_1} < {I_2} < {I_3} < {I_4}$.
Therefore,${I_4}$ is the greatest integral.

(C) Given $I_1 = \int_{1 - k}^k x f\{x(1 - x)\} dx$ and $I_2 = \int_{1 - k}^k f\{x(1 - x)\} dx$.
Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a + b - x) dx$,we have $a + b = 1 - k + k = 1$.
Thus,$I_1 = \int_{1 - k}^k (1 - x) f\{(1 - x)(1 - (1 - x))\} dx$.
$I_1 = \int_{1 - k}^k (1 - x) f\{(1 - x)x\} dx$.
$I_1 = \int_{1 - k}^k f\{x(1 - x)\} dx - \int_{1 - k}^k x f\{x(1 - x)\} dx$.
$I_1 = I_2 - I_1$.
$2I_1 = I_2$.
Therefore,$I_1/I_2 = 1/2$.

(C) Let $I = \int_0^\pi {e^{\cos^2 x} \cos^3((2n + 1)x)} \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we have:
$I = \int_0^\pi {e^{\cos^2(\pi - x)} \cos^3((2n + 1)(\pi - x))} \, dx$.
Since $\cos(\pi - x) = -\cos x$,we have $\cos^2(\pi - x) = \cos^2 x$.
Also,$\cos((2n + 1)(\pi - x)) = \cos((2n + 1)\pi - (2n + 1)x) = \cos((2n + 1)\pi) \cos((2n + 1)x) + \sin((2n + 1)\pi) \sin((2n + 1)x)$.
Since $(2n + 1)$ is an odd integer,$\cos((2n + 1)\pi) = -1$ and $\sin((2n + 1)\pi) = 0$.
Thus,$\cos((2n + 1)(\pi - x)) = -\cos((2n + 1)x)$.
Therefore,$\cos^3((2n + 1)(\pi - x)) = (-\cos((2n + 1)x))^3 = -\cos^3((2n + 1)x)$.
Substituting this back into the integral,we get $I = \int_0^\pi e^{\cos^2 x} (-\cos^3((2n + 1)x)) \, dx = -I$.
$2I = 0 \implies I = 0$.

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

(D) Given $f(x) = \frac{e^x}{1 + e^x}$. Note that $f(-a) = \frac{e^{-a}}{1 + e^{-a}} = \frac{1}{e^a + 1}$.
Thus,$f(a) + f(-a) = \frac{e^a}{1 + e^a} + \frac{1}{1 + e^a} = 1$.
Let $I_1 = \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$. Using the property $\int_{A}^{B} h(x) dx = \int_{A}^{B} h(A + B - x) dx$,where $A + B = 1$:
$I_1 = \int_{f(-a)}^{f(a)} (1 - x) g\{(1 - x)(1 - (1 - x))\} dx = \int_{f(-a)}^{f(a)} (1 - x) g\{(1 - x)x\} dx$.
$I_1 = \int_{f(-a)}^{f(a)} g\{x(1 - x)\} dx - \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$.
$I_1 = I_2 - I_1$.
$2I_1 = I_2$.
Therefore,$\frac{I_2}{I_1} = 2$.

(D) Let $h(x) = [f(x) + f(-x)][g(x) - g(-x)]$.
Now,calculate $h(-x)$:
$h(-x) = [f(-x) + f(-(-x))][g(-x) - g(-(-x))] = [f(-x) + f(x)][g(-x) - g(x)]$.
We can rewrite this as:
$h(-x) = -[f(x) + f(-x)][g(x) - g(-x)] = -h(x)$.
Since $h(-x) = -h(x)$,the function $h(x)$ is an odd function.
For any odd function $h(x)$,the integral over a symmetric interval $[-a, a]$ is zero:
$\int_{-a}^{a} h(x) \, dx = 0$.
Therefore,$\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)][g(x) - g(-x)] \, dx = 0$.

(D) Let the given expression be $S = \sum\limits_{n = 1}^{10} \int_{ - 2n - 1}^{2n} \sin^{27}x \, dx + \sum\limits_{n = 1}^{10} \int_{2n}^{2n + 1} \sin^{27}x \, dx$.
Consider the first summation: $I_1 = \sum\limits_{n = 1}^{10} \int_{ - 2n - 1}^{2n} \sin^{27}x \, dx$.
Let $x = -t$,then $dx = -dt$. When $x = -2n-1$,$t = 2n+1$. When $x = 2n$,$t = -2n$.
Thus,$\int_{ - 2n - 1}^{2n} \sin^{27}x \, dx = \int_{2n+1}^{-2n} \sin^{27}(-t) (-dt) = \int_{2n+1}^{-2n} -\sin^{27}t (-dt) = \int_{2n+1}^{-2n} \sin^{27}t \, dt = -\int_{-2n}^{2n+1} \sin^{27}t \, dt$.
This approach is complex. Let's simplify the sum: $S = \sum\limits_{n = 1}^{10} \left( \int_{ - 2n - 1}^{2n} \sin^{27}x \, dx + \int_{2n}^{2n + 1} \sin^{27}x \, dx \right)$.
By the property of definite integrals,$\int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx = \int_{a}^{c} f(x) dx$.
So,$S = \sum\limits_{n = 1}^{10} \int_{ - 2n - 1}^{2n + 1} \sin^{27}x \, dx$.
Since $\sin^{27}x$ is an odd function,$\int_{-a}^{a} \sin^{27}x \, dx = 0$.
Therefore,$\int_{ - (2n + 1)}^{(2n + 1)} \sin^{27}x \, dx = 0$ for each $n$.
Thus,$S = \sum\limits_{n = 1}^{10} 0 = 0$.

(B) Given ${I_n} = \int_{0}^{\pi /4} {\tan^n x} \,dx$.
We know that $\tan^2 x = \sec^2 x - 1$.
So,${I_n} = \int_{0}^{\pi /4} {\tan^{n-2} x (\sec^2 x - 1)} \,dx$.
${I_n} = \int_{0}^{\pi /4} {\tan^{n-2} x \sec^2 x} \,dx - \int_{0}^{\pi /4} {\tan^{n-2} x} \,dx$.
${I_n} = \left[ \frac{\tan^{n-1} x}{n-1} \right]_{0}^{\pi /4} - {I_{n-2}}$.
Since $\tan(\pi/4) = 1$ and $\tan(0) = 0$,we get ${I_n} = \frac{1}{n-1} - {I_{n-2}}$.
Thus,${I_n} + {I_{n-2}} = \frac{1}{n-1}$.
Now,we need to find $\lim_{n \to \infty} n[{I_n} + {I_{n-2}}]$.
$\lim_{n \to \infty} n \left( \frac{1}{n-1} \right) = \lim_{n \to \infty} \frac{n}{n-1} = \lim_{n \to \infty} \frac{1}{1 - 1/n} = 1$.

(C) We know that $2\sin \frac{x}{2} \left( {\frac{1}{2} + \cos x + \cos 2x + \dots + \cos nx} \right) = \sin \frac{x}{2} + \sum_{k=1}^n 2\sin \frac{x}{2} \cos kx$.
Using the identity $2\sin A \cos B = \sin(A+B) + \sin(A-B)$,we get:
$2\sin \frac{x}{2} \cos kx = \sin \left( k + \frac{1}{2} \right)x - \sin \left( k - \frac{1}{2} \right)x$.
Summing this from $k=1$ to $n$,we get a telescoping series:
$\sin \frac{x}{2} + \left( \sin \frac{3x}{2} - \sin \frac{x}{2} \right) + \left( \sin \frac{5x}{2} - \sin \frac{3x}{2} \right) + \dots + \left( \sin \left( n + \frac{1}{2} \right)x - \sin \left( n - \frac{1}{2} \right)x \right) = \sin \left( n + \frac{1}{2} \right)x$.
Thus,$\frac{\sin \left( n + \frac{1}{2} \right)x}{\sin \frac{x}{2}} = 1 + 2\cos x + 2\cos 2x + \dots + 2\cos nx$.
Now,integrate from $0$ to $\pi$:
$\int_0^\pi \left( 1 + 2\sum_{k=1}^n \cos kx \right) dx = \left[ x + 2\sum_{k=1}^n \frac{\sin kx}{k} \right]_0^\pi$.
Since $\sin(k\pi) = 0$ for all integers $k$,the integral evaluates to $\pi - 0 = \pi$.

(C) Given $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0}$.
This can be rewritten as $\int_0^1 {x{e^{{x^2}}}dx = \alpha \int_0^1 {{e^{{x^2}}}dx} }$.
For the left side,let $u = x^2$,then $du = 2x \, dx$,so $x \, dx = \frac{1}{2} du$. When $x=0, u=0$ and when $x=1, u=1$.
Thus,$\frac{1}{2} \int_0^1 {e^u du} = \alpha \int_0^1 {{e^{{x^2}}}dx}$.
$\frac{1}{2} [e^u]_0^1 = \alpha \int_0^1 {{e^{{x^2}}}dx}$.
$\frac{1}{2} (e - 1) = \alpha \int_0^1 {{e^{{x^2}}}dx}$.
Therefore,$\alpha = \frac{e - 1}{2 \int_0^1 {{e^{{x^2}}}dx}}$.
Since $e^x$ is an increasing function on $[0, 1]$,$1 \le e^{x^2} \le e$. Thus,$\int_0^1 1 \, dx < \int_0^1 e^{x^2} \, dx < \int_0^1 e \, dx$,which means $1 < \int_0^1 e^{x^2} \, dx < e$.
Since $e \approx 2.718$,$\frac{e-1}{2} \approx 0.859$. Since $\int_0^1 e^{x^2} \, dx > 1$,$\alpha = \frac{0.859}{\text{value} > 1} < 0.859 < 1$.
Also,since $e > 1$,$\alpha > 0$. Thus,$0 < \alpha < 1$.

(D) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$.
Using the property $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$,where $T$ is the period,we have:
$\int_{\,\pi }^{\,10\pi } {|\sin x|dx} = \int_{\,0}^{\,9\pi } {|\sin x|dx}$ (by shifting the limits by $-\pi$).
Since the period is $\pi$,we can write this as:
$9 \int_{\,0}^{\,\pi } {|\sin x|dx} = 9 \int_{\,0}^{\,\pi } {\sin x dx}$ (since $\sin x \ge 0$ in $[0, \pi]$).
Evaluating the integral:
$9 [-\cos x]_{0}^{\pi} = 9 (-(\cos \pi) + \cos 0) = 9 (-(-1) + 1) = 9(1 + 1) = 18$.

(B) Let $I = \int_{ - \pi }^\pi {\frac{{2x(1 + \sin x)}}{{1 + {{\cos }^2}x}}dx} $.
We can split the integral into two parts:
$I = \int_{ - \pi }^\pi {\frac{{2x}}{{1 + {{\cos }^2}x}}dx} + \int_{ - \pi }^\pi {\frac{{2x\sin x}}{{1 + {{\cos }^2}x}}dx} $.
For the first part,let $f(x) = \frac{2x}{1 + \cos^2 x}$. Since $f(-x) = \frac{-2x}{1 + \cos^2(-x)} = -f(x)$,the function is odd. Thus,$\int_{-\pi}^{\pi} f(x) dx = 0$.
For the second part,let $g(x) = \frac{2x\sin x}{1 + \cos^2 x}$. Since $g(-x) = \frac{2(-x)\sin(-x)}{1 + \cos^2(-x)} = \frac{2x\sin x}{1 + \cos^2 x} = g(x)$,the function is even. Thus,$\int_{-\pi}^{\pi} g(x) dx = 2 \int_0^{\pi} g(x) dx$.
So,$I = 2 \int_0^{\pi} \frac{2x\sin x}{1 + \cos^2 x} dx = 4 \int_0^{\pi} \frac{x\sin x}{1 + \cos^2 x} dx$ ... $(i)$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = 4 \int_0^{\pi} \frac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)} dx = 4 \int_0^{\pi} \frac{(\pi - x)\sin x}{1 + \cos^2 x} dx$ ... $(ii)$.
Adding $(i)$ and $(ii)$:
$2I = 4 \int_0^{\pi} \frac{x\sin x + (\pi - x)\sin x}{1 + \cos^2 x} dx = 4\pi \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$.
$I = 2\pi \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$.
Let $t = \cos x$,then $dt = -\sin x dx$. When $x=0, t=1$; when $x=\pi, t=-1$.
$I = 2\pi \int_1^{-1} \frac{-dt}{1 + t^2} = 2\pi \int_{-1}^1 \frac{dt}{1 + t^2} = 2\pi [\tan^{-1} t]_{-1}^1$.
$I = 2\pi (\tan^{-1}(1) - \tan^{-1}(-1)) = 2\pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = 2\pi (\frac{\pi}{2}) = \pi^2$.

(D) Let $I = \int_{0}^{\pi} x f(\sin x) dx$ $(1)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\pi} (\pi - x) f(\sin(\pi - x)) dx$
Since $\sin(\pi - x) = \sin x$,we have:
$I = \int_{0}^{\pi} (\pi - x) f(\sin x) dx$
$I = \pi \int_{0}^{\pi} f(\sin x) dx - \int_{0}^{\pi} x f(\sin x) dx$
$I = \pi \int_{0}^{\pi} f(\sin x) dx - I$
$2I = \pi \int_{0}^{\pi} f(\sin x) dx$
$I = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) dx$
Using the property $\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(2a-x) = f(x)$:
Here $f(\sin(\pi - x)) = f(\sin x)$,so $\int_{0}^{\pi} f(\sin x) dx = 2 \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$
$I = \frac{\pi}{2} \times 2 \int_{0}^{\frac{\pi}{2}} f(\sin x) dx = \pi \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$
Using $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get $\int_{0}^{\frac{\pi}{2}} f(\sin x) dx = \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$
Therefore,$I = \pi \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$.

(C) Let $I = \int_{-\frac{3\pi}{2}}^{-\frac{\pi}{2}} \left[ (x+\pi)^3 + \cos^2(x+3\pi) \right] dx$.
Substitute $t = x + \pi$,then $dt = dx$.
When $x = -\frac{3\pi}{2}$,$t = -\frac{\pi}{2}$.
When $x = -\frac{\pi}{2}$,$t = \frac{\pi}{2}$.
Since $\cos(x+3\pi) = \cos(x+\pi) = -\cos x$,we have $\cos^2(x+3\pi) = \cos^2 x$.
Thus,$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (t^3 + \cos^2 t) dt$.
Since $t^3$ is an odd function and $\cos^2 t$ is an even function,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t^3 dt = 0$.
Therefore,$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 t dt = 2 \int_{0}^{\frac{\pi}{2}} \cos^2 t dt$.
Using the identity $\cos^2 t = \frac{1 + \cos 2t}{2}$,we get $I = 2 \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos 2t}{2} dt = \int_{0}^{\frac{\pi}{2}} (1 + \cos 2t) dt$.
Evaluating the integral: $[t + \frac{\sin 2t}{2}]_{0}^{\frac{\pi}{2}} = (\frac{\pi}{2} + 0) - (0 + 0) = \frac{\pi}{2}$.

(C) Given $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt$.
Consider $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt$.
Let $t = \frac{1}{u}$,then $dt = -\frac{1}{u^2} du$.
When $t=1, u=1$. When $t=1/x, u=x$.
$f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_{e}(1/u)}{1+(1/u)} \left(-\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\log_{e} u}{\frac{u+1}{u}} \left(\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\log_{e} u}{u(u+1)} du$.
Now,$F(x) = f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt - \int_{1}^{x} \frac{\log_{e} t}{t(1+t)} dt$.
$F(x) = \int_{1}^{x} \log_{e} t \left( \frac{1}{1+t} - \frac{1}{t(1+t)} \right) dt = \int_{1}^{x} \log_{e} t \left( \frac{t-1}{t(1+t)} \right) dt$.
Wait,let's re-evaluate: $F(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} \left( 1 - \frac{1}{t} \right) dt = \int_{1}^{x} \frac{\log_{e} t}{1+t} \left( \frac{t-1}{t} \right) dt$.
Actually,$f(x) + f(1/x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt + \int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt$.
Using the property $\int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt = \int_{1}^{x} \frac{\log_{e}(1/u)}{1+(1/u)} (-du/u^2) = \int_{1}^{x} \frac{\log_{e} u}{u(u+1)} du$.
$F(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} (1 + 1/t) dt = \int_{1}^{x} \frac{\log_{e} t}{t} dt$.
$F(x) = \left[ \frac{(\log_{e} t)^2}{2} \right]_{1}^{x} = \frac{(\log_{e} x)^2}{2}$.
For $x=e$,$F(e) = \frac{(\log_{e} e)^2}{2} = \frac{1}{2} = 0.5$.

(C) Let $I = \int_{0}^{\pi} [\cot x] dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\pi} [\cot(\pi - x)] dx = \int_{0}^{\pi} [-\cot x] dx$.
Adding the two expressions for $I$:
$2I = \int_{0}^{\pi} ([\cot x] + [-\cot x]) dx$.
We know that for any $y \notin \mathbb{Z}$,$[y] + [-y] = -1$,and for $y \in \mathbb{Z}$,$[y] + [-y] = 0$.
The function $\cot x$ is not defined at $x=0, \pi/2, \pi$. At $x=\pi/2$,$\cot x = 0$,so $[0] + [0] = 0$.
For all other $x \in (0, \pi)$,$\cot x$ is not an integer.
Thus,$[\cot x] + [-\cot x] = -1$ almost everywhere in $(0, \pi)$.
$2I = \int_{0}^{\pi} (-1) dx = -\pi$.
Therefore,$I = -\frac{\pi}{2}$.

(A) Given $p'(x) = p'(1 - x)$.
Integrating both sides with respect to $x$,we get $p(x) = -p(1 - x) + C$.
At $x = 0$,$p(0) = -p(1) + C$.
Substituting the given values $p(0) = 1$ and $p(1) = 41$,we have $1 = -41 + C$,which implies $C = 42$.
Thus,$p(x) + p(1 - x) = 42$.
Let $I = \int_{0}^{1} p(x) dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have $I = \int_{0}^{1} p(1 - x) dx$.
Adding the two expressions for $I$,we get $2I = \int_{0}^{1} (p(x) + p(1 - x)) dx$.
Substituting $p(x) + p(1 - x) = 42$,we get $2I = \int_{0}^{1} 42 dx = 42[x]_{0}^{1} = 42$.
Therefore,$I = 21$.

(D) 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 $x=0, \theta=0$ and when $x=1, \theta=\frac{\pi}{4}$.
$I = 8 \int_{0}^{\pi/4} \frac{\log(1+\tan \theta)}{1+\tan^{2} \theta} \sec^{2} \theta d\theta = 8 \int_{0}^{\pi/4} \log(1+\tan \theta) d\theta$ ... $(i)$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$:
$I = 8 \int_{0}^{\pi/4} \log(1+\tan(\frac{\pi}{4}-\theta)) d\theta$.
Since $\tan(\frac{\pi}{4}-\theta) = \frac{1-\tan \theta}{1+\tan \theta}$,we have:
$I = 8 \int_{0}^{\pi/4} \log(1 + \frac{1-\tan \theta}{1+\tan \theta}) d\theta = 8 \int_{0}^{\pi/4} \log(\frac{2}{1+\tan \theta}) d\theta$.
$I = 8 \int_{0}^{\pi/4} (\log 2 - \log(1+\tan \theta)) d\theta$.
$I = 8 \log 2 [\theta]_{0}^{\pi/4} - 8 \int_{0}^{\pi/4} \log(1+\tan \theta) d\theta$.
$I = 8 \log 2 (\frac{\pi}{4}) - I$.
$2I = 2\pi \log 2 \Rightarrow I = \pi \log 2$.

(D) Let $I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan x}}$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we have:
$I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan(\frac{\pi}{6} + \frac{\pi}{3} - x)}} = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan(\frac{\pi}{2} - x)}} = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\cot x}}$.
$I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \frac{1}{\sqrt{\tan x}}} = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + 1} dx$.
Adding the two expressions for $I$:
$2I = \int_{\pi/6}^{\pi/3} \left( \frac{1}{1 + \sqrt{\tan x}} + \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \right) dx = \int_{\pi/6}^{\pi/3} 1 dx = [x]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Thus,$I = \frac{\pi}{12}$.
Statement $-1$ claims the value is $\frac{\pi}{6}$,which is false. Statement $-2$ is a standard property of definite integrals,which is true.

(D) Let $I = \int_{2}^{4} \frac{\log(x^2)}{\log(x^2) + \log((6-x)^2)} \, dx$.
Using the property $\log(a^2) = 2\log|a|$,we have:
$I = \int_{2}^{4} \frac{2\log x}{2\log x + 2\log(6-x)} \, dx = \int_{2}^{4} \frac{\log x}{\log x + \log(6-x)} \, dx \quad \dots(1)$.
Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,where $a+b = 2+4 = 6$:
$I = \int_{2}^{4} \frac{\log(6-x)}{\log(6-x) + \log(6-(6-x))} \, dx = \int_{2}^{4} \frac{\log(6-x)}{\log(6-x) + \log x} \, dx \quad \dots(2)$.
Adding $(1)$ and $(2)$:
$2I = \int_{2}^{4} \frac{\log x + \log(6-x)}{\log x + \log(6-x)} \, dx = \int_{2}^{4} 1 \, dx$.
$2I = [x]_{2}^{4} = 4 - 2 = 2$.
$I = 1$.

(C) Let $I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} \quad (i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi$:
$I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos(\pi - x)} = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 - \cos x} \quad (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \left( \frac{1}{1 + \cos x} + \frac{1}{1 - \cos x} \right) dx$
$2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1 - \cos x + 1 + \cos x}{1 - \cos^2 x} dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{2}{\sin^2 x} dx$
$2I = 2 \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \csc^2 x dx$
$I = [-\cot x]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$
$I = -(\cot \frac{3\pi}{4} - \cot \frac{\pi}{4}) = -(-1 - 1) = 2$

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

(A) Given ${I_n} = \int_0^{\pi /4} {{\tan ^n}\theta \,d\theta }$.
Consider ${I_{n + 1}} + {I_{n - 1}} = \int_0^{\pi /4} {{\tan ^{n + 1}}\theta \,d\theta } + \int_0^{\pi /4} {{\tan ^{n - 1}}\theta \,d\theta }$.
$= \int_0^{\pi /4} {{\tan ^{n - 1}}\theta (\tan ^2\theta + 1)d\theta }$.
$= \int_0^{\pi /4} {{\tan ^{n - 1}}\theta \sec ^2\theta \,d\theta }$.
Let $u = \tan \theta$,then $du = \sec ^2\theta \,d\theta$. When $\theta = 0, u = 0$ and when $\theta = \pi /4, u = 1$.
$= \int_0^1 {{u^{n - 1}}du} = \left[ \frac{u^n}{n} \right]_0^1 = \frac{1}{n}$.
Therefore,$n({I_{n - 1}} + {I_{n + 1}}) = 1$.

(A) Let $u = h(x)$. Then $du = h'(x) \, dx$.
When $x = a$,$u = h(a)$. When $x = b$,$u = h(b)$.
The integral becomes $\int_{h(a)}^{h(b)} [f(g(u))]^{-1} f'(g(u)) g'(u) \, du$.
Let $v = g(u)$. Then $dv = g'(u) \, du$.
When $u = h(a)$,$v = g(h(a))$. When $u = h(b)$,$v = g(h(b))$.
The integral becomes $\int_{g(h(a))}^{g(h(b))} [f(v)]^{-1} f'(v) \, dv$.
This is $\int_{g(h(a))}^{g(h(b))} \frac{f'(v)}{f(v)} \, dv = [\ln|f(v)|]_{g(h(a))}^{g(h(b))}$.
Since $h(a) = h(b)$,the upper and lower limits are equal: $g(h(a)) = g(h(b))$.
Therefore,the value of the integral is $0$.

(A) 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:
$I = \int_0^a f(a - x)g(a - x) dx$.
Given $f(x) = f(a - x)$ and $g(a - x) = 2 - g(x)$,we substitute these into the integral:
$I = \int_0^a f(x)[2 - g(x)] dx$.
$I = 2\int_0^a f(x) dx - \int_0^a f(x)g(x) dx$.
$I = 2\int_0^a f(x) dx - I$.
$2I = 2\int_0^a f(x) dx$.
$I = \int_0^a f(x) dx$.

(C) Let $I = \int_1^3 \sqrt{3 + x^3} \,dx$.
Since $f(x) = \sqrt{3 + x^3}$ is an increasing function on the interval $[1, 3]$,the minimum value occurs at $x = 1$ and the maximum value occurs at $x = 3$.
$f(1) = \sqrt{3 + 1^3} = \sqrt{4} = 2$.
$f(3) = \sqrt{3 + 3^3} = \sqrt{3 + 27} = \sqrt{30}$.
Using the property of definite integrals,$(b - a)f(x)_{\min} \le I \le (b - a)f(x)_{\max}$.
Here,$a = 1$ and $b = 3$,so $(b - a) = 2$.
Thus,$2 \times 2 \le I \le 2 \times \sqrt{30}$.
$4 \le I \le 2\sqrt{30}$.
Therefore,the value of the integral lies in the interval $(4, 2\sqrt{30})$.

(D) Let $I = \int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $.
Using the property $\int_0^a {f(x)dx = \int_0^a {f(a - x)dx} }$,we have:
$I = \int_0^{2\pi } {\frac{{(2\pi - x){{\sin }^{2n}}(2\pi - x)}}{{{{\sin }^{2n}}(2\pi - x) + {{\cos }^{2n}}(2\pi - x)}}dx} $
$I = \int_0^{2\pi } {\frac{{(2\pi - x){{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
Adding the two expressions for $I$:
$2I = \int_0^{2\pi } {\frac{{2\pi {{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
$I = \pi \int_0^{2\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
Using the property $\int_0^{2a} {f(x)dx = 2\int_0^a {f(x)dx} }$ if $f(2a-x) = f(x)$:
$I = 2\pi \int_0^{\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
$I = 4\pi \int_0^{\pi /2} {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $ (Equation $i$)
Similarly,$I = 4\pi \int_0^{\pi /2} {\frac{{{{\cos }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $ (Equation $ii$)
Adding $(i)$ and $(ii)$:
$2I = 4\pi \int_0^{\pi /2} {1 dx} = 4\pi \left( \frac{\pi }{2} \right) = 2{\pi ^2}$
Therefore,$I = {\pi ^2}$.

(C) Let $J = \int_{3}^{3 + 3T} f(2x) dx$. Substitute $u = 2x$,so $du = 2 dx$ or $dx = \frac{1}{2} du$.
When $x = 3, u = 6$. When $x = 3 + 3T, u = 6 + 6T$.
Thus,$J = \frac{1}{2} \int_{6}^{6 + 6T} f(u) du$.
Since $f$ is periodic with period $T$,$\int_{a}^{a + nT} f(x) dx = n \int_{0}^{T} f(x) dx$ for any $a \in \mathbb{R}$ and $n \in \mathbb{Z}$.
Here,the integral is over an interval of length $6T$,which is $6$ times the period $T$.
Therefore,$\int_{6}^{6 + 6T} f(u) du = 6 \int_{0}^{T} f(u) du = 6I$.
Substituting this back,$J = \frac{1}{2} \times 6I = 3I$.

(D) Let $I = \int_0^\pi \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} \dots (i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:
$I = \int_0^\pi \frac{(\pi - x) \, dx}{a^2 \cos^2(\pi - x) + b^2 \sin^2(\pi - x)} = \int_0^\pi \frac{(\pi - x) \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^\pi \frac{\pi \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} = \pi \int_0^\pi \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$
Since the integrand is symmetric about $\frac{\pi}{2}$,$I = \pi \int_0^{\pi/2} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$
Dividing numerator and denominator by $\cos^2 x$:
$I = \pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{a^2 + b^2 \tan^2 x}$
Let $b \tan x = t$,then $b \sec^2 x \, dx = dt \implies \sec^2 x \, dx = \frac{dt}{b}$.
As $x \to 0, t \to 0$ and as $x \to \frac{\pi}{2}, t \to \infty$.
$I = \frac{\pi}{b} \int_0^\infty \frac{dt}{a^2 + t^2} = \frac{\pi}{b} \left[ \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) \right]_0^\infty = \frac{\pi}{ab} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi^2}{2ab}$.

(B) Let $f(x) = \sin^{-1}(3x - 4x^3) - \cos^{-1}(4x^3 - 3x)$.
We know that $\cos^{-1}(y) = \pi - \sin^{-1}(y)$ for $y \in [-1, 1]$.
Thus,$f(x) = \sin^{-1}(3x - 4x^3) - [\pi - \sin^{-1}(4x^3 - 3x)]$.
Since $4x^3 - 3x = -(3x - 4x^3)$,we have $\sin^{-1}(4x^3 - 3x) = \sin^{-1}(-(3x - 4x^3)) = -\sin^{-1}(3x - 4x^3)$.
Substituting this into the expression for $f(x)$:
$f(x) = \sin^{-1}(3x - 4x^3) - \pi + \sin^{-1}(4x^3 - 3x) = \sin^{-1}(3x - 4x^3) - \pi - \sin^{-1}(3x - 4x^3) = -\pi$.
Wait,let us re-evaluate using the identity $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x$ for $x \in [-\frac{1}{2}, \frac{1}{2}]$.
And $\cos^{-1}(4x^3 - 3x) = \cos^{-1}(-(3x - 4x^3)) = \pi - \cos^{-1}(3x - 4x^3)$.
Using $\sin^{-1}u + \cos^{-1}u = \frac{\pi}{2}$,we have $\cos^{-1}(3x - 4x^3) = \frac{\pi}{2} - \sin^{-1}(3x - 4x^3)$.
So,$\cos^{-1}(4x^3 - 3x) = \pi - (\frac{\pi}{2} - \sin^{-1}(3x - 4x^3)) = \frac{\pi}{2} + \sin^{-1}(3x - 4x^3)$.
Then $f(x) = \sin^{-1}(3x - 4x^3) - (\frac{\pi}{2} + \sin^{-1}(3x - 4x^3)) = -\frac{\pi}{2}$.
Therefore,$I = \int_{-1/2}^{1/2} -\frac{\pi}{2} dx = -\frac{\pi}{2} [x]_{-1/2}^{1/2} = -\frac{\pi}{2} (\frac{1}{2} - (-\frac{1}{2})) = -\frac{\pi}{2}$.

(A) Let $I = \int\limits_0^\pi x \sin^2 x \cos x \, dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^\pi (\pi - x) \sin^2(\pi - x) \cos(\pi - x) \, dx$
Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we get:
$I = \int_0^\pi (\pi - x) \sin^2 x (-\cos x) \, dx$
$I = -\pi \int_0^\pi \sin^2 x \cos x \, dx + \int_0^\pi x \sin^2 x \cos x \, dx$
$I = -\pi \int_0^\pi \sin^2 x \cos x \, dx + I$
This implies $\pi \int_0^\pi \sin^2 x \cos x \, dx = 0$.
Let $u = \sin x$,then $du = \cos x \, dx$.
When $x=0, u=0$ and when $x=\pi, u=0$.
Thus,$\int_0^\pi \sin^2 x \cos x \, dx = \int_0^0 u^2 \, du = 0$.
Therefore,$I = 0$.

(B) Let $I = \int\limits_0^{\pi / 2n} \frac{dx}{1 + \tan^n(nx)}$.
Substitute $nx = t$,then $n \, dx = dt$,so $dx = \frac{dt}{n}$.
When $x = 0, t = 0$. When $x = \frac{\pi}{2n}, t = \frac{\pi}{2}$.
$I = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{dt}{1 + \tan^n(t)} = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{\cos^n(t)}{\sin^n(t) + \cos^n(t)} \, dt$.
Using the property $\int\limits_0^a f(t) \, dt = \int\limits_0^a f(a-t) \, dt$,we get:
$I = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{\sin^n(t)}{\cos^n(t) + \sin^n(t)} \, dt$.
Adding the two expressions for $I$:
$2I = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{\cos^n(t) + \sin^n(t)}{\cos^n(t) + \sin^n(t)} \, dt = \frac{1}{n} \int\limits_0^{\pi / 2} 1 \, dt = \frac{1}{n} \left[ t \right]_0^{\pi / 2} = \frac{\pi}{2n}$.
Therefore,$I = \frac{\pi}{4n}$.

(A) Let $I = \int\limits_{-a}^a f(x)\,dx$.
Using the property $\int\limits_a^b f(x)\,dx = \int\limits_a^b f(a+b-x)\,dx$,we have $I = \int\limits_{-a}^a f(-x)\,dx$.
Adding the two expressions for $I$:
$2I = \int\limits_{-a}^a [f(x) + f(-x)]\,dx$.
Since the integrand $g(x) = f(x) + f(-x)$ is an even function (because $g(-x) = f(-x) + f(x) = g(x)$),we can write:
$2I = 2\int\limits_0^a [f(x) + f(-x)]\,dx$.
Dividing by $2$,we get $I = \int\limits_0^a [f(x) + f(-x)]\,dx$.
Thus,the correct option is $A$.

(C) Let $f(x) = \frac{1}{2}(|x - 3| + |1 - x| - 4)$.
For $x \in [\frac{1}{2}, 1]$,$f(x) = \frac{1}{2}(3 - x + 1 - x - 4) = \frac{1}{2}(-2x) = -x$. Thus,$\{f(x)\} = \{-x\} = 1 - x$ (since $x \in [\frac{1}{2}, 1]$,$-x \in [-1, -\frac{1}{2}]$,so $\{-x\} = -x - (-1) = 1 - x$).
For $x \in [1, 3]$,$f(x) = \frac{1}{2}(3 - x + x - 1 - 4) = \frac{1}{2}(-2) = -1$. Thus,$\{f(x)\} = \{-1\} = 0$.
For $x \in [3, \frac{7}{2}]$,$f(x) = \frac{1}{2}(x - 3 + x - 1 - 4) = \frac{1}{2}(2x - 8) = x - 4$. Thus,$\{f(x)\} = \{x - 4\} = x - 4$ (since $x \in [3, \frac{7}{2}]$,$x - 4 \in [-1, -\frac{1}{2}]$,so $\{x - 4\} = x - 4 - (-1) = x - 3$).
Now,the integral is $\int_{\frac{1}{2}}^{1} (1 - x) dx + \int_{1}^{3} 0 dx + \int_{3}^{\frac{7}{2}} (x - 3) dx$.
$= [x - \frac{x^2}{2}]_{\frac{1}{2}}^{1} + 0 + [\frac{x^2}{2} - 3x]_{3}^{\frac{7}{2}}$.
$= (1 - \frac{1}{2}) - (\frac{1}{2} - \frac{1}{8}) + ((\frac{49}{8} - \frac{21}{2}) - (\frac{9}{2} - 9))$.
$= \frac{1}{2} - \frac{3}{8} + (\frac{49 - 84}{8} - (-\frac{9}{2})) = \frac{1}{8} + (-\frac{35}{8} + \frac{36}{8}) = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$.

(C) Let $I = \int_{a}^{b} x \cdot f(a + b - x) \, dx$.
Using the property $\int_{a}^{b} g(x) \, dx = \int_{a}^{b} g(a + b - x) \, dx$,we get:
$I = \int_{a}^{b} (a + b - x) \cdot f(a + b - (a + b - x)) \, dx$
$I = \int_{a}^{b} (a + b - x) \cdot f(x) \, dx$
$I = (a + b) \int_{a}^{b} f(x) \, dx - \int_{a}^{b} x \cdot f(x) \, dx$.
Since $f(a + b - x) = f(x)$,the original integral is $I = \int_{a}^{b} x \cdot f(x) \, dx$.
Substituting this into the equation:
$I = (a + b) \int_{a}^{b} f(x) \, dx - I$
$2I = (a + b) \int_{a}^{b} f(x) \, dx$
$I = \frac{a + b}{2} \int_{a}^{b} f(x) \, dx$.

(B) Let $I = \int_{0}^{a} f(x) g(x) dx$ ....$(1)$
Using the property $\int_{0}^{a} h(x) dx = \int_{0}^{a} h(a - x) dx$,we get:
$I = \int_{0}^{a} f(a - x) g(a - x) dx$
Since $f(x) = f(a - x)$ and $g(a - x) = 4 - g(x)$,we have:
$I = \int_{0}^{a} f(x) (4 - g(x)) dx$
$I = 4 \int_{0}^{a} f(x) dx - \int_{0}^{a} f(x) g(x) dx$
$I = 4 \int_{0}^{a} f(x) dx - I$
Adding $I$ to both sides:
$2I = 4 \int_{0}^{a} f(x) dx$
$I = 2 \int_{0}^{a} f(x) dx$

(A) Let $I = \int\limits_{\frac{1}{2}}^2 \frac{1}{x} \sin \left( x - \frac{1}{x} \right) dx$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a = \frac{1}{2}$ and $b = 2$,we have $a+b = \frac{5}{2}$.
Substituting $x = \frac{1}{t}$,we get $dx = -\frac{1}{t^2} dt$.
When $x = \frac{1}{2}$,$t = 2$. When $x = 2$,$t = \frac{1}{2}$.
$I = \int\limits_2^{\frac{1}{2}} t \sin \left( \frac{1}{t} - t \right) \left( -\frac{1}{t^2} \right) dt$
$I = \int\limits_2^{\frac{1}{2}} \frac{1}{t} \sin \left( t - \frac{1}{t} \right) dt$
$I = -\int\limits_{\frac{1}{2}}^2 \frac{1}{t} \sin \left( t - \frac{1}{t} \right) dt$
$I = -I$
Therefore,$2I = 0$,which implies $I = 0$.

(C) Given $I_n = \int_{1}^{e} (\ln x)^n dx$.
Using integration by parts,let $u = (\ln x)^n$ and $dv = dx$.
Then $du = n(\ln x)^{n-1} \cdot \frac{1}{x} dx$ and $v = x$.
$I_n = [x(\ln x)^n]_{1}^{e} - \int_{1}^{e} x \cdot n(\ln x)^{n-1} \cdot \frac{1}{x} dx$.
$I_n = (e(\ln e)^n - 1(\ln 1)^n) - n \int_{1}^{e} (\ln x)^{n-1} dx$.
Since $\ln e = 1$ and $\ln 1 = 0$,we have $I_n = e - n I_{n-1}$.
Replacing $n$ with $n+1$,we get $I_{n+1} = e - (n+1) I_n$.
Rearranging the terms,we get $I_{n+1} + (n+1) I_n = e$.