Properties of definite integration Questions in English

(A) Let $f(x) = a x^3 + b x$.
Since $f(-x) = a(-x)^3 + b(-x) = -(a x^3 + b x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^a f(x) dx = 0$.
Alternatively,evaluating the integral:
$\int_{-1}^1 (a x^3 + b x) dx = \left[ a \frac{x^4}{4} + b \frac{x^2}{2} \right]_{-1}^1$
$= \left( \frac{a(1)^4}{4} + \frac{b(1)^2}{2} \right) - \left( \frac{a(-1)^4}{4} + \frac{b(-1)^2}{2} \right)$
$= \left( \frac{a}{4} + \frac{b}{2} \right) - \left( \frac{a}{4} + \frac{b}{2} \right) = 0$.
This result holds true for any real values of $a$ and $b$.

(C) Given $9 > m > l > s > n > r > 2$ ... $(i)$
For $\int_{-2 \pi}^{2 \pi} \sin ^m x \cos ^n x \, dx = 4 \int_0^\pi \sin ^m x \cos ^n x \, dx$,the integral over $[-2\pi, 2\pi]$ is $4 \int_0^{\pi} f(x) dx$ only if $m$ is even. Thus,$m = 8$.
For $\int_{-\pi}^\pi \sin ^r x \cos ^s x \, dx = 4 \int_0^{\pi / 2} \sin ^r x \cos ^s x \, dx$,this holds if $r$ is even and $s$ is even. Given the constraints $9 > m > l > s > n > r > 2$ and $m=8$,we have $8 > l > s > n > r > 2$.
For $\int_{-\pi / 2}^{\pi / 2} \sin ^l x \cos ^m x \, dx = 0$,the integrand must be an odd function,which implies $l$ is odd.
With $m=8$,the remaining integers are $7, 6, 5, 4, 3$.
Since $l$ is odd and $l < 8$,$l=7$.
Since $s$ is even and $s < 7$,$s=6$.
Since $n < 6$ and $n > r > 2$,we have $n=5, r=4$.
Checking the options:
$(s-2)(s+2) = (6-2)(6+2) = 4 \times 8 = 32$.
$ln - 3 = (7 \times 5) - 3 = 35 - 3 = 32$.
Thus,$(s-2)(s+2) = ln - 3$ is correct.

(B) Let $I = \frac{3}{25} \int_0^{25 \pi} \sqrt{|\cos x - \cos^3 x|} \, dx$.
Since the function $f(x) = \sqrt{|\cos x - \cos^3 x|}$ has a period of $\pi$,we can write:
$I = \frac{3}{25} \times 25 \int_0^{\pi} \sqrt{|\cos x(1 - \cos^2 x)|} \, dx = 3 \int_0^{\pi} \sqrt{|\cos x| \sin^2 x} \, dx = 3 \int_0^{\pi} \sin x \sqrt{|\cos x|} \, dx$.
Splitting the integral at $x = \frac{\pi}{2}$:
$I = 3 \left( \int_0^{\frac{\pi}{2}} \sin x \sqrt{\cos x} \, dx + \int_{\frac{\pi}{2}}^{\pi} \sin x \sqrt{-\cos x} \, dx \right)$.
For the first part,let $u = \cos x$,$du = -\sin x \, dx$:
$3 \int_0^1 u^{1/2} \, du = 3 \left[ \frac{u^{3/2}}{3/2} \right]_0^1 = 3 \times \frac{2}{3} = 2$.
For the second part,let $u = \cos x$,$du = -\sin x \, dx$:
$3 \int_{-1}^0 \sqrt{-u} \, du = 3 \int_0^1 \sqrt{v} \, dv = 3 \left[ \frac{v^{3/2}}{3/2} \right]_0^1 = 2$.
Thus,$I = 2 + 2 = 4$.

(C) Given the integral $I = \int_{-1}^1 \frac{\log (1+x)}{1+x^2} d x$.
Split the integral at $x=0$:
$I = \int_{-1}^0 \frac{\log (1+x)}{1+x^2} d x + \int_0^1 \frac{\log (1+x)}{1+x^2} d x$.
Let $I_1 = \int_{-1}^0 \frac{\log (1+x)}{1+x^2} d x$.
Substitute $x = -t$,so $d x = -d t$. When $x = -1, t = 1$ and when $x = 0, t = 0$.
$I_1 = \int_1^0 \frac{\log (1-t)}{1+(-t)^2} (-d t) = \int_0^1 \frac{\log (1-t)}{1+t^2} d t$.
By the property of definite integrals $\int_a^b f(t) d t = \int_a^b f(x) d x$,we have $I_1 = \int_0^1 \frac{\log (1-x)}{1+x^2} d x$.
Comparing this with the given equation $\int_{-1}^1 \frac{\log (1+x)}{1+x^2} d x = \int_0^1 \frac{\log (1+x)}{1+x^2} d x + \int_0^1 f(x) d x$,we identify $f(x) = \frac{\log (1-x)}{1+x^2}$.

(D) Let $I = \int_{-1}^{3/2} |x \sin(\pi x)| dx$.
Since $x \sin(\pi x) \ge 0$ for $x \in [-1, 0]$ and $x \in [1, 3/2]$,and $x \sin(\pi x) \le 0$ for $x \in [0, 1]$,we split the integral:
$I = \int_{-1}^{0} -x \sin(\pi x) dx + \int_{0}^{1} x \sin(\pi x) dx + \int_{1}^{3/2} -x \sin(\pi x) dx$.
Using the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ for even functions,note that $f(x) = x \sin(\pi x)$ is an even function.
Thus,$I = 2 \int_{0}^{1} x \sin(\pi x) dx - \int_{1}^{3/2} x \sin(\pi x) dx$.
Using integration by parts $\int u dv = uv - \int v du$:
$\int x \sin(\pi x) dx = -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2}$.
Evaluating the first part: $2 \left[ -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{0}^{1} = 2 \left[ (\frac{1}{\pi} + 0) - (0 + 0) \right] = \frac{2}{\pi}$.
Evaluating the second part: $-\left[ -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{1}^{3/2} = -\left[ (0 - \frac{1}{\pi^2}) - (\frac{1}{\pi} + 0) \right] = \frac{1}{\pi^2} + \frac{1}{\pi}$.
Adding them together: $I = \frac{2}{\pi} + \frac{1}{\pi} + \frac{1}{\pi^2} = \frac{3}{\pi} + \frac{1}{\pi^2}$.

(C) Let $I = \int_0^{1/2} |\sin(4\pi x)| \, dx$.
Since the period of $|\sin(4\pi x)|$ is $\frac{\pi}{4\pi} = \frac{1}{4}$,we can use the property of periodic functions.
$I = 2 \int_0^{1/4} |\sin(4\pi x)| \, dx$.
In the interval $[0, 1/4]$,$\sin(4\pi x) \ge 0$,so $|\sin(4\pi x)| = \sin(4\pi x)$.
$I = 2 \int_0^{1/4} \sin(4\pi x) \, dx$.
$I = 2 \left[ -\frac{\cos(4\pi x)}{4\pi} \right]_0^{1/4}$.
$I = -\frac{1}{2\pi} [\cos(\pi) - \cos(0)]$.
$I = -\frac{1}{2\pi} [-1 - 1] = -\frac{1}{2\pi} (-2) = \frac{1}{\pi}$.

(C) Let $I = \int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x$.
We can rewrite the numerator as $100(\sin x + \cos x) + 100 \sin x$.
So,$I = \int_0^{\pi / 2} \frac{100(\sin x + \cos x) + 100 \sin x}{\sin x + \cos x} d x = 100 \int_0^{\pi / 2} 1 d x + 100 \int_0^{\pi / 2} \frac{\sin x}{\sin x + \cos x} d x$.
Let $I_1 = \int_0^{\pi / 2} \frac{\sin x}{\sin x + \cos x} d x$ ... $(i)$.
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get:
$I_1 = \int_0^{\pi / 2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} d x = \int_0^{\pi / 2} \frac{\cos x}{\cos x + \sin x} d x$ ... (ii).
Adding $(i)$ and (ii):
$2I_1 = \int_0^{\pi / 2} \frac{\sin x + \cos x}{\sin x + \cos x} d x = \int_0^{\pi / 2} 1 d x = [x]_0^{\pi / 2} = \frac{\pi}{2}$.
Thus,$I_1 = \frac{\pi}{4}$.
Now,substituting back into the expression for $I$:
$I = 100 \times [x]_0^{\pi / 2} + 100 \times I_1 = 100 \times \frac{\pi}{2} + 100 \times \frac{\pi}{4} = 50\pi + 25\pi = 75\pi$.

(B) Let $I = \int_{\frac{-3}{4}}^{\frac{\pi-6}{8}} \log (\sin (4 x+3)) \, dx$.
Substitute $t = 4x + 3$,then $dt = 4 \, dx$,which implies $dx = \frac{dt}{4}$.
When $x = \frac{-3}{4}$,$t = 4(\frac{-3}{4}) + 3 = 0$.
When $x = \frac{\pi-6}{8}$,$t = 4(\frac{\pi-6}{8}) + 3 = \frac{\pi-6}{2} + 3 = \frac{\pi}{2} - 3 + 3 = \frac{\pi}{2}$.
Thus,the integral becomes $I = \int_{0}^{\frac{\pi}{2}} \log (\sin t) \frac{dt}{4} = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \log (\sin t) \, dt$.
Using the standard definite integral property $\int_{0}^{\frac{\pi}{2}} \log (\sin t) \, dt = -\frac{\pi}{2} \log 2$,we get:
$I = \frac{1}{4} \times (-\frac{\pi}{2} \log 2) = -\frac{\pi}{8} \log 2$.

(A) Let $I = \int_{\pi/5}^{3\pi/10} \frac{dx}{\sec^2 x + (\tan^{2022} x - 1)(\sec^2 x - 1)}$.
Since $\sec^2 x = 1 + \tan^2 x$ and $\sec^2 x - 1 = \tan^2 x$,we have:
$I = \int_{\pi/5}^{3\pi/10} \frac{dx}{1 + \tan^2 x + (\tan^{2022} x - 1)\tan^2 x}$
$I = \int_{\pi/5}^{3\pi/10} \frac{dx}{1 + \tan^2 x + \tan^{2024} x - \tan^2 x} = \int_{\pi/5}^{3\pi/10} \frac{dx}{1 + \tan^{2024} x}$ ... $(i)$
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a = \pi/5$ and $b = 3\pi/10$,$a+b = \pi/2$:
$I = \int_{\pi/5}^{3\pi/10} \frac{dx}{1 + \tan^{2024}(\pi/2 - x)} = \int_{\pi/5}^{3\pi/10} \frac{dx}{1 + \cot^{2024} x}$
$I = \int_{\pi/5}^{3\pi/10} \frac{\tan^{2024} x}{1 + \tan^{2024} x} dx$ ... (ii)
Adding $(i)$ and (ii):
$2I = \int_{\pi/5}^{3\pi/10} \frac{1 + \tan^{2024} x}{1 + \tan^{2024} x} dx = \int_{\pi/5}^{3\pi/10} dx = \frac{3\pi}{10} - \frac{\pi}{5} = \frac{\pi}{10}$
Therefore,$I = \frac{\pi}{20}$.

(C) Let $I = \int_0^{\pi / 2} \frac{16 x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$ $(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^{\pi / 2} \frac{16(\frac{\pi}{2}-x) \sin(\frac{\pi}{2}-x) \cos(\frac{\pi}{2}-x)}{\sin^4(\frac{\pi}{2}-x)+\cos^4(\frac{\pi}{2}-x)} dx$
$I = \int_0^{\pi / 2} \frac{(8\pi - 16x) \cos x \sin x}{\cos^4 x + \sin^4 x} dx$ (ii)
Adding $(i)$ and (ii):
$2I = \int_0^{\pi / 2} \frac{8\pi \sin x \cos x}{\sin^4 x + \cos^4 x} dx$
$I = 4\pi \int_0^{\pi / 2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx$
Divide numerator and denominator by $\cos^4 x$:
$I = 4\pi \int_0^{\pi / 2} \frac{\tan x \sec^2 x}{\tan^4 x + 1} dx$
Let $u = \tan^2 x$,then $du = 2 \tan x \sec^2 x dx$:
$I = 2\pi \int_0^{\infty} \frac{du}{u^2 + 1} = 2\pi [\tan^{-1}(u)]_0^{\infty}$
$I = 2\pi [\frac{\pi}{2} - 0] = \pi^2$

(A) We have,$f(x) = \int_1^x \frac{1}{2+t^4} dt$.
So,$f(2) = \int_1^2 \frac{1}{2+t^4} dt$.
For $t \in [1, 2]$,the function $g(t) = \frac{1}{2+t^4}$ is a decreasing function.
Therefore,the minimum value occurs at $t = 2$ and the maximum value occurs at $t = 1$.
Thus,$\frac{1}{2+2^4} \leq \frac{1}{2+t^4} \leq \frac{1}{2+1^4}$.
$\frac{1}{18} \leq \frac{1}{2+t^4} \leq \frac{1}{3}$.
Integrating from $1$ to $2$:
$\int_1^2 \frac{1}{18} dt < \int_1^2 \frac{1}{2+t^4} dt < \int_1^2 \frac{1}{3} dt$.
$\frac{1}{18}(2-1) < f(2) < \frac{1}{3}(2-1)$.
$\frac{1}{18} < f(2) < \frac{1}{3}$.

(B) Let $I = \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5 x}{1+e^{5 x}} d x$.
Using the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx$,we have:
$I = \int_{0}^{\frac{\pi}{15}} \left( \frac{\cos 5x}{1+e^{5x}} + \frac{\cos(-5x)}{1+e^{-5x}} \right) dx$.
Since $\cos(-5x) = \cos(5x)$ and $\frac{1}{1+e^{-5x}} = \frac{e^{5x}}{e^{5x}+1}$,the expression becomes:
$I = \int_{0}^{\frac{\pi}{15}} \left( \frac{\cos 5x}{1+e^{5x}} + \frac{e^{5x} \cos 5x}{1+e^{5x}} \right) dx$.
$I = \int_{0}^{\frac{\pi}{15}} \frac{\cos 5x (1+e^{5x})}{1+e^{5x}} dx = \int_{0}^{\frac{\pi}{15}} \cos 5x dx$.
$I = \left[ \frac{\sin 5x}{5} \right]_{0}^{\frac{\pi}{15}} = \frac{1}{5} \left( \sin \frac{\pi}{3} - \sin 0 \right)$.
$I = \frac{1}{5} \left( \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{10}$.

(B) $I = \int_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx \quad ... (i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:
$I = \int_0^{\pi/2} \frac{\cos^2 x}{\cos x + \sin x} dx \quad ... (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\pi/2} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} dx = \int_0^{\pi/2} \frac{1}{\sin x + \cos x} dx$
$2I = \frac{1}{\sqrt{2}} \int_0^{\pi/2} \frac{1}{\sin(x + \pi/4)} dx = \frac{1}{\sqrt{2}} \int_0^{\pi/2} \operatorname{cosec}(x + \pi/4) dx$
$2I = \frac{1}{\sqrt{2}} [\log|\operatorname{cosec}(x + \pi/4) - \cot(x + \pi/4)|]_0^{\pi/2}$
$2I = \frac{1}{\sqrt{2}} [\log|\operatorname{cosec}(3\pi/4) - \cot(3\pi/4)| - \log|\operatorname{cosec}(\pi/4) - \cot(\pi/4)|]$
$2I = \frac{1}{\sqrt{2}} [\log(\sqrt{2} + 1) - \log(\sqrt{2} - 1)] = \frac{1}{\sqrt{2}} \log\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)$
Since $\frac{\sqrt{2} + 1}{\sqrt{2} - 1} = (\sqrt{2} + 1)^2$,we have $2I = \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1)^2 = \frac{2}{\sqrt{2}} \log(\sqrt{2} + 1)$
$I = \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1)$

(I-D, II-A, III-E, IV-C) The correct matches are as follows:
$I. \int_{-1}^1 x|x| dx = 0$ (since $f(x) = x|x|$ is an odd function,i.e.,$f(-x) = -x|-x| = -x|x| = -f(x)$). Thus,$I \rightarrow (d)$.
$II. \text{Let } I = \int_0^{\pi/2} \left(1 + \log \frac{4+3\sin x}{4+3\cos x}\right) dx$.
Using $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\pi/2} \left(1 + \log \frac{4+3\cos x}{4+3\sin x}\right) dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi/2} \left(2 + \log \left(\frac{4+3\sin x}{4+3\cos x} \cdot \frac{4+3\cos x}{4+3\sin x}\right)\right) dx = \int_0^{\pi/2} (2 + \log 1) dx = \int_0^{\pi/2} 2 dx = \pi$.
Therefore,$I = \frac{\pi}{2}$. Thus,$II \rightarrow (a)$.
$III. \int_0^a f(x) dx = \int_0^a f(a-x) dx$ is a standard property of definite integrals. Thus,$III \rightarrow (e)$.
$IV. \int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$. Let $x = -t$ in the first integral,$dx = -dt$.
$int_{-a}^0 f(x) dx = \int_a^0 f(-t) (-dt) = \int_0^a f(-t) dt = \int_0^a f(-x) dx$.
So,$\int_{-a}^a f(x) dx = \int_0^a f(-x) dx + \int_0^a f(x) dx = \int_0^a [f(x) + f(-x)] dx$. Thus,$IV \rightarrow (c)$.

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

(D) Consider the function $f(x) = \sin x$. On the interval $[0, \pi/2]$,we know that $\sin x \geq \frac{2x}{\pi}$ by Jordan's inequality.
Since $R > 0$,we have $-R \sin x \leq -R \frac{2x}{\pi}$.
Thus,$e^{-R \sin x} \leq e^{-2Rx/\pi}$.
Integrating both sides from $0$ to $\pi/2$ gives $\int_0^{\pi/2} e^{-R \sin x} dx \leq \int_0^{\pi/2} e^{-2Rx/\pi} dx = \frac{\pi}{2R}(1 - e^{-R})$.
However,the integral $I(R)$ is defined from $0$ to $R$. For large $R$,the behavior of the integral is dominated by the region near $x=0$ where $\sin x \approx x$.
By comparing the growth rates and the bounds of the integral,it is observed that for general $R > 0$,the values of $I(R)$ and the expression $\frac{\pi}{2R}(1 - e^{-R})$ do not maintain a consistent inequality for all $R$,making them not directly comparable in a simple global sense.

(C) First,observe that $f(a) + f(-a) = \frac{e^a}{1+e^a} + \frac{e^{-a}}{1+e^{-a}} = \frac{e^a}{1+e^a} + \frac{1}{e^a+1} = \frac{e^a+1}{e^a+1} = 1$.
Let $t = f(-a)$,then $f(a) = 1-t$.
Now,$l_1 = \int_{t}^{1-t} x g(x(1-x)) dx$.
Using the property $\int_{A}^{B} h(x) dx = \int_{A}^{B} h(A+B-x) dx$,we have:
$l_1 = \int_{t}^{1-t} (t + (1-t) - x) g((t + (1-t) - x)(1 - (t + (1-t) - x))) dx$
$l_1 = \int_{t}^{1-t} (1-x) g((1-x)(1-(1-x))) dx = \int_{t}^{1-t} (1-x) g(x(1-x)) dx$.
Adding the two expressions for $l_1$:
$2l_1 = \int_{t}^{1-t} (x + 1 - x) g(x(1-x)) dx = \int_{t}^{1-t} g(x(1-x)) dx$.
Since $l_2 = \int_{t}^{1-t} g(x(1-x)) dx$,we get $2l_1 = l_2$.
Therefore,$\frac{l_2}{l_1} = 2$.

(C) Let $I = \int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$.
Substitute $t = 2a - 1 - x$,then $dt = -dx$.
When $t = a-1$,$x = a$. When $t = a$,$x = a-1$.
$I = \int_{a}^{a-1} \frac{e^{-(2a-1-x)}}{(2a-1-x)-a-1} (-dx) = \int_{a-1}^{a} \frac{e^{x-2a+1}}{a-x-2} dx$.
This substitution does not simplify directly to the form of $b$.
Alternatively,let $u = t - a + 1$,then $t = u + a - 1$ and $dt = du$.
When $t = a-1$,$u = 0$. When $t = a$,$u = 1$.
$I = \int_{0}^{1} \frac{e^{-(u+a-1)}}{(u+a-1)-a-1} du = \int_{0}^{1} \frac{e^{-u-a+1}}{u-2} du$.
Given $b = \int_{0}^{1} \frac{e^{t}}{t+1} dt$.
Evaluating the integral $\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$ using substitution $u = t - a + 1$ leads to $-b e^{-a}$.

(B) Let $I = \int_{1/2014}^{2014} \frac{\tan^{-1} x}{x} dx$ $(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(\frac{ab}{x}) dx$ is not directly applicable,but we use the substitution $x = \frac{1}{t}$,so $dx = -\frac{1}{t^2} dt$.
When $x = 1/2014$,$t = 2014$. When $x = 2014$,$t = 1/2014$.
$I = \int_{2014}^{1/2014} \frac{\tan^{-1}(1/t)}{1/t} (-\frac{1}{t^2}) dt = \int_{1/2014}^{2014} \frac{\cot^{-1} t}{t} dt = \int_{1/2014}^{2014} \frac{\cot^{-1} x}{x} dx$ $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{1/2014}^{2014} \frac{\tan^{-1} x + \cot^{-1} x}{x} dx$
Since $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$:
$2I = \int_{1/2014}^{2014} \frac{\pi/2}{x} dx = \frac{\pi}{2} [\ln x]_{1/2014}^{2014}$
$2I = \frac{\pi}{2} (\ln 2014 - \ln(1/2014)) = \frac{\pi}{2} (\ln 2014 + \ln 2014) = \frac{\pi}{2} (2 \ln 2014) = \pi \ln 2014$
$I = \frac{\pi}{2} \log 2014$.

(B) Let $I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)} + e^{\tan^{-1}(\cos x)}} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = \frac{\pi}{2} + \frac{5\pi}{2} = 3\pi$.
$I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan^{-1}(\sin(3\pi - x))}}{e^{\tan^{-1}(\sin(3\pi - x))} + e^{\tan^{-1}(\cos(3\pi - x))}} dx$.
Since $\sin(3\pi - x) = \sin x$ and $\cos(3\pi - x) = -\cos x$,this does not simplify directly.
Let us use the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ on the interval $[0, 2\pi]$.
The integrand $f(x) = \frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)} + e^{\tan^{-1}(\cos x)}}$ has a period of $\frac{\pi}{2}$.
$I = \int_{\pi/2}^{5\pi/2} f(x) dx = \int_{0}^{2\pi} f(x) dx$.
Since $f(x)$ is periodic with period $\frac{\pi}{2}$,$\int_{0}^{2\pi} f(x) dx = 4 \int_{0}^{\pi/2} f(x) dx$.
Let $J = \int_{0}^{\pi/2} \frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)} + e^{\tan^{-1}(\cos x)}} dx$.
Using $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,$J = \int_{0}^{\pi/2} \frac{e^{\tan^{-1}(\cos x)}}{e^{\tan^{-1}(\cos x)} + e^{\tan^{-1}(\sin x)}} dx$.
$2J = \int_{0}^{\pi/2} 1 dx = \frac{\pi}{2} \implies J = \frac{\pi}{4}$.
Thus,$I = 4 \times \frac{\pi}{4} = \pi$.

(B) Let $I = \int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x$.
Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,we get:
$I = \int_{0}^{1} \log \left(\frac{1-(1-x)}{1-x}\right) d x$
$I = \int_{0}^{1} \log \left(\frac{x}{1-x}\right) d x$
$I = \int_{0}^{1} \log \left(\left(\frac{1-x}{x}\right)^{-1}\right) d x$
$I = -\int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x$
$I = -I$
$2I = 0 \implies I = 0$.

(D) Let $I = \int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\pi / 2} \frac{1}{1+(\tan(\frac{\pi}{2}-x))^{-101}} dx$
$I = \int_{0}^{\pi / 2} \frac{1}{1+(\cot x)^{-101}} dx$
$I = \int_{0}^{\pi / 2} \frac{1}{1+(\frac{1}{\tan x})^{-101}} dx = \int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{101}} dx$.
Adding the two expressions for $I$:
$2I = \int_{0}^{\pi / 2} (\frac{1}{1+(\tan x)^{-101}} + \frac{1}{1+(\tan x)^{101}}) dx$
$2I = \int_{0}^{\pi / 2} (\frac{(\tan x)^{101}}{(\tan x)^{101}+1} + \frac{1}{1+(\tan x)^{101}}) dx$
$2I = \int_{0}^{\pi / 2} \frac{1+(\tan x)^{101}}{1+(\tan x)^{101}} dx = \int_{0}^{\pi / 2} 1 dx = \frac{\pi}{2}$.
Therefore,$I = \frac{\pi}{4}$.

(C) Let $I = \int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\pi / 2} \log \left(\frac{4+3 \sin(\pi/2 - x)}{4+3 \cos(\pi/2 - x)}\right) d x$
$I = \int_0^{\pi / 2} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi / 2} \left[ \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) + \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) \right] d x$
$2I = \int_0^{\pi / 2} \log \left( \frac{4+3 \sin x}{4+3 \cos x} \times \frac{4+3 \cos x}{4+3 \sin x} \right) d x$
$2I = \int_0^{\pi / 2} \log(1) d x = \int_0^{\pi / 2} 0 d x = 0$.
Therefore,$I = 0$.

(B) Let $I = \int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} dx$.
We can split the integral into two parts: $I = \int_{-1}^1 \frac{x^3}{x^2+2|x|+1} dx + \int_{-1}^1 \frac{|x|+1}{x^2+2|x|+1} dx$.
Let $f(x) = \frac{x^3}{x^2+2|x|+1}$. Since $f(-x) = \frac{(-x)^3}{(-x)^2+2|-x|+1} = -\frac{x^3}{x^2+2|x|+1} = -f(x)$,the function is odd. Therefore,$\int_{-1}^1 f(x) dx = 0$.
Now,consider the second part $I_2 = \int_{-1}^1 \frac{|x|+1}{x^2+2|x|+1} dx$. Since the integrand is an even function,$I_2 = 2 \int_0^1 \frac{x+1}{x^2+2x+1} dx$.
Simplifying the integrand: $I_2 = 2 \int_0^1 \frac{x+1}{(x+1)^2} dx = 2 \int_0^1 \frac{1}{x+1} dx$.
Evaluating the integral: $I_2 = 2 [\ln |x+1|]_0^1 = 2 (\ln 2 - \ln 1) = 2 \ln 2$.
Thus,$I = 0 + 2 \ln 2 = 2 \ln 2$.

(B) Let $I = \int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ $(1)$
Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,we get:
$I = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{9-(9-x)}+\sqrt{9-x}} d x$
$I = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$ $(2)$
Adding equations $(1)$ and $(2)$:
$2I = \int_3^6 \frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} d x$
$2I = \int_3^6 1 d x$
$2I = [x]_3^6 = 6-3 = 3$
$I = \frac{3}{2}$

(C) Let $f(x) = \frac{x+x^3+x^5}{1+x^2+x^4+x^6}$.
Check if $f(x)$ is an odd function by evaluating $f(-x)$:
$f(-x) = \frac{(-x)+(-x)^3+(-x)^5}{1+(-x)^2+(-x)^4+(-x)^6} = \frac{-x-x^3-x^5}{1+x^2+x^4+x^6} = -\left(\frac{x+x^3+x^5}{1+x^2+x^4+x^6}\right) = -f(x)$.
Since $f(-x) = -f(x)$,the function is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) dx = 0$.
Therefore,$\int_{-100}^{100} \frac{x+x^3+x^5}{1+x^2+x^4+x^6} dx = 0$.

(B) Using the Leibniz rule for differentiation under the integral sign:
$f'(x) = \sin^{-1}(\sqrt{\sin^2 x}) \cdot \frac{d}{dx}(\sin^2 x) = x \cdot (2 \sin x \cos x) = x \sin(2x)$
$g'(x) = \cos^{-1}(\sqrt{\cos^2 x}) \cdot \frac{d}{dx}(\cos^2 x) = x \cdot (-2 \cos x \sin x) = -x \sin(2x)$
Thus,$f'(x) + g'(x) = x \sin(2x) - x \sin(2x) = 0$.
Since the derivative is zero,$f(x) + g(x) = C$ (a constant).
To find $C$,substitute $x = \frac{\pi}{4}$:
$f(\frac{\pi}{4}) = \int_0^{1/2} \sin^{-1} \sqrt{t} \, dt$
$g(\frac{\pi}{4}) = \int_0^{1/2} \cos^{-1} \sqrt{t} \, dt$
$f(\frac{\pi}{4}) + g(\frac{\pi}{4}) = \int_0^{1/2} (\sin^{-1} \sqrt{t} + \cos^{-1} \sqrt{t}) \, dt$
Since $\sin^{-1} \sqrt{t} + \cos^{-1} \sqrt{t} = \frac{\pi}{2}$ for $t \in [0, 1]$:
$C = \int_0^{1/2} \frac{\pi}{2} \, dt = \frac{\pi}{2} [t]_0^{1/2} = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}$.

(A) Let $I = \int_0^{\pi / 2} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x} + (\sin x)^{\cos x}} dx$ --- $(1)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\pi / 2} \frac{(\cos(\pi/2 - x))^{\sin(\pi/2 - x)}}{(\cos(\pi/2 - x))^{\sin(\pi/2 - x)} + (\sin(\pi/2 - x))^{\cos(\pi/2 - x)}} dx$
$I = \int_0^{\pi / 2} \frac{(\sin x)^{\cos x}}{(\sin x)^{\cos x} + (\cos x)^{\sin x}} dx$ --- $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_0^{\pi / 2} \frac{(\cos x)^{\sin x} + (\sin x)^{\cos x}}{(\cos x)^{\sin x} + (\sin x)^{\cos x}} dx$
$2I = \int_0^{\pi / 2} 1 dx$
$2I = [x]_0^{\pi / 2} = \pi / 2$
$I = \pi / 4$

(C) Let $I = \int_{-1 / 2}^{1 / 2} \sqrt{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2} d x$.
Using the identity $a^2 + b^2 - 2 = (a-b)^2$,we have $\sqrt{(a-b)^2} = |a-b|$.
Thus,$I = \int_{-1 / 2}^{1 / 2} \left| \frac{x+1}{x-1} - \frac{x-1}{x+1} \right| d x$.
Simplifying the expression inside the absolute value: $\frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} = \frac{4x}{x^2-1}$.
So,$I = \int_{-1 / 2}^{1 / 2} \left| \frac{4x}{x^2-1} \right| d x$.
Since the integrand is an even function,$I = 2 \int_{0}^{1 / 2} \left| \frac{4x}{x^2-1} \right| d x$.
For $x \in [0, 1/2]$,$x^2-1 < 0$,so $|\frac{4x}{x^2-1}| = -\frac{4x}{x^2-1} = \frac{4x}{1-x^2}$.
$I = 2 \int_{0}^{1 / 2} \frac{4x}{1-x^2} d x = 4 \int_{0}^{1 / 2} \frac{2x}{1-x^2} d x$.
Let $u = 1-x^2$,then $du = -2x dx$.
$I = 4 [-\ln|1-x^2|]_{0}^{1/2} = 4 [-\ln(3/4) + \ln(1)] = 4 \ln(4/3)$.

(D) Let $I = \sum_{n=1}^{10} \int_{-2n-1}^{-2n} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$.
Consider the first integral term: $J_n = \int_{-2n-1}^{-2n} \sin^{27} x \, dx$.
Let $x = -t$,then $dx = -dt$. When $x = -2n-1$,$t = 2n+1$. When $x = -2n$,$t = 2n$.
So,$J_n = \int_{2n+1}^{2n} \sin^{27}(-t) (-dt) = \int_{2n}^{2n+1} -\sin^{27} t \, dt = -\int_{2n}^{2n+1} \sin^{27} t \, dt$.
Substituting this back into the sum,we get:
$I = \sum_{n=1}^{10} (-\int_{2n}^{2n+1} \sin^{27} x \, dx) + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$.
$I = -\sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx = 0$.

(B) Let $I = \int_{-\pi/4}^{\pi/4} (\lambda|\sin x| + \frac{\mu \sin x}{1+\cos x} + \gamma) \, dx$.
We can split the integral into three parts:
$I = \lambda \int_{-\pi/4}^{\pi/4} |\sin x| \, dx + \mu \int_{-\pi/4}^{\pi/4} \frac{\sin x}{1+\cos x} \, dx + \gamma \int_{-\pi/4}^{\pi/4} 1 \, dx$.
Consider the second integral $J = \int_{-\pi/4}^{\pi/4} \frac{\sin x}{1+\cos x} \, dx$.
Let $f(x) = \frac{\sin x}{1+\cos x}$.
Then $f(-x) = \frac{\sin(-x)}{1+\cos(-x)} = \frac{-\sin x}{1+\cos x} = -f(x)$.
Since $f(x)$ is an odd function and the interval $[-\pi/4, \pi/4]$ is symmetric about $0$,the integral $J = 0$.
Thus,$I = \lambda \int_{-\pi/4}^{\pi/4} |\sin x| \, dx + \gamma \int_{-\pi/4}^{\pi/4} 1 \, dx$.
Since the term involving $\mu$ vanishes,the value of the integral is independent of $\mu$.

(D) Let $I = \int_{-1}^{1} \left\{ \frac{x^{2015}}{e^{|x|}(x^2 + \cos x)} + \frac{1}{e^{|x|}} \right\} dx$.
Split the integral into two parts: $I = \int_{-1}^{1} \frac{x^{2015}}{e^{|x|}(x^2 + \cos x)} dx + \int_{-1}^{1} \frac{1}{e^{|x|}} dx$.
Let $f(x) = \frac{x^{2015}}{e^{|x|}(x^2 + \cos x)}$ and $g(x) = \frac{1}{e^{|x|}}$.
Check for symmetry: $f(-x) = \frac{(-x)^{2015}}{e^{|-x|}((-x)^2 + \cos(-x))} = \frac{-x^{2015}}{e^{|x|}(x^2 + \cos x)} = -f(x)$. Thus,$f(x)$ is an odd function.
$g(-x) = \frac{1}{e^{|-x|}} = \frac{1}{e^{|x|}} = g(x)$. Thus,$g(x)$ is an even function.
Since $f(x)$ is odd,$\int_{-1}^{1} f(x) dx = 0$.
Since $g(x)$ is even,$\int_{-1}^{1} g(x) dx = 2 \int_{0}^{1} g(x) dx$.
Therefore,$I = 0 + 2 \int_{0}^{1} e^{-x} dx = 2 [-e^{-x}]_{0}^{1}$.
$I = 2 (-e^{-1} - (-e^{0})) = 2(1 - e^{-1})$.

(D) Given,$M = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{(x+1)^{2}} dx$.
Using the identity $\sin x \cos x = \frac{1}{2} \sin 2x$,we have $N = \int_{0}^{\frac{\pi}{4}} \frac{\sin 2x}{2(x+1)^{2}} dx$.
Let $2x = t$,then $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=\frac{\pi}{2}$.
Substituting these into $N$,we get $N = \int_{0}^{\frac{\pi}{2}} \frac{\sin t}{2(\frac{t}{2}+1)^{2}} \cdot \frac{dt}{2} = \int_{0}^{\frac{\pi}{2}} \frac{\sin t}{(t+2)^{2}} dt$.
Now,$M - N = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{x+2} dx - \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(x+2)^{2}} dx$.
Using integration by parts on the first integral,let $u = \frac{1}{x+2}$ and $dv = \cos x dx$. Then $du = -\frac{1}{(x+2)^{2}} dx$ and $v = \sin x$.
$M = \left[ \frac{\sin x}{x+2} \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \sin x \left( -\frac{1}{(x+2)^{2}} \right) dx = \left[ \frac{\sin x}{x+2} \right]_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(x+2)^{2}} dx$.
Thus,$M - N = \left[ \frac{\sin x}{x+2} \right]_{0}^{\frac{\pi}{2}} = \frac{\sin(\pi/2)}{\pi/2 + 2} - \frac{\sin(0)}{0+2} = \frac{1}{\frac{\pi+4}{2}} = \frac{2}{\pi+4}$.

(D) Given,$M = \int_{0}^{\pi / 2} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} dx$.
Using the identity $\sin x \cos x = \frac{1}{2} \sin 2x$,we have $N = \int_{0}^{\pi / 4} \frac{\sin 2x}{2(x+1)^{2}} dx$.
Let $2x = t$,then $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=\pi/4, t=\pi/2$.
$N = \int_{0}^{\pi / 2} \frac{\sin t}{2(t/2 + 1)^{2}} \cdot \frac{dt}{2} = \int_{0}^{\pi / 2} \frac{\sin t}{2(\frac{t+2}{2})^{2}} dt = \int_{0}^{\pi / 2} \frac{\sin t}{(t+2)^{2}} dt = \int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} dx$.
Now,$M - N = \int_{0}^{\pi / 2} \left( \frac{\cos x}{x+2} - \frac{\sin x}{(x+2)^{2}} \right) dx$.
Using integration by parts on $\int \frac{\cos x}{x+2} dx$ with $u = \frac{1}{x+2}$ and $dv = \cos x dx$,we get $du = -\frac{1}{(x+2)^{2}} dx$ and $v = \sin x$.
$M = \left[ \frac{\sin x}{x+2} \right]_{0}^{\pi / 2} - \int_{0}^{\pi / 2} \sin x \left( -\frac{1}{(x+2)^{2}} \right) dx = \left[ \frac{\sin x}{x+2} \right]_{0}^{\pi / 2} + \int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} dx$.
Thus,$M - \int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} dx = \left[ \frac{\sin x}{x+2} \right]_{0}^{\pi / 2} = \frac{\sin(\pi/2)}{\pi/2 + 2} - \frac{\sin(0)}{0+2} = \frac{1}{\frac{\pi+4}{2}} = \frac{2}{\pi+4}$.

(D) Let $I = \int_{-1}^{1} \left\{ \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)} + \frac{1}{e^{|x|}} \right\} dx$.
We can split the integral into two parts:
$I = \int_{-1}^{1} \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)} dx + \int_{-1}^{1} \frac{1}{e^{|x|}} dx$.
Let $f(x) = \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)}$. Since $f(-x) = \frac{(-x)^{2013}}{e^{|-x|}((-x)^{2}+\cos(-x))} = \frac{-x^{2013}}{e^{|x|}(x^{2}+\cos x)} = -f(x)$,the function $f(x)$ is an odd function. Therefore,$\int_{-1}^{1} f(x) dx = 0$.
Let $g(x) = \frac{1}{e^{|x|}} = e^{-|x|}$. Since $g(-x) = e^{-|-x|} = e^{-|x|} = g(x)$,the function $g(x)$ is an even function. Therefore,$\int_{-1}^{1} g(x) dx = 2 \int_{0}^{1} e^{-x} dx$.
Calculating the integral: $2 \int_{0}^{1} e^{-x} dx = 2 [-e^{-x}]_{0}^{1} = 2 (-e^{-1} - (-e^{0})) = 2(1 - e^{-1})$.
Thus,$I = 0 + 2(1 - e^{-1}) = 2(1 - e^{-1})$.

(D) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [\sin x \cos x] dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [\frac{1}{2} \sin 2x] dx$.
Substitute $\theta = 2x$,so $d\theta = 2dx$ or $dx = \frac{1}{2} d\theta$.
When $x = -\frac{\pi}{2}$,$\theta = -\pi$. When $x = \frac{\pi}{2}$,$\theta = \pi$.
Thus,$I = \frac{1}{2} \int_{-\pi}^{\pi} [\frac{1}{2} \sin \theta] d\theta$.
Since $[\frac{1}{2} \sin \theta]$ is an odd function,we check the intervals:
For $\theta \in [-\pi, 0]$,$\sin \theta \in [-1, 0]$,so $\frac{1}{2} \sin \theta \in [-\frac{1}{2}, 0]$,hence $[\frac{1}{2} \sin \theta] = -1$ for $\theta \in [-\pi, 0)$.
For $\theta \in [0, \pi]$,$\sin \theta \in [0, 1]$,so $\frac{1}{2} \sin \theta \in [0, \frac{1}{2}]$,hence $[\frac{1}{2} \sin \theta] = 0$ for $\theta \in [0, \pi]$.
Therefore,$I = \frac{1}{2} [\int_{-\pi}^{0} (-1) d\theta + \int_{0}^{\pi} (0) d\theta] = \frac{1}{2} [-\theta]_{-\pi}^{0} = \frac{1}{2} [0 - (\pi)] = -\frac{\pi}{2}$.

(C) Let $I = \int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$.
We can split the integral as:
$I = \int_{-2}^{2} e^{|x|} d x + 2 \int_{-2}^{2} \sin x e^{|x|} d x$.
Consider the first part: $f(x) = e^{|x|}$. Since $f(-x) = e^{|-x|} = e^{|x|} = f(x)$,$f(x)$ is an even function.
Thus,$\int_{-2}^{2} e^{|x|} d x = 2 \int_{0}^{2} e^{x} d x = 2[e^{x}]_{0}^{2} = 2(e^{2}-1)$.
Consider the second part: $g(x) = \sin x e^{|x|}$. Since $g(-x) = \sin(-x) e^{|-x|} = -\sin x e^{|x|} = -g(x)$,$g(x)$ is an odd function.
Thus,$\int_{-2}^{2} \sin x e^{|x|} d x = 0$.
Therefore,$I = 2(e^{2}-1) + 2(0) = 2(e^{2}-1)$.

(D) Let $I = \int_{-2}^2(x \cos x+\sin x+1) d x$.
We can split the integral as $I = \int_{-2}^2 x \cos x d x + \int_{-2}^2 \sin x d x + \int_{-2}^2 1 d x$.
Recall the property of definite integrals: $\int_{-a}^a f(x) d x = 0$ if $f(x)$ is an odd function,and $2 \int_0^a f(x) d x$ if $f(x)$ is an even function.
Let $f_1(x) = x \cos x$. Since $f_1(-x) = (-x) \cos(-x) = -x \cos x = -f_1(x)$,$f_1(x)$ is an odd function. Thus,$\int_{-2}^2 x \cos x d x = 0$.
Let $f_2(x) = \sin x$. Since $f_2(-x) = \sin(-x) = -\sin x = -f_2(x)$,$f_2(x)$ is an odd function. Thus,$\int_{-2}^2 \sin x d x = 0$.
Therefore,$I = 0 + 0 + \int_{-2}^2 1 d x = [x]_{-2}^2 = 2 - (-2) = 4$.

(A) Let $I = \int_0^\pi \sin^{50} x \cos^{49} x \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:
$I = \int_0^\pi \sin^{50}(\pi - x) \cos^{49}(\pi - x) \, dx$.
Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we get:
$I = \int_0^\pi \sin^{50} x (-\cos x)^{49} \, dx$.
$I = -\int_0^\pi \sin^{50} x \cos^{49} x \, dx$.
$I = -I$.
Adding $I$ to both sides,we get $2I = 0$,which implies $I = 0$.

(B) Given the integral $I = \int_{-\pi / 2}^{\pi / 2} |\sin x| \, dx$.
Since $|\sin x|$ is an even function,we can use the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$.
Thus,$I = 2 \int_{0}^{\pi / 2} |\sin x| \, dx$.
In the interval $[0, \pi / 2]$,$\sin x \geq 0$,so $|\sin x| = \sin x$.
Therefore,$I = 2 \int_{0}^{\pi / 2} \sin x \, dx$.
Evaluating the integral: $I = 2 [-\cos x]_{0}^{\pi / 2}$.
$I = 2 [-\cos(\pi / 2) - (-\cos(0))]$.
$I = 2 [0 - (-1)] = 2(1) = 2$.

(C) Let $I = \int_0^a x f(x) dx$.
Using the property $\int_0^a g(x) dx = \int_0^a g(a-x) dx$,we get:
$I = \int_0^a (a-x) f(a-x) dx$.
Since $f(a-x) = f(x)$,this becomes:
$I = \int_0^a (a-x) f(x) dx = \int_0^a a f(x) dx - \int_0^a x f(x) dx$.
$I = a \int_0^a f(x) dx - I$.
Adding $I$ to both sides,we get:
$2I = a \int_0^a f(x) dx$.
Therefore,$I = \frac{a}{2} \int_0^a f(x) dx$.

(D) Given that $\int_{-1}^4 f(x) dx = 4$ and $\int_2^4 \{3 - f(x)\} dx = 7$.
First,solve the second integral:
$\int_2^4 3 dx - \int_2^4 f(x) dx = 7$
$[3x]_2^4 - \int_2^4 f(x) dx = 7$
$3(4 - 2) - \int_2^4 f(x) dx = 7$
$6 - \int_2^4 f(x) dx = 7$
$\int_2^4 f(x) dx = 6 - 7 = -1$.
Now,use the property of definite integrals:
$\int_{-1}^4 f(x) dx = \int_{-1}^2 f(x) dx + \int_2^4 f(x) dx$
$4 = \int_{-1}^2 f(x) dx + (-1)$
$\int_{-1}^2 f(x) dx = 4 + 1 = 5$.

(B, D) For a periodic function $f(x)$ with period $T$,the integral over any interval of length $T$ is constant,i.e.,$\int_a^{a+T} f(x) dx = \int_0^T f(x) dx$. Thus,statement $(B)$ is true and $(A)$ is false.
For the Dirichlet function $f(x) = \begin{cases} 1, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$,$f(x+T) = f(x)$ holds for any rational $T$ because if $x$ is rational,$x+T$ is rational,and if $x$ is irrational,$x+T$ is irrational. Thus,$f$ is periodic for all rational $T$. Statement $(D)$ is true.

(B) For a periodic function $f(x)$ with period $T$,the integral over any interval of length $T$ is constant.
Let $I(a) = \int_{a}^{a+T} f(x) \, dx$.
Differentiating with respect to $a$ using the Leibniz rule:
$\frac{dI}{da} = f(a+T) \cdot \frac{d}{da}(a+T) - f(a) \cdot \frac{d}{da}(a)$
Since $f(x)$ is periodic with period $T$,$f(a+T) = f(a)$.
Thus,$\frac{dI}{da} = f(a) - f(a) = 0$.
Since the derivative is $0$,$I$ is independent of $a$ and $I = \int_{0}^{T} f(x) \, dx$.

(C) Let $I = \int_{0}^{100} e^{x-[x]} d x$.
Since $f(x) = x - [x]$ is a periodic function with period $T = 1$,we can use the property $\int_{0}^{nT} f(x) d x = n \int_{0}^{T} f(x) d x$.
Here,$n = 100$ and $T = 1$,so $I = 100 \int_{0}^{1} e^{x-[x]} d x$.
For $0 < x < 1$,the greatest integer function $[x] = 0$,so $x - [x] = x$.
Thus,$I = 100 \int_{0}^{1} e^{x} d x$.
Evaluating the integral,we get $I = 100 [e^{x}]_{0}^{1}$.
$I = 100 (e^{1} - e^{0}) = 100 (e - 1)$.

(B) Given $I = \int_{0}^{100 \pi} \sqrt{1 - \cos 2x} \, dx$.
Using the identity $1 - \cos 2x = 2 \sin^2 x$,we have $I = \int_{0}^{100 \pi} \sqrt{2 \sin^2 x} \, dx$.
$I = \sqrt{2} \int_{0}^{100 \pi} |\sin x| \, dx$.
Since $|\sin x|$ is a periodic function with period $\pi$,we can write $I = \sqrt{2} \times 100 \int_{0}^{\pi} |\sin x| \, dx$.
In the interval $[0, \pi]$,$\sin x \geq 0$,so $|\sin x| = \sin x$.
$I = 100 \sqrt{2} \int_{0}^{\pi} \sin x \, dx$.
$I = 100 \sqrt{2} [-\cos x]_{0}^{\pi}$.
$I = 100 \sqrt{2} [-\cos \pi - (-\cos 0)]$.
$I = 100 \sqrt{2} [-(-1) - (-1)]$.
$I = 100 \sqrt{2} [1 + 1] = 100 \sqrt{2} \times 2 = 200 \sqrt{2}$.

(C) The function $f(x) = |\sin x|$ is a periodic function with a period of $\pi$.
We know that $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$ for a periodic function with period $T$.
Here,the interval length is $16\pi - \pi = 15\pi$.
Thus,$\int_{\pi}^{16\pi} |\sin x| dx = 15 \int_{0}^{\pi} |\sin x| dx$.
Since $\sin x \ge 0$ for $x \in [0, \pi]$,we have $|\sin x| = \sin x$.
Therefore,$15 \int_{0}^{\pi} \sin x dx = 15 [-\cos x]_{0}^{\pi}$.
$= 15 [-\cos(\pi) - (-\cos(0))] = 15 [-(-1) - (-1)] = 15 [1 + 1] = 15 \times 2 = 30$.

(C) Let $g(x) = f(\cos^2 x)$.
Since $\cos^2(x + \pi) = (-\cos x)^2 = \cos^2 x$,the function $g(x)$ is periodic with period $\pi$.
Using the property of definite integrals for periodic functions,$\int_0^{n T} g(x) dx = n \int_0^T g(x) dx$,where $T$ is the period.
Here,$T = \pi$ and $n = 3$.
Therefore,$I_1 = \int_0^{3 \pi} f(\cos^2 x) dx = 3 \int_0^\pi f(\cos^2 x) dx$.
Since $I_2 = \int_0^\pi f(\cos^2 x) dx$,we have $I_1 = 3 I_2$.