Properties of definite integration Questions in English

(D) Let $I = \int_{0}^{1} \tan^{-1}\left(\frac{\tan x - 2\cot x}{3}\right) dx$.
Using the identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we can rewrite the integrand.
Note that $\frac{\tan x - 2\cot x}{3} = \frac{\tan x - 2/\tan x}{1 + (\tan x)(2/\tan x)} = \frac{\tan x - 2/\tan x}{1 + 2}$.
Thus,$\tan^{-1}\left(\frac{\tan x - 2\cot x}{3}\right) = \tan^{-1}(\tan x) - \tan^{-1}(2\cot x) = x - \tan^{-1}(2\cot x)$.
However,a simpler approach is to use $\tan^{-1}(\frac{\tan x}{2}) - \tan^{-1}(\cot x) = tan^{-1}\left(\frac{\frac{\tan x}{2} - \cot x}{1 + \frac{\tan x}{2} \cot x}\right) = \tan^{-1}\left(\frac{\tan x - 2\cot x}{2(1 + 1/2)}\right) = \tan^{-1}\left(\frac{\tan x - 2\cot x}{3}\right)$.
Therefore,the integral is $\int_{0}^{1} \tan^{-1}\left(\frac{\tan x}{2}\right) dx - \int_{0}^{1} \tan^{-1}(\cot x) dx$.
Given $\int_{0}^{1} \tan^{-1}\left(\frac{\tan x}{2}\right) dx = \alpha$,we have $I = \alpha - \int_{0}^{1} (\frac{\pi}{2} - x) dx$.
$I = \alpha - [\frac{\pi}{2}x - \frac{x^2}{2}]_{0}^{1} = \alpha - (\frac{\pi}{2} - \frac{1}{2}) = \alpha - \frac{\pi}{2} + \frac{1}{2}$.

(A) Let $I = \int_{-3}^{1} (2(t+1)^5 - 5(t+1)^3 + t + 3) dt$.
Substitute $z = t+1$,then $dz = dt$.
When $t = -3$,$z = -2$. When $t = 1$,$z = 2$.
$I = \int_{-2}^{2} (2z^5 - 5z^3 + (z-1) + 3) dz$
$I = \int_{-2}^{2} (2z^5 - 5z^3 + z + 2) dz$
Since $f(z) = 2z^5 - 5z^3 + z$ is an odd function,$\int_{-2}^{2} (2z^5 - 5z^3 + z) dz = 0$.
Therefore,$I = \int_{-2}^{2} 2 dz = [2z]_{-2}^{2} = 2(2 - (-2)) = 2(4) = 8$.

(B) Given $P_n = \int_1^e (\ln x)^n dx$.
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 $v = x$.
$P_n = [x(\ln x)^n]_1^e - \int_1^e x \cdot n(\ln x)^{n-1} \cdot \frac{1}{x} dx$.
$P_n = e(1)^n - 0 - n \int_1^e (\ln x)^{n-1} dx$.
$P_n = e - n P_{n-1}$.
Now,we need to find $P_{10} - 90P_8$.
From the recurrence relation:
$P_{10} = e - 10P_9$.
$P_9 = e - 9P_8$.
Substitute $P_9$ into the expression for $P_{10}$:
$P_{10} = e - 10(e - 9P_8)$.
$P_{10} = e - 10e + 90P_8$.
$P_{10} = -9e + 90P_8$.
Therefore,$P_{10} - 90P_8 = -9e$.

(B) We are given the expression $I = \int_{0}^{\sqrt{3}} (x+4)^2 e^{x^2} dx + \int_{\sqrt{3}}^{0} (x-4)^2 e^{x^2} dx$.
Using the property $\int_{b}^{a} f(x) dx = -\int_{a}^{b} f(x) dx$,we can rewrite the second integral as $-\int_{0}^{\sqrt{3}} (x-4)^2 e^{x^2} dx$.
Thus,$I = \int_{0}^{\sqrt{3}} e^{x^2} \{(x+4)^2 - (x-4)^2\} dx$.
Expanding the terms inside the bracket: $(x+4)^2 - (x-4)^2 = (x^2 + 8x + 16) - (x^2 - 8x + 16) = 16x$.
So,$I = \int_{0}^{\sqrt{3}} e^{x^2} (16x) dx$.
Let $u = x^2$,then $du = 2x dx$,which implies $8 du = 16x dx$.
When $x=0, u=0$. When $x=\sqrt{3}, u=3$.
$I = \int_{0}^{3} 8 e^u du = 8 [e^u]_{0}^{3} = 8(e^3 - e^0) = 8(e^3 - 1)$.

(B) Let $I = \int_{-\infty}^{\infty} f\left(x - \frac{1}{x}\right) dx$.
We use the property $\int_{-\infty}^{\infty} g\left(x - \frac{1}{x}\right) dx = \int_{-\infty}^{\infty} g(u) du$ for a suitable transformation.
Let $u = x - \frac{1}{x}$. Then $du = (1 + \frac{1}{x^2}) dx$.
This integral is a standard result in definite integration involving the substitution $x - \frac{1}{x} = u$.
Specifically,$\int_{-\infty}^{\infty} f\left(x - \frac{1}{x}\right) dx = \int_{-\infty}^{\infty} f(u) du$.
Given $\int_{-\infty}^{\infty} f(x) dx = 1$,it follows that $I = 1$.

(A) Given the equation: $\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx = \int_{0}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$.
We can rewrite the right side as: $\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx + \int_{1}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$.
Subtracting the integral from $0$ to $1$ from both sides,we get: $\int_{1}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx = 0$.
Let $f(x) = ax^2 + bx + c$ and $g(x) = 1 + |\sin x|$. Since $g(x) > 0$ for all $x \in [1, 2]$,for the integral of the product $f(x)g(x)$ to be zero over the interval $[1, 2]$,the function $f(x)$ must change its sign at least once in the interval $(1, 2)$.
By the Intermediate Value Theorem,if a continuous function changes sign in an interval,it must have at least one root in that interval.
Therefore,$ax^2 + bx + c = 0$ has at least one real root between $(1, 2)$.

(C) Let $I = \int\limits_{\frac{1}{2}}^2 {\frac{1}{x}{{\tan }^{2015}}\left( {x - \frac{1}{x}} \right)dx} $.
Substitute $x = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$.
When $x = \frac{1}{2}$,$t = 2$. When $x = 2$,$t = \frac{1}{2}$.
So,$I = \int\limits_2^{\frac{1}{2}} {t \cdot {{\tan }^{2015}}\left( {\frac{1}{t} - t} \right) \cdot \left( -\frac{1}{t^2} \right) dt} $.
$I = \int\limits_{\frac{1}{2}}^2 {\frac{1}{t} \cdot {{\tan }^{2015}}\left( {-(t - \frac{1}{t})} \right) dt} $.
Since $\tan(- \theta) = -\tan(\theta)$,we have $\tan^{2015}(-(t - \frac{1}{t})) = -\tan^{2015}(t - \frac{1}{t})$.
Thus,$I = - \int\limits_{\frac{1}{2}}^2 {\frac{1}{t} {{\tan }^{2015}}\left( {t - \frac{1}{t}} \right) dt} = -I$.
$2I = 0 \Rightarrow I = 0$.

(C) Let $I = \int_{2}^{3} \frac{(x+2)^2}{2x^2-10x+53} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = 2+3 = 5$.
So,$I = \int_{2}^{3} \frac{(5-x+2)^2}{2(5-x)^2-10(5-x)+53} dx = \int_{2}^{3} \frac{(7-x)^2}{2(25-10x+x^2)-50+10x+53} dx$.
Simplifying the denominator: $2x^2-20x+50-50+10x+53 = 2x^2-10x+53$.
Thus,$I = \int_{2}^{3} \frac{(7-x)^2}{2x^2-10x+53} dx$.
Adding the two expressions for $I$: $2I = \int_{2}^{3} \frac{(x+2)^2+(7-x)^2}{2x^2-10x+53} dx = \int_{2}^{3} \frac{x^2+4x+4+49-14x+x^2}{2x^2-10x+53} dx = \int_{2}^{3} \frac{2x^2-10x+53}{2x^2-10x+53} dx = \int_{2}^{3} 1 dx = [x]_{2}^{3} = 3-2 = 1$.
Therefore,$I = \frac{1}{2}$.

(A) Let $I = \int_{1/2}^{2} \frac{1}{x} \sin \left( x - \frac{1}{x} \right) dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f\left( \frac{ab}{x} \right) \frac{ab}{x^2} dx$,where $a = 1/2$ and $b = 2$,we have $ab = 1$.
So,$I = \int_{1/2}^{2} \frac{1}{(1/x)} \sin \left( \frac{1}{x} - x \right) \cdot \frac{1}{x^2} dx$.
$I = \int_{1/2}^{2} x \sin \left( -\left( x - \frac{1}{x} \right) \right) \cdot \frac{1}{x^2} dx$.
Since $\sin(-\theta) = -\sin(\theta)$,we get $I = -\int_{1/2}^{2} \frac{1}{x} \sin \left( x - \frac{1}{x} \right) dx$.
Thus,$I = -I$,which implies $2I = 0$,so $I = 0$.

(C) We know that $\sqrt{\frac{1 + \cos 2x}{2}} = \sqrt{\frac{2 \cos^2 x}{2}} = |\cos x|$.
Thus,the integral becomes $I = \int_{0}^{n\pi + V} |\cos x| dx$.
Since the period of $|\cos x|$ is $\pi$,we can write $I = \int_{0}^{n\pi} |\cos x| dx + \int_{n\pi}^{n\pi + V} |\cos x| dx$.
$I = n \int_{0}^{\pi} |\cos x| dx + \int_{0}^{V} |\cos x| dx$.
Since $\int_{0}^{\pi} |\cos x| dx = \int_{0}^{\pi/2} \cos x dx + \int_{\pi/2}^{\pi} (-\cos x) dx = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{\pi} = 1 - (0 - 1) = 2$.
For the second part,since $\frac{\pi}{2} < V < \pi$,$|\cos x| = \cos x$ for $x \in [0, \pi/2]$ and $|\cos x| = -\cos x$ for $x \in [\pi/2, V]$.
So,$\int_{0}^{V} |\cos x| dx = \int_{0}^{\pi/2} \cos x dx + \int_{\pi/2}^{V} (-\cos x) dx = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{V} = 1 - (\sin V - 1) = 2 - \sin V$.
Therefore,$I = n(2) + 2 - \sin V = 2n + 2 - \sin V$.

(C) Given $f(x) = x^4 + |x|$. Since $f(-x) = (-x)^4 + |-x| = x^4 + |x| = f(x)$,$f(x)$ is an even function.
$I_1 = \int_{0}^{\pi} f(\cos x) dx$.
Using the property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$,we get:
$I_1 = \int_{0}^{\pi} f(\cos(\pi - x)) dx = \int_{0}^{\pi} f(-\cos x) dx$.
Since $f(x)$ is an even function,$f(-\cos x) = f(\cos x)$.
Thus,$I_1 = \int_{0}^{\pi} f(\cos x) dx = 2 \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$.
Now,consider $I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$. Using the property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$:
$I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin(\frac{\pi}{2} - x)) dx = \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$.
Therefore,$I_1 = 2 I_2$,which implies $\frac{I_1}{I_2} = 2$.

(C) Let $I = \int_{-1}^{0} \frac{4x^2 + 4x + 3}{1 + e^{2x + 1}} dx$.
Notice that $4x^2 + 4x + 3 = (2x + 1)^2 + 2$.
Substitute $t = 2x + 1$,then $dt = 2dx$ or $dx = \frac{dt}{2}$.
When $x = -1$,$t = -1$. When $x = 0$,$t = 1$.
$I = \int_{-1}^{1} \frac{t^2 + 2}{1 + e^t} \cdot \frac{dt}{2} = \frac{1}{2} \int_{-1}^{1} \frac{t^2 + 2}{1 + e^t} dt$.
Using the property $\int_{a}^{b} f(t) dt = \int_{a}^{b} f(a+b-t) dt$,here $a+b = -1+1 = 0$,so $I = \frac{1}{2} \int_{-1}^{1} \frac{(-t)^2 + 2}{1 + e^{-t}} dt = \frac{1}{2} \int_{-1}^{1} \frac{t^2 + 2}{1 + e^{-t}} \cdot \frac{e^t}{e^t} dt = \frac{1}{2} \int_{-1}^{1} \frac{(t^2 + 2)e^t}{e^t + 1} dt$.
Adding the two expressions for $I$:
$2I = \frac{1}{2} \int_{-1}^{1} \frac{t^2 + 2}{1 + e^t} dt + \frac{1}{2} \int_{-1}^{1} \frac{(t^2 + 2)e^t}{1 + e^t} dt = \frac{1}{2} \int_{-1}^{1} (t^2 + 2) \frac{1 + e^t}{1 + e^t} dt = \frac{1}{2} \int_{-1}^{1} (t^2 + 2) dt$.
Since $t^2 + 2$ is an even function,$\int_{-1}^{1} (t^2 + 2) dt = 2 \int_{0}^{1} (t^2 + 2) dt$.
$2I = \frac{1}{2} \cdot 2 \int_{0}^{1} (t^2 + 2) dt = \left[ \frac{t^3}{3} + 2t \right]_{0}^{1} = \frac{1}{3} + 2 = \frac{7}{3}$.
Thus,$2I = \frac{7}{3} \implies I = \frac{7}{6}$.

(D) Let $I = \int\limits_0^\pi e^{\cos^4 x} \cdot \cos^5(2n + 1)x \,dx$.
Using the property $\int_0^a f(x) \,dx = \int_0^a f(a - x) \,dx$,we have:
$I = \int_0^\pi e^{\cos^4(\pi - x)} \cdot \cos^5(2n + 1)(\pi - x) \,dx$.
Since $\cos(\pi - x) = -\cos x$,then $\cos^4(\pi - x) = \cos^4 x$.
Also,$\cos(2n + 1)(\pi - x) = \cos((2n + 1)\pi - (2n + 1)x) = \cos((2n + 1)\pi) \cdot \cos(2n + 1)x + \sin((2n + 1)\pi) \cdot \sin(2n + 1)x$.
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,$\cos^5(2n + 1)(\pi - x) = (-\cos(2n + 1)x)^5 = -\cos^5(2n + 1)x$.
Substituting this back into the integral:
$I = \int_0^\pi e^{\cos^4 x} \cdot (-\cos^5(2n + 1)x) \,dx = -I$.
$2I = 0 \implies I = 0$.

(B) We know that $x - [x] = \{x\}$,where $\{x\}$ is the fractional part function.
Thus,$\frac{3^x}{3^{[x]}} = 3^{x-[x]} = 3^{\{x\}}$.
The integral becomes $I = \int_{-10}^{10} 3^{\{x\}} \, dx$.
Since $\{x\}$ is a periodic function with period $1$,and the interval length is $20$,we have:
$I = 20 \int_{0}^{1} 3^x \, dx$.
Evaluating the integral:
$I = 20 \left[ \frac{3^x}{\ln 3} \right]_{0}^{1} = \frac{20}{\ln 3} (3^1 - 3^0) = \frac{20}{\ln 3} (3 - 1) = \frac{40}{\ln 3}$.

(A) Let $I = \int_{a}^{b} \frac{f(x)}{f(x) + f(a+b-x)} \, dx$.
Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we know that $I = \frac{b-a}{2}$.
Here,$a = 2 - \log 3$ and $b = 3 + \log 3$.
Also,$f(x) = \log(4+x)$,so $f(a+b-x) = \log(4 + (2 - \log 3 + 3 + \log 3 - x)) = \log(4 + 5 - x) = \log(9 - x)$.
Thus,$I = \frac{(3 + \log 3) - (2 - \log 3)}{2} = \frac{1 + 2 \log 3}{2} = \frac{1}{2} + \log 3$.

(B) We are given ${I_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,dx}$.
Consider the sum ${I_n} + {I_{n - 2}} = \int\limits_0^{\frac{\pi }{4}} {(\tan^n x + \tan^{n-2} x) dx}$.
Factoring out ${\tan^{n-2} x}$,we get ${I_n} + {I_{n - 2}} = \int\limits_0^{\frac{\pi }{4}} {\tan^{n-2} x (1 + \tan^2 x) dx}$.
Since $1 + \tan^2 x = \sec^2 x$,we have ${I_n} + {I_{n - 2}} = \int\limits_0^{\frac{\pi }{4}} {\tan^{n-2} x \sec^2 x dx}$.
Let $u = \tan x$,then $du = \sec^2 x dx$. When $x=0, u=0$ and when $x=\pi/4, u=1$.
Thus,${I_n} + {I_{n - 2}} = \int\limits_0^1 {u^{n-2} du} = \left[ \frac{u^{n-1}}{n-1} \right]_0^1 = \frac{1}{n-1}$.
Now,we evaluate the limit: $\mathop {\lim }\limits_{n \to \infty } n({I_n} + {I_{n - 2}}) = \mathop {\lim }\limits_{n \to \infty } \frac{n}{n-1}$.
Dividing numerator and denominator by $n$,we get $\mathop {\lim }\limits_{n \to \infty } \frac{1}{1 - 1/n} = 1$.

(C) Let $I = \int_{-1}^{1} \frac{x^4}{1 + e^{x^7}} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get:
$I = \int_{-1}^{1} \frac{(-x)^4}{1 + e^{(-x)^7}} dx = \int_{-1}^{1} \frac{x^4}{1 + e^{-x^7}} dx$.
Multiplying the numerator and denominator by $e^{x^7}$:
$I = \int_{-1}^{1} \frac{x^4 e^{x^7}}{e^{x^7} + 1} dx$.
Adding the two expressions for $I$:
$2I = \int_{-1}^{1} \frac{x^4}{1 + e^{x^7}} dx + \int_{-1}^{1} \frac{x^4 e^{x^7}}{1 + e^{x^7}} dx = \int_{-1}^{1} \frac{x^4(1 + e^{x^7})}{1 + e^{x^7}} dx$.
$2I = \int_{-1}^{1} x^4 dx$.
Since $x^4$ is an even function,$2I = 2 \int_{0}^{1} x^4 dx$.
$I = \int_{0}^{1} x^4 dx = \left[ \frac{x^5}{5} \right]_{0}^{1} = \frac{1}{5} - 0 = \frac{1}{5}$.

(C) Let $I(x) = \int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt + \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt$.
Applying Leibniz's rule to differentiate with respect to $x$:
$I'(x) = \sin^{-1}(\sqrt{\sin^2 x}) \cdot \frac{d}{dx}(\sin^2 x) + \cos^{-1}(\sqrt{\cos^2 x}) \cdot \frac{d}{dx}(\cos^2 x)$.
Since $x \in (0, \pi/2)$,$\sin x > 0$ and $\cos x > 0$,so $\sqrt{\sin^2 x} = \sin x$ and $\sqrt{\cos^2 x} = \cos x$.
$I'(x) = (x) \cdot (2 \sin x \cos x) + (x) \cdot (-2 \cos x \sin x)$.
$I'(x) = x \sin(2x) - x \sin(2x) = 0$.
Since the derivative is $0$,$I(x)$ is a constant.
To find the constant,evaluate at $x = \pi/4$:
$I(\pi/4) = \int_{0}^{1/2} \sin^{-1} \sqrt{t} \, dt + \int_{0}^{1/2} \cos^{-1} \sqrt{t} \, dt$.
Using the identity $\sin^{-1} \sqrt{t} + \cos^{-1} \sqrt{t} = \pi/2$:
$I(\pi/4) = \int_{0}^{1/2} (\sin^{-1} \sqrt{t} + \cos^{-1} \sqrt{t}) \, dt = \int_{0}^{1/2} \frac{\pi}{2} \, dt = \frac{\pi}{2} [t]_{0}^{1/2} = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}$.

(B) For $u = \int_0^1 \frac{\ln(x+1)}{x^2+1} dx$,let $x = \tan \theta$,then $dx = \sec^2 \theta d\theta$. When $x=0, \theta=0$ and when $x=1, \theta=\frac{\pi}{4}$.
$u = \int_0^{\pi/4} \ln(1+\tan \theta) d\theta$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$u = \int_0^{\pi/4} \ln(1+\tan(\frac{\pi}{4}-\theta)) d\theta = \int_0^{\pi/4} \ln(1+\frac{1-\tan \theta}{1+\tan \theta}) d\theta = \int_0^{\pi/4} \ln(\frac{2}{1+\tan \theta}) d\theta$.
$u = \int_0^{\pi/4} \ln 2 d\theta - \int_0^{\pi/4} \ln(1+\tan \theta) d\theta = \frac{\pi}{4} \ln 2 - u$.
$2u = \frac{\pi}{4} \ln 2 \implies u = \frac{\pi}{8} \ln 2 \implies 4u = \frac{\pi}{2} \ln 2 \dots (1)$.
For $v = \int_0^{\pi/2} \ln(\sin 2x) dx$,let $2x = t$,then $dx = \frac{1}{2} dt$. When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\pi$.
$v = \frac{1}{2} \int_0^{\pi} \ln(\sin t) dt = \int_0^{\pi/2} \ln(\sin t) dt$.
Using the standard integral $\int_0^{\pi/2} \ln(\sin t) dt = -\frac{\pi}{2} \ln 2$,we get $v = -\frac{\pi}{2} \ln 2 \dots (2)$.
From $(1)$ and $(2)$,$4u = -v \implies 4u + v = 0$.

(B) Let $I = \int_{-2}^{\pi} \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}} \,dx$.
We split the integral based on the value of $[\frac{x}{\pi}]$.
For $x \in [-2, 0)$,$\frac{x}{\pi} \in [-\frac{2}{\pi}, 0)$,so $[\frac{x}{\pi}] = -1$.
For $x \in [0, \pi)$,$\frac{x}{\pi} \in [0, 1)$,so $[\frac{x}{\pi}] = 0$.
Thus,$I = \int_{-2}^{0} \frac{\sin^2 x}{-1 + 1/2} \,dx + \int_{0}^{\pi} \frac{\sin^2 x}{0 + 1/2} \,dx$.
$I = -2 \int_{-2}^{0} \sin^2 x \,dx + 2 \int_{0}^{\pi} \sin^2 x \,dx$.
Using $\sin^2 x = \frac{1 - \cos 2x}{2}$,we get:
$I = -2 \int_{-2}^{0} \frac{1 - \cos 2x}{2} \,dx + 2 \int_{0}^{\pi} \frac{1 - \cos 2x}{2} \,dx$.
$I = -[x - \frac{\sin 2x}{2}]_{-2}^{0} + [x - \frac{\sin 2x}{2}]_{0}^{\pi}$.
$I = -[0 - (-2 - \frac{\sin(-4)}{2})] + [(\pi - \frac{\sin 2\pi}{2}) - 0]$.
$I = -[2 - \frac{\sin 4}{2}] + \pi = \pi - 2 + \frac{\sin 4}{2}$.
Since $\sin 4 = 2 \sin 2 \cos 2$,we have $I = \pi - 2 + \sin 2 \cos 2$.

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

(D) For $x \in (0, 1)$,we have $x^3 < x^2 < x$.
Multiplying by $-1$,we get $-x^3 > -x^2 > -x$.
Since the exponential function $f(t) = e^t$ is strictly increasing,we have $e^{-x^3} > e^{-x^2} > e^{-x}$.
Now,consider the integrands:
For $I_3$,the integrand is $f_3(x) = e^{-x^3}$.
For $I_2$,the integrand is $f_2(x) = e^{-x^2} \cos^2 x$. Since $0 \le \cos^2 x \le 1$,we have $e^{-x^2} \cos^2 x \le e^{-x^2}$.
For $I_1$,the integrand is $f_1(x) = e^{-x} \cos^2 x$. Since $0 \le \cos^2 x \le 1$,we have $e^{-x} \cos^2 x \le e^{-x}$.
Comparing the functions on $(0, 1)$:
$e^{-x^3} > e^{-x^2} \ge e^{-x^2} \cos^2 x$ implies $I_3 > I_2$.
Also,$e^{-x^2} \cos^2 x > e^{-x} \cos^2 x$ because $e^{-x^2} > e^{-x}$ for $x \in (0, 1)$.
Thus,$I_3 > I_2 > I_1$.

(B) We are given the identity $2 \int_{0}^{1} \tan^{-1} x \, dx = \int_{0}^{1} \cot^{-1} (1 - x + x^2) \, dx$.
Using the property $\cot^{-1} u = \frac{\pi}{2} - \tan^{-1} u$,we can rewrite the right side:
$2 \int_{0}^{1} \tan^{-1} x \, dx = \int_{0}^{1} \left( \frac{\pi}{2} - \tan^{-1} (1 - x + x^2) \right) dx$.
$2 \int_{0}^{1} \tan^{-1} x \, dx = \frac{\pi}{2} - \int_{0}^{1} \tan^{-1} (1 - x + x^2) \, dx$.
Rearranging the terms,we get:
$\int_{0}^{1} \tan^{-1} (1 - x + x^2) \, dx = \frac{\pi}{2} - 2 \int_{0}^{1} \tan^{-1} x \, dx$.
Now,we evaluate $I = \int_{0}^{1} \tan^{-1} x \, dx$ using integration by parts:
$I = [x \tan^{-1} x]_0^1 - \int_0^1 \frac{x}{1+x^2} \, dx = \frac{\pi}{4} - \frac{1}{2} [\ln(1+x^2)]_0^1 = \frac{\pi}{4} - \frac{1}{2} \ln 2$.
Substituting this back into our expression:
$\int_{0}^{1} \tan^{-1} (1 - x + x^2) \, dx = \frac{\pi}{2} - 2 \left( \frac{\pi}{4} - \frac{1}{2} \ln 2 \right) = \frac{\pi}{2} - \frac{\pi}{2} + \ln 2 = \ln 2$.

(D) Let $I = \int_{4}^{10} \frac{[x^2]}{[(x-14)^2] + [x^2]} dx$ ... $(1)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a=4$ and $b=10$,we have $a+b-x = 14-x$.
Substituting $x$ with $14-x$ in the integral:
$I = \int_{4}^{10} \frac{[(14-x)^2]}{[(14-x-14)^2] + [(14-x)^2]} dx$
$I = \int_{4}^{10} \frac{[(14-x)^2]}{[(-x)^2] + [(14-x)^2]} dx$
$I = \int_{4}^{10} \frac{[(14-x)^2]}{[x^2] + [(14-x)^2]} dx$ ... $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{4}^{10} \frac{[x^2] + [(14-x)^2]}{[x^2] + [(14-x)^2]} dx$
$2I = \int_{4}^{10} 1 dx$
$2I = [x]_{4}^{10} = 10 - 4 = 6$
$I = 3$

(C) Given $f(x) = \int_{1}^{x} \frac{\ln t}{1+t} dt$.
Consider $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\ln t}{1+t} dt$.
Let $t = \frac{1}{u}$,then $dt = -\frac{1}{u^2} du$. When $t=1, u=1$ and when $t=1/x, u=x$.
$f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\ln(1/u)}{1 + 1/u} \left(-\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\ln u}{\frac{u+1}{u}} \left(-\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{\ln u}{u(u+1)} du$.
Now,$f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\ln t}{1+t} dt + \int_{1}^{x} \frac{\ln t}{t(1+t)} dt = \int_{1}^{x} \ln t \left( \frac{1}{1+t} + \frac{1}{t(1+t)} \right) dt$.
Simplifying the integrand: $\frac{1}{1+t} + \frac{1}{t(1+t)} = \frac{t+1}{t(1+t)} = \frac{1}{t}$.
Thus,$f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\ln t}{t} dt$.
Let $\ln t = v$,then $\frac{1}{t} dt = dv$. When $t=1, v=0$ and when $t=x, v=\ln x$.
$f(x) + f\left(\frac{1}{x}\right) = \int_{0}^{\ln x} v dv = \left[ \frac{v^2}{2} \right]_{0}^{\ln x} = \frac{1}{2}(\ln x)^2$.

(C) Given $f(2 - x) = f(2 + x)$,the function is symmetric about $x = 2$.
Given $f(4 - x) = f(4 + x)$,the function is symmetric about $x = 4$.
Replacing $x$ with $x - 2$ in the second equation: $f(4 - (x - 2)) = f(4 + (x - 2)) \Rightarrow f(6 - x) = f(2 + x)$.
Since $f(2 - x) = f(2 + x)$,we have $f(6 - x) = f(2 - x)$.
Let $t = 2 - x$,then $f(4 + t) = f(t)$,which shows that $f(x)$ is a periodic function with period $T = 4$.
We are given $\int_{0}^{2} f(x) dx = 5$.
Since $f(2 - x) = f(2 + x)$,$\int_{0}^{4} f(x) dx = \int_{0}^{2} f(x) dx + \int_{2}^{4} f(x) dx = 2 \int_{0}^{2} f(x) dx = 2 \times 5 = 10$.
Now,$\int_{10}^{50} f(x) dx = \frac{50 - 10}{4} \int_{0}^{4} f(x) dx = 10 \times 10 = 100$.

(C) Let $I = \int\limits_0^{1/2} \frac{\ln(1 + 2x)}{1 + (2x)^2} dx$.
Substitute $2x = \tan \theta$,then $2 dx = \sec^2 \theta d\theta$,which implies $dx = \frac{1}{2} \sec^2 \theta d\theta$.
When $x = 0$,$\theta = 0$. When $x = 1/2$,$\theta = \pi/4$.
Substituting these into the integral:
$I = \int\limits_0^{\pi/4} \frac{\ln(1 + \tan \theta)}{1 + \tan^2 \theta} \cdot \frac{1}{2} \sec^2 \theta d\theta$
Since $1 + \tan^2 \theta = \sec^2 \theta$,the expression simplifies to:
$I = \frac{1}{2} \int\limits_0^{\pi/4} \ln(1 + \tan \theta) d\theta$ --- $(1)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:
$I = \frac{1}{2} \int\limits_0^{\pi/4} \ln(1 + \tan(\pi/4 - \theta)) d\theta$
Using $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:
$I = \frac{1}{2} \int\limits_0^{\pi/4} \ln\left(1 + \frac{1 - \tan \theta}{1 + \tan \theta}\right) d\theta$
$I = \frac{1}{2} \int\limits_0^{\pi/4} \ln\left(\frac{1 + \tan \theta + 1 - \tan \theta}{1 + \tan \theta}\right) d\theta = \frac{1}{2} \int\limits_0^{\pi/4} \ln\left(\frac{2}{1 + \tan \theta}\right) d\theta$
$I = \frac{1}{2} \int\limits_0^{\pi/4} (\ln 2 - \ln(1 + \tan \theta)) d\theta$
$I = \frac{1}{2} \ln 2 \cdot [\theta]_0^{\pi/4} - \frac{1}{2} \int\limits_0^{\pi/4} \ln(1 + \tan \theta) d\theta$
$I = \frac{1}{2} \ln 2 \cdot \frac{\pi}{4} - I$
$2I = \frac{\pi}{8} \ln 2 \implies I = \frac{\pi}{16} \ln 2$.

(D) Given $P_n = \int\limits_1^e (\log x)^n \, dx$.
Let $\log x = t$,then $x = e^t$ and $dx = e^t \, dt$.
When $x = 1$,$t = 0$ and when $x = e$,$t = 1$.
Thus,$P_n = \int\limits_0^1 t^n e^t \, dt$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$.
Let $u = t^n$ and $dv = e^t \, dt$,then $du = nt^{n-1} \, dt$ and $v = e^t$.
$P_n = [t^n e^t]_0^1 - n \int\limits_0^1 t^{n-1} e^t \, dt = e - n P_{n-1}$.
So,$P_{10} = e - 10 P_9$.
Also,$P_9 = e - 9 P_8$.
Substituting $P_9$ into the equation for $P_{10}$:
$P_{10} = e - 10(e - 9 P_8) = e - 10e + 90 P_8$.
$P_{10} = -9e + 90 P_8$.
Therefore,$P_{10} - 90 P_8 = -9e$.

(D) Let $I = \int_{0}^{\pi} [\cos x] \, dx \quad \dots(1)$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get:
$I = \int_{0}^{\pi} [\cos(\pi - x)] \, dx = \int_{0}^{\pi} [-\cos x] \, dx \quad \dots(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{0}^{\pi} ([\cos x] + [-\cos x]) \, dx$
Since $[x] + [-x] = -1$ if $x \notin \mathbb{Z}$ and $0$ if $x \in \mathbb{Z}$,and the set of points where $\cos x \in \mathbb{Z}$ in $[0, \pi]$ is finite (measure zero),we have:
$2I = \int_{0}^{\pi} (-1) \, dx$
$2I = -[x]_{0}^{\pi} = -\pi$
$I = -\frac{\pi}{2}$

(D) Given $F(x) = \int_{1}^{x} \frac{e^{t}}{t} dt$ for $x > 0$.
Let $I = \int_{1}^{x} \frac{e^{t}}{t+a} dt$.
Substitute $t+a = z$,so $t = z-a$ and $dt = dz$.
When $t = 1$,$z = 1+a$. When $t = x$,$z = x+a$.
Substituting these into the integral:
$I = \int_{1+a}^{x+a} \frac{e^{z-a}}{z} dz = e^{-a} \int_{1+a}^{x+a} \frac{e^{z}}{z} dz$.
Using the property of definite integrals $\int_{b}^{c} f(z) dz = \int_{1}^{c} f(z) dz - \int_{1}^{b} f(z) dz$:
$I = e^{-a} \left[ \int_{1}^{x+a} \frac{e^{z}}{z} dz - \int_{1}^{1+a} \frac{e^{z}}{z} dz \right]$.
By the definition of $F(x)$,this becomes:
$I = e^{-a} [F(x+a) - F(1+a)]$.
Thus,the correct option is $D$.

(D) Let $I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} \quad ......(1)$
Using the property $\int\limits_a^b f(x)dx = \int\limits_a^b f(a+b-x)dx$,we replace $x$ with $-\frac{\pi}{2} + \frac{\pi}{2} - x = -x$:
$I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}(-x)}}{{1 + {2^{-x}}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + \frac{1}{2^x}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{2^x}{{\sin }^2}x}}{{2^x + 1}}} dx \quad ......(2)$
Adding $(1)$ and $(2)$:
$2I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x + {2^x}{{\sin }^2}x}}{{1 + {2^x}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}x dx}$
Since $\sin^2 x$ is an even function,$2I = 2 \int\limits_0^{\pi /2} {\sin^2 x dx} = 2 \int\limits_0^{\pi /2} {\frac{1 - \cos 2x}{2} dx}$
$2I = \int\limits_0^{\pi /2} {(1 - \cos 2x) dx} = [x - \frac{\sin 2x}{2}]_0^{\pi /2} = (\frac{\pi}{2} - 0) - (0 - 0) = \frac{\pi}{2}$
Therefore,$I = \frac{\pi}{4}$.

(A) Let $f(x) = \int_{0}^{\sin^{2}x} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{\cos^{2}x} \cos^{-1}(\sqrt{t}) \, dt$.
Using the Leibniz integral rule,we differentiate $f(x)$ with respect to $x$:
$f'(x) = \sin^{-1}(\sqrt{\sin^{2}x}) \cdot \frac{d}{dx}(\sin^{2}x) + \cos^{-1}(\sqrt{\cos^{2}x}) \cdot \frac{d}{dx}(\cos^{2}x)$
$f'(x) = x \cdot (2 \sin x \cos x) + (\frac{\pi}{2} - x) \cdot (-2 \cos x \sin x)$
$f'(x) = x \sin(2x) - (\frac{\pi}{2} - x) \sin(2x)$
$f'(x) = x \sin(2x) - \frac{\pi}{2} \sin(2x) + x \sin(2x) = (2x - \frac{\pi}{2}) \sin(2x)$.
Since $f'(x)$ is not zero,let us evaluate $f(x)$ at $x = \frac{\pi}{4}$:
$f(\frac{\pi}{4}) = \int_{0}^{1/2} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{1/2} \cos^{-1}(\sqrt{t}) \, dt$
$f(\frac{\pi}{4}) = \int_{0}^{1/2} (\sin^{-1}(\sqrt{t}) + \cos^{-1}(\sqrt{t})) \, dt$
Since $\sin^{-1}(\sqrt{t}) + \cos^{-1}(\sqrt{t}) = \frac{\pi}{2}$ for $t \in [0, 1]$:
$f(\frac{\pi}{4}) = \int_{0}^{1/2} \frac{\pi}{2} \, dt = \frac{\pi}{2} \cdot [t]_{0}^{1/2} = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}$.

(D) We are given $R_1 = \int_{-2}^3 f(x) dx$ and $R_2 = \int_{-2}^3 x f(x) dx$.
Using the property $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$,we have $a+b = -2+3 = 1$.
Thus,$R_2 = \int_{-2}^3 x f(x) dx = \int_{-2}^3 (1-x) f(1-x) dx$.
Since $f(1-x) = f(x)$,we get $R_2 = \int_{-2}^3 (1-x) f(x) dx$.
$R_2 = \int_{-2}^3 f(x) dx - \int_{-2}^3 x f(x) dx$.
$R_2 = R_1 - R_2$.
$2R_2 = R_1$ or $R_1 = 2R_2$.

(D) Let $I = \int_{-0.9}^{0.9} \left( [x^2] + \log \left( \frac{2-x}{2+x} \right) \right) dx$.
We can split this into two integrals: $I = \int_{-0.9}^{0.9} [x^2] dx + \int_{-0.9}^{0.9} \log \left( \frac{2-x}{2+x} \right) dx$.
For the first part,since $-0.9 < x < 0.9$,we have $0 \leq x^2 < 0.81$. Thus,$[x^2] = 0$ for all $x \in (-0.9, 0.9)$. Therefore,$\int_{-0.9}^{0.9} [x^2] dx = 0$.
For the second part,let $f(x) = \log \left( \frac{2-x}{2+x} \right)$.
Then $f(-x) = \log \left( \frac{2-(-x)}{2+(-x)} \right) = \log \left( \frac{2+x}{2-x} \right) = \log \left( \left( \frac{2-x}{2+x} \right)^{-1} \right) = -\log \left( \frac{2-x}{2+x} \right) = -f(x)$.
Since $f(x)$ is an odd function and the interval $[-0.9, 0.9]$ is symmetric about the origin,$\int_{-0.9}^{0.9} f(x) dx = 0$.
Thus,$I = 0 + 0 = 0$.

(C) Let $I = \int_{-\pi/2}^{\pi/2} \frac{dx}{[x] + [\sin x] + 4}$.
We split the integral at $x=0$:
$I = \int_{-\pi/2}^{0} \frac{dx}{[x] + [\sin x] + 4} + \int_{0}^{\pi/2} \frac{dx}{[x] + [\sin x] + 4}$.
For $x \in [-\pi/2, 0)$,$[\sin x] = -1$ (except at $x=0$). For $x \in [0, \pi/2]$,$[\sin x] = 0$.
$I = \int_{-\pi/2}^{-1} \frac{dx}{[x] - 1 + 4} + \int_{-1}^{0} \frac{dx}{[x] - 1 + 4} + \int_{0}^{1} \frac{dx}{[x] + 0 + 4} + \int_{1}^{\pi/2} \frac{dx}{[x] + 0 + 4}$.
$I = \int_{-\pi/2}^{-1} \frac{dx}{-2+3} + \int_{-1}^{0} \frac{dx}{-1+3} + \int_{0}^{1} \frac{dx}{0+4} + \int_{1}^{\pi/2} \frac{dx}{1+4}$.
$I = \int_{-\pi/2}^{-1} 1 dx + \int_{-1}^{0} \frac{1}{2} dx + \int_{0}^{1} \frac{1}{4} dx + \int_{1}^{\pi/2} \frac{1}{5} dx$.
$I = (-1 - (-\pi/2)) + \frac{1}{2}(0 - (-1)) + \frac{1}{4}(1 - 0) + \frac{1}{5}(\pi/2 - 1)$.
$I = \frac{\pi}{2} - 1 + \frac{1}{2} + \frac{1}{4} + \frac{\pi}{10} - \frac{1}{5}$.
$I = \pi(\frac{1}{2} + \frac{1}{10}) + (-1 + 0.5 + 0.25 - 0.2)$.
$I = \frac{6\pi}{10} - 0.45 = \frac{3\pi}{5} - \frac{9}{20} = \frac{12\pi - 9}{20} = \frac{3(4\pi - 3)}{20}$.

(A) Let $I = \int_{-2}^{2} f(x) \, dx$,where $f(x) = \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}}$.
Since the interval is $[-2, 2]$ and $\pi \approx 3.14$,the value of $\frac{x}{\pi}$ lies in the range $(-\frac{2}{\pi}, \frac{2}{\pi}) \approx (-0.63, 0.63)$.
Thus,for $x \in (-2, 0)$,$[\frac{x}{\pi}] = -1$,and for $x \in [0, 2)$,$[\frac{x}{\pi}] = 0$.
We split the integral: $I = \int_{-2}^{0} \frac{\sin^2 x}{-1 + 0.5} \, dx + \int_{0}^{2} \frac{\sin^2 x}{0 + 0.5} \, dx$.
$I = \int_{-2}^{0} \frac{\sin^2 x}{-0.5} \, dx + \int_{0}^{2} \frac{\sin^2 x}{0.5} \, dx$.
$I = -2 \int_{-2}^{0} \sin^2 x \, dx + 2 \int_{0}^{2} \sin^2 x \, dx$.
Using the property $\int_{-a}^{0} f(x) \, dx = \int_{0}^{a} f(-x) \, dx$ and since $\sin^2(-x) = \sin^2 x$:
$I = -2 \int_{0}^{2} \sin^2 x \, dx + 2 \int_{0}^{2} \sin^2 x \, dx = 0$.

(A) Let $I = \int_{-\pi/4}^{\pi/4} g(f(x)) dx$.
Since $f(-x) = \frac{2 - (-x) \cos(-x)}{2 + (-x) \cos(-x)} = \frac{2 + x \cos x}{2 - x \cos x} = \frac{1}{f(x)}$,
we have $g(f(-x)) = \ln(f(-x)) = \ln(1/f(x)) = -\ln(f(x)) = -g(f(x))$.
Thus,$g(f(x))$ is an odd function.
For any odd function $h(x)$,the integral $\int_{-a}^{a} h(x) dx = 0$.
Therefore,$I = 0$.
Since $\ln 1 = 0$,the correct option is $\ln 1$.

(A) Given $f(x) = \int_{0}^{x} g(t) dt$. Since $g(t)$ is an even function,$f(x)$ is an odd function because $f(-x) = \int_{0}^{-x} g(t) dt$. Let $t = -u$,then $dt = -du$. So $f(-x) = \int_{0}^{x} g(-u) (-du) = -\int_{0}^{x} g(u) du = -f(x)$.
Given $f(x+5) = g(x)$. Since $g(x)$ is even,$g(x) = g(-x)$,so $f(x+5) = g(-x)$.
Also,since $f$ is odd,$f(x+5) = -f(-x-5)$.
We need to evaluate $I = \int_{0}^{x} f(t) dt$.
From $f(x+5) = g(x)$,we have $g(t) = f(t+5)$.
Since $g$ is even,$g(t) = g(-t)$,thus $f(t+5) = f(-t+5)$.
Using the derivative,$f'(x) = g(x)$.
Integrating $f(x+5) = g(x)$ from $0$ to $x$:
$\int_{0}^{x} f(t+5) dt = \int_{0}^{x} g(t) dt = f(x)$.
Let $u = t+5$,then $du = dt$. When $t=0, u=5$; when $t=x, u=x+5$.
So,$\int_{5}^{x+5} f(u) du = f(x)$.
This does not directly give the integral of $f$. Let's use $f'(x) = g(x)$.
Since $f(x+5) = g(x)$,then $f'(x+5) = g'(x)$.
Integrating $f(x+5) = g(x)$ with respect to $x$:
$\int_{0}^{x} f(t+5) dt = \int_{0}^{x} g(t) dt = f(x)$.
Using the property $\int_{0}^{x} f(t) dt = -\int_{0}^{x} f(-t) dt$,and $f$ is odd,$f(-t) = -f(t)$.
Actually,from $f(x+5) = g(x)$,we have $f(x) = \int_{0}^{x} g(t) dt = \int_{0}^{x} f(t+5) dt$.
Let $u = t+5$,then $\int_{5}^{x+5} f(u) du = f(x)$.
Thus $\int_{0}^{x} f(t) dt = \int_{x+5}^{5} g(t) dt$ is the correct transformation.

(C) Let $I = \int_{0}^{\pi / 2} \frac{\sin^3 x}{\sin x + \cos x} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\pi / 2} \frac{\cos^3 x}{\cos x + \sin x} dx$.
Adding the two expressions for $I$:
$2I = \int_{0}^{\pi / 2} \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} dx$.
Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$:
$2I = \int_{0}^{\pi / 2} \frac{(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)}{\sin x + \cos x} dx$.
$2I = \int_{0}^{\pi / 2} (1 - \sin x \cos x) dx$.
$2I = \int_{0}^{\pi / 2} (1 - \frac{1}{2} \sin 2x) dx$.
$2I = [x + \frac{1}{4} \cos 2x]_{0}^{\pi / 2}$.
$2I = (\frac{\pi}{2} + \frac{1}{4} \cos \pi) - (0 + \frac{1}{4} \cos 0)$.
$2I = (\frac{\pi}{2} - \frac{1}{4}) - (0 + \frac{1}{4}) = \frac{\pi}{2} - \frac{1}{2}$.
$I = \frac{\pi - 1}{4}$.

(A) Let $I = \int_{0}^{1} x \cot^{-1}(1 - x^2 + x^4) dx$.
Using the property $\cot^{-1}(u) = \tan^{-1}(\frac{1}{u})$ for $u > 0$,we get:
$I = \int_{0}^{1} x \tan^{-1}\left(\frac{1}{1 - x^2 + x^4}\right) dx$.
We can write the argument as $\frac{x^2 - (x^2 - 1)}{1 + x^2(x^2 - 1)}$.
Thus,$I = \int_{0}^{1} x \left[ \tan^{-1}(x^2) - \tan^{-1}(x^2 - 1) \right] dx$.
Let $x^2 = t$,then $2x dx = dt$,so $x dx = \frac{1}{2} dt$.
When $x=0, t=0$ and when $x=1, t=1$.
$I = \frac{1}{2} \int_{0}^{1} \tan^{-1}(t) dt - \frac{1}{2} \int_{0}^{1} \tan^{-1}(t - 1) dt$.
Using the property $\int_{0}^{a} f(t) dt = \int_{0}^{a} f(a - t) dt$ on the second integral:
$I = \frac{1}{2} \int_{0}^{1} \tan^{-1}(t) dt - \frac{1}{2} \int_{0}^{1} \tan^{-1}(1 - t - 1) dt = \frac{1}{2} \int_{0}^{1} \tan^{-1}(t) dt - \frac{1}{2} \int_{0}^{1} \tan^{-1}(-t) dt$.
Since $\tan^{-1}(-t) = -\tan^{-1}(t)$:
$I = \frac{1}{2} \int_{0}^{1} \tan^{-1}(t) dt + \frac{1}{2} \int_{0}^{1} \tan^{-1}(t) dt = \int_{0}^{1} \tan^{-1}(t) dt$.
Integrating by parts: $\int \tan^{-1}(t) dt = t \tan^{-1}(t) - \int \frac{t}{1+t^2} dt = t \tan^{-1}(t) - \frac{1}{2} \ln(1+t^2)$.
Evaluating from $0$ to $1$:
$I = [1 \cdot \tan^{-1}(1) - \frac{1}{2} \ln(2)] - [0 - 0] = \frac{\pi}{4} - \frac{1}{2} \ln 2$.

(D) Let $I = \int_{0}^{2\pi} [\sin 2x(1 + \cos 3x)] \,dx$.
Using the property $\int_{0}^{a} f(x) \,dx = \int_{0}^{a} f(a-x) \,dx$,we observe that the integrand $f(x) = [\sin 2x(1 + \cos 3x)]$ satisfies $f(2\pi - x) = [\sin(4\pi - 2x)(1 + \cos(6\pi - 3x))] = [-\sin 2x(1 + \cos 3x)]$.
Using the property $\int_{0}^{a} [f(x) + f(a-x)] \,dx$,we know that for any $x$ such that $\sin 2x(1 + \cos 3x)$ is not an integer,$[t] + [-t] = -1$.
Since the function is not an integer almost everywhere in the interval $[0, 2\pi]$,we have:
$2I = \int_{0}^{2\pi} ( [\sin 2x(1 + \cos 3x)] + [-\sin 2x(1 + \cos 3x)] ) \,dx$
$2I = \int_{0}^{2\pi} -1 \,dx = -2\pi$.
Therefore,$I = -\pi$.

(D) Let $I = \frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) dx \quad \dots(1)$
Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,we get:
$I = \frac{1}{a+b} \int_{a}^{b} (a+b-x)(f(a+b-x)+f(a+b-x+1)) dx$
Given $f(a+b+1-x) = f(x)$,we have $f(a+b-x) = f(x+1)$.
Thus,$I = \frac{1}{a+b} \int_{a}^{b} (a+b-x)(f(x+1)+f(x)) dx \quad \dots(2)$
Adding $(1)$ and $(2)$:
$2I = \frac{1}{a+b} \int_{a}^{b} (a+b)(f(x)+f(x+1)) dx = \int_{a}^{b} (f(x)+f(x+1)) dx$
Since $f(a+b+1-x) = f(x)$,let $t = a+b+1-x$,then $dt = -dx$. When $x=a, t=b+1$; when $x=b, t=a+1$.
$\int_{a}^{b} f(x) dx = \int_{a+1}^{b+1} f(a+b+1-t) dt = \int_{a+1}^{b+1} f(t) dt$.
Thus,$2I = \int_{a}^{b} f(x) dx + \int_{a}^{b} f(x+1) dx = \int_{a+1}^{b+1} f(x+1) dx + \int_{a}^{b} f(x+1) dx$. This simplifies to $I = \int_{a}^{b} f(x+1) dx = \int_{a-1}^{b-1} f(x) dx$.

(D) Let $I = \int_{0}^{2 \pi} \frac{x \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have:
$I = \int_{0}^{2 \pi} \frac{(2 \pi - x) \sin^{8}(2 \pi - x)}{\sin^{8}(2 \pi - x) + \cos^{8}(2 \pi - x)} dx = \int_{0}^{2 \pi} \frac{(2 \pi - x) \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$.
Adding the two expressions for $I$:
$2I = \int_{0}^{2 \pi} \frac{2 \pi \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx = 2 \pi \int_{0}^{2 \pi} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$.
$I = \pi \int_{0}^{2 \pi} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$.
Using the property $\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(2a-x) = f(x)$:
$I = 2 \pi \int_{0}^{\pi} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx = 4 \pi \int_{0}^{\pi/2} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$.
Using $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$:
$I = 4 \pi \int_{0}^{\pi/2} \frac{\cos^{8} x}{\cos^{8} x + \sin^{8} x} dx$.
Adding these two: $2I = 4 \pi \int_{0}^{\pi/2} 1 dx = 4 \pi \cdot \frac{\pi}{2} = 2 \pi^{2}$.
Thus,$I = \pi^{2}$.

(A) We observe that $f(x) = \sin^{2} x$ is an even function because $f(-x) = \sin^{2}(-x) = (-\sin x)^{2} = \sin^{2} x = f(x)$.
Therefore,by the property of definite integrals $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$,we get:
$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^{2} x \, dx = 2 \int_{0}^{\frac{\pi}{4}} \sin^{2} x \, dx$
Using the identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$,we have:
$= 2 \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos 2x}{2} \, dx = \int_{0}^{\frac{\pi}{4}} (1 - \cos 2x) \, dx$
$= \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{4}}$
$= \left( \frac{\pi}{4} - \frac{\sin(\frac{\pi}{2})}{2} \right) - (0 - 0)$
$= \frac{\pi}{4} - \frac{1}{2} = \frac{\pi - 2}{4}$

Let $I = \int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$.
By using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have:
$I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} dx = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} dx$.
$I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx - \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx$.
$I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx - I$.
$2I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$.
$I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$.
Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$ and when $x=\pi, t=-1$.
$I = \frac{\pi}{2} \int_{1}^{-1} \frac{-dt}{1 + t^2} = \frac{\pi}{2} \int_{-1}^{1} \frac{dt}{1 + t^2}$.
Since $\frac{1}{1+t^2}$ is an even function,$I = \frac{\pi}{2} \times 2 \int_{0}^{1} \frac{dt}{1 + t^2} = \pi [\tan^{-1} t]_{0}^{1}$.
$I = \pi (\tan^{-1}(1) - \tan^{-1}(0)) = \pi (\frac{\pi}{4} - 0) = \frac{\pi^2}{4}$.

(A) Let $I = \int_{-1}^{1} \sin^{5} x \cos^{4} x \, dx$.
Define $f(x) = \sin^{5} x \cos^{4} x$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \sin^{5}(-x) \cos^{4}(-x) = (-\sin x)^{5} (\cos x)^{4} = -\sin^{5} x \cos^{4} x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$I = 0$.

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

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

(C) Let $I = \int_{0}^{\frac{\pi}{2}} \log \sin x \, dx$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we have:
$I = \int_{0}^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-x\right) \, dx = \int_{0}^{\frac{\pi}{2}} \log \cos x \, dx$
Adding the two expressions for $I$:
$2I = \int_{0}^{\frac{\pi}{2}} (\log \sin x + \log \cos x) \, dx = \int_{0}^{\frac{\pi}{2}} \log(\sin x \cos x) \, dx$
Multiply and divide by $2$ inside the logarithm:
$2I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{\sin 2x}{2}\right) \, dx = \int_{0}^{\frac{\pi}{2}} \log \sin 2x \, dx - \int_{0}^{\frac{\pi}{2}} \log 2 \, dx$
$2I = \int_{0}^{\frac{\pi}{2}} \log \sin 2x \, dx - \frac{\pi}{2} \log 2$
For the first integral,let $2x = t$,then $2 \, dx = dt$. When $x=0, t=0$; when $x=\frac{\pi}{2}, t=\pi$:
$\int_{0}^{\frac{\pi}{2}} \log \sin 2x \, dx = \frac{1}{2} \int_{0}^{\pi} \log \sin t \, dt = \frac{1}{2} \times 2 \int_{0}^{\frac{\pi}{2}} \log \sin t \, dt = I$ (using $\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(2a-x) = f(x)$)
Substituting back:
$2I = I - \frac{\pi}{2} \log 2$
$I = -\frac{\pi}{2} \log 2$

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