Properties of definite integration Questions in English

(C) We evaluate the definite integral $\int_{-2}^{2} (ax^3 + bx + c) dx$.
Using the property of linearity,we can split the integral: $\int_{-2}^{2} ax^3 dx + \int_{-2}^{2} bx dx + \int_{-2}^{2} c dx$.
Since $ax^3$ and $bx$ are odd functions and the interval $[-2, 2]$ is symmetric about the origin,$\int_{-2}^{2} ax^3 dx = 0$ and $\int_{-2}^{2} bx dx = 0$.
Thus,the integral simplifies to $\int_{-2}^{2} c dx = [cx]_{-2}^{2} = c(2) - c(-2) = 2c + 2c = 4c$.
Since the result is $4c$,the value of the integral depends only on the value of $c$.

(C) Let $I = \int_0^{\pi /2} \frac{\cos x}{1 + \cos x + \sin x} \,dx$.
Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get:
$I = \int_0^{\pi /2} \frac{\cos(\pi/2 - x)}{1 + \cos(\pi/2 - x) + \sin(\pi/2 - x)} \,dx = \int_0^{\pi /2} \frac{\sin x}{1 + \sin x + \cos x} \,dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi /2} \frac{\cos x + \sin x}{1 + \cos x + \sin x} \,dx$.
Let $f(x) = 1 + \cos x + \sin x$,then $f'(x) = \cos x - \sin x$. This does not match the numerator directly.
Alternatively,$2I = \int_0^{\pi /2} \frac{(1 + \cos x + \sin x) - 1}{1 + \cos x + \sin x} \,dx = \int_0^{\pi /2} 1 \,dx - \int_0^{\pi /2} \frac{1}{1 + \cos x + \sin x} \,dx$.
$2I = \frac{\pi}{2} - \int_0^{\pi /2} \frac{1}{2\cos^2(x/2) + 2\sin(x/2)\cos(x/2)} \,dx = \frac{\pi}{2} - \frac{1}{2} \int_0^{\pi /2} \sec^2(x/2) \frac{1}{1 + \tan(x/2)} \,dx$.
Let $u = \tan(x/2)$,then $du = \frac{1}{2} \sec^2(x/2) \,dx$. Limits: $x=0 \implies u=0$,$x=\pi/2 \implies u=1$.
$2I = \frac{\pi}{2} - \int_0^1 \frac{1}{1+u} \,du = \frac{\pi}{2} - [\ln(1+u)]_0^1 = \frac{\pi}{2} - \ln 2$.
$I = \frac{\pi}{4} - \frac{1}{2} \ln 2$.

(A) Let $I = \int_{-\pi}^{\pi} \sin(mx) \sin(nx) \, dx$.
Since the integrand $f(x) = \sin(mx) \sin(nx)$ is an even function,we can write:
$I = 2 \int_{0}^{\pi} \sin(mx) \sin(nx) \, dx$.
Using the trigonometric identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$,we get:
$I = \int_{0}^{\pi} [\cos((m - n)x) - \cos((m + n)x)] \, dx$.
Integrating term by term:
$I = \left[ \frac{\sin((m - n)x)}{m - n} - \frac{\sin((m + n)x)}{m + n} \right]_{0}^{\pi}$.
Evaluating at the limits:
$I = \left( \frac{\sin((m - n)\pi)}{m - n} - \frac{\sin((m + n)\pi)}{m + n} \right) - (0 - 0)$.
Since $m, n \in I$ (integers),$\sin(k\pi) = 0$ for any integer $k$.
Therefore,$I = 0 - 0 = 0$.

(A) We know that for $0 < x \le \frac{\pi}{2}$,the inequality $\sin x < x$ holds true.
Dividing both sides by $x$ (since $x > 0$),we get $\frac{\sin x}{x} < 1$.
Integrating both sides from $0$ to $\frac{\pi}{2}$ with respect to $x$:
$\int_0^{\pi /2} \frac{\sin x}{x} \, dx < \int_0^{\pi /2} 1 \, dx$.
Evaluating the right side: $\int_0^{\pi /2} 1 \, dx = [x]_0^{\pi /2} = \frac{\pi}{2}$.
Therefore,$\int_0^{\pi /2} \frac{\sin x}{x} \, dx < \frac{\pi}{2}$.
Thus,the greater value is $\frac{\pi}{2}$.

(B) Let $I = \int_{-1}^{3} \left( \tan^{-1} \frac{x}{x^2+1} + \tan^{-1} \frac{x^2+1}{x} \right) dx$.
Using the identity $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for $u > 0$,we note that $\tan^{-1} \frac{x^2+1}{x} = \cot^{-1} \frac{x}{x^2+1}$.
Thus,the integrand becomes $\tan^{-1} \left( \frac{x}{x^2+1} \right) + \cot^{-1} \left( \frac{x}{x^2+1} \right) = \frac{\pi}{2}$.
Therefore,$I = \int_{-1}^{3} \frac{\pi}{2} dx$.
$I = \frac{\pi}{2} [x]_{-1}^{3} = \frac{\pi}{2} (3 - (-1)) = \frac{\pi}{2} (4) = 2\pi$.

(C) Let $I = \int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \,dt + \int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \,dt$.
For the first integral,let $t = \sin^2 u$,then $dt = 2 \sin u \cos u \,du = \sin 2u \,du$. When $t=0, u=0$ and when $t=\sin^2 x, u=x$.
So,$\int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \,dt = \int_0^x u \sin 2u \,du$.
For the second integral,let $t = \cos^2 v$,then $dt = -2 \cos v \sin v \,dv = -\sin 2v \,dv$. When $t=0, v=\pi/2$ and when $t=\cos^2 x, v=x$.
So,$\int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \,dt = \int_{\pi/2}^x v (-\sin 2v) \,dv = \int_x^{\pi/2} v \sin 2v \,dv$.
Thus,$I = \int_0^x u \sin 2u \,du + \int_x^{\pi/2} u \sin 2u \,du = \int_0^{\pi/2} u \sin 2u \,du$.
Using integration by parts: $\int u \sin 2u \,du = u \left( -\frac{\cos 2u}{2} \right) - \int 1 \cdot \left( -\frac{\cos 2u}{2} \right) \,du = -\frac{u \cos 2u}{2} + \frac{\sin 2u}{4}$.
Evaluating from $0$ to $\pi/2$: $\left[ -\frac{u \cos 2u}{2} + \frac{\sin 2u}{4} \right]_0^{\pi/2} = \left( -\frac{\pi/2 \cos \pi}{2} + 0 \right) - (0 + 0) = -\frac{\pi/2 (-1)}{2} = \frac{\pi}{4}$.

(B) Given the equation: $af(x) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 5$ ... $(i)$
Replacing $x$ with $\frac{1}{x}$ in $(i)$,we get: $af\left( {\frac{1}{x}} \right) + bf(x) = x - 5$ ... $(ii)$
To eliminate $f\left( {\frac{1}{x}} \right)$,multiply $(i)$ by $a$ and $(ii)$ by $b$:
$a^2 f(x) + ab f\left( {\frac{1}{x}} \right) = \frac{a}{x} - 5a$
$b^2 f(x) + ab f\left( {\frac{1}{x}} \right) = bx - 5b$
Subtracting the second from the first: $(a^2 - b^2) f(x) = \frac{a}{x} - bx - 5a + 5b$
Integrating both sides from $1$ to $2$:
$(a^2 - b^2) \int_1^2 f(x) dx = \int_1^2 \left( \frac{a}{x} - bx - 5(a - b) \right) dx$
$= \left[ a \log |x| - \frac{b x^2}{2} - 5(a - b)x \right]_1^2$
$= \left( a \log 2 - \frac{b(4)}{2} - 5(a - b)(2) \right) - \left( a \log 1 - \frac{b(1)}{2} - 5(a - b)(1) \right)$
$= a \log 2 - 2b - 10a + 10b - 0 + \frac{b}{2} + 5a - 5b$
$= a \log 2 - 5a + \frac{7}{2}b$
Thus,$\int_1^2 f(x) dx = \frac{1}{a^2 - b^2} \left[ a \log 2 - 5a + \frac{7}{2}b \right]$.

(D) We are given ${I_n} = \int_0^{\pi /4} {{\tan ^n}\theta \,d\theta }$.
Consider ${I_n} + {I_{n-2}} = \int_0^{\pi /4} {{\tan ^n}\theta \,d\theta } + \int_0^{\pi /4} {{\tan ^{n-2}}\theta \,d\theta }$.
${I_n} + {I_{n-2}} = \int_0^{\pi /4} {{\tan ^{n-2}}\theta (\tan ^2\theta + 1) \,d\theta }$.
Since $1 + \tan ^2\theta = \sec ^2\theta$,we have:
${I_n} + {I_{n-2}} = \int_0^{\pi /4} {{\tan ^{n-2}}\theta \sec ^2\theta \,d\theta }$.
Let $u = \tan \theta$,then $du = \sec ^2\theta \,d\theta$.
When $\theta = 0, u = 0$ and when $\theta = \pi /4, u = 1$.
Thus,${I_n} + {I_{n-2}} = \int_0^1 {u^{n-2}} \,du = \left[ \frac{u^{n-1}}{n-1} \right]_0^1 = \frac{1}{n-1}$.
For $n = 8$,we have ${I_8} + {I_6} = \frac{1}{8-1} = \frac{1}{7}$.

(A) We know that $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for any $u > 0$.
Also,$\tan^{-1} \left( \frac{1}{u} \right) = \cot^{-1}(u)$ for $u > 0$.
Let $I = \int_{-1}^{3} \left( \tan^{-1} \left( \frac{x}{x^2+1} \right) + \tan^{-1} \left( \frac{x^2+1}{x} \right) \right) dx$.
Note that the integrand is defined for $x \neq 0$.
Using the property $\tan^{-1} \left( \frac{1}{u} \right) = \cot^{-1}(u)$,the expression becomes:
$I = \int_{-1}^{3} \left( \tan^{-1} \left( \frac{x}{x^2+1} \right) + \cot^{-1} \left( \frac{x}{x^2+1} \right) \right) dx$.
Since $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$,the integrand simplifies to $\frac{\pi}{2}$.
Thus,$I = \int_{-1}^{3} \frac{\pi}{2} dx = \frac{\pi}{2} [x]_{-1}^{3} = \frac{\pi}{2} (3 - (-1)) = \frac{\pi}{2} (4) = 2\pi$.

(B) Let $I = \int_{0}^{1} \sin \left( 2 \tan^{-1} \sqrt{\frac{1+x}{1-x}} \right) \, dx$.
Substitute $x = \cos \theta$,then $dx = -\sin \theta \, d\theta$.
When $x = 0$,$\theta = \pi / 2$. When $x = 1$,$\theta = 0$.
The expression inside the sine function becomes:
$2 \tan^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}} = 2 \tan^{-1} \sqrt{\frac{2 \cos^2(\theta/2)}{2 \sin^2(\theta/2)}} = 2 \tan^{-1} (\cot(\theta/2)) = 2 \tan^{-1} \left( \tan \left( \frac{\pi}{2} - \frac{\theta}{2} \right) \right) = 2 \left( \frac{\pi}{2} - \frac{\theta}{2} \right) = \pi - \theta$.
Thus,the integral becomes:
$I = \int_{\pi/2}^{0} \sin(\pi - \theta) (- \sin \theta) \, d\theta = \int_{0}^{\pi/2} \sin^2 \theta \, d\theta$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$:
$I = \int_{0}^{\pi/2} \frac{1 - \cos 2\theta}{2} \, d\theta = \left[ \frac{\theta}{2} - \frac{\sin 2\theta}{4} \right]_{0}^{\pi/2} = \left( \frac{\pi}{4} - 0 \right) - (0 - 0) = \frac{\pi}{4}$.

(B) Let $I = \int_0^\pi x f(\sin x) dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^\pi (\pi - x) f(\sin(\pi - x)) dx$.
Since $\sin(\pi - x) = \sin x$,this becomes:
$I = \int_0^\pi (\pi - x) f(\sin x) dx = \pi \int_0^\pi f(\sin x) dx - \int_0^\pi x f(\sin x) dx$.
$I = \pi \int_0^\pi f(\sin x) dx - I$.
$2I = \pi \int_0^\pi f(\sin x) dx$.
$I = \frac{\pi}{2} \int_0^\pi f(\sin x) dx$.

(C) Let $I = \int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx$ ..... $(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^{\pi /2} \frac{\sqrt{\cot(\pi/2 - x)}}{\sqrt{\cot(\pi/2 - x)} + \sqrt{\tan(\pi/2 - x)}} \, dx$
Since $\cot(\pi/2 - x) = \tan x$ and $\tan(\pi/2 - x) = \cot x$,we have:
$I = \int_0^{\pi /2} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} \, dx$ ..... $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\pi /2} \frac{\sqrt{\cot x} + \sqrt{\tan x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx$
$2I = \int_0^{\pi /2} 1 \, dx$
$2I = [x]_0^{\pi /2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$
$I = \frac{\pi}{4}$

(D) Let $I = \int_0^{\pi /2} \frac{d\theta}{1 + \tan \theta}$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^{\pi /2} \frac{d\theta}{1 + \tan(\frac{\pi}{2} - \theta)} = \int_0^{\pi /2} \frac{d\theta}{1 + \cot \theta}$.
Since $\cot \theta = \frac{1}{\tan \theta}$,this becomes:
$I = \int_0^{\pi /2} \frac{d\theta}{1 + \frac{1}{\tan \theta}} = \int_0^{\pi /2} \frac{\tan \theta}{1 + \tan \theta} d\theta$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi /2} \left( \frac{1}{1 + \tan \theta} + \frac{\tan \theta}{1 + \tan \theta} \right) d\theta$.
$2I = \int_0^{\pi /2} \frac{1 + \tan \theta}{1 + \tan \theta} d\theta = \int_0^{\pi /2} 1 d\theta$.
$2I = [\theta]_0^{\pi /2} = \frac{\pi}{2}$.
Therefore,$I = \frac{\pi}{4}$.

(B) Let $f(x) = x|x|$.
We check if the function is even or odd:
$f(-x) = (-x)|-x| = -x|x| = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x) = x|x|$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-1}^{1} x|x| \, dx = 0$.

(B) Let $I = \int_0^\pi x \log(\sin x) \, dx$ ..... $(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^\pi (\pi - x) \log(\sin(\pi - x)) \, dx = \int_0^\pi (\pi - x) \log(\sin x) \, dx$ ..... $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^\pi (x + \pi - x) \log(\sin x) \, dx = \pi \int_0^\pi \log(\sin x) \, dx$
Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$:
$2I = 2\pi \int_0^{\pi/2} \log(\sin x) \, dx$
$I = \pi \int_0^{\pi/2} \log(\sin x) \, dx$
We know that $\int_0^{\pi/2} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2 = \frac{\pi}{2} \log(\frac{1}{2})$.
Therefore,$I = \pi \left( \frac{\pi}{2} \log(\frac{1}{2}) \right) = \frac{\pi^2}{2} \log(\frac{1}{2})$.

(D) Let $I = \int_0^{\pi /2} \log(\tan x) \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^{\pi /2} \log(\tan(\frac{\pi}{2} - x)) \, dx$
$I = \int_0^{\pi /2} \log(\cot x) \, dx$
$I = \int_0^{\pi /2} \log(\frac{1}{\tan x}) \, dx$
$I = \int_0^{\pi /2} (\log 1 - \log(\tan x)) \, dx$
Since $\log 1 = 0$,we have:
$I = - \int_0^{\pi /2} \log(\tan x) \, dx$
$I = -I$
$2I = 0 \implies I = 0$.

(A) Let $I = \int_0^{\pi /2} \log(\sin x) \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have $I = \int_0^{\pi /2} \log(\cos x) \, dx$.
Adding these two equations:
$2I = \int_0^{\pi /2} (\log(\sin x) + \log(\cos x)) \, dx = \int_0^{\pi /2} \log(\sin x \cos x) \, dx$.
Multiplying and dividing by $2$ inside the log:
$2I = \int_0^{\pi /2} \log(\frac{\sin 2x}{2}) \, dx = \int_0^{\pi /2} \log(\sin 2x) \, dx - \int_0^{\pi /2} \log 2 \, dx$.
$2I = \int_0^{\pi /2} \log(\sin 2x) \, dx - \frac{\pi}{2} \log 2$.
Let $2x = t$,then $2 \, dx = dt$. When $x=0, t=0$ and when $x=\pi/2, t=\pi$.
$2I = \frac{1}{2} \int_0^{\pi} \log(\sin t) \, dt - \frac{\pi}{2} \log 2$.
Using $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$:
$2I = \frac{1}{2} \cdot 2 \int_0^{\pi /2} \log(\sin t) \, dt - \frac{\pi}{2} \log 2$.
$2I = I - \frac{\pi}{2} \log 2$.
$I = -\frac{\pi}{2} \log 2$.

(C) Let $I = \int_0^{\pi /2} \frac{\cos x - \sin x}{1 + \sin x \cos x} \,dx$ .... $(i)$
Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get:
$I = \int_0^{\pi /2} \frac{\cos(\pi/2 - x) - \sin(\pi/2 - x)}{1 + \sin(\pi/2 - x)\cos(\pi/2 - x)} \,dx$
Since $\cos(\pi/2 - x) = \sin x$ and $\sin(\pi/2 - x) = \cos x$,we have:
$I = \int_0^{\pi /2} \frac{\sin x - \cos x}{1 + \cos x \sin x} \,dx$ .... $(ii)$
Adding equations $(i)$ and $(ii)$:
$I + I = \int_0^{\pi /2} \frac{\cos x - \sin x + \sin x - \cos x}{1 + \sin x \cos x} \,dx$
$2I = \int_0^{\pi /2} \frac{0}{1 + \sin x \cos x} \,dx = 0$
Therefore,$I = 0$.

(D) Let $f(x) = \log \left( \frac{2 - x}{2 + x} \right)$.
Now,calculate $f(-x)$:
$f(-x) = \log \left( \frac{2 - (-x)}{2 + (-x)} \right) = \log \left( \frac{2 + x}{2 - x} \right)$.
We know that $\log \left( \frac{1}{a} \right) = -\log(a)$,so:
$f(-x) = \log \left( \left( \frac{2 - x}{2 + x} \right)^{-1} \right) = -\log \left( \frac{2 - x}{2 + x} \right) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-1}^{1} \log \left( \frac{2 - x}{2 + x} \right) \, dx = 0$.

(C) Let $f(x) = x^{17} \cos^{4} x$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = (-x)^{17} \cos^{4}(-x) = -x^{17} \cos^{4} x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-1}^{1} x^{17} \cos^{4} x \, dx = 0$.

(D) Let $I = \int_0^{\pi /2} \frac{\sin^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx$ ..... $(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\pi /2} \frac{\sin^{3/2} (\pi /2 - x)}{\cos^{3/2} (\pi /2 - x) + \sin^{3/2} (\pi /2 - x)} dx$
Since $\sin(\pi /2 - x) = \cos x$ and $\cos(\pi /2 - x) = \sin x$,we have:
$I = \int_0^{\pi /2} \frac{\cos^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx$ ..... $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\pi /2} \frac{\sin^{3/2} x + \cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx$
$2I = \int_0^{\pi /2} 1 dx$
$2I = [x]_0^{\pi /2} = \pi /2$
$I = \pi /4$

(A) Let $f(\theta) = \log \left( \frac{2 - \sin \theta}{2 + \sin \theta} \right)$.
We check if the function is odd or even by evaluating $f(-\theta)$:
$f(-\theta) = \log \left( \frac{2 - \sin(-\theta)}{2 + \sin(-\theta)} \right) = \log \left( \frac{2 + \sin \theta}{2 - \sin \theta} \right)$.
Since $\log \left( \frac{2 + \sin \theta}{2 - \sin \theta} \right) = \log \left( \left( \frac{2 - \sin \theta}{2 + \sin \theta} \right)^{-1} \right) = -\log \left( \frac{2 - \sin \theta}{2 + \sin \theta} \right) = -f(\theta)$.
Thus,$f(\theta)$ is an odd function.
By the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) dx = 0$.
Therefore,$\int_{-\pi/2}^{\pi/2} \log \left( \frac{2 - \sin \theta}{2 + \sin \theta} \right) d\theta = 0$.

(C) Let $I = \int_0^{\pi /4} \log (1 + \tan \theta )\,d\theta$
Using the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$,we get:
$I = \int_0^{\pi /4} \log \left( 1 + \tan \left( \frac{\pi }{4} - \theta \right) \right) d\theta$
Since $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we have $\tan(\frac{\pi}{4} - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta}$
$I = \int_0^{\pi /4} \log \left( 1 + \frac{1 - \tan \theta}{1 + \tan \theta} \right) d\theta$
$I = \int_0^{\pi /4} \log \left( \frac{1 + \tan \theta + 1 - \tan \theta}{1 + \tan \theta} \right) d\theta$
$I = \int_0^{\pi /4} \log \left( \frac{2}{1 + \tan \theta} \right) d\theta$
$I = \int_0^{\pi /4} (\log 2 - \log (1 + \tan \theta )) d\theta$
$I = \int_0^{\pi /4} \log 2 \, d\theta - \int_0^{\pi /4} \log (1 + \tan \theta ) d\theta$
$I = \log 2 [\theta]_0^{\pi /4} - I$
$2I = \frac{\pi}{4} \log 2$
$I = \frac{\pi}{8} \log 2$

(D) Let $I = \int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}d\theta } $.
Using the property $\int_0^{2a} f(x) dx = \int_0^{2a} f(2a-x) dx$,we have:
$I = \int_0^{2\pi } {\frac{{\sin (2(2\pi - \theta ))}}{{a - b\cos (2\pi - \theta )}}d\theta } $
$I = \int_0^{2\pi } {\frac{{\sin (4\pi - 2\theta )}}{{a - b\cos \theta }}d\theta } $
Since $\sin(4\pi - 2\theta) = -\sin 2\theta$,we get:
$I = - \int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}d\theta } $
$I = -I$
$2I = 0$
$I = 0$.

(A) Let $I = \int_0^1 f(1 - x) \, dx$.
Substitute $1 - x = t$,which implies $x = 1 - t$ and $dx = -dt$.
When $x = 0$,$t = 1$. When $x = 1$,$t = 0$.
Substituting these into the integral:
$I = \int_1^0 f(t) (-dt) = -\int_1^0 f(t) \, dt = \int_0^1 f(t) \, dt$.
Since the variable of integration is a dummy variable,$\int_0^1 f(t) \, dt = \int_0^1 f(x) \, dx$.
Thus,$\int_0^1 f(1 - x) \, dx = \int_0^1 f(x) \, dx$.

(A) Let $I = \int_{-1/2}^{1/2} f(x) dx$,where $f(x) = (\cos x) \left[ \log \left( \frac{1-x}{1+x} \right) \right]$.
Check if the function is odd or even:
$f(-x) = \cos(-x) \left[ \log \left( \frac{1-(-x)}{1+(-x)} \right) \right]$
Since $\cos(-x) = \cos x$ and $\log \left( \frac{1+x}{1-x} \right) = \log \left( \left( \frac{1-x}{1+x} \right)^{-1} \right) = -\log \left( \frac{1-x}{1+x} \right)$,
We have $f(-x) = \cos x \left[ -\log \left( \frac{1-x}{1+x} \right) \right] = -f(x)$.
Since $f(x)$ is an odd function and the limits of integration are symmetric $[-a, a]$,the integral is $0$.

(D) Let $I = \int_{0}^{1} \frac{dx}{x + \sqrt{1 - x^2}}$.
Substitute $x = \sin \theta$,then $dx = \cos \theta \, d\theta$.
When $x = 0$,$\theta = 0$. When $x = 1$,$\theta = \frac{\pi}{2}$.
$I = \int_{0}^{\pi/2} \frac{\cos \theta \, d\theta}{\sin \theta + \cos \theta}$.
Using the property $\int_{0}^{a} f(\theta) \, d\theta = \int_{0}^{a} f(a - \theta) \, d\theta$,we get:
$I = \int_{0}^{\pi/2} \frac{\cos(\pi/2 - \theta) \, d\theta}{\sin(\pi/2 - \theta) + \cos(\pi/2 - \theta)} = \int_{0}^{\pi/2} \frac{\sin \theta \, d\theta}{\cos \theta + \sin \theta}$.
Adding the two expressions for $I$:
$2I = \int_{0}^{\pi/2} \frac{\cos \theta + \sin \theta}{\sin \theta + \cos \theta} \, d\theta = \int_{0}^{\pi/2} 1 \, d\theta = [\theta]_{0}^{\pi/2} = \frac{\pi}{2}$.
Therefore,$I = \frac{\pi}{4}$.

(B) The property of definite integrals states that for an odd function $f(x)$,where $f(-x) = -f(x)$,the integral over a symmetric interval $[-a, a]$ is zero.
Specifically,$\int_{-a}^{a} f(x) \, dx = 0$ if $f(x)$ is an odd function.
In this case,the interval is $[-1, 1]$,so $\int_{-1}^{1} f(x) \, dx = 0$ implies that $f(x)$ is an odd function,which satisfies the condition $f(-x) = -f(x)$.
Therefore,the correct option is $(b)$.

(C) The expression ${x - [x]}$ represents the fractional part of $x$,denoted as $\{x\}$.
Since the fractional part function $f(x) = \{x\} = x - [x]$ is a periodic function with period $T = 1$,we can use the property $\int_0^{nT} f(x) \,dx = n \int_0^T f(x) \,dx$.
Therefore,$\int_0^n \{x\} \,dx = n \int_0^1 \{x\} \,dx$.
In the interval $[0, 1)$,$[x] = 0$,so $\{x\} = x - 0 = x$.
Thus,$\int_0^n \{x\} \,dx = n \int_0^1 x \,dx$.
Evaluating the integral: $n \left[ \frac{x^2}{2} \right]_0^1 = n \left( \frac{1}{2} - 0 \right) = \frac{n}{2}$.

(B) Let $I = \int_0^\pi x \sin^3 x \, dx$ ..... $(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:
$I = \int_0^\pi (\pi - x) \sin^3(\pi - x) \, dx = \int_0^\pi (\pi - x) \sin^3 x \, dx$ ..... $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^\pi (x + \pi - x) \sin^3 x \, dx = \pi \int_0^\pi \sin^3 x \, dx$
Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we get $\sin^3 x = \frac{3 \sin x - \sin 3x}{4}$.
$2I = \pi \int_0^\pi \frac{3 \sin x - \sin 3x}{4} \, dx = \frac{\pi}{4} \left[ -3 \cos x + \frac{\cos 3x}{3} \right]_0^\pi$
$2I = \frac{\pi}{4} \left[ (-3(-1) + \frac{-1}{3}) - (-3(1) + \frac{1}{3}) \right] = \frac{\pi}{4} \left[ (3 - \frac{1}{3}) - (-3 + \frac{1}{3}) \right]$
$2I = \frac{\pi}{4} \left[ \frac{8}{3} + \frac{8}{3} \right] = \frac{\pi}{4} \left( \frac{16}{3} \right) = \frac{4\pi}{3}$
Therefore,$I = \frac{2\pi}{3}$.

(C) Let $I = \int_0^{\pi /2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx$ .....$(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:
$I = \int_0^{\pi /2} \frac{\sqrt{\cos(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} \, dx$
Since $\cos(\pi/2 - x) = \sin x$ and $\sin(\pi/2 - x) = \cos x$,we get:
$I = \int_0^{\pi /2} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx$ .....$(ii)$
Adding equations $(i)$ and $(ii)$:
$2I = \int_0^{\pi /2} \left( \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) \, dx$
$2I = \int_0^{\pi /2} 1 \, dx$
$2I = [x]_0^{\pi /2} = \frac{\pi}{2}$
$I = \frac{\pi}{4}$

(D) Let $I = \int_0^{\pi /2} \frac{x \sin x \cos x}{\cos^4 x + \sin^4 x} \, dx$ .....$(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^{\pi /2} \frac{(\frac{\pi}{2} - x) \cos x \sin x}{\sin^4 x + \cos^4 x} \, dx$ .....$(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\pi /2} \frac{\frac{\pi}{2} \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx$
$I = \frac{\pi}{4} \int_0^{\pi /2} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \, dx$
Divide numerator and denominator by $\cos^4 x$:
$I = \frac{\pi}{4} \int_0^{\pi /2} \frac{\tan x \sec^2 x}{1 + \tan^4 x} \, dx$
Let $\tan^2 x = t$,then $2 \tan x \sec^2 x \, dx = dt$,so $\tan x \sec^2 x \, dx = \frac{1}{2} dt$. When $x=0, t=0$; when $x=\frac{\pi}{2}, t \to \infty$.
$I = \frac{\pi}{4} \int_0^{\infty} \frac{1}{1 + t^2} \cdot \frac{1}{2} \, dt = \frac{\pi}{8} [\tan^{-1} t]_0^{\infty} = \frac{\pi}{8} \cdot \frac{\pi}{2} = \frac{\pi^2}{16}$.

(D) The property of definite integrals states that $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$.
This is a standard property used in evaluating definite integrals where the variable $x$ is replaced by $(a - x)$.

(B) We need to evaluate the integral $I = \int_0^{\pi /2} |\sin x - \cos x| \, dx$.
Since $\sin x - \cos x \le 0$ for $x \in [0, \pi/4]$ and $\sin x - \cos x \ge 0$ for $x \in [\pi/4, \pi/2]$,we split the integral at $x = \pi/4$:
$I = \int_0^{\pi /4} -(\sin x - \cos x) \, dx + \int_{\pi /4}^{\pi /2} (\sin x - \cos x) \, dx$
$I = \int_0^{\pi /4} (\cos x - \sin x) \, dx + \int_{\pi /4}^{\pi /2} (\sin x - \cos x) \, dx$
Evaluating the first part: $[\sin x + \cos x]_0^{\pi /4} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.
Evaluating the second part: $[-\cos x - \sin x]_{\pi /4}^{\pi /2} = (0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 + \sqrt{2} = \sqrt{2} - 1$.
Adding both parts: $I = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)$.

(D) Let $I = \int_0^\pi \frac{x \tan x}{\sec x + \tan x} \,dx$.
Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get:
$I = \int_0^\pi \frac{(\pi - x) \tan(\pi - x)}{\sec(\pi - x) + \tan(\pi - x)} \,dx = \int_0^\pi \frac{(\pi - x)(-\tan x)}{-\sec x - \tan x} \,dx = \int_0^\pi \frac{(\pi - x) \tan x}{\sec x + \tan x} \,dx$.
Adding the two expressions for $I$:
$2I = \int_0^\pi \frac{\pi \tan x}{\sec x + \tan x} \,dx = \pi \int_0^\pi \frac{\sin x}{1 + \sin x} \,dx$.
$2I = \pi \int_0^\pi \frac{1 + \sin x - 1}{1 + \sin x} \,dx = \pi \left( \int_0^\pi 1 \,dx - \int_0^\pi \frac{1}{1 + \sin x} \,dx \right)$.
Since $\int_0^\pi \frac{1}{1 + \sin x} \,dx = \int_0^\pi \frac{1 - \sin x}{\cos^2 x} \,dx = \int_0^\pi (\sec^2 x - \sec x \tan x) \,dx = [\tan x - \sec x]_0^\pi = (0 - (-1)) - (0 - 1) = 2$.
Thus,$2I = \pi (\pi - 2) = \pi^2 - 2\pi$.
Therefore,$I = \frac{\pi^2}{2} - \pi = \pi \left( \frac{\pi}{2} - 1 \right)$.

(A) Let $I = \int_0^\pi \frac{x \tan x}{\sec x + \cos x} \,dx$.
Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get:
$I = \int_0^\pi \frac{(\pi - x) \tan(\pi - x)}{\sec(\pi - x) + \cos(\pi - x)} \,dx = \int_0^\pi \frac{(\pi - x)(-\tan x)}{-\sec x - \cos x} \,dx = \int_0^\pi \frac{(\pi - x) \tan x}{\sec x + \cos x} \,dx$.
Adding the two expressions for $I$:
$2I = \int_0^\pi \frac{\pi \tan x}{\sec x + \cos x} \,dx = \pi \int_0^\pi \frac{\sin x / \cos x}{1/\cos x + \cos x} \,dx = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \,dx$.
Let $t = \cos x$,then $dt = -\sin x \,dx$. When $x=0, t=1$ and when $x=\pi, t=-1$.
$2I = \pi \int_1^{-1} \frac{-dt}{1 + t^2} = \pi \int_{-1}^1 \frac{dt}{1 + t^2} = \pi [\tan^{-1} t]_{-1}^1$.
$2I = \pi (\tan^{-1}(1) - \tan^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$.
Therefore,$I = \frac{\pi^2}{4}$.

(A) Let $f(x) = \sin^3 x \cos^2 x$.
Check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \sin^3(-x) \cos^2(-x) = (-\sin x)^3 (\cos x)^2 = -\sin^3 x \cos^2 x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-1}^{1} \sin^3 x \cos^2 x \, dx = 0$.

(B) Let $I = \int_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}(2n + 1)x\,dx}$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^\pi {{e^{{{\sin }^2}(\pi-x)}}{{\cos }^3}(2n + 1)(\pi-x)\,dx}$.
Since $\sin(\pi-x) = \sin x$,we have $\sin^2(\pi-x) = \sin^2 x$.
Also,$\cos((2n+1)(\pi-x)) = \cos((2n+1)\pi - (2n+1)x) = -\cos((2n+1)x)$ because $(2n+1)$ is an odd integer.
Therefore,$\cos^3((2n+1)(\pi-x)) = (-\cos((2n+1)x))^3 = -\cos^3((2n+1)x)$.
Substituting these into the integral:
$I = \int_0^\pi {{e^{{{\sin }^2}x}} \cdot [-\cos^3(2n + 1)x]\,dx} = -I$.
Thus,$2I = 0$,which implies $I = 0$.

(C) Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have:
$I = \int_{0}^{1} (1 - x)(1 - (1 - x))^n dx$
$I = \int_{0}^{1} (1 - x)x^n dx$
$I = \int_{0}^{1} (x^n - x^{n+1}) dx$
$I = \left[ \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_{0}^{1}$
$I = \left( \frac{1}{n+1} - \frac{1}{n+2} \right) - (0 - 0)$
$I = \frac{1}{n+1} - \frac{1}{n+2}$

(C) Let $g(a) = \int_a^{a + T} {f(x)dx}$.
Differentiating with respect to $a$ using the Leibniz rule:
$g'(a) = \frac{d}{da} \int_a^{a + T} {f(x)dx} = f(a + T) \cdot \frac{d}{da}(a + T) - f(a) \cdot \frac{d}{da}(a)$.
Since $f(x)$ is a periodic function with period $T$,we have $f(a + T) = f(a)$.
Thus,$g'(a) = f(a + T) - f(a) = f(a) - f(a) = 0$.
Since the derivative $g'(a) = 0$,the function $g(a)$ is a constant value.
Therefore,the integral $I = \int_a^{a + T} {f(x)dx}$ is independent of $a$.

(B) Let $I = \int_0^\pi x f(\cos^2 x + \tan^4 x) dx$ ... $(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^\pi (\pi - x) f(\cos^2(\pi - x) + \tan^4(\pi - x)) dx$
Since $\cos(\pi - x) = -\cos x$ and $\tan(\pi - x) = -\tan x$,we have $\cos^2(\pi - x) = \cos^2 x$ and $\tan^4(\pi - x) = \tan^4 x$.
Thus,$I = \int_0^\pi (\pi - x) f(\cos^2 x + \tan^4 x) dx$ ... $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^\pi (x + \pi - x) f(\cos^2 x + \tan^4 x) dx$
$2I = \pi \int_0^\pi f(\cos^2 x + \tan^4 x) dx$
Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$:
Since $f(\cos^2(\pi-x) + \tan^4(\pi-x)) = f(\cos^2 x + \tan^4 x)$,the integral satisfies this condition.
$2I = \pi \cdot 2 \int_0^{\pi/2} f(\cos^2 x + \tan^4 x) dx$
$I = \pi \int_0^{\pi/2} f(\cos^2 x + \tan^4 x) dx$
Comparing this with the given equation $I = k \int_0^{\pi/2} f(\cos^2 x + \tan^4 x) dx$,we find $k = \pi$.

(C) Let $f(x) = \frac{x^2 \sin x}{1 + x^6}$.
Check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \frac{(-x)^2 \sin(-x)}{1 + (-x)^6} = \frac{x^2 (-\sin x)}{1 + x^6} = -\frac{x^2 \sin x}{1 + x^6} = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
Using the property of definite integrals,$\int_{-a}^{a} f(x) \, dx = 0$ if $f(x)$ is an odd function.
Therefore,$\int_{-3}^{3} \frac{x^2 \sin x}{1 + x^6} \, dx = 0$.

(D) Let $I = \int_0^{\pi /2} \frac{dx}{1 + \tan^3 x}$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^{\pi /2} \frac{dx}{1 + \tan^3(\pi/2 - x)} = \int_0^{\pi /2} \frac{dx}{1 + \cot^3 x}$.
$I = \int_0^{\pi /2} \frac{\tan^3 x}{1 + \tan^3 x} dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi /2} \left( \frac{1}{1 + \tan^3 x} + \frac{\tan^3 x}{1 + \tan^3 x} \right) dx$.
$2I = \int_0^{\pi /2} \frac{1 + \tan^3 x}{1 + \tan^3 x} dx = \int_0^{\pi /2} 1 dx$.
$2I = [x]_0^{\pi /2} = \frac{\pi}{2}$.
Therefore,$I = \frac{\pi}{4}$.

(A) Let $I = \int_{\pi /4}^{3\pi /4} \frac{\phi}{1 + \sin \phi} \, d\phi$.
Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we have:
$I = \int_{\pi /4}^{3\pi /4} \frac{\pi - \phi}{1 + \sin(\pi - \phi)} \, d\phi = \int_{\pi /4}^{3\pi /4} \frac{\pi - \phi}{1 + \sin \phi} \, d\phi$.
Adding the two expressions for $I$:
$2I = \int_{\pi /4}^{3\pi /4} \frac{\phi + \pi - \phi}{1 + \sin \phi} \, d\phi = \pi \int_{\pi /4}^{3\pi /4} \frac{1}{1 + \sin \phi} \, d\phi$.
$2I = \pi \int_{\pi /4}^{3\pi /4} \frac{1 - \sin \phi}{\cos^2 \phi} \, d\phi = \pi \int_{\pi /4}^{3\pi /4} (\sec^2 \phi - \sec \phi \tan \phi) \, d\phi$.
$2I = \pi [\tan \phi - \sec \phi]_{\pi /4}^{3\pi /4}$.
$2I = \pi [(\tan(3\pi/4) - \sec(3\pi/4)) - (\tan(\pi/4) - \sec(\pi/4))]$.
$2I = \pi [(-1 - (-\sqrt{2})) - (1 - \sqrt{2})] = \pi [\sqrt{2} - 1 - 1 + \sqrt{2}] = \pi [2\sqrt{2} - 2] = 2\pi(\sqrt{2} - 1)$.
$I = \pi(\sqrt{2} - 1)$.
Since $\tan(\pi/8) = \sqrt{2} - 1$,the value is $\pi \tan(\pi/8)$.

(B) Let $I = \int_a^b x f(x) dx$.
Using the property $\int_a^b g(x) dx = \int_a^b g(a + b - x) dx$,we get:
$I = \int_a^b (a + b - x) f(a + b - x) dx$.
Since it is given that $f(a + b - x) = f(x)$,we substitute this into the integral:
$I = \int_a^b (a + b - x) f(x) dx$.
$I = (a + b) \int_a^b f(x) dx - \int_a^b x f(x) dx$.
$I = (a + b) \int_a^b f(x) dx - I$.
$2I = (a + b) \int_a^b f(x) dx$.
$I = \frac{a + b}{2} \int_a^b f(x) dx$.

(A) Let $I = \int_0^\pi x \sin x \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^\pi (\pi - x) \sin(\pi - x) \, dx$
Since $\sin(\pi - x) = \sin x$,we have:
$I = \int_0^\pi (\pi - x) \sin x \, dx = \pi \int_0^\pi \sin x \, dx - \int_0^\pi x \sin x \, dx$
$I = \pi \int_0^\pi \sin x \, dx - I$
$2I = \pi \int_0^\pi \sin x \, dx$
$2I = \pi [-\cos x]_0^\pi$
$2I = \pi [-\cos \pi - (-\cos 0)]$
$2I = \pi [-(-1) - (-1)] = \pi [1 + 1] = 2\pi$
$I = \pi$.

(D) Let $I = \int_{ - a}^a {\sqrt {\frac{{a - x}}{{a + x}}} dx} $.
To evaluate this,we substitute $x = a \cos \theta$.
Then $dx = -a \sin \theta \, d\theta$.
When $x = -a$,$\cos \theta = -1$,so $\theta = \pi$.
When $x = a$,$\cos \theta = 1$,so $\theta = 0$.
$I = \int_{\pi}^0 \sqrt{\frac{a - a \cos \theta}{a + a \cos \theta}} (-a \sin \theta) \, d\theta$.
$I = \int_0^{\pi} \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} (a \sin \theta) \, d\theta$.
Using half-angle formulas,$\sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \sqrt{\frac{2 \sin^2(\theta/2)}{2 \cos^2(\theta/2)}} = \tan(\theta/2)$.
$I = a \int_0^{\pi} \tan(\theta/2) \cdot 2 \sin(\theta/2) \cos(\theta/2) \, d\theta$.
$I = 2a \int_0^{\pi} \frac{\sin(\theta/2)}{\cos(\theta/2)} \cdot \sin(\theta/2) \cos(\theta/2) \, d\theta$.
$I = 2a \int_0^{\pi} \sin^2(\theta/2) \, d\theta$.
Using $\sin^2(\theta/2) = \frac{1 - \cos \theta}{2}$,we get $I = 2a \int_0^{\pi} \frac{1 - \cos \theta}{2} \, d\theta = a [\theta - \sin \theta]_0^{\pi} = a(\pi - 0) = a\pi$.
Comparing $a\pi$ with $k\pi$,we find $k = a$.

(B) We are given the property $\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_a^{2a} f(x) \, dx$.
By substituting $x = 2a - t$ in the second integral $\int_a^{2a} f(x) \, dx$,we get $dx = -dt$. When $x = a$,$t = a$,and when $x = 2a$,$t = 0$.
Thus,$\int_a^{2a} f(x) \, dx = \int_a^0 f(2a - t) (-dt) = \int_0^a f(2a - t) \, dt = \int_0^a f(2a - x) \, dx$.
Substituting this back,we have $\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_0^a f(2a - x) \, dx = \int_0^a [f(x) + f(2a - x)] \, dx$.
Given that $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$,it follows that $\int_0^a [f(x) + f(2a - x)] \, dx = \int_0^a 2f(x) \, dx$.
This equality holds if $f(2a - x) = f(x)$.

(A) Given $I = \int_0^{\pi /4} \sin^2 x \, dx$ and $J = \int_0^{\pi /4} \cos^2 x \, dx$.
Adding the two integrals:
$I + J = \int_0^{\pi /4} (\sin^2 x + \cos^2 x) \, dx$.
Using the trigonometric identity $\sin^2 x + \cos^2 x = 1$,we get:
$I + J = \int_0^{\pi /4} 1 \, dx$.
Evaluating the integral:
$I + J = [x]_0^{\pi /4} = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.
Therefore,$I = \frac{\pi}{4} - J$.

(D) Let $I = \int_2^3 {\frac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} \,dx$ .....$(i)$
Using the property $\int_a^b f(x) \,dx = \int_a^b f(a + b - x) \,dx$,we replace $x$ with $(2 + 3 - x) = (5 - x)$.
$\therefore I = \int_2^3 {\frac{{\sqrt {5 - x} }}{{\sqrt {5 - (5 - x)} + \sqrt {5 - x} }}} \,dx$
$I = \int_2^3 {\frac{{\sqrt {5 - x} }}{{\sqrt x + \sqrt {5 - x} }}} \,dx$ .....$(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_2^3 {\frac{{\sqrt x + \sqrt {5 - x} }}{{\sqrt {5 - x} + \sqrt x }}} \,dx$
$2I = \int_2^3 1 \,dx$
$2I = [x]_2^3 = 3 - 2 = 1$
$I = \frac{1}{2}$.