Maths Solve the Following Question.(4 Marks)

Let $I=\int_a^b f(x) d x$
Put x= a + b - t
∴ dx = -dt
When x = a, t = b and when x = b, t = a
$\therefore I=\int_b^a f(a+b-t)(-d t)$
$\therefore I=-\int_b^a f(a+b-t) d t$
$\therefore I=\int_a^b f(a+b-t) d t \ldots\left[\because \int_a^b f(x) d x=-\int_b^a f(x) d x\right]$
$\therefore \int_a^b f(x) d x=\int_a^b f(a+b-x) d x \ldots\left[\because \int_a^b f(x) d x=\int_a^b f(t) d t\right]$
Let $I=\int_a^b \frac{f(x)}{f(x)+f(a+b-x)} d x \ldots(i)$
$\therefore I=\int_a^b \frac{f(a+b-x)}{f(a+b-x)+f(a+b-(a+b-x))} d x$
$\therefore I=\int_a^b \frac{f(a+b-x)}{f(a+b-x)+f(x)} d x \ldots(i i)$
Adding (i) and (ii) we get
$\begin{aligned} & 2 I=\int_a^b \frac{f(x)+f(a+b-x)}{f(x)+f(a+b-x)} d x \\ & \therefore 2 I=\int_a^b 1 d x \\ & \therefore 2 I=[x]_a^b \\ & \therefore I=\frac{b-a}{2}\end{aligned}$

$\int_{-a}^a \sqrt{\frac{a-x}{a+x}} d x$
Let $I=\int_{-a}^a \sqrt{\frac{a-x}{a+x}} d x$
$=\int_{-a}^a \sqrt{\frac{(a-x)(a-x)}{(a+x)(a-x)}} d x$
$=\int_{-a}^a \frac{a-x}{\sqrt{a^2-x^2}} d x$
$=\int_{-a}^a \frac{a}{\sqrt{a^2-x^2}} d x-\int_{-a}^a \frac{x}{\sqrt{a^2-x^2}} d x$
[but $\frac{a}{\sqrt{a^2-x^2}}$ is an is an even function and $\frac{x}{\sqrt{a^2-x^2}}$ is an odd function]
$=2 a \cdot\left[\sin ^{-1}\left(\frac{x}{a}\right)\right]_0^a$
$=2 a \cdot\left[\sin ^{-1} 1-\sin ^{-1} 0\right]$
$=2 a\left[\frac{\pi}{2}-0\right]$
$\int_{-a}^a \sqrt{\frac{a-x}{a+x}} \cdot d x=\pi a$

$\begin{aligned} & \text { Let } I=\int_0^\pi \frac{x \sin x}{1+\sin x} d x \\ & =\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\sin (\pi-x)} d x\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right]\end{aligned}$$\begin{aligned} & =\int_0^\pi \frac{(\pi-x) \sin x}{1+\sin x} d x \\ & =\int_0^\pi \frac{\pi \sin x}{1+\sin x} d x-I \\ & I=\int_0^\pi \frac{\pi \sin x}{1+\sin x} d x-I\end{aligned}$$\begin{aligned} & 2 I=\int_0^\pi \frac{\pi \sin x \cdot(1-\sin x)}{(1+\sin x)(1-\sin x)} d x \\ & 2 I=\int_0^\pi \frac{\pi \sin x \cdot(1-\sin x)}{1-\sin ^2 x} d x \\ & \frac{2 I}{\pi}=\int_0^\pi \frac{\sin x \cdot(1-\sin x)}{\cos ^2 x} d x\end{aligned}$$\begin{aligned} \frac{2 I}{\pi} & =\int_0^\pi \frac{\sin x \cdot-\sin ^2 x}{\cos ^2 x} d x \\ \frac{2 I}{\pi} & =\int_0^\pi \frac{\sin x}{\cos ^2 x} d x-\int_0^\pi \frac{\sin ^2 x}{\cos ^2 x} d x\end{aligned}$
$\begin{aligned} & \frac{2 I}{\pi}=\int_0^\pi \sec x \cdot \tan x d x-\int_0^\pi \tan ^2 x d x \\ & \frac{2 I}{\pi}=[\sec x]_0^\pi-\int_0^\pi\left(\sec ^2 x-1\right) d x\end{aligned}$$\begin{aligned} & \frac{2 I}{\pi}=[\sec \pi-\sec 0]-\int_0^\pi \sec ^2 x \cdot d x+\int_0^\pi 1 d x \\ & \frac{2 I}{\pi}=[-1-1]-[\tan x]_0^\pi-[x]_0^\pi \\ & \frac{2 I}{\pi}=[-2]-[\tan \pi-\tan 0]+\pi \\ & \frac{2 I}{\pi}=[-2]-0+\pi \\ & \therefore I=\frac{(\pi-2) \pi}{2}\end{aligned}$

L.H.S becomes
$\int_{-a}^a f(x) d x=\int_{-a}^0 f(x) d x+\int_0^a f(x) d x$...(i)
Consider $\int_{-a}^0 f(x) d x$
Put x = – t
∴ dx = – dt
When x = – a, t = a
and when x = 0, t = 0
$\therefore \int_{-a}^0 f(x) d x=\int_a^0 f(-t)(-d t)$
$=-\int_a^0 f(-t) d t$
$\begin{aligned} & =\int_0^a f(-t) d t \quad \ldots . . .\left[\because \int_a^b f(x) d x=-\int_b^a f(x) d x\right] \\ & =\int_0^a f(-x) d x \quad \ldots . .\left[\because \int_a^b f(x) d x=\int_a^b f(t) d t\right]\end{aligned}$
Equation (i) becomes
$\begin{aligned} & \int_{-a}^a f(x) d x=\int_0^a f(-x) d x+\int_0^a f(x) d x \\ & =\int_0^a[f(-x)+f(x)] d x \ldots . . \text { (ii) }\end{aligned}$
Case I:
If f(x) is an even function, then f(– x) = f(x),
Equation (ii) becomes
$\int_{-a}^a f(x) d x=2 \cdot \int_0^a f(x) d x$
Case II:
If f(x) is an odd function, then f(– x) = – f(x),
Equation (ii) becomes
$\begin{array}{rlrl}\int_{-a}^a f(x) d x & =0 & \\ \int_{-a}^a f(x) d x & =2 \cdot \int_0^a f(x) d x, & & \text { If } f(x) \text { is an even function } \\ & =0, & & \text { if } f(x) \text { is an odd function }\end{array}$

Let $I =\int_0^a f(x) d x$
Put x = a – t
∴ dx = – dt
When x = 0, t = a - 0 = a
When x = a, t = a - a = 0
$\begin{aligned} & I=\int_0^a f(x) d x=\int_a^0 f(a-t)(-d t) \\ & =-\int_a^0 f(a-t) d t \quad \ldots \ldots .\left[\because \int_a^b f(x) d x=-\int_b^a f(x) d x\right] \\ & =\int_0^a f(a-x) d x \quad \ldots \ldots .\left[\because \int_a^b f(x) d x=-\int_b^a f(t) d x\right] \\ & \therefore \int_0^a f(x) d x=\int_0^a f(a-x) d x\end{aligned}$
Let $I =\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x}$......(i)
$I=\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} \quad \ldots \ldots .\left[\because \int_0^a f(x) d x=-\int_0^a f(a-x) d x\right]$
$=\int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x$......(ii)
Adding (i) and (ii), we get
$\begin{aligned} & 2 I=\int_0^{\frac{\pi}{2}} \frac{\sin x+\cos x}{\sin x+\cos x} d x \\ & =\int_0^{\frac{\pi}{2}} 1 d x\end{aligned}$
$\begin{aligned} & =[x]_0^{\frac{\pi}{2}} \\ & =\frac{\pi}{2}-0 \\ & 2 I=\frac{\pi}{2}\end{aligned}$
$\begin{aligned} & I=\frac{\pi}{4} \\ & \therefore \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x=\frac{\pi}{4}\end{aligned}$