Maths Solve the Following Question.(3 Marks)

Let $I =\int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x$...(1)
$=\int_1^2 \frac{\sqrt{3-x}}{\sqrt{3-(3-x)}+\sqrt{3-x}} d x$
$\ldots \ldots\left\{\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right\}$
$I=\int_1^2 \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} d x$...(2)
Adding (1) and (2)
$2 I =\int_1^2 \frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} d x$
$=\int_1^2 1 d x=[x]_1^2$
$\therefore \quad 2 I =2-1=1$
$\therefore \quad I=\frac{1}{2}$

$\begin{aligned} & I=\int_0^\pi \frac{x}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x \ldots \ldots \ldots \ldots(i) \\ & I=\int_0^\pi \frac{\pi-x}{a^2 \cos ^2(\pi-x)+b^2 \sin ^2(\pi-x)} d x \\ & I=\int_0^\pi \frac{\pi-x}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x \ldots \ldots \ldots .(i i) \\ & \int_0^a f(x) d x=\int_0^a f(a-x) d x\end{aligned}$
Adding (i) and (ii), we get
$2 I =\int_0^\pi \frac{x+\pi-x}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x$
$2 I =\int_0^\pi \frac{\pi}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x$
$2 I =\int_0^\pi \frac{\pi \sec ^2 x}{a^2+b^2 \tan ^2 x} \quad \cdots \cdots \ldots . . \frac{1}{b^2} \int_0^\pi \frac{\pi \sec ^2 x d x}{\left(\frac{a}{b}\right)^2+\tan ^2 x}$
$2 I =\frac{\pi}{b^2} \int \frac{d t}{\left(\frac{a}{b}\right)^2}+t^2 \quad \ldots \ldots \ldots . .\left[\tan x=t \rightarrow \sec ^2 x d x=d t\right]$
$2 I =\frac{\pi}{b^2}\left[\left(\frac{b}{a}\right) \tan ^{-1}\left(b \frac{t}{a}\right)\right]_0^\pi$
$2 l =\frac{\pi}{a b}\left[\tan ^{-1}\left(\frac{b}{a} \tan x\right)\right]_0^\pi$
$2 I =\frac{\pi}{a b}(0-0)=0$
2 I = 0
I = 0