Answer:
Option A
Explanation:
We have,
$I=\int_{0}^{\pi} \frac{xdx}{4 \cos^{2}x+9 \sin^{2}x}$
$\Rightarrow$ $I=\int_{0}^{\pi} \frac{(\pi-x)dx}{4 \cos^{2}(\pi-x)+9 \sin^{2}(\pi-x)}$
$\Rightarrow$ $I=\int_{0}^{\pi} \frac{(\pi-x)dx}{4 \cos^{2}x+9 \sin^{2}x}$
$\Rightarrow$ $2I=\int_{0}^{\pi} \frac{\pi dx}{4 \cos^{2}x+9 \sin^{2}x}$
$\Rightarrow$ $2I=\int_{0}^{\pi} \frac{\pi \sec^{2}x dx}{4 +9 \tan^{2}x}$
$\Rightarrow$ $2I=\int_{0}^{\pi/2} \frac{2\pi \sec^{2} x dx}{4 +9 \tan^{2}x}$
$\left[\because \int_{0}^{2a} f(x) dx-2\int_{0}^{a} f(x) dx\Rightarrow f(2a-x)=f(x)\right]$
$\Rightarrow$ $I= \frac{\pi}{9}\int_{0}^{\pi/2} \frac{\sec^{2} x dx}{\frac{4}{9}+\tan^{2}x }$
Put $\tan x=t \Rightarrow \sec^{2} x dx=dt$
$x=0, t=0, x=\frac{\pi}{2},t=\infty$
$\Rightarrow $ $I= \frac{\pi}{9} \int_{0}^{\infty} \frac{dt}{\left(\frac{2}{3}\right)^{2}+t^{2}}$
$\Rightarrow $ $I= \frac{\pi}{9} \times \frac{3}{2}\left[ \tan ^{-1} \frac{3t}{2}\right]_{0}^{\infty}$
$\Rightarrow $ $I= \frac{\pi}{9} \times \frac{3}{2}\times \frac{\pi}{2}= \frac{\pi^{2}}{12}$
