We have,
$$
\begin{aligned}
& I=\int_0^\pi \frac{x d x}{4 \cos ^2 x+9 \sin ^2 x} \\
\Rightarrow & I=\int_0^\pi \frac{(\pi-x) d x}{4 \cos ^2(\pi-x)+9 \sin ^2(\pi-x)} \\
\Rightarrow & I=\int_0^\pi \frac{(\pi-x) d x}{4 \cos ^2 x+9 \sin ^2 x} \\
\Rightarrow & I=\int_0^\pi \frac{\pi d x}{4 \cos ^2 x+9 \sin ^2 x} \\
\Rightarrow & 2 I=\int_0^\pi \frac{\pi \sec ^2 x d x}{4+9 \tan ^2 x}
\end{aligned}
$$


$$
\begin{aligned}
& \Rightarrow 2 I=\int_0^{\pi / 2} \frac{2 \pi \sec ^2 x d x}{4+9 \tan ^2 x} \\
& {\left[\because \int_0^{2 a} f(x) d x-2 \int_0^a f(x) d x \Rightarrow f(2 a-x)=f(x)\right]} \\
& \Rightarrow I=\frac{\pi}{9} \int_0^{\pi / 2} \frac{\sec ^2 x d x}{\frac{4}{9}+\tan ^2 x}
\end{aligned}
$$
Put $\tan x=t \Rightarrow \sec ^2 x d x=d t$
$$
\begin{aligned}
& x=0, t=0, x=\frac{\pi}{2}, t=\infty \\
\Rightarrow & I=\frac{\pi}{9} \int_0^{\infty} \frac{d t}{\left(\frac{2}{3}\right)^2+t^2} \\
\Rightarrow & I=\frac{\pi}{9} \times \frac{3}{2}\left[\tan ^{-1} \frac{3 t}{2}\right]_0^{\infty} \\
\Rightarrow & I=\frac{\pi}{9} \times \frac{3}{2} \times \frac{\pi}{2}=\frac{\pi^2}{12}
\end{aligned}
$$