The integral $\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x}$ is equal to

<p>$$\begin{aligned} & I=\int_0^\pi \frac{8 x}{4 \cos ^2 x+\sin ^2 x} d x \ldots(1) \\ & I=\int_0^\pi \frac{8(\pi-x)}{4 \cos ^2(\pi-x)+\sin ^2(\pi-x)} d x \\ & I=\int_0^\pi \frac{8(\pi-x)}{4 \cos ^2 x+\sin ^2 x} d x \ldots(2) \end{aligned}$$</p> <p>Adding (1) and (2)</p> <p>$$\begin{aligned} & 2 I=8 \pi \int_0^\pi \frac{1}{4 \cos ^2 x+\sin ^2 x} d x \\ & I=4 \pi \times 2 \int_0^\pi \frac{\sec ^2 x}{4 \tan ^2 x} d x \end{aligned}$$</p> <p>Put $\tan x=t$</p> <p>$\sec ^2 x d x=d t$</p> <p>$I=8 \pi \int_0^{\infty} \frac{d t}{4+t^2}$</p> <p>$$\begin{aligned} & I=8 \pi \frac{1}{2}\left(\tan ^{-1} \frac{t}{2}\right)_0^{\infty} \\ & I=4 \pi\left(\frac{\pi}{2}\right) \\ & I=2 \pi^2 \end{aligned}$$</p>