How to evaluate $\int_{0}^{\pi}\frac{x}{1+\cos^2x}dx$

Question

How to evaluate:

$$\int_{0}^{\pi}\frac{x}{1+\cos^2x}dx$$

(maybe the question is wrong)

Answer 1

Hint...try substituting $u=\pi-x$.....

Answer 2

Let $u=\pi-x$.

$$\int_0^\pi \frac{x}{1+\cos^2x}dx=-\int_\pi^0 \frac{\pi-u}{1+\cos^2(\pi-u)}du=\pi\int_0^\pi \frac{1}{1+\cos^2u}dx-\int_0^\pi \frac{u}{1+\cos^2u}du$$

So,

$$\int_0^\pi \frac{x}{1+\cos^2x}dx=\frac{\pi}{2}\int_0^\pi \frac{1}{1+\cos^2x}dx=\frac{\pi}{2}\int_0^\pi \frac{\sec^2x}{2+\tan^2x}dx$$

Then let $\tan x=\sqrt{2}\tan \theta$.