Evaluation of integral involving trigonometric functions

Question

Evaluate $$\int_0^{\pi}\frac{x\sin x}{1 + \cos^2x}\,dx$$

I've tried many substitutions like $u = x\sin x $, $u = \cos x$ and $u = 1+\cos^2 x$ but they didn't work.

Answer 1

Since we've $$ \int_0^a f(x) \rm dx = \int_0^a f(a-x) \rm dx$$ $$ \rm I= \color{blue}{\int_0^{\pi}\frac{x\sin x}{1 + cos^2x}} \tag 1$$

$$\implies \rm I = \int_0^{\pi}\frac{(\pi-x) \sin (\pi -x)}{1 + cos^2 (\pi - x)} \tag 2 = \color{blue}{\int_0^{\pi}\frac{(\pi - x)\sin x}{1 + cos^2x}} $$

Adding $(1)$ and $(2)$

$$2 \rm I = \pi \int_0^{\pi}\frac{\sin x}{1 + cos^2x} \rm dx$$

Now subtsitute $\cos x = t$ to get

$$ 2 \rm I = -\pi \int_0^{\pi} \frac{1}{1+t^2} \rm dt$$