I need assistance in integrating $ \frac{x \sin x}{1+(\cos x)^2}$ Find the integral
$$ \int_0^{\pi} \frac{x \sin x}{1+(\cos x)^2}$$
 A: Integrate by parts:
$$\begin{align}\int_0^{\pi} dx\: \frac{x \sin x}{1+(\cos x)^2} &= -\int_0^{\pi} d(\cos{x})\: \frac{x}{1+(\cos x)^2}\\ &= -[x \arctan{\cos{x}}]_0^{\pi} + \underbrace{\int_0^{\pi} dx \:\arctan{\cos{x}}}_{\mathrm{this} = 0} \\ &= \frac{\pi^2}{4} \end{align}$$
Keep in mind that I used the principal branch of the arctangent.
A: Solution 1
Let the variable change $x=\pi-y$, and then
$$I=\int_0^{\pi} \frac{x \sin x}{1+(\cos x)^2}\mathrm{dx}=-\int_{0}^{\pi} \frac{ y\sin y}{1+(\cos y)^2}\mathrm{dy}+\pi\int_0^{\pi}\frac{\sin y}{1+(\cos y)^2}\mathrm{dy}$$
$$I=-\frac{\pi}{2}\int_0^{\pi}\frac{(\cos x)'}{1+(\cos x)^2}\mathrm{dx}$$
$$I=\frac{\pi}{2}[-\arctan(\cos x)]_0^{\pi}=\frac{\pi^2}{4}$$  
Solution 2 (the fast way)
We recall and employ the formula 
$$\int_0^\pi xf(\sin x )\mathrm{dx}=\frac \pi2\int_0^\pi f(\sin x )\mathrm{dx}$$  
that I used in another answer you may see here. Then
$$\int_0^{\pi} \frac{x \sin x}{1+(\cos x)^2}\mathrm{dx}=-\frac{\pi}{2}\int_0^{\pi} \frac{(\cos x)'}{1+(\cos x)^2}\mathrm{dx}=\frac{\pi}{2}[-\arctan(\cos x)]_0^{\pi/2}=\frac{\pi^2}{4}$$
$\quad$
