Evaluate $\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}\,\mathrm dx$ I can't seem to get any kind of result which I find useful when I use the fact that $$\int_0^af(x)\,\mathrm dx=\int_0^af(a-x)\,\mathrm dx$$
After using trigonometry, I end up getting:
$$\int_0^{\pi}\frac{(\pi-x)\sin x}{1+\cos^2x}\,\mathrm dx$$
I don't see how this is useful. Please help.
 A: It follows from what you've done so far that
$$2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx = \int_0^\pi \frac{\pi\sin x}{1+\cos^2 x} dx$$
The integral on the left is twice the one you're after. The integral on the right can be evaluated using the 'obvious' substitution of $u = \cos x$ and then hopefully you recognize the integral.

Added (in response to comment below)
You have shown that
$$\int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx  = \int_0^\pi \frac{(\pi - x) \sin x}{1+\cos^2 x} dx $$
In other words,
$$\int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx  = \int_0^\pi \frac{\pi \sin x}{1+\cos^2 x} dx - \int_0^\pi \frac{x\sin x}{1+\cos^2 x} dx $$
and hence
$$2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx = \int_0^\pi \frac{\pi\sin x}{1+\cos^2 x} dx$$
So we can write the original integral as being equal to this integral:
$$ \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx =  \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$$
Clear now?
A: $\color{red}{\hat{I}=\displaystyle\int_{0}^\pi\dfrac{x\sin x}{1+\cos^2x}dx}=\color{blue}{\displaystyle\int_{0}^\pi\dfrac{(\pi-x)\sin x}{1+\cos^2x}dx=\hat{I}}$
$\therefore 2\hat{I}=\color{red}{\hat{I}}+\color{blue}{\hat{I}}=\color{green}{\pi\displaystyle\int_{0}^\pi\dfrac{\sin x}{1+\cos^2x}dx}$
$I=\displaystyle\int\dfrac{\sin x}{1+\cos^2x}dx=\displaystyle\int\dfrac{\sec x\tan x}{\sec^2x+1}dx$
$\therefore\sec x=z \implies \displaystyle\int\dfrac{\sec x\tan x}{\sec^2x+1}dx=\displaystyle\int\dfrac{dz}{z^2+1}$
