<s>[INST] Integrating $\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$ Could someone help with the following integration:
  $$\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$$
So far I have done the following, but I am stuck:
I denoted $ y=-\cos x $ then:
$$\begin{align*}&\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&= \arccos(-1) \arctan 1+\arccos 1 \arctan(-1) - \int^1_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\\&=\frac{\pi^2}{4}-\int^{1}_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\end{align*}$$
Then I am really stuck. Could someone help me?
 [/INST] Let $y=x-\frac{\pi}{2}$.
\begin{array}{l}\displaystyle  \int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x&=\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(y+\frac{\pi}{2}\right) \cos y}{1+\sin ^{2} y} d y\\&=\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{y \cos y}{1+\sin ^{2} y} d y +\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos y}{1+\sin ^{2} y} d y
\\\displaystyle  &=\displaystyle \pi \int_{0}^{\frac{\pi}{2}} \frac{d(\sin y)}{1+\sin ^{2} y}
\\&=\displaystyle 0+\pi\left[\tan ^{-1}(\sin y)\right]_{0}^{\frac{\pi}{2}}\\&=\displaystyle \frac{\pi^{2}}{4} \end{array}
Hence
$$\int_{\pi}^{0} \frac{x \sin x}{1+\cos ^{2} x} d x =-\frac{\pi^2}{4}$$
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