# 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?

$$I=\int_0^{\pi} \frac{-x\sin x}{1+\cos^2 x}\,dx=\int_0^{\pi} \frac{(x-\pi)\sin x}{1+\cos^2 x}dx\quad(x\to \pi-x)$$
$$\Rightarrow I=\frac{\pi}{2}\int_0^{\pi}\frac{-\sin x}{1+\cos^2 x}\,dx$$
Let $t=\cos x:$
$$I=\frac{\pi}{2}\int_{-1}^{1}-\frac{1}{1+t^2}\,dt=-\frac{\pi^2}{4}$$
Let $$I = \int_0^{\pi} \dfrac{x \sin(x)}{1+\cos^2(x)} dx = \int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \sin(x+\pi/2)}{1 + \cos^2(x+\pi/2)} dx = \int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \cos(x)}{1 + \sin^2(x)} dx$$ Now $$\int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \cos(x)}{1 + \sin^2(x)} dx = \int_{-\pi/2}^{\pi/2} \underbrace{\dfrac{x \cos(x)}{1 + \sin^2(x)}}_{\text{Odd function}} dx + \dfrac{\pi}2 \cdot \int_{-\pi/2}^{\pi/2} \dfrac{\cos(x)}{1 + \sin^2(x)} dx$$ Hence, we get that $$I = \dfrac{\pi}2 \cdot \int_{-\pi/2}^{\pi/2} \dfrac{\cos(x)}{1 + \sin^2(x)} dx = \dfrac{\pi}2 \cdot \int_{-1}^1 \dfrac{dt}{1+t^2} = \dfrac{\pi}2 \cdot \left( \dfrac{\pi}4 - \dfrac{-\pi}4\right) = \dfrac{\pi^2}4$$ The integral you are after is $-I$ and hence the answer is $-\dfrac{\pi^2}4$.