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

Question

Please could you help me calculate the next integral:

$$\int_0^\pi\dfrac{x\sin x}{1+\cos^2x}dx$$

Thank you.

Answer 1

Note that $$\int_0^{\pi} xf(\sin x) \, \mathrm{d}x = \frac{\pi}{2}\int_0^{\pi} f(\sin x) \, \mathrm{d}x$$ via $x \mapsto \pi-x$.

So here you have (remember that $\cos^2 x = 1-\sin^2 x)$ $$\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x} \, \mathrm{d}x = \frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \, \mathrm{d}x.$$

Now use $x \mapsto \cos x$ to get $$\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x}\, \mathrm{d}x = \frac{\pi}{2}\int_{-1}^1 \frac{\mathrm{d}x}{1+x^2} = \frac{\pi}{2}\cdot \frac{\pi}{2} = \frac{\pi^2}{4}.$$