How to evaluate $\int_{0}^{\pi }\frac{x\sin x}{1+\cos^{2}x}dx$

Question

How can I evaluate this integral?

$$\int_{0}^{\pi }\frac{x\sin x}{1+\cos^{2}x}dx$$

Answer 1

As $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$

$$I=\int_{0}^{\pi }\frac{x\sin x dx}{1+\cos^{2}x}$$ $$=\int_{0}^{\pi }\frac{(\pi+0-x) \sin\left (\pi+0-x\right ) dx}{1+\cos^{2}(\pi+0-x)}$$

$$=\int_{0}^{\pi }\frac{(\pi-x) \sin x dx}{1+\cos^{2}(x)}$$

So, $$I+I=\int_{0}^{\pi }\frac{x \sin x dx}{1+\cos^{2}x}+\int_{0}^{\pi }\frac{(\pi-x)\sin xdx}{1+\cos^{2}(x)}$$ as $\sin(\pi-x)=\sin x,\cos(\pi-x)=-\cos x$

$$2I=\pi\int_{0}^{\pi }\frac{ \sin x dx}{1+\cos^{2}x}$$

Put $\cos x=t, -\sin xdx=dt$

$x=0\implies t=\cos 0=1, x=\pi\implies t=\cos\pi=-1$

So, $$2I=\pi\int_1^{-1}\frac{-dt}{1+t^2}=\pi\int_{-1}^1\frac{dt}{1+t^2}$$ as $\int_a^bf(x)dx=-\int_b^af(x)dx$

$$2I=\pi(\arctan t)_{-1}^1=\pi\{\frac\pi4-(-\frac\pi4)\}=\frac{\pi^2}2$$

Answer 2

You can do something like (I think in english it's called integration by parts):

$\int\frac{xsin(x)}{1+cos^2(x)} = x\int\frac{sin(x)}{1+cos^2(x)} + \int\int\frac{sin(x)}{1+cos^2(x)}$

now, just put $cos(x) = t$ and you should be done