Let I=∫0^π(xsin x)/(1+cos ^2x) dx. Then the number I is equal to

Use the property $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ with $a=\pi$. Since $\sin(\pi-x)=\sin x$ and $\cos^2(\pi-x)=\cos^2x$, we get $I=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}dx$. Adding the two forms, $2I=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx$. Now put $u=\cos x$, $du=-\sin x\,dx$: the integral becomes $\int_{-1}^{1}\frac{du}{1+u^2}=\tan^{-1}1-\tan^{-1}(-1)=\frac\pi2$. Therefore $2I=\pi\cdot\frac\pi2$ and $I=\dfrac{\pi^2}{4}$. Forgetting to halve gives $\pi^2/2$, and a wrong limit change gives $\pi^2/8$. The same King-property trick, replacing the variable by the upper limit minus itself, is the standard method for integrals with a factor $x$ multiplying a function symmetric about $x=\pi/2$.