Show by substitution that $\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x} \,\mathrm dx = \frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x}  \,\mathrm dx$ How do you show 
$$\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x} \,\mathrm{d}x = \frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \,\mathrm{d}x$$
without integrating by parts, but only using substitution?
 A: For any integral of the form
$$I = \int_0^{\pi} x f(\sin x) \mathrm{d}x $$
since $\sin(\pi-x) = \sin(x)$, using substitution $x = \pi-u$, or $u=\pi- x$, one may reduce this integral to the following using
when $x=0, u=\pi$, and when $x=\pi, u=0$, and also $ \mathrm{d}x = \mathrm{-d}u$
$$
\begin{align*}
I = \int_0^{\pi} (\pi - u) f(\sin u) \mathrm{d}u 
&= \pi \int_0^{\pi}  f(\sin u) \mathrm{d}u -  \int_{\pi}^{0} u f(\sin u) \mathrm{(-d}u)  \\
&= \pi \int_0^{\pi}  f(\sin u) \mathrm{d}u -  \int_0^{\pi} u f(\sin u) \mathrm{d}u  \\
&= \pi \int_0^{\pi}  f(\sin u) \mathrm{d}u - I\\
2I &= \pi \int_0^{\pi}  f(\sin u) \mathrm{d}u\\
I & = \frac{\pi}{2}\int_0^{\pi}  f(\sin u)  \mathrm{d}u = \frac{\pi}{2}\int_0^{\pi}  f(\sin x) \mathrm{d}x  \tag{A}
\end{align*}
$$
Apply $(A)$ for $\displaystyle{f(\sin x) = \frac{\sin x}{2-\sin^2 x} = \frac{\sin x}{1+\cos^2 x}}$ 
Although it was not asked, to further notice what is the value of the integral
by substituting further $\cos x = t, -\sin x \mathrm{d}x = \mathrm{d}t \hspace{5pt} cos(\pi) = -1, cos(0) = 1$
$$
\begin{align*}
\frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \,\mathrm{d}x &= \frac{\pi}{2} \int_{1}^{-1} \frac{\mathrm{-d}t}{1+t^2}\\
&= \frac{\pi}{2} \int_{-1}^{1} \frac{\mathrm{d}t}{1+t^2}\\
&= \frac{\pi}{2} \left( \tan^{-1}(1) - \tan^{-1}(-1) \right) \\
&= \frac{\pi}{2} \left( \frac{\pi}{4} - \frac{-\pi}{4} \right) = \frac{\pi}{2} \left(\frac{\pi}{2}\right) = \frac{{\pi}^2}{4}\\

\end{align*}
$$
