Problem source

By making the change of variable $t=\pi-x$ in the integral 
\[
\int_{0}^{\pi}x\mathrm{f}(\sin x)\,\mathrm{d}x,
\]
or otherwise, show that, for any function $\mathrm{f},$ 
\[
\int_{0}^{\pi}x\mathrm{f}(\sin x)\,\mathrm{d}x=\frac{\pi}{2}\int_{0}^{\pi}\mathrm{f}(\sin x)\,\mathrm{d}x\,.
\]
Evaluate 
\[
\int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x\quad\mbox{ and }\quad\int_{0}^{2\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x\,.
\]

Solution source

\begin{align*}
 && I &= \int_0^{\pi} x f(\sin x) \d x \\
t = \pi - x, \d t = -\d t : &&&= \int_{t = \pi}^{t = 0} (\pi - t) f(\sin (\pi - t)) -\d t \\
&&&= \int_0^{\pi} (\pi - t) f(\sin t) \d t  \\
\Rightarrow && 2 I &= \pi \int_0^\pi f(\sin t) \d t \\
\Rightarrow && I &= \frac{\pi}{2} \int_0^{\pi} f(\sin x) \d x
\end{align*}

\begin{align*}
&& I &= \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} \d x \\
&&&= \frac{\pi}{2}\left [ -\tan^{-1} \cos x\right]_0^{\pi} \\
&&&= \tan 1 - \tan (-1) = \frac{\pi^2}{4} \\
\\
&& I &= \int_{0}^{2\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\
&&&= \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x + \int_{\pi}^{2\pi} \frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\
u = x - \pi, \d u = \d x: &&&= \frac{\pi^2}{4} + \int_{0}^{\pi} \frac{(u+\pi)(-\sin u)}{1 + \cos^2 u}\d u \\
&&&= \frac{\pi^2}{4}  -\frac{3\pi}{2} \int_0^{\pi} \frac{\sin u}{1+\cos^2 u} \d u \\
&&&= - \frac{\pi^2}2
  \end{align*}