Problem source

By making the substitution $x=\pi-t\,$, show that
\[
 \! \int_0^\pi x\f(\sin x) \d x = \tfrac12 \pi \! \int_0^\pi \f(\sin x) \d x\,,
\]
where $\f(\sin x)$ is a given function of $\sin x$.
Evaluate the following integrals:
\begin{questionparts}
 \item $\displaystyle \int_0^\pi \frac {x \sin x}{3+\sin^2 x}\,\d x\,$;
\item  $\displaystyle \int_0^{2\pi} 
\frac {x \sin x}{3+\sin^2 x}\,\d x\,$;
\item   $\displaystyle \int_{0}^{\pi} 
\frac {x \big\vert\sin 2x\big\vert}{3+\sin^2 x}\,\d x\,$.
\end{questionparts}