prove property of the definite integral of a function of sine. Show that 
$$\int_0^{\pi}x\;f(\sin x)dx=\frac{\pi}{2}\int_0^{\pi}f(\sin x)dx$$
I tried to prove this using integration by parts and u-subs but I can't work it out.
 A: HINT:
Use  $\displaystyle I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$
and $\displaystyle\sin(\pi+0-x)=\sin x$
So, $\displaystyle I+I=\int_a^bf(x)dx+\int_a^bf(a+b-x)dx=\int_a^b\left[f(x)+f(a+b-x)\right]dx$
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\color{#00f}{\large\int_{0}^{\pi}x\fermi\pars{\sin\pars{x}}\,\dd x}
&=
\half\bracks{%
\int_{0}^{\pi}x\fermi\pars{\sin\pars{x}}\,\dd x
+
\int_{0}^{\pi}\pars{\pi - x}\fermi\pars{\sin\pars{\pi - x}}\,\dd x}
\\[3mm]&=
\half\bracks{%
\int_{0}^{\pi}x\fermi\pars{\sin\pars{x}}\,\dd x
+
\int_{0}^{\pi}\pars{\pi - x}\fermi\pars{\sin\pars{x}}\,\dd x}
\\[3mm]&=
\color{#00f}{\large{\pi \over 2}\int_{0}^{\pi}\fermi\pars{\sin\pars{x}}\,\dd x}
\end{align}
