How to show this integral 
Show that
  $$
\int_{0}^{\pi} xf(\sin(x))\,dx = \frac{\pi}{2}\int_{0}^{\pi} f(\sin(x))\,dx
$$

My progress: nothing cause I don't know where start. 
I accept hints, (explicits). And the exercise is from Apostol's book.
Ahh... The chapter I'm reading is about "integration by substitution", the book gives this Hint: $u=\pi-x$.
 A: If you accept the book's hint, you have
\begin{align}
\int_0^{\pi}xf(\sin(x))\,dx
&=\int_{\pi}^0(\pi-u)f(\sin(\pi-u))\cdot(-1)\,du\\
&=\int_{0}^\pi(\pi-u)f(\sin(u))\,du\\
&=\int_{0}^\pi\pi f(\sin(u))\,du-\int_{0}^\pi uf(\sin(u))\,du\\
&=\pi\int_{0}^\pi f(\sin(u))\,du-\int_{0}^\pi uf(\sin(u))\,du\\
\end{align}
Now set
$$
I=\int_0^{\pi}xf(\sin(x))\,dx
$$
and…
A: The key to solving this is to notice that since $\sin x$ is symmetric about $x=\pi/2$, then so is $ f(\sin x)$. This means that $$\int_0^\pi xf(\sin x) dx=\int_0^\pi (\pi-x)f(\sin (\pi-x)) dx=\int_0^\pi (\pi-x)f(\sin x) dx,$$ after which you can proceed as in the other answers.
A: If you make the substitution that the book suggests, you have $x = \pi-u$ and $dx = -du$, so the integral becomes
\begin{align}
  \int_{0}^{\pi} x\,f(\sin(x))\,dx
    &= -\int_{\pi}^0 (\pi-u)f(\sin(\pi-u))\,du \\
    &= \int_0^{\pi} (\pi-u)f(\sin u)\,du \\
    &= \pi\int_0^{\pi} f(\sin u)\,du - \int_0^{\pi} u\,f(\sin u)\,du.
\end{align}
Collect terms and simplify.
