Evaluate $\int_0^\pi xf(\sin x)dx$

Question

Let $f(\sin x)$ be a given function of $\sin x$.

How would I show that $\int_0^\pi xf(\sin x)dx=\frac{1}{2}\pi\int_0^\pi f(\sin x)dx$?

Answer 1

If you make the substitution $w = \pi-x$, so that $dw = -dx$, you get

\begin{align} \int_0^\pi xf(\sin x)dx &= -\int_\pi^0 (\pi-w)f(\sin(\pi-w))dw \\ &= \int_0^\pi (\pi-x)f(\sin(x))dx \\ &= \pi\int_0^\pi f(\sin(x))dx - \int_0^\pi xf(\sin(x))dx \end{align} which gives the result you want.

Answer 2

$\int_0^\pi xf(\sin x)dx$

=$\int_0^\pi (\pi-x)f(\sin (\pi - x))dx$

= $\int_0^\pi (\pi-x)f(\sin x)dx$

= $\pi\int_0^\pi f(\sin x)dx - \int_0^\pi xf(\sin x)dx$

$2\int_0^\pi xf(\sin x)d$ = $\pi\int_0^\pi f(\sin x)dx$

$\int_0^\pi xf(\sin x)d$ = $\frac{\pi}{2}\int_0^\pi f(\sin x)dx$

I also used: $\int_a^b f(x)dx$ =$\int_a^b f(a+b-x)dx$