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

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

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

## 2 Answers

If you make the substitution $w = \pi-x$, so that $dw = -dx$, you get $$\int_0^\pi xf(\sin x)dx=-\int_\pi^0 (\pi-w)f(\sin(\pi-w))dw$$ $$= \int_0^\pi (\pi-x)f(\sin(x))dx$$ $$= \pi\int_0^\pi f(\sin(x))dx - \int_0^\pi xf(\sin(x))dx$$ which gives the result you want.

• Thanks, most appreciated. :) – Steven Jun 17 '12 at 16:01

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

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

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

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

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

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

I am using this formula, $\int_a^b f(x)dx$ =$\int_a^b f(a+b-x)dx$

• Its ok, thanks for your help! – Steven Jun 17 '12 at 16:01
• there should be $\frac{\pi}{2}\int_0^\pi f(\sin x)dx$ instead of $\frac{\pi}{2}\int_0^\pi \sin xdx$. – Aang Jul 12 '12 at 8:55
• @avatar: Thanks – Prasad G Jul 12 '12 at 8:59