Let $f(\sin x)$ be a given function of $\sin x$.
How would I show that $\int_0^\pi xf(\sin x)dx=\frac{1}{2}\pi\int_0^\pi f(\sin x)dx$?
Let $f(\sin x)$ be a given function of $\sin x$.
How would I show that $\int_0^\pi xf(\sin x)dx=\frac{1}{2}\pi\int_0^\pi f(\sin x)dx$?
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.
$\int_0^\pi xf(\sin x)d$
=$\int_0^\pi (\pi-x)f(\sin (\pi - x))dx$
= $\int_0^\pi (\pi-x)f(\sin x)dx$
= $\pi\int_0^\pi f(\sin x)dx$ - $\int_0^\pi xf(\sin x)dx$
$2\int_0^\pi xf(\sin x)d$ = $\pi\int_0^\pi f(\sin x)dx$
$\int_0^\pi xf(\sin x)d$ = $\frac{\pi}{2}\int_0^\pi f(\sin x)dx$
I am using this formula, $\int_a^b f(x)dx$ =$\int_a^b f(a+b-x)dx$