Let $f(t)$ be a continuous function.
so-$$\frac{\pi}{2} \int_{0}^{\pi}f(\sin x)dx=\int_{0}^{\pi}xf(\sin x)dx$$ I tried many times Integration in parts but didn't succeed.
Let $f(t)$ be a continuous function.
so-$$\frac{\pi}{2} \int_{0}^{\pi}f(\sin x)dx=\int_{0}^{\pi}xf(\sin x)dx$$ I tried many times Integration in parts but didn't succeed.
Copyright © 2021 JogjaFile Inc.
By the change of variable $u=\pi-x$, one has $$ \int_{0}^{\pi}xf(\sin x)dx=\int_{0}^{\pi}(\pi-u)f(\sin (\pi-u))du=\pi\int_{0}^{\pi}f(\sin u)du-\int_{0}^{\pi}\!uf(\sin u)du $$ using $\sin (\pi-u)=\sin u$.