Mathematics - Definite Integration Question with Solution | TestHub

$I={\int }_0^{\pi }xf(\sin x)dx={\int }_0^{\pi }(\pi -x)f(\sin x)dx$ $=\pi {\int }_0^{\pi }f(\sin x)dx-1$ $2I=\pi {\int }_0^{\pi }f(\sin x)dx$ $I=\frac{\pi }{2}{\int }_0^{\pi }f(\sin x)dx=\pi {\int }_0^{\pi /2}f(\sin x)dx$ $=\pi {\int }_0^{\pi /2}f(\cos x)dx$