$\int_0^\pi x f(\sin x) d x$ is equal to

$\int_0^\pi x f(\sin x) d x$ is equal to
  1. $\pi \int_0^\pi f(\cos x) d x$
  2. $\pi \int_0^\pi f(\sin x) d x$
  3. $\frac{\pi}{2} \int_0^{\pi / 2} f(\sin x) d x$
  4. $\pi \int_0^{\pi / 2} f(\cos x) d x$

Solution

$I=\int_0^\pi x f(\sin x) d x=\int_0^\pi(\pi-x) f(\sin x) d x$ $=\pi \int_0^\pi f(\sin x) d x-1$ $2 I=\pi \int_0^\pi f(\sin x) d x$ $I=\frac{\pi}{2} \int_0^\pi f(\sin x) d x=\pi \int_0^{\pi / 2} f(\sin x) d x$ $=\pi \int_0^{\pi / 2} f(\cos x) d x$

Asked in: JEE Main 2006

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