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Q. The value of the integral $\int\limits_0^\pi x \ell n \sin x d x$ is

Integrals

Solution:

$I=\int\limits_0^\pi x \ell n \sin x d x$
$I=\int\limits_0^\pi(\pi-x) \ell n \sin (\pi-x) d x$
$=\int\limits_0^\pi(\pi-x) \ell n \sin x d x$
Add $2 I=\pi \int\limits_0^\pi \ell n \sin x d x$
$I=\frac{\pi}{2} \int\limits_0^\pi \ell n \sin x d x$
$=-\frac{\pi^2}{2} \ln 2$