Question

Evaluate: $I=\int \log \sin xdx$

Answer 1

$=log\sin x\cdot \int dx-\int (\int dx)\cdot \frac{d}{dx}\left(log\sin x\right)dx+c$
$=log\sin x\cdot x-\int x\cdot \frac{1}{\sin x}\cdot \cos xdx+c$
$=xlog\sin x-\int x\cdot \frac{\cos x}{\sin x}dx+c$
$=xlog\sin x-\int x\cot xdx+c$
$=xlog\sin x-x\cdot \int \cot xdx-\int (\int \cot xdx)\frac{d}{dx}\left(x\right)dx+c+$
$=xlog\sin x-x\cdot log|\sin x|-\int log|\sin x|dx+c^{\prime }$
$\because I=\int log|\sin x|$
$I=-I+c^{\prime }$

$2I=c^{\prime }$

$I=c^{\prime \prime }$