Integrals — Class 12 Maths Solution

: Let$I = \int {\cot x\log \sin x} dx$
Put $\log \sin x = t$ $\Rightarrow$ $\cfrac{1}{{\sin x}}\cos xdx = dt$

$\Rightarrow$ $\cot x\,dx = dt$
$\therefore$ $I = \int {t\,dt} = \cfrac{{{t^2}}}{2} + C = \cfrac{1}{2}{\left( {\log \sin x} \right)^2} + C$