Answer
$\ln |\sin x|-\frac{1}{2}\sin^2 x+C$
Work Step by Step
$\int\cot x\cos^2 x\ dx$
$=\int\frac{\cos x}{\sin x}*\cos^2 x\ dx$
$=\int\frac{\cos^3 x}{\sin x}\ dx$
$=\int\frac{\cos x*\cos^2 x}{\sin x}\ dx$
$=\int\frac{\cos x(1-\sin^2 x)}{\sin x}\ dx$
Let $u=\sin x$. Then $du=\cos x\ dx$.
$=\int\frac{1-u^2}{u}du$
$=\int(\frac{1}{u}-\frac{u^2}{u})du$
$=\int(u^{-1}-u)du$
$=\ln |u|-\frac{u^2}{2}+C$
$=\boxed{\ln |\sin x|-\frac{1}{2}\sin^2 x+C}$
