Answer
$\frac{\sin^3 x}{3}+C$
Work Step by Step
$\int \tan^2 x\cos^3 x\ dx$
$=\int\frac{\sin^2 x}{\cos^2 x}*\cos^3 x\ dx$
$=\int\sin^2 x\cos x\ dx$
We use u substitution. Let $u=\sin x$. Then, $du=\cos x\ dx$.
$=\int u^2 du$
$=\frac{u^3}{3}+C$
$=\boxed{\frac{\sin^3 x}{3}+C}$
