## Calculus: Early Transcendentals 8th Edition

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