## Calculus, 10th Edition (Anton)

$\frac{\sin^6 2t}{12}+C$
Let $u=\sin 2t$. Then $du=2\cos 2tdt$, so $\cos 2tdt=\frac{1}{2}du$. $\int\cos 2t\sin^5 2tdt$ $=\int\sin^5 2t\cos 2tdt$ $=\int u^5*\frac{1}{2}du$ $=\frac{1}{2}*\frac{u^6}{6}+C$ $=\frac{u^6}{12}+C$ $=\frac{\sin^6 2t}{12}+C$