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