Answer
$$\int\sin^42x\cos2xdx=\frac{(\sin 2x)^5}{10}+C$$
Work Step by Step
$$A=\int\sin^42x\cos2xdx$$
Take $a= 2x$, which means $$da=2dx$$ $$dx=\frac{1}{2}da$$
Therefore, $$A=\frac{1}{2}\int\sin^4a\cos ada$$
Take $u=\sin a$, we have $$du=\cos ada$$
So, $$A=\frac{1}{2}\int u^4du$$ $$A=\frac{1}{2}\times\frac{u^5}{5}+C=\frac{u^5}{10}+C$$ $$A=\frac{(\sin 2x)^5}{10}+C$$
