Answer
$$\int\sin^5\frac{x}{3}\cos\frac{x}{3}dx=\frac{1}{2}\sin^6\frac{x}{3}+C$$
Work Step by Step
$$A=\int\sin^5\frac{x}{3}\cos\frac{x}{3}dx$$
We set $u=\sin\frac{x}{3}$
Then $$du=\frac{1}{3}\cos\frac{x}{3}dx$$
That means $$\cos\frac{x}{3}dx=3du$$
Therefore, $$A=3\int u^5du=\frac{3u^6}{6}+C=\frac{u^6}{2}+C$$ $$A=\frac{1}{2}\sin^6\frac{x}{3}+C$$
