Answer
$$\frac{{{{\sin }^6}3x}}{{18}} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^5}3x\cos 3x} dx \cr
& {\text{substitute }}u = \sin 3x,{\text{ }} \cr
& du = 3\cos 3xdx \cr
& \frac{1}{3}du = \cos 3xdx \cr
& \int {{{\sin }^5}3x\cos 3x} dx = \int {{u^5}\left( {\frac{1}{3}du} \right)} \cr
& = \frac{1}{3}\int {{u^5}du} \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \frac{{{u^6}}}{{18}} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \sin 3x,{\text{ }} \cr
& = \frac{{{{\sin }^6}3x}}{{18}} + C \cr} $$
