Answer
$\frac{1}{9} \sin^3(3\theta)+c.$
Work Step by Step
We have
\begin{align*}
\int \sin^2(3\theta)\cos(3\theta) d\theta &=\frac{1}{3}\int 3\sin^2(3\theta)\cos(3\theta) d\theta \\
&=\frac{1}{3}\int \sin^2(3\theta) d(\sin(3\theta)) \\
&=\frac{1}{3}\frac{1}{3} \sin^3(3\theta)+c\\
&=\frac{1}{9} \sin^3(3\theta)+c.
\end{align*}
