Answer
$$\frac{{{{\cos }^3}2\theta }}{6} - \frac{{\cos 2\theta }}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^3}2\theta } d\theta \cr
& {\text{split }} \cr
& = \int {{{\sin }^2}2\theta } \sin 2\theta d\theta \cr
& {\text{identity co}}{{\text{s}}^2}\phi + {\sin ^2}\phi = 1 \cr
& = \int {\left( {1 - {{\cos }^2}2\theta } \right)} \sin 2\theta d\theta \cr
& {\text{substitute }}u = \cos 2\theta ,{\text{ }}du = - 2\sin 2\theta d\theta \cr
& = \int {\left( {1 - {u^2}} \right)} \left( { - \frac{1}{2}} \right)du \cr
& = \frac{1}{2}\int {\left( {{u^2} - 1} \right)du} \cr
& {\text{find antiderivative }} \cr
& = \frac{1}{2}\left( {\frac{{{u^3}}}{3} - u} \right) + C \cr
& = \frac{{{u^3}}}{6} - \frac{u}{2} + C \cr
& {\text{write in terms of }}\theta ,{\text{ replace }}u = \cos 2\theta \cr
& = \frac{{{{\cos }^3}2\theta }}{6} - \frac{{\cos 2\theta }}{2} + C \cr} $$
