Answer
$$\frac{1}{6}\sin 3\theta - \frac{1}{4}\sin \theta - \frac{1}{{20}}\sin 5\theta + C $$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^2}\theta \cos 3\theta } d\theta \cr
& {\text{Use the identity si}}{{\text{n}}^2}\theta = \frac{1}{2}\left( {1 - \cos 2\theta } \right) \cr
& = \int {\frac{1}{2}\left( {1 - \cos 2\theta } \right)\cos 3\theta } d\theta \cr
& {\text{distribute}} \cr
& = \int {\frac{1}{2}\cos 3\theta } d\theta - \int {\frac{1}{2}\cos 2\theta \cos 3\theta } d\theta \cr
& = \frac{1}{2}\int {\cos 3\theta } d\theta - \frac{1}{2}\int {\cos 2\theta \cos 3\theta } d\theta \cr
& {\text{Use the identity }}\cos mx\cos xnx = \frac{1}{2}\left[ {\cos \left( {m - n} \right)x + \cos \left( {m + n} \right)x} \right]\,\,\left( {{\text{see page 463}}} \right) \cr
& = \frac{1}{2}\int {\cos 3\theta } d\theta - \frac{1}{2}\int {\frac{1}{2}\left[ {\cos \left( {2 - 3} \right)\theta + \cos \left( {2 + 3} \right)\theta } \right]} d\theta \cr
& = \frac{1}{2}\int {\cos 3\theta } d\theta - \frac{1}{4}\int {\left( {\cos \theta + \cos 5\theta } \right)} d\theta \cr
& = \frac{1}{2}\int {\cos 3\theta } d\theta - \frac{1}{4}\int {\cos \theta } d\theta - \frac{1}{4}\int {\cos 5\theta } d\theta \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\left( {\frac{1}{3}\sin 3\theta } \right) - \frac{1}{4}\sin \theta - \frac{1}{4}\left( {\frac{1}{5}\sin 5\theta } \right) + C \cr
& = \frac{1}{6}\sin 3\theta - \frac{1}{4}\sin \theta - \frac{1}{{20}}\sin 5\theta + C \cr} $$
