Answer
$$ - \frac{1}{2}\cos \theta - \frac{1}{{10}}\cos 5\theta + C$$
Work Step by Step
$$\eqalign{
& \int {\sin 3\theta } cos2\theta d\theta \cr
& {\text{product }}\sin mx\cos nx = \frac{1}{2}\left( {\sin \left( {\left( {m - n} \right)x} \right) + \sin \left( {\left( {m + n} \right)x} \right)} \right) \cr
& \sin 3\theta \cos 2\theta = \frac{1}{2}\left( {\sin \left( {\left( {3 - 2} \right)\theta } \right) + \sin \left( {\left( {3 + 2} \right)\theta } \right)} \right) \cr
& \sin 3\theta \cos 2\theta = \frac{1}{2}\left( {\sin \theta + \sin 5\theta } \right) \cr
& \int {\sin 3\theta } cos2\theta d\theta = \frac{1}{2}\int {\left( {\sin \theta + \sin 5\theta } \right)} dx \cr
& {\text{sum rule}} \cr
& = \frac{1}{2}\int {\sin \theta } d\theta + \frac{1}{2}\int {\sin 5\theta } d\theta \cr
& {\text{find antiderivatives}} \cr
& = \frac{1}{2}\left( { - \cos \theta } \right) + \frac{1}{2}\left( { - \frac{1}{5}\cos 5\theta } \right) + C \cr
& = - \frac{1}{2}\cos \theta - \frac{1}{{10}}\cos 5\theta + C \cr} $$
