Answer
$$\frac{1}{2}\cos x - \frac{1}{{10}}\cos 5x + C $$
Work Step by Step
$$\eqalign{
& \int {\sin 2x} \cos 3xdx \cr
& {\text{Use the identity }}\sin mx\cos nx = \frac{1}{2}\left[ {\sin \left( {m - n} \right)x + \sin \left( {m + n} \right)x} \right]\,\,\left( {{\text{see page 463}}} \right) \cr
& \int {\sin 3x} \cos 2xdx = \int {\frac{1}{2}} \left[ {\sin \left( {2 - 3} \right)x + \sin \left( {2 + 3} \right)x} \right]dx \cr
& = \int {\frac{1}{2}} \left( {\sin \left( { - x} \right) + \sin 5x} \right)dx \cr
& = \frac{1}{2}\int {\left( { - \sin x + \sin 5x} \right)} dx \cr
& {\text{distribute}} \cr
& = - \frac{1}{2}\int {\sin x} dx + \frac{1}{2}\int {\sin 5x} dx \cr
& {\text{integrating}} \cr
& = - \frac{1}{2}\left( { - \cos x} \right) + \frac{1}{2}\left( { - \frac{1}{5}\cos 5x} \right) + C \cr
& = \frac{1}{2}\cos x - \frac{1}{{10}}\cos 5x + C \cr} $$
