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