Answer
$$ - \frac{1}{6}\cos 5x + \frac{1}{2}\cos x + C$$
Work Step by Step
$$\eqalign{
& \int {\sin x\cos 2x} dx \cr
& {\text{product }}\sin mx\cos nx = \frac{1}{2}\left( {\sin \left( {\left( {m - n} \right)x} \right) + \sin \left( {\left( {m + n} \right)x} \right)} \right) \cr
& \sin x\cos 2x = \frac{1}{2}\left( {\sin \left( {\left( {1 - 2} \right)x} \right) + \sin \left( {\left( {1 + 2} \right)x} \right)} \right) \cr
& \sin x\cos 2x = \frac{1}{2}\left( { - \sin x + \sin 3x} \right) \cr
& \int {\sin x\cos 2x} dx = \frac{1}{2}\int {\left( {\sin 3x - \sin x} \right)} dx \cr
& {\text{sum rule}} \cr
& = \frac{1}{2}\int {\sin 3x} dx - \frac{1}{2}\int {\sin x} dx \cr
& {\text{find antiderivatives}} \cr
& = \frac{1}{2}\left( { - \frac{1}{3}\cos 5x} \right) - \frac{1}{2}\left( { - \cos x} \right) + C \cr
& = - \frac{1}{6}\cos 5x + \frac{1}{2}\cos x + C \cr} $$
