Answer
$$\frac{1}{8}\cos 4x - \frac{1}{{20}}\cos 10x + C$$
Work Step by Step
$$\eqalign{
& \int {\sin 3x\cos 7x} dx \cr
& \sin mx\cos nx = \frac{1}{2}\left( {\sin \left( {\left( {m - n} \right)x} \right) + \sin \left( {\left( {m + n} \right)x} \right)} \right) \cr
& = \frac{1}{2}\int {\left( {\sin \left( {\left( { - 4} \right)x} \right) + \sin \left( {\left( {10} \right)x} \right)} \right)} dx \cr
& = \frac{1}{2}\int {\left( { - \sin 4x + \sin 10x} \right)} dx \cr
& {\text{sum rule}} \cr
& = - \frac{1}{2}\int {\sin 4x} dx + \frac{1}{2}\int {\sin 10x} dx \cr
& {\text{integrating}} \cr
& = - \frac{1}{2}\left( { - \frac{1}{4}\cos 4x} \right) + \frac{1}{2}\left( { - \frac{1}{{10}}\cos 10x} \right) + C \cr
& {\text{simplify}} \cr
& = \frac{1}{8}\cos 4x - \frac{1}{{20}}\cos 10x + C \cr} $$