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