Answer
$$\frac{1}{4}\sin 2x - \frac{1}{{24}}\sin 12x + C$$
Work Step by Step
$$\eqalign{
& \int {\sin 5x\sin 7x} dx \cr
& \sin mx\sin nx = \frac{1}{2}\left( {\cos \left( {\left( {m - n} \right)x} \right) - \cos \left( {\left( {m + n} \right)x} \right)} \right) \cr
& = \frac{1}{2}\int {\left( {\cos \left( {\left( { - 2} \right)x} \right) - \cos \left( {\left( {12} \right)x} \right)} \right)} dx \cr
& = \frac{1}{2}\int {\left( {\cos \left( {2x} \right) - \cos \left( {12x} \right)} \right)} dx \cr
& {\text{sum rule}} \cr
& = \frac{1}{2}\int {\left( {\cos \left( {2x} \right)} \right)} dx - \frac{1}{2}\int {\left( {\cos \left( {12x} \right)} \right)} dx \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\left( {\frac{1}{2}\sin 2x} \right) - \frac{1}{2}\left( {\frac{1}{{12}}\sin 12x} \right) + C \cr
& {\text{simplify}} \cr
& = \frac{1}{4}\sin 2x - \frac{1}{{24}}\sin 12x + C \cr} $$
