Answer
$$ - \frac{{\sin 7x}}{{14}} + \frac{{\sin x}}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\sin 3x\sin 4x} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{The integrand has a product of trigonometric functions}} \cr
& {\text{Use formula 38}} \cr
& \left( {38} \right):\,\,\,\,\,\int {\sin mu\sin nu} du = - \frac{{\sin \left( {m + n} \right)u}}{{2\left( {m + n} \right)}} + \frac{{\sin \left( {m - n} \right)u}}{{2\left( {m - n} \right)}} + C \cr
& \int {\sin 3x\sin 4x} dx\,\, \Rightarrow \,\,\,\,m = 3,\,\,\,n = 4 \cr
& {\text{Then by formula 38}} \cr
& \int {\sin 3x\sin 4x} dx = - \frac{{\sin \left( {3 + 4} \right)x}}{{2\left( {3 + 4} \right)}} + \frac{{\sin \left( {3 - 4} \right)x}}{{2\left( {3 - 4} \right)}} + C \cr
& {\text{simplifying}} \cr
& \int {\sin 3x\sin 4x} dx = - \frac{{\sin 7x}}{{14}} + \frac{{\sin \left( { - x} \right)}}{{ - 2}} + C \cr
& \int {\sin 3x\sin 4x} dx = - \frac{{\sin 7x}}{{14}} + \frac{{\sin x}}{2} + C \cr} $$
