Answer
$$ - \frac{1}{{10}}\cos 5x - \frac{1}{2}\cos x + C$$
Work Step by Step
$$\eqalign{
& \int {\sin 3x} \cos 2xdx \cr
& {\text{using the special product }}\sin Ax\cos Bx = \frac{1}{2}\left[ {\sin \left( {A + B} \right)x + \sin \left( {A - B} \right)x} \right] \cr
& \int {\sin 3x} \cos 2xdx = \int {\frac{1}{2}\left[ {\sin \left( {3 + 2} \right)x + \sin \left( {3 - 2} \right)x} \right]} dx \cr
& = \frac{1}{2}\int {\left( {\sin 5x + \sin x} \right)} dx \cr
& = \frac{1}{2}\int {\sin 5x} dx + \frac{1}{2}\int {\sin x} dx \cr
& \cr
& {\text{integrating by the formula }}\int {\sin ax} dx = - \frac{1}{a}\cos ax + C \cr
& = \frac{1}{2}\left( { - \frac{1}{5}\cos 5x} \right) + \frac{1}{2}\left( { - \cos x} \right) + C \cr
& \cr
& {\text{simplifying, we get}} \cr
& = - \frac{1}{{10}}\cos 5x - \frac{1}{2}\cos x + C \cr} $$
