Answer
$$ - \frac{{\cos 5x}}{{10}} - \frac{{\cos x}}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\sin 3x\cos 2x} dx \cr
& {\text{A matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{formula }}48 \cr
& \int {\sin mx\cos nx} dx = - \frac{{\cos \left( {m + n} \right)x}}{{2\left( {m + 2} \right)}} - \frac{{\cos \left( {m - n} \right)x}}{{2\left( {m - n} \right)}} + C \cr
& {\text{with }}m = 3{\text{ and }}n = 2 \cr
& \int {\sin 3x\cos 2x} dx = - \frac{{\cos \left( {3 + 2} \right)x}}{{2\left( {3 + 2} \right)}} - \frac{{\cos \left( {3 - 2} \right)x}}{{2\left( {3 - 2} \right)}} + C \cr
& \int {\sin 3x\cos 2x} dx = - \frac{{\cos 5x}}{{10}} - \frac{{\cos x}}{2} + C \cr} $$
