Answer
$$ - \frac{{\cos 5x}}{{10}} - \frac{{\cos x}}{2} + C $$
Work Step by Step
$$\eqalign{
& \int {\sin 3x\cos 2x} dx \cr
& {\text{Integrate using the table of integrals in the book}} \cr
& {\text{We use formula 69a:}}\cr
& \,\,\,\int {\sin ax\cos bx} dx = - \frac{{\cos \left( {a + b} \right)x}}{{2\left( {a + b} \right)}} - \frac{{\cos \left( {a - b} \right)x}}{{2\left( {a - b} \right)}} + C \cr
& {\text{setting }}a = 3{\text{ and }}b = 2{\text{, then}} \cr
& \int {\sin 3x\cos 2x} dx = - \frac{{\cos \left( {3 + 2} \right)x}}{{2\left( {3 + 2} \right)}} - \frac{{\cos \left( {3 - 2} \right)x}}{{2\left( {3 - 2} \right)}} + C \cr
& \int {\sin 3x\cos 2x} dx = - \frac{{\cos 5x}}{{10}} - \frac{{\cos x}}{2} + C \cr} $$
