Answer
$-\frac{1}{2}cos x -\frac{1}{10}cos 5x +C$
Work Step by Step
Applying the product-to-sum identity, we have
$\int sin 3x cos 2xdx=\int\frac{1}{2}(sinx+sin5x)dx$
$= \frac{1}{2}\int sinxdx + \frac{1}{2}\int sin 5xdx$
$=\frac{1}{2}\times-cosx + \frac{1}{2}\times\frac{-cos5x}{5}+C$
$= -\frac{1}{2}cos x -\frac{1}{10}cos 5x +C$
