Answer
$\frac{1}{2}(\sin 5x-\sin x)$
Work Step by Step
Using the product-to-sum formulas on page 559, $\sin u\cos v=\frac{1}{2}[\sin(u+v)+\sin(u-v)]$:
$\sin 2x\cos 3x$
=$\frac{1}{2}[\sin(2x+3x)+\sin(2x-3x)]$
=$\frac{1}{2}[\sin 5x+\sin (-x)]$
=$\frac{1}{2}(\sin 5x-\sin x)$