## University Calculus: Early Transcendentals (3rd Edition)

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