Question

Find:$\int \sin 2x\cos 3xdx$

Answer 1

We know that, $2\sin A\cos B=\sin (A+B)+\sin (A-B)$

$\therefore \int \sin 2x\cos 3xdx=\frac{1}{2}\int 2\sin 2x\cos 3xdx=\frac{1}{2}\int \{\sin (2x+3x)+\sin (2x-3x)\}dx=\frac{1}{2}\int \left\{\sin 5x+\sin (-x)\right\}dx=\frac{1}{2}\int \left\{\sin 5x-\sin x\right\}dx(\because \sin (-x)=-\sin x)=\frac{1}{2}\int \sin 5xdx-\frac{1}{2}\int \sin xdx=\frac{1}{10}(-\cos 5x)+\frac{1}{2}\cos x=\frac{1}{2}\cos x-\frac{1}{10}\cos 5x$