Question

Find the integrals of the function $\sin 3x\cos 4x$

Answer 1

It is known that, $\sin A\cos B=\frac{1}{2}\left\{\sin \right(A+B)+\sin (A-B\left)\right\}$
$\therefore \int \sin 3x\cos 4xdx=\frac{1}{2}\int \left\{\sin \right(3x+4x)+\sin (3x-4x\left)\right\}dx$
$=\frac{1}{2}\int \{\sin 7x+\sin (-x\left)\right\}dx$
$=\frac{1}{2}\int \{\sin 7x-\sin x\}dx$
$=\frac{1}{2}\int \sin 7xdx-\frac{1}{2}\int \sin xdx$
$=\frac{1}{2}\left(\frac{-\cos 7x}{7}\right)-\frac{1}{2}(-\cos x)+C$
$=\frac{-\cos 7x}{14}+\frac{\cos x}{2}+C$