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