Question

Find the integral of the function $\sin 3x\cos 4x$.

Answer 1

for integrals of type $\sin ax\cos bx$

$\sin ax\cos bx=\frac{1}{2}\left(\sin \right(ax+bx)+\sin (ax-bx\left)\right)$

If $a=3$ $b=4$

$\sin 3x\cos 4x=\frac{1}{2}\left(\sin \right(7x)+\sin (-x\left)\right)$

$=\frac{1}{2}(\sin 7x-\sin x)$

$\int \sin 3x\cos 4xdx=\frac{1}{2}\int (\sin 7x-\sin x)dx$

$=\frac{1}{2}(\frac{-1}{7}\cos 7x+\cos x)+c$

$=\frac{\cos x}{2}-\frac{-\cos 7x}{14}+c$