Question

Find the integrals of the function :
$\sin 4x\sin 8x$

Answer 1

$\int \sin 4x\sin 8xdx$

$=\int \frac{-\cos (4x+8x)+\cos (4x-8x)}{2}dx$

$=\frac{1}{2}\int (-\cos (4x+8x)+\cos (4x-8x\left)\right)dx$

substitute $u=4x+8x\Rightarrow du=12dx,v=4x-8x\Rightarrow dv=-4dx$

$=\frac{1}{2}[\int \left(\frac{1}{12}\cos u\right)du+\int (-\frac{1}{4}\cos v)dv]$

$=\frac{1}{2}(\frac{1}{12}\sin u-\frac{1}{4}\sin v)$

$=\frac{1}{2}(-\frac{1}{12}\sin (12x)+\frac{1}{4}\sin (4x\left)\right)+C$