Question

Find the integrals of the functions.
$\int sin4xsin8x$ dx

Answer 1

$\int sin4xsin8xdx=\frac{1}{2}\int 2sin8xsin4xdx=\frac{1}{2}\int (cos4x-cos12x)dx[\therefore 2sinAsinB=cos(A-B)-cos(A+B\left)\right]=\frac{1}{2}\{\frac{sin4x}{4}-\frac{sin12x}{12}\}+C$