Question

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

Answer 1

The correct answer is: $\frac{1}{2}[\frac{sin4x}{4}-\frac{sin12x}{12}]$
It is known that, $sin A\, sin B = \frac{1}{2}cos(A-B)-cos(A+B)$
$∴ ∫sin4x\,sin8x. dx = ∫[\frac{1}{2}cos(4x-8x)-cos(4x+8x)]dx$
$= \frac{1}{2} ∫(cos(-4x)-cos12x) dx$
$= \frac{1}{2} ∫(cos4x-cos12x) dx$
$= \frac{1}{2}[\frac{sin4x}{4}-\frac{sin12x}{12}]$