Question

Find the integrals of the function: $sin\, x\,\, sin\, 2x\,\, sin\, 3x$

Answer 1

The correct answer is: $= \frac{1}{8}[\frac{cos6x}{3}-\frac{cos4x}{2}-cos2x]+C$
It is known that, $sin A\, sin B = \frac{1}{2}{cos(A-B)-cos(A+B)}$
$∴ ∫sin\, x\,\, sin\, 2x\,\, sin\, 3x. dx = ∫[sin x.\frac{1}{2}{cos(2x-3x)-cos(2x+3x)}]dx$
$= \frac{1}{2}∫(sinx\, cos(-x)-sinx\, cos\,5x)dx$
$= \frac{1}{2} ∫(sinx\,cosx-sinx\,cos5x)dx$
$= \frac{1}{2} ∫\frac{sin2x}{2}. dx -\frac{1}{2} ∫sin x\, cos5x. dx$
$= \frac{1}{4}[\frac{-cos2x}{2}]-\frac{1}{2} ∫{\frac{1}{2}sin(x+5x)+sin(x-5x)}dx$
$= \frac{-cos2x}{8}-\frac{1}{4} ∫(sin6x+sin(-4x))dx$
$=\frac{-cos2x}{8}-\frac{1}{4}[\frac{-cos6x}{3}+\frac{cos4x}{4}]+C$
$=\frac{-cos2x}{8}-\frac{1}{8}[\frac{-cos6x}{3}+\frac{cos4x}{2}]+C$
$= \frac{1}{8}[\frac{cos6x}{3}-\frac{cos4x}{2}-cos2x]+C$