Question

Evaluate: $\int \sin x\sin 2x\sin 3xdx.$

Answer 1

Let $I=\int sinx.sin2x.sin3xdx$

$=\frac{1}{2}\int 2sinx.sin2x.sin3xdx$

$=\frac{1}{2}\int sinx.(2sin2x.sin3x)dx$

We know that $[2sinA.sinB=cos(A-B)-cos(A+B\left)\right]$

So, $=\frac{1}{2}\int sinx.(cosx-cos5x)dx$

$=\frac{1}{2\times 2}\int 2sinx.cosxdx-\frac{1}{2\times 2}\int 2sinx.cos5xdx$

$=\frac{1}{4}\int sin2xdx-\frac{1}{4}\int (sin6x-sin4x)dx$

= $-\frac{cos2x}{8}+\frac{cos6x}{24}-\frac{cos4x}{16}+c$