Question

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

Answer 1

$\int sinxsin2xsin3xdx$

we know $2sinasinb=\left[cos\right(a+b)-cos(a+b\left)\right]$

then

$\int sinxsin2xsin3x=\frac{1}{2}\int sin2x2sinxsin3xdx$

$\Rightarrow \frac{1}{2}\int sin2x[cos2x-cos4x]dx$

$\Rightarrow \frac{1}{2}\int sin2xcosx-sin2xcos4xdx$

$\Rightarrow \frac{1}{2}\int \frac{sin4x}{2}-sin2xcos4xdx$

$\Rightarrow \frac{1}{2}\int \frac{sin4x}{2}dx-\frac{1}{4}\int 2sin2xcos4xdx$

we know $2sinacosb=sin(a+b)+sin(a-b)$

$\Rightarrow \frac{1}{2}\int \frac{sin4x}{2}dx-\frac{1}{4}\int sin6x+sin(-2x)dx$

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