`int sin x sin 2x sin 3x dx`

Explanation

`int sin x sin 2x sin 3x dx`
`=(1)/(2)int(2sin 3x sin x)sin 2x dx`
`=(1)/(2)int(cos 2x-cos 4x)sin 2x dx`
`=(1)/(4) int(2sin 2x cos 2x- 2 cos 4x sin 2x)dx`
`=(1)/(4)int[sin 4x-(sin6x-sin2x)]dx`
`=(1)/(4)int(sin4x-sin6x+sin2x)dx`
`=(1)/(4)[-(1)/(4)cos 4x+(1)/(6)cos 6x-(1)/(2)cos 2x]+c`
`=-(1)/(16)cos 4x+(1)/(24)cos 6x-(1)/(8)cos 2x+c`