Evaluate: ∫sinxsin2xsin3x dx

Let I = ∫sinx.sin2x.sin3x dx = 1/2 ∫2sinx.sin2x.sin3x dx = 1/2 ∫sinx (2sin2x.sin3x)dx
We know that [2sinA.sinB = cos(A-B) - cos(A+B)]
So, = 1/2 ∫sinx (cos x - cos5x) dx = 1/2 * 2 ∫sinx.cosx dx - 1/2 * 2 ∫sinx.cos5x dx
= 1/4 ∫sin2x dx - ∫(sin6x - sin4x) dx = -cos2x/8 + cos6x/24 - cos4x/16 + c