I don't know the integral of sin3x either...but I looked it up.
The integral of sin3x = (-1/3) cos x (sin2x + 2) + C.
But to be fair I think you integrate sin3x this way.
Start with cos2x - sin2x = cos 2x
Then sin2x = (1-cos 2x)/2
sin3x = (1/2)(sin x - sin x cos 2x)
and you should be able to integrate sin x cos 2x by parts.
Sorry, not quite right:
The integral is correct, but you get it by integrating sin3x as sin x sin2x dx by parts.