Another brave attempt on xSin[x]Sin[2x]dx

[PrudensOptimus]

∫xSin[x]Sin[2x]dx = ∫xsinxsin2xdx

= ∫2xsin^2xcosx dx
= 2∫x(1-cos^2x)cosx dx
= 2(∫xcosx dx - ∫xcos^3x dx)

∫xcosx dx = xsinx + cosx + C
∫xcos^3x dx = x(sinx - sin^3x/3) + cosx - cosx/3 + cos^3x/9 + C


so, 2(∫xcosx dx - ∫xcos^3x dx)

= 2(xsinx + cosx - [x(sinx - sin^3x/3) + cosx - cosx/3 + cos3x/9]) + C

:)