Integral of (sin2x)(cos3x)

Answer 1

Int[sin(2x)*cos(3x)]dx = Int[(2sinx*cosx)*(4cos^3x - 3cosx)]dx= Int[(8sinx*cos^4x - 6sinx*cos^2x)]dx

Let cosx = u then du/dx = -sinx

So, the integral is

Int[-8*u^4 + 6*u^2]du

= -8/5*u^5 + 2u^3 + c where c is a constant of integration

= -8/5*cos^5x + 2cos^3x + c

Answer 2

(1/10)(5 cos x - cos (5x)) + Caccording to Wolfram Alpha