cos2 2x sin3 2x dx Evaluate the integral using trig substitu

cos2 2x sin3 2x dx Evaluate the integral using trig substitution or integration by parts or u substitution. (Please show steps) The answer is 2/15

Solution

=integral(cos^22xsin^32x)dx =integral(cos^22x*(1-cos^22x)sin2xdx let cos2x=t -sin2x 2dx=dt =integral(t^2*(1-t^2)*-dt/2 =integral(t^2-t^4)*-dt/2 =(t^5/10 -t^3/6) =(cos^52x/10 -cos^22x/6) with limits 0 to pi/2 =(cos^5(pi)/10- 1/10)-(cos^2(pi)/6-1/6) =(-2/10)-(5/6) =-1/5 -5/6 =-4/30