How do you integrate #int x^2sin^2x#?

1 Answer
Nov 8, 2017

#(x^3)/6+(x*cos2x)/4-(x^2-2)*(sin2x)/4+C#

Explanation:

#int x^2*(sinx)^2*dx#

#=1/2*int x^2*(1-cos2x)*dx#

#=int (x^2*dx)/2#-#1/2*int x^2*cos2x*dx#

After using tabular integration for second integral.

#int x^2*cos2x*dx#

=#x^2*(sin2x)/2-2x*(cos2x)/(-4)+2*(sin2x)/(-8)#

=#(2x^2-4)*(sin2x)/4+(x*cos2x)/2#

Thus,

#int x^2*(sinx)^2*dx#

=#(x^3)/6+C#-#1/2*[(2x^2-4)*(sin2x)/4+(x*cos2x)/2]#

=#(x^3)/6+(x*cos2x)/4-(x^2-2)*(sin2x)/4+C#