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

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Answer 1 · 2017-11-08T12:03:04

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

#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#