How do you find the integral of #x(sinx)^2#?

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Answer 1 · 2016-09-06T19:12:30

#1/8(2x^2-2xsin2x-cos2x)+C#.

Let #I=intxsin^2xdx#.

Then, #I=int{(x(1-cos2x))/2}dx=1/2intxdx-1/2intxsin2xdx=1/4x^2-1/2J,#

where, #J=intxcos2xdx#

To find #J#, we use the Rule of Integration by Parts (IBP) :

(IBP) : #intuvdx=uintvdx-int((du)/dx*intvdx)dx#.

Taking, #u=x", so that, "(du)/dx=1, &, v=cos2x", so "intvdx=sin(2x)/2#.

#:. J=x/2sin2x-1/2intsin2xdx#

#=x/2sin2x-1/2(-1/2cos2x)#

#=x/2sin2x+1/4cos2x#

Hence, #I=x^2/4-x/4sin2x-1/8cos2x#

#=1/8(2x^2-2xsin2x-cos2x)+C#.

Enjoy Maths.!