Let I=intxsin^2xdxI=∫xsin2xdx.
Then, I=int{(x(1-cos2x))/2}dx=1/2intxdx-1/2intxsin2xdx=1/4x^2-1/2J,I=∫{x(1−cos2x)2}dx=12∫xdx−12∫xsin2xdx=14x2−12J,
where, J=intxcos2xdxJ=∫xcos2xdx
To find JJ, we use the Rule of Integration by Parts (IBP) :
(IBP) : intuvdx=uintvdx-int((du)/dx*intvdx)dx∫uvdx=u∫vdx−∫(dudx⋅∫vdx)dx.
Taking, u=x", so that, "(du)/dx=1, &, v=cos2x", so "intvdx=sin(2x)/2u=x, so that, dudx=1,&,v=cos2x, so ∫vdx=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.!