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

1 Answer
Sep 6, 2016

1/8(2x^2-2xsin2x-cos2x)+C18(2x22xsin2xcos2x)+C.

Explanation:

Let I=intxsin^2xdxI=xsin2xdx.

Then, I=int{(x(1-cos2x))/2}dx=1/2intxdx-1/2intxsin2xdx=1/4x^2-1/2J,I={x(1cos2x)2}dx=12xdx12xsin2xdx=14x212J,

where, J=intxcos2xdxJ=xcos2xdx

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

(IBP) : intuvdx=uintvdx-int((du)/dx*intvdx)dxuvdx=uvdx(dudxvdx)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.!