What is the integral of #int x sin^2(x) dx#?

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Answer 1 · 2018-04-03T16:07:07

#x^2/4 -xsin(2x)/4 - cos(2x)/8#

The solution is really simple if you do it by parties.

We start by integrating the function #sin^2(x)#

so #int sin^2(x)dx = int 1/2 (1-cos(2x)) = 1/2 (x- sin(2x)/2) + C#

so then you do the original integral by party.
you take:
#u = x rArr dot u = 1#
#dot v = sin^2x rArr v=1/2 (x- sin(2x)/2)#

and we know that #int u dot v = uv - int dot u v#
and so

#intxsin^2xdx = x/2(x-sin(2x)/2) - int 1/2 (x- sin(2x)/2)#

#= x/2(x-sin(2x)/2) - 1/2 (x^2 + cos(2x)/4)#

Which then can be simplified to

#x^2/4 -xsin(2x)/4 - cos(2x)/8#