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

1 Answer
Apr 3, 2018

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

Explanation:

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