How do you integrate #int xsin^2x dx#?

Question

1484 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-17T01:26:46

#int x sin^2x dx = x^2/4 -1/4 xsin2x -1/8cos2x + C#

Use the identity:

#sin^2x = (1-cos2x)/2#

#int x sin^2x dx = int x((1-cos2x)/2 )dx = 1/2int xdx - 1/2int xcos2xdx#

The first integral is:

#int xdx = x^2/2+C_1#

the second can be integrated by part:

#int xcos2xdx = 1/2 int x d(sin2x) = 1/2xsin2x - 1/2 int sin2x dx = 1/2xsin2x +1/4 cos2x+C_2#

Putting it together:

#int x sin^2x dx = x^2/4 -1/4 xsin2x -1/8cos2x + C#