How do you find the antiderivative of #int sin^2xdx#?

Question

32392 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-20T12:43:54

I got: #1/2[x-sin(x)cos(x)]+c#

I tried this:
enter image source here

Answer 2 · 2017-06-20T15:55:12

We can start with the cosine double angle formula to derive another expression equivalent to #sin^2x#:

#cos2x=cos^2x-sin^2x=1-2sin^2x#

Then:

#2sin^2x=1-cos2x#

So:

#intsin^2xdx=1/2int(1-cos2x)dx#

Integrating both of these term-by-term, with a substitution in the integration of #cos2x# gives:

#=1/2(x-1/2sin2x)#

Which can be rewritten using #sin2x=2sinxcosx#:

#=1/2(x-sinxcosx)+C#