What is the antiderivative of #sin^2(x)#?

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Euan S. Share
Jul 4, 2016

Answer:

#= 1/2[x - 1/2sin2x] + C#

Explanation:

We're going to use the trig identity

#cos2theta = 1 -2sin^2theta#

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

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

#= 1/2[x - 1/2sin2x] + C#

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