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How do you find the integral of #sin^2(2x) dx#?

1 Answer

Jan 21, 2017

The answer is #=x/2-(sin4x)/8+C#

Explanation:

We use

#cos4x=1-2sin^2(x)#

#sin^2 2x=1/2(1-cos4x)#

Therefore,

#int(sin^2 2x)dx=1/2int(1-cos4x)dx#

#=1/2(x-(sin4x)/4)+C#

#=x/2-(sin4x)/8+C#

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