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What is the integral of #cos^2(2x)dx#?

1 Answer

Jan 15, 2017

#int cos^2(2x)dx = x/2 + 1/8sin(4x) + C#

Explanation:

Use the identity:

#cos^2theta= (1+cos2theta)/2#

so that:

#int cos^2(2x)dx = int (1+cos(4x))/2dx = 1/2intdx + 1/8 int cos(4x)d(4x)= x/2 + 1/8sin(4x) + C#

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