What is #int cos^2(4x)dx #?

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Answer 1 · 2016-02-07T01:53:48

#1/16sin(8x)+x/2+C#

In this instance we need to use a trig identity to re-express #cos^2(4x)# in a form which can be integrated. We will use:

#cos^2(a) = 1/2cos(2a)+1/2#

from the double angle formulae.

So re expressing the integral gives us:

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

And now we can integrate easily to get:

#int 1/2cos(8x)+1/2dx = 1/16sin(8x)+x/2+C#