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How do you find the integral of #int [cos^3 (2x)] dx #?

1 Answer

Sep 21, 2015

#1/2sin 2x-1/6sin^3 2x+C#

Explanation:

#I=int cos^3 2xdx = int cos 2x cos^2 2x dx#
#I=int cos 2x (1-sin^2 2x)dx#

#sin 2x=t => 2cos 2xdx = dt => cos 2xdx = dt/2#

#I=int(1-t^2)dt/2=1/2(t-t^3/3)+C#

#I=1/2sin 2x-1/6sin^3 2x+C#

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