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How do you find the integral of #cos^(2)2xdx#?

1 Answer

Jul 18, 2016

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

Explanation:

Using the fact

#cos(a+b)=cos(a) cos(b)-sin( a) sin(b)# we know that
#cos(2a) = 2cos^2(a)-1# then
#cos^2(a)=(cos(2a)+1)/2#

so

#cos^2(2x) =(cos(4x)+1)/2# and finally

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

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