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

Question

1279 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-09T15:35:30

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

We use

#cos2theta=2cos^2theta-1#

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

Here #theta=2x#

So,

#cos^2(2x)=(1+cos4x)/2#

#intcos^2(2x)dx=int((1+cos4x)dx)/2#

#=1/2(x+sin(4x)/4)+C#

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