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

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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,

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

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

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

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

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