How do you find the integral of $(cosx)^2 dx$ ?

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Answer 1 · 2015-10-15T13:50:55

Use the power reduction identity, then substitution.

$cos(2x) = cos^2x-sin^2x$

$= 1-2sin^2x$
$= 2cos^2x-1$

To reduce an even power of cosine, use
$cos(2x) = 2cos^2x-1$

to see that $cos^2x=1/2(1+cos(2x))$

So,

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

$= 1/2[intdx+int cos(2x)dx]$

$= 1/2[x+1/2sin(2x)]+C$

You may choose to rewrite the last line as

$= 1/2x+1/4sin(2x)+C$
or as
$= 1/4[2x+sin(2x)]+C$
or in some other way.

How do you find the integral of (cosx)^2 dx(cosx)^2 dx?