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.