What is the antiderivative of #cos^2x#?

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Answer 1 · 2015-06-08T00:16:42

The antiderivative is pretty much the same as the integral, except it's more general, so I'll do the indefinite integral.

#intcos^2xdx#

An identity for #cos^2x# is:

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

#=> 1/2int 1+cos(2x)dx#

Since #d/(dx)[sin(2x)] = 2cos(2x)#, #intcos(2x)dx = 1/2sin(2x)#;

#sin(2x) = 2sinxcosx#, so #1/2sin(2x) = sinxcosx#

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

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

#= x/2 + 1/2sinxcosx + C#