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solve integral cos^2(2x)
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\[\int\limits \cos^2(2x)\]
dx
you know the relation for cos 2x ?
i use: 1+cos2x/2
that is how it's being solved in my textbook
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ok, so cos^2 2x will be (1+cos 4x)/2 right ?
then the answer should be \[\frac{ x }{ 2 } + \frac{ \sin2x }{ 4 }\] + k
oh, it's cos4x/2?
2nd term = cos 4x/2 its integral = (1/2) sin 4x /4 ok ?
yes, 2nd term is (cos 4x )/2 any doubts ?
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yes, i'm checking
it appears my professor wrote it wrong
and i've been stuck for hours
2nd term = cos 4x/2 its integral = (1/2) sin 4x /4 got this ?
yes yes, i got it now
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good :) ask if any more doubts anywhere ...
nope, i'll try and solve it again and thank you!
welcome ^_^
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