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Integral (cos x)^2 dx

  1. Apr 28, 2004 #1
    Do any one have an idea how to calculate integral of (cos x)^2 ? Or is it even possible? I tried some substitutions and/or rules of trigonometry, like cosxcosx+sinxsinx=1, but it didn't help. Thank you!
  2. jcsd
  3. Apr 28, 2004 #2


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    Therefore cos2x=(1+cos2x)/2

    I'll let you finish.
  4. Apr 28, 2004 #3
    Thank you. :) integral (cos x)^2 dx
    Last edited by a moderator: Feb 17, 2008
  5. Mar 3, 2008 #4
    dont you have to use half angle identities to get integral of cos^2 ?
  6. Mar 3, 2008 #5


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    No, double angle formulas as mathman said.
  7. Mar 5, 2008 #6
    an easy way to remember the solution to this common integral, when integrating over a whole period:

    cos^2 x + sin ^2 x =1
    [tex] \int cos^2 x = \int sin^2 x [/tex]
    , at least when you integrate over a whole period

    [tex] \int cos^2 x + \int sin^2 x =[/tex] length of a period

    so the integral gives length of a period divided by 2
  8. Mar 5, 2008 #7
    Why does this thread have over 16,000 views?

    edit: Oh, it's four years old.
  9. Sep 13, 2009 #8
    First use the half-angle formula to change the cos(x)^2 to (1+cos(2x))/2...
    This will allow you to break the integral into two seperate problems much easier to solve
    integral{ 1/2dx + integral{ cos(2x)dx
    Then you will have x/2 + (sin(2x)/2) + C
  10. Sep 13, 2009 #9
    What the, that's not even correct. If you're gonna revive a 5-year old thread, at least make sure you don't have arithmetic errors.
  11. Sep 14, 2009 #10
    sin(2x)/4 ;)
  12. Feb 22, 2010 #11
    use the euler's formula

    cos x= [e^ix+e^-ix ]
    [ 2 ]
    Last edited by a moderator: Mar 8, 2012
  13. Feb 22, 2010 #12
  14. Feb 22, 2010 #13


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    This is crazy. The very first reply, post #2, answered the question. Six years ago!
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