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Trig Integrals

  1. Nov 8, 2006 #1
    Quick question: What is the integral of

    [tex]\int^{\frac{1}{2}}_{0}cos^2((\pi)x)[/tex]?

    Is it

    [tex]\frac{1}{3}(cos^3((\pi)x))\frac{1}{\pi}sin((\pi)x)[/tex]

    and then plug in or is there something wrong with that?
     
    Last edited: Nov 8, 2006
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  3. Nov 8, 2006 #2

    acm

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    Cos^2(pix) = 1/2 (1 + Cos(2pix))
    This is simply integrated.
     
  4. Nov 8, 2006 #3
    Do you know where I can get a proof of that?
     
  5. Nov 8, 2006 #4

    quasar987

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    [tex]\cos x=\frac{e^{ix}+e^{-ix}}{2}[/tex]

    Square that and you have your result.
     
  6. Nov 8, 2006 #5

    quasar987

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    Or square the power series of cosx and rearange but that's more difficult.
     
    Last edited: Nov 8, 2006
  7. Nov 9, 2006 #6

    dextercioby

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    Or simply use that

    [tex] \cos 2x=\cos^{2} x -\sin^{2} x [/tex]

    and

    [tex] 1=\cos^{2} x +\sin^{2} x [/tex]

    Daniel.
     
  8. Nov 9, 2006 #7

    VietDao29

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    Or basically, we can use the Power Reduction Fomulae:
    [tex]\cos ^ 2 x = \frac{1 + \cos (2x)}{2}[/tex]
    [tex]\sin ^ 2 x = \frac{1 - \cos (2x)}{2}[/tex]
    -------------
    Now, back to your problem:
    [tex]\int_{0} ^ {\frac{\pi}{2}} \cos ^ 2 (\pi x) dx = \int_{0} ^ {\frac{\pi}{2}} \left( \frac{1 + \cos (2x)}{2} \right) dx[/tex]
    Now, all you need to do is to use a u-substitution to solve it.
    Can you go from here? :)
     
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