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Just an integral

  1. Sep 28, 2006 #1

    quasar987

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    How could I compute

    [tex]\int \sqrt{1-cost} \ \ dt [/tex]

    ?
     
  2. jcsd
  3. Sep 28, 2006 #2
    just throwing this out there... could it have anything to do with a hanfle angle formula??

    because i do rmember that
    [tex] \sin t/2 = \pm \sqrt{\frac{1 - \cos (t)}{2}} [/tex]

    that plus or minus does make things a bit... iffy though
     
  4. Sep 28, 2006 #3

    acm

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    Sqrt ( 1- cost)
    Sqrt ( 1 - cos^2t) / Sqrt (1 + Cost)
    Sqrt ( Sin^2t) / Sqrt (1 + Cost)
    Sint / Sqrt (1 + Cost)
    Integral becomes: I = Int ( sint/ sqrt( 1+ Cost) ) dt
    Let u = 1 + Cost
    du/dt= -Sint
    du = -Sintdt
    Integral becomes: I = -Int ( -du/Sqrt (u) )
    I = -2 Sqrt(u) + C
    I = -2 Sqrt (1 + Cost) + C
     
  5. Sep 28, 2006 #4

    quasar987

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    Excellent! Thanks you very much acm!
     
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