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Integral of the function?

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    ∫(sin x)√(1+cos2x) dx
     
    Last edited: Oct 12, 2011
  2. jcsd
  3. Oct 12, 2011 #2

    HallsofIvy

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    What you have written makes no sense (even if the latex were working). you have \sqrt{} with nothing inside the square root and have a subscript that makes no sense.

    Did you mean [itex]\int sin(x)\sqrt{1+ cos^2(x)}dx[/itex]?

    If so, start by letting y= cos(x).
     
  4. Oct 12, 2011 #3
    Yes, that's exactly what I was try to make. Well, this is my first time using this so sorry..
     
  5. Oct 12, 2011 #4
    Well, I got myself until:

    [itex]\large - [/itex] [itex]\int \sqrt{1+y^2}[/itex] [itex]\large dy [/itex]

    then I stucked again.
     
  6. Oct 12, 2011 #5

    Dick

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    Try and think of a trig (or hyperbolic trig) substitution that will turn 1+y^2 into the square of something so you can get rid of the square root.
     
  7. Oct 12, 2011 #6
    Argh... I don't seem to get it after cracking my head for a while. Can I have a little bit more tips?
     
  8. Oct 12, 2011 #7

    SammyS

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    Let's see, ...

    1+tan2(θ) = sec2(θ)

    1+sinh2(u) = cosh2(u)
     
  9. Oct 14, 2011 #8
    using 1+tan2v=sec2v , I got it until :

    -∫sec3v dv

    am I suppose to do by parts or is there other ways?
     
  10. Oct 14, 2011 #9

    Dick

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    Parts. Split it into sec(v)^2*dv and sec(v).
     
  11. Oct 14, 2011 #10
    Okay, I think I got it. Thanks for the help from everyone.^^
     
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