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How to integrate the following

  1. Jul 19, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫[itex]\frac{sin\Theta}{1+sin^2\Theta}[/itex] d[itex]\Theta[/itex] on [0,[itex]\frac{\pi}{2}[/itex]]


    2. Relevant equations
    I substituted:
    u=cos[itex]\Theta[/itex]
    du=-sin[itex]\Theta[/itex]


    3. The attempt at a solution
    = ∫ [itex]\frac{sin\Theta}{1+1-cos^2\Theta}[/itex]

    = ∫ [itex]\frac{sin\Theta}{2-cos\Theta}[/itex]

    = ∫ [itex]\frac{sin\Theta}{2-u^2}[/itex]

    = [itex]\frac{1}{(u+\sqrt{2})(u-\sqrt{2})}[/itex] <--- I'm pretty sure I shouldn't have done this.

    And that's as far as I've gotten.

    Any tips greatly appreciated.
     
  2. jcsd
  3. Jul 19, 2012 #2

    vela

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    You can use partial fractions or a trig substitution.
     
  4. Jul 19, 2012 #3
    We haven't done trig substitution, I think partial fractions is covered in the next section of my notes though so I'll check it out. Thanks :)
     
  5. Jul 19, 2012 #4

    HallsofIvy

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    You've dropped the "[itex]d\Theta[/itex]" in each of those- don't do that!

    You need, of course, "du" and you have the sign wrong in the denominator.

     
  6. Jul 19, 2012 #5

    eumyang

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    (The LaTeX was a little messy, so I cleaned them up for you.)
    Don't forget the dθ. Also, from here, move the negative sign:
    [itex]-du = \sin \theta \; d\theta[/itex]

    Again, don't forget the dθ!
    Substitute in -du for sin θ dθ:
    [tex]= \int \frac{-du}{2-u^2}[/tex]
    Can you take it from there?

    EDIT: Beaten to it. :wink:
     
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