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Help with an integral, yet again.

  1. Jan 23, 2007 #1
    Hi, the integral I'm trying to solve is:

    [tex] \int \frac{1}{1+\sin^2{\theta}}d\theta [/tex]

    From 0 to pi, which is the same as half the same integral over the unit circle.

    I changed the sine squared into 1-cos[2x] and then expressed cos[2x] as half z^2 +(z-1)^2 and i finally got it in this form:

    [tex] \frac{1}{i} \int \frac{z}{-z^4+z^2-1} dz [/tex]

    I found out the poles to be -1/2 +- i sqrt(3)/2 which happen to be on the countour path, and in that case i don't know how to apply the residue theorem, if it could be applied that is.

    Edit: The answer is given in the book to be pi/sqrt(2)
     
    Last edited: Jan 23, 2007
  2. jcsd
  3. Jan 23, 2007 #2

    Gib Z

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    Let sin(x)=u and express the entire integral in terms of u. the denominator is just 1+u^2, and solve and sub to dtheta to du, we get

    [tex]\int \frac{\sqrt{1-u^2}}{1+u^2} du[/tex]. Trig sub.
     
    Last edited: Jan 23, 2007
  4. Jan 23, 2007 #3

    Gib Z

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    Ahh nvm my last post..I must have done something wrong cuz when you check it on www.calc101.com it dusnt gimme the same thing...But I dont undersatnd how you textbook gives you a constant for an indefinite integral..

    BTW: The answer isnt nice..
    Your textbook is correct if this is pi:
     

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  5. Jan 23, 2007 #4
    Okay I see where you're going here, it should be solved using that method. However, i took the integral from a chapter on the residue method in integration. So i'm trying to solve it using that method, and explained how i got stuck.
     
  6. Jan 23, 2007 #5

    Gib Z

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    I have no idea what the theorem is :D

    Edit: Method, not theorem
     
  7. Jan 23, 2007 #6
    your sub to cos(2x) is fine, and then exp's. what you then want to do is substitute [tex] z=e^{2 i \theta} thus dz=2i z d\theta [/tex]. Your substitution was a little funky. thus [tex]cos(2 \theta) = \frac{z+ (1/z)}{2} [/tex].
     
  8. Jan 23, 2007 #7
    youll get a slightly easier integral which you can just take the residue of for the answer
     
  9. Jan 23, 2007 #8
    Oh my God that was so trivial. Thanks for the help guys.

    Edit: Actually it IS the residue THEOREM not method :P
     
  10. Jan 23, 2007 #9
    cool problem, shows the awesome power of the residue theorem!
     
  11. Jan 24, 2007 #10
    Yes awesome indeed. I just started studying this and it looks really nice, although some aspects of it are hard to grasp. I guess i should be practicing even more...
     
  12. Jan 24, 2007 #11
    Practice makes permanant!
     
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