Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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:

    Attached Files:

    • OMg.JPG
      File size:
      9.8 KB
    Last edited: Jan 23, 2007
  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

    User Avatar
    Homework Helper

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook