1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Application of Cauchy's formula for trigonometric integrals.

  1. Oct 15, 2008 #1
    1. The problem statement, all variables and given/known data
    ∫dθ/(1 + (sinθ)^2 ), [0, π]


    2. Relevant equations

    Cauchy's Formula, perhaps Cauchy's Thm.
    http://mathworld.wolfram.com/CauchyIntegralFormula.html
    http://mathworld.wolfram.com/CauchyIntegralTheorem.html

    3. The attempt at a solution
    I first substituted sinθ = (1/(2i))(z - 1/z), where z = e^(2iθ) to get the expression
    ∫4iz/(z^4 - 6z^2 + 1)dz with the path |z| = 1.

    I found the zeros for the denominator, which are z^2 = 3 ± 2√(2), and so z = ±√(3 ± 2√(2)). Only the points ±√(3 - 2√(2)) are in |z|= 1.

    I proceeded to use Cauchy's formula to get 0, however, I am not confident with my answer. There are several things that I did as a result of my unfamiliarity of domains other than [0, 2π] when it comes to application of Cauchy's formula. I am not sure I am allowed to use the substitute of z but if I don't then I do not believe we are considering a closed curve at all.

    By the way, I am trying to find an answer in a specific way, I know there are other ways to solve this problem but my book says it can be done using something Cauchy related. I hope my question is well stated and thank you.
     
  2. jcsd
  3. Oct 15, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The answer obviously isn't zero. The integrand is positive. The problem is that if z=exp(2i*theta), (z-1/z)/(2i)=sin(2*theta), not sin(theta). Start by using sin(x)^2=(1-cos(2x))/2. Now express cos(2x) in terms of z.
     
  4. Oct 15, 2008 #3
    That doesn't actually help because you end with the same results. The denominator becomes (3 - cos(2θ)), which you can derive ∫4iz/(z^4 - 6z^2 + 1)dz from.

    Thanks for the suggestion though.
     
  5. Oct 15, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I DOES help. You just aren't being careful. You don't get a quartic in the denominator anymore. Where is the z^4 coming from? I get z^2-6z+1 in the denominator. The roots are now z=3+/-2*sqrt(2). And only ONE of them is in the unit circle.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Application of Cauchy's formula for trigonometric integrals.
Loading...