Application of Cauchy's formula for trigonometric integrals.

  • #1

Homework Statement


∫dθ/(1 + (sinθ)^2 ), [0, π]


Homework Equations



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

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.
 

Answers and Replies

  • #2
Dick
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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.
 
  • #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.
 
  • #4
Dick
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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.
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.
 

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