Application of Cauchy's formula for trigonometric integrals.

In summary, the conversation discusses solving the integral ∫dθ/(1 + (sinθ)^2) from 0 to π using Cauchy's Formula. The conversation mentions substituting sinθ = (1/(2i))(z - 1/z) and finding the zeros of the denominator, z^2 = 3 ± 2√(2). The conversation also mentions using Cauchy's Formula and obtaining an answer of 0, but questioning the validity of this answer. It is suggested to use sin(x)^2=(1-cos(2x))/2 and express cos(2x) in terms of z. The result is a denominator of (3 - cos(2θ)) and the integral
  • #1
StumpedPupil
11
0

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.
 
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  • #2
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
StumpedPupil said:
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.
 

1. What is Cauchy's formula for trigonometric integrals?

Cauchy's formula for trigonometric integrals is a method for solving integrals involving trigonometric functions. It states that if f(x) is a continuous function on the interval [a, b], then the integral of f(x) over that interval can be calculated by taking the average value of f(x) over the interval and multiplying it by the length of the interval.

2. How is Cauchy's formula used to solve trigonometric integrals?

To use Cauchy's formula, we first need to find the average value of the function over the interval. This can be done by taking the integral of the function and dividing it by the length of the interval. Then, we multiply this average value by the length of the interval to get the value of the integral.

3. What are some common applications of Cauchy's formula for trigonometric integrals?

Cauchy's formula can be used to solve various types of integrals involving trigonometric functions, such as those involving sine, cosine, and tangent. It is also commonly used in physics and engineering to solve problems involving periodic functions.

4. Are there any limitations to Cauchy's formula for trigonometric integrals?

Cauchy's formula can only be used for integrals over a finite interval. It also assumes that the function is continuous over that interval. If the function is not continuous or if the interval is infinite, Cauchy's formula cannot be applied.

5. Can Cauchy's formula be applied to integrals involving multiple trigonometric functions?

Yes, Cauchy's formula can be applied to integrals involving multiple trigonometric functions. In this case, the average value of the function over the interval is calculated separately for each trigonometric function and then multiplied together to get the final result.

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