Help with an integral, yet again.

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Discussion Overview

The discussion revolves around solving the integral \(\int \frac{1}{1+\sin^2{\theta}}d\theta\) from 0 to \(\pi\). Participants explore various methods of integration, including trigonometric substitution and the residue theorem, while addressing challenges and uncertainties in their approaches.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes changing \(\sin^2{\theta}\) into \(1 - \cos(2\theta)\) and expresses the integral in terms of a complex variable, leading to a form involving poles.
  • Another suggests substituting \(\sin(x) = u\) to express the integral in terms of \(u\), leading to a different integral that may be easier to solve.
  • A participant expresses confusion about the nature of indefinite integrals and the provided answer in the textbook, questioning the validity of constants in such contexts.
  • Some participants discuss the application of the residue theorem, with one noting a misunderstanding of terminology between method and theorem.
  • Another participant mentions that the substitution to \(z = e^{2i\theta}\) could simplify the integral further.
  • Several participants express enthusiasm about the residue theorem and its applications, noting its complexity and the need for practice.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to solve the integral, with multiple competing approaches and some uncertainty about the application of the residue theorem.

Contextual Notes

There are unresolved mathematical steps and assumptions regarding the application of the residue theorem and the validity of various substitutions. The discussion reflects differing levels of familiarity with the techniques involved.

Who May Find This Useful

Readers interested in advanced integration techniques, particularly those involving complex analysis and the residue theorem, may find this discussion beneficial.

DrKareem
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Hi, the integral I'm trying to solve is:

\int \frac{1}{1+\sin^2{\theta}}d\theta

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:

\frac{1}{i} \int \frac{z}{-z^4+z^2-1} dz

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)
 
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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

\int \frac{\sqrt{1-u^2}}{1+u^2} du. Trig sub.
 
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Ahh nvm my last post..I must have done something wrong because when you check it on www.calc101.com it dusnt gimme the same thing...But I don't undersatnd how you textbook gives you a constant for an indefinite integral..

BTW: The answer isn't nice..
Your textbook is correct if this is pi:
 

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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.
 
I have no idea what the theorem is :D

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

Edit: Actually it IS the residue THEOREM not method :P
 
cool problem, shows the awesome power of the residue theorem!
 
  • #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...
 
  • #11
Practice makes permanant!
 

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