I was wondering how one could break the integral


into other elliptic integrals of the first, second, and third kind. I came upon a thread that said all elliptic integrals can be broken down into elliptic integrals of the first, second, and third kind, and I noticed that another similar integral


was equal (or at least so I think). I feel like if I could break them down, equivalent elliptic factors could cancel out and I could prove their equivalence, but I don't know how to break them down. How do you break both integrals into their decompositions? Also, sorry this is my first question on Stack Exchange, so there are probably flaws everywhere in my question. Thanks!

To clarify, I'm asking if

$$\int_{0}^{2\pi}\frac{1}{(b^2sin^2(\theta)+a^2cos^2(\theta))^{1/4}}d\theta = \int_{0}^{2\pi}\frac{1}{(a^2sin^2(\theta)+b^2cos^2(\theta))^{1/4}}d\theta $$

is true or not for all REAL a and b.

Oh yeah, and here is a source I found that might clarify: https://pdfs.semanticscholar.org/3a41/3ed6779db174ff01c0debea7f41e12a215d0.pdf

And here is someone on Quora decomposing another elliptic integral: https://www.quora.com/How-do-I-integrate-displaystyle-int_0-frac-pi-2-frac-dx-sqrt-1+-sin-3x/answer/Luke-Gustafson

  • $\begingroup$ The two integrals are literally essentially the same, the only difference at all is that $b$ and $a$ have been exchanged places, and those look to be constants unless there's a detail missing here. So if you have an expression for the second, simply interchange $a$ and $b$ in it, and that will be the expression for the first. $\endgroup$ – The_Sympathizer Jun 24 at 3:00
  • $\begingroup$ I understand what you are saying, but I think I should have clarified by mentioning that the second integral should use the same b and the same a from the first integral, and that they can't be changed. I'm not trying to rename the same variable, I'm trying to use the same constant but in different places, if that makes sense. $\endgroup$ – Manu Gupta Jun 24 at 3:04
  • $\begingroup$ The point is, it's immaterial which value you label "b" and "a", but the values themselves, so long as they go in the right spots in the formula - so if you have the values labeled opposite, then simply changing the labels and making the corresponding change in the formula so that the right names go with the right places, and it will work just same. $\endgroup$ – The_Sympathizer Jun 24 at 3:12
  • $\begingroup$ I understand, but in this case one cannot make the corresponding change in the formula so that the right names go with the right places. For example, if one were to plug in a = 1 and b = 2, I'm asking if $\int_{0}^{2\pi}\frac{1}{sin^2(\theta)+4cos^2(\theta))^{1/4})}d\theta = \int_{0}^{2\pi}\frac{1}{(4sin^2(\theta)+ cos^2(\theta))^{1/4})}d\theta $ is true. The integrands are completely different. Maybe I didn't clarify that I am focusing on the values and not the names, or I said it wrong. $\endgroup$ – Manu Gupta Jun 24 at 3:16
  • $\begingroup$ To find the integral with the inner part looking like $4 \sin^2 \theta + \cos^2 \theta$, then take $a = 4$ and $b = 1$. For it the other way around, take $a = 1$ and $b = 4$, in the same formula. $\endgroup$ – The_Sympathizer Jun 24 at 3:32

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