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The incomplete elliptic integral of the first kind is written (using trigonometric form) : $ F(\varphi,k)=\int_{0}^{\varphi} \frac{1}{\sqrt{1-k^2 \sin^2(\theta)}} \mathrm{d}\theta $.

Then, it is noted everywhere that if we make the change of variable $t=\sin(\theta)$, then the integral can be re-written in the so-called Jacobi's form : $F(x,k)=\int_{0}^{x} \frac{1}{\sqrt{ (1-t^2)(1-k^2 t^2) }} \mathrm{d}t$, where $x$ is used instead of $\sin(\varphi)$...

So good up to there, but, $t=\sin(\theta) \Longrightarrow \mathrm{d}t=\cos(\theta)\mathrm{d}\theta$, and, depending on $\theta$, $\cos(\theta)=\pm\sqrt{1-\sin^2(\theta)}=\pm\sqrt{1-t^2}$

So I wonder why, in the form of Jacobi, we use the positive writing of $\cos(\theta)$ ?

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Because the original integral makes sense for $\theta$ in a neighborhood of $0$. There, $\cos$ is positive.

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  • $\begingroup$ Hi Giuseppe, I don't understand your "makes sense" .... $\theta$ is supposed to vary between $0$ and $\varphi$ ; is there, by definition, any limitation imposed to $\varphi$ ? $\endgroup$ – Andrew Mar 9 at 18:32
  • $\begingroup$ I confess I still do not understand why changing the variable $ t = \sin(\theta) $ leads to choosing $ \cos(\theta) = \sqrt {1-t ^ 2} $ to express the Jacobi form of the elliptical integral ... $\endgroup$ – Andrew Mar 9 at 21:15
  • $\begingroup$ Oh yes I get it. The true change of variable is $dt=|\cos \theta|\, d\theta$. If you change variable this way, in the style of multiple integrals (absolute value of the Jacobian determinant), you don't have to worry about this kind of things. $\endgroup$ – Giuseppe Negro Mar 11 at 20:59

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