# Incomplete elliptic integral and Jacobi's form

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 $$\cos(\theta)$$ ?

Because the original integral makes sense for $$\theta$$ in a neighborhood of $$0$$. There, $$\cos$$ is positive.
• 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 $\varphi$ ? – Andrew Mar 9 at 18:32
• 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 ... – Andrew Mar 9 at 21:15
• 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. – Giuseppe Negro Mar 11 at 20:59