Re: help rewriting an integral
I am not totally sure of what you mean other than writing it in the form E(a,x).
But, is this what you're getting at?.
We get the Jacobi form of the second kind by making the subs:
I am going to use phi instead of x.
\(\displaystyle x=sin{\phi}, \;\ dx=cos{\phi}d{\phi}, \;\ or \;\ d{\phi}=\frac{dx}{cos{\phi}}=\frac{dx}{\sqrt{1-x^{2}}}\)
\(\displaystyle {\phi}=\frac{\pi}{2}\) corresponds to x=1.
\(\displaystyle F(a,{\phi})=\int_{0}^{\phi}\frac{d{\phi}}{\sqrt{1-a^{2}sin^{2}{\phi}}}=\int_{0}^{x}\frac{dx}{\sqrt{(1-x^{2})(1-a^{2}x^{2})}}\)
Say we have \(\displaystyle E(a,{\phi})=\int_{0}^{\phi}\sqrt{1-a^{2}sin^{2}(\phi)}d{\phi}\)
\(\displaystyle =\int_{0}^{x}\sqrt{\frac{1-a^{2}x^{2}}{1-x^{2}}}dx\)
\(\displaystyle K=F(a,\frac{\pi}{2})=\int_{0}^{1}\frac{dx}{\sqrt{(1-x^{2})(1-a^{2}x^{2})}}\)
\(\displaystyle E=\int_{0}^{1}\sqrt{\frac{1-a^{2}x^{2}}{1-x^{2}}}dx\)
Many integrals are reduced to one or more combinations of these forms.
For instance:
\(\displaystyle \int_{0}^{x}\sqrt{\frac{10-5x^{2}}{1-x^{2}}}dx\)
can be found by factoring out \(\displaystyle \sqrt{10}\) and then the integral is \(\displaystyle \sqrt{10}E(a{\phi})\) in the above with
\(\displaystyle a^{2}=\frac{1}{2}\)
Also have you been familiarized with 'sn u' and 'cn u'?.
The function 'sn u' is an elliptic function. They have some resemblance to the regular trig functions.
\(\displaystyle sn(u)=sin{\phi}=sin(amp(u))\)
\(\displaystyle cn(u)=cos{\phi}=cos(amp \;\ u)=\sqrt{1-sin^{2}(amp \;\ u)}=\sqrt{1-sn^{2}(u)}=\sqrt{1-x^{2}}\)
\(\displaystyle dn(u) = \frac{d{\phi}}{du}=\frac{1}{\frac{du}{d{\phi}}}=\sqrt{1-a^{2}sin^{2}{\phi}}=\sqrt{1-a^{2}sn^{2}(u)}=\sqrt{1-a^{2}x^{2}}\)
Also, \(\displaystyle \frac{d}{du}(sn(u))=\frac{d}{du}(sin{\phi})=cos{\phi}\frac{d{\phi}}{du}=cn(u)du\)
I hope this small tutorial was of some help. I am out of time now. Have to leave shortly. I will not be back for several days.
Elliptic integrals are wacky things. They were first devised to find the arc length/circumferences of ellipses if I am not mistaken.
