help rewriting elliptic integral: int[sqrt(1 + (asinx)^2)]dx

[dts5044]

hey...so my problem is to rewrite the [integral from 0 to pi/2] SQRT(1 + (asinx)^2)dx in terms of the elliptic integral E(k, phi) = [integral from 0 to phi] SQRT(1 - (ksinx)^2)dx.

please note i am aware elliptic integrals are unsolvable in closed form...i am not looking for a solution, I am just attempting to rewrite the integral to make it look like the form of the elliptic integral E

using the substitution tanu = asinx, I have worked it out to be: [integral from 0 to arctan(asinx)] du/((cosx)^2 * SQRT(a^2 -(a^2 + 1)(sinx)^2))

but I can't figure out where to go from here. Please help!

 

[galactus]

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.

 

[dts5044]

Re: help rewriting an integral

well thank you for taking the time to write out that response! I'm not sure it explicitly answers my question but hopefully it will be of some help as I work on it. Thanks again!