elliptic integral of SQRT(cosx)

**minivan15** · September 10th 2008, 10:14 AM

hey..we are studying elliptic integrals in my class and I did some additional research online....where i came across the formula for the integral of SQRT(cosx) = 2E (1/2 *x, SQRT(2)). I couldn't figure out how to get there myself...would someone mind writing out the proof of this?

**shawsend** · September 10th 2008, 10:42 AM

If we define the elliptic integral of the second kind as:

$\textbf{E}(\phi,k)=\int_{0}^{\phi} \sqrt{1-k^2\sin(\theta)}d\theta$

and we can write (using half-angle formula, then sub $u=\theta/2$ ):

$\int_0^x \sqrt{\cos(\theta)}d\theta=2\int_{0}^{x/2}\sqrt{1-2\sin^2(u)} du$

then:

$2\int_{0}^{x/2}\sqrt{1-2\sin^2(u)} du=2\textbf{E}(x/2,\sqrt{2})$

**minivan15** · September 10th 2008, 07:58 PM

is there a way to write it in the forms of both the first and second kinds of elliptic integrals?

(F is the same as E but d(theta) is on the top of a fraction in the integral, with the rest on the bottom of the fraction

**shawsend** · September 11th 2008, 04:18 AM

Originally Posted by minivan15

is there a way to write it in the forms of both the first and second kinds of elliptic integrals?

(F is the same as E but d(theta) is on the top of a fraction in the integral, with the rest on the bottom of the fraction

I don't see how it can be written in terms of elliptical integrals of the first kind:

$\textbf{F}(\phi,k)=\int_0^{\phi}\frac{d\theta}{\sq rt{1-k^2\sin^2(\theta)}}$

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