calculating the circumference of an ellipse

[bwanaaa]

I have searched for the answer and always find that there is an open solution (integral).

C1 = INTEGRAL(0->2pi) sqrt(a^2 * cos^2[t]+b^2 *sin^2[t]) dt.

(a nicer page is here:
http://mathforum.org/dr.math/faq/formulas/faq.ellipse.circumference.html)

plugging this into wolfram alpha runs out of computation time but a similar integral here
C2 = INTEGRAL(0->2pi) (a^2 * cos^2[t]+b^2 *sin^2[t]) dt.

(here:
http://www.wolframalpha.com/input/?i=integrate+(a^2*cos^2[t]+b^2*sin^2[t])+dt+++from+t=0+to+2pi
)

gives
pi * (a^2 + b^2)

I assume that the first equation (c1) is intregate-able as well, but do not have the resources (mathematica) to calculate it. Is there anyone here with math-ability?

 

[Turtle.Man]

My "mathability" has about 25 years of rust on it, but all of the methods that I've seen for solving this problem are approximations. I don't believe that there is a (known) closed form solution.

 

[Essence_of_War]

C1 is an elliptic integral of the 2nd kind, and it cannot be expressed in terms of elementary functions. If you just want the arc length, it can be evaluated numerically using your favorite numerical integration scheme.

You can also find an exact infinite series solution if you're so inclined. The wikipedia article on elliptic integrals is probably useful there:

https://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_second_kind