I am asked to give the length of the curve:
$\gamma: [0,2\pi]\to\mathbb{R}^2$, $\gamma(t)=(a\cos(t), b\sin(t))$ where $a,b\in(0,\infty)$
We get to solving the following:
$\int_0^{2\pi} \sqrt{a^2\sin(t)^2+b^2\cos(t)^2}\, dt$
Which should not be possible to integrate probably.
For $a=b$ I can solve it, but thats it. We would get:
$2\pi\cdot a$
Is there any way, which does not involve using the form above and its solution with "second kind elliptic integrals" which wolframalpha refers to?
Thanks in advance.
Quoting the Wikipedia page, the circumference $C$ of an ellipse is given by $$C = 4 a\, E(e)\qquad \text{with} \qquad e=\sqrt{1-\frac{b^2}{a^2}}$$ The exact infinite series is given by $$C=2\pi a \left[1 - \sum_{n=1}^\infty \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{e^{2n}}{2n-1}\right]$$ but a much faster convergen series is $$C = \pi (a+b) \left[1 + \sum_{n=1}^\infty \left(\frac{(2n-1)!!}{2^n(2n-1) n!}\right)^2 {h^n}\right]\qquad \text{with} \qquad h= \frac {(a-b)^2 } {(a+b)^2}$$
In the linked page, you will find a series of approximations wich can be more than sufficient for practical usage.