<s>[INST] Arc Length of an Ellipse using integration I was thinking about what the arc length of an ellipse is, but throughout my calculations I got stuck. Here is how I approached the problem:$$$$We have an ellipse in the form:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow y=\pm \frac{b}{a}\sqrt{a^2-x^2}$$By applying the formula of the arc length of a function, we get:$$L=4\int_0^a\sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}}dx=4\int_0^a\sqrt{\frac{a^4+(b^2-a^2)x^2}{a^2(a^2-x^2)}}dx$$Now I made a little subsitution recalling trigonometry: $$x=a\sin(u)\\dx=a\cos(u)du$$So the Integral now can be expressed as:$$L=4\int_0^{\frac{\pi}{2}}a\cos(u)\sqrt{\frac{a^4+(b^2-a^2)a^2\sin^2(u)}{a^2(a^2-a^2\sin^2(u))}}du=\\4\int_0^{\frac{\pi}{2}}a\cos(u)\sqrt{\frac{a^4+(b^2-a^2)a^2\sin^2(u)}{a^2(a^2\cos^2(u)+a^2\sin^2(u)-a^2\sin^2(u))}}du=\\4\int_0^{\frac{\pi}{2}}\sqrt{a^2+(b^2-a^2)\sin^2(u)}du$$So we have:$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+\frac{(b^2-a^2)}{a^2}\sin^2(u)}du$$
Letting $m=\frac{(b^2-a^2)}{a^2}$ we finally get:$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+m\sin^2(u)}du$$
However, at this point I do not know any way on how to integrate this function because the $m$ is 'in the way'. Does anyone have any hints?
 [/INST] This is an elliptical integral. 
You can use the elliptical integral tables to look up a value for the integral- much like the standard normal distribution. 
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