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?

I suppose that you missed an $a$ somewhere and have in fact, for your last expression,$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+m\sin^2(u)}\,du=4a\,E(-m)$$ where appears  the complete elliptic integral.
This cannot be expressed in terms of elementary functions but some approximations are available by the great Ramanujan. In particular
$$L=\pi  \left(3 (a+b)-\sqrt{(3 a+b) (a+3 b)}\right)\tag 1$$ and  $$L=\pi  (a+b) \left(1+\frac{3\frac{
   (a-b)^2}{(a+b)^2}}{10  \sqrt{4-3\frac{
   (a-b)^2}{(a+b)^2}}}\right)\tag 2$$
Appliead to your example $(a=4,b=8)$, $(1)$ would give $4\pi \left(9-\sqrt{35}\right)\approx 38.7537 $ and  $(2)$ would give $\frac{2\pi}{55} \left(330+\sqrt{33}\right)\approx 38.3554$ while the exact value should be $16 E(-3)\approx 38.7538$.
You also could use an infinite sum formulation $$L=\pi(a+b)\sum_{n=0}^\infty \binom{\frac{1}{2}}{n}^2 h^n \qquad \qquad \text{where} \qquad h=\frac{
   (a-b)^2}{(a+b)^2}$$ Limiting to $p$ terms, the convergence is quite fast 
$$\left(
\begin{array}{cc}
 p & L_p \\
 0 & 37.69911184 \\
 1 & 38.74630939 \\
 2 & 38.75358160 \\
 3 & 38.75378361 \\
 4 & 38.75379237 \\
 5 & 38.75379285 \\
 6 & 38.75379288 
\end{array}
\right)$$
