<s>[INST] Perimeter of an ellipse solution and elliptic integrals. I developed a formula for finding the length of a curve; let's call it $L(x)$:
$$L(x)=\int_{0}^x ((\frac{dy}{dx})^2+1)^\frac{1}{2}dx$$
Applying this to an ellipse of equation 
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
and therefore $\frac {dy}{dx}={-xb\over a\sqrt{a^2-x^2}}$ 
so $((\frac{dy}{dx})^2+1)={(b^2-a^2)x^2+a^4\over a^2({a^2-x^2})}$
This means a quarter of an ellipse's perimeter is 
$$\int_{0}^a\sqrt{(b^2-a^2)x^2+a^4\over a^2({a^2-x^2})}dx$$
which is
$$\int_{0}^a\frac{1}{a}\sqrt{(b^2-a^2)x^2+a^4\over ({a^2-x^2})}dx $$
This means the perimeter of an ellipse is equal to 
$$\frac{4}{a}\int_{0}^a\sqrt{(b^2-a^2)x^2+a^4\over ({a^2-x^2})}dx $$
Can anyone integrate this for me please? I don't have the necessary skills yet. I've been told that to integrate this I'd need elliptic integrals, but I don't understand what they are. If you do, could you please explain what they are and how they work? Thanks a lot.
 [/INST] Your integral is correct and there're alternative representations for the arclength:
\begin{align}
  (x,y) &= (a\sin \theta,b\cos \theta) \\
  k &= \sqrt{1-\frac{b^2}{a^2}} \\
  k' &= \frac{b}{a} \\
  s &= \int_0^\theta \sqrt{a^2\cos^2 \phi+b^2\sin^2 \phi} \, d\phi \\
  &= a\int_0^{\frac{x}{a}} \sqrt{\frac{1-k^2 t^2}{1-t^2}} \, dt \\
  &= b\int_{\frac{y}{b}}^1 \sqrt{\frac{k'^2+k^2 t^2}{1-t^2}} \, dt \\
  &= a\int_{b\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}}}^1
  \frac{t^2 dt}{\sqrt{(1-t^2)(t^2-k'^2)}} \\
  &= aE(\theta,k)
\end{align}
where $E(.,.)$ is called incomplete elliptic integral of the second kind.
May compare these with Jacobi elliptic functions:
\begin{align}
  (x,y) &= (a\operatorname{sn} u,b\operatorname{cn} u) \\
  u &= \int_0^{\operatorname{sn} u} \frac{1}{\sqrt{(1-t^2)(1-k^2 t^2)}} \, dt \\
  &= \int_{\operatorname{cn} u}^1 \frac{1}{\sqrt{(1-t^2)(k'^2+k^2 t^2)}} \, dt \\
  &= \int_{\operatorname{dn} u}^1 \frac{dt}{\sqrt{(1-t^2)(t^2-k'^2)}} \\
  &= F(\operatorname{sn u},k)
\end{align}
where $F(.,.)$ is called incomplete elliptic integral of the first kind.
See the link here in other ways for calculating the perimeter.

Further points to be noticed
  
  
*
  
*$k$ is elliptic modulus of elliptic integrals/functions which also equals to the eccentricity of the ellipse.
  
*The integrals are the arclength measured between the fixed point $(0,b)$ and an arbitrary point $(x,y)$ on the ellipse within one quadrant.
  
*A quarter of perimeter is between $(x,y)=(0,b)$ and $(x,y)=(a,0)$.
  
*The perimeter is $$4aE\left( \dfrac{\pi}{2}, k \right)=4aE(k)$$  where $E(k)$ is a complete elliptic integral of the second kind.
  
*We can derive the above integrals by change of variables
  
  
*
  
*$t=\dfrac{x}{a}=\sin \phi=\operatorname{sn} (u,k)$
  
*$t=\dfrac{y}{b}=\cos \phi=\operatorname{cn} (u,k)$
  
*$t=b\sqrt{\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}}
      =\sqrt{1-k^2\sin^2 \phi}=\operatorname{dn} (u,k)$
  
  
*If $p$ is the perpendicular distance of a tangent, at $(x,y)$ on the ellipse, from the origin
  $$p=\frac{1}{\sqrt{\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}}}$$
  we have arclength element $$ds=\frac{ab\, d\phi}{p}$$ and curvature at $(x,y)$ $$\kappa=\frac{p^3}{a^2 b^2}$$

</s>