Length of Lemniscate of Bernoulli

Theorem
The total length of the lemniscate of Bernoulli given in polar coordinates as:
 * $r^2 = a^2 \cos 2 \theta$

is given by:
 * $L = 4 a F \left({\sqrt 2, \dfrac \pi 4}\right)$

where $F$ denotes the incomplete elliptic integral of the first kind.

Proof
The arc length of a small length increment $\mathrm d s$ is given in polar co-ordinates by:
 * $\left({\mathrm d s}\right)^2 = \left({r \mathrm d \theta}\right)^2 + \left({\mathrm d r}\right)^2$

from which:
 * $\dfrac {\mathrm d s} {\mathrm d \theta} = \sqrt {r^2 + \left({\dfrac {\mathrm d r} {\mathrm d \theta} }\right)^2}$

Half of one lobe of the lemniscate is achieved when $\theta$ goes from $0$ to $\pi / 4$.

Therefore the total length of the lemniscate of Bernoulli is given by:
 * $\displaystyle L = 4 \int_0^{\pi/4} \sqrt{r^2 + \left({\dfrac {\mathrm d r} {\mathrm d \theta} }\right)^2} \, \mathrm d \theta$

First we show:

So:

Thus: