Finding out the perimeter of the ellipse?

Good work so far. The perimeter of an ellipse with semi major and semi minor axes $a,b$ should be $$P(a,b)=\int_0^{2\pi}\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}~\mathrm d\theta$$

Which we'll rewrite a bit, by adding and subtracting $a^2\sin^2$ $$P(a,b)=\int_0^{2\pi}\sqrt{a^2\cos^2\theta+a^2\sin^2\theta+b^2\sin^2\theta-a^2\sin^2\theta}~\mathrm{d}\theta$$ Now using $\cos^2+\sin^2=1$, we have removed the cosine term: $$P(a,b)=\int_0^{2\pi}\sqrt{a^2+(b^2-a^2)\sin^2\theta}$$ We'll rewrite this again as $$P(a,b)=a\int_0^{2\pi}\sqrt{1-\left(1-\frac{b^2}{a^2}\right)\sin^2\theta}$$ However due to symmetry of the ellipse, we know this is just four times the integral taken from $0$ to $\pi/2$, so $$P(a,b)=4a\cdot \int_0^{\pi/2}\sqrt{1-\left(1-\frac{b^2}{a^2}\right)\sin^2\theta}~\mathrm d\theta$$ But this is just the complete elliptic integral of the second kind Which I'll denote as $$P(a,b)=4a\cdot \operatorname{Eli}_2\left(\sqrt{1-\frac{b^2}{a^2}}\right)$$

Approximations

The complete elliptic integral of the second kind has a series expansion: $$\operatorname{Eli}_2(z)=\frac{\pi}{2}\left[1-\sum_{n=1}^\infty\left(\frac{(2n-1)!_2}{(2n)!_2}\right)^2\frac{1}{2n-1}z^{2n}\right]$$ Where $!_2$ is a double factorial. (I dislike the "$!!$" notation as I find it misleading.) Take $z=\sqrt{1-b^2/a^2}$ and use as many terms as you wish for as good an approximation as you wish.