Converting $\intop_{0}^{a}\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}dx$ to elliptic integral Tried using $x=a\sin\left(\theta\right)$ $$\rightarrow \intop_{0}^{\pi/2}\sqrt{\frac{a^{2}-\left(a\sin\left(\theta\right)\right)^{2}}{1-\left(a\sin\left(\theta\right)\right)^{2}}}a\cos{\left(\theta\right)d\theta}$$
$$\iff \intop_{0}^{\pi/2}\frac{a^{2}\cos^{2}\left(\theta\right)}{\sqrt{1-\left(a\sin\left(\theta\right)\right)^{2}}}{d\theta}$$
Which looks similar to the complete elliptical of first or second kind, is there a way to make the conversion? Thanks
 A: You can convert this integral in terms of elliptic functions. You need to make an assumption about the parameter $a$:
If $0<a<1$ we can continue with your approach:
First, note
$$ J=\intop_{0}^{\pi/2} \frac{\sin^{2}\left(\theta\right)}{\sqrt{1-k^2\sin^2\left(\theta\right)}}{d\theta} = \frac{K(k)-E(k)}{k^2}$$
To show this recall the definition of the complete elliptic integrals of the first and second kind, respectively:
$$K(k) = \intop_{0}^{\pi/2} \frac{1}{\sqrt{1-k^2\sin^2\left(\theta\right)}}{d\theta}$$
$$E(k) = \intop_{0}^{\pi/2} \sqrt{1-k^2\sin^2\left(\theta\right)}{d\theta}$$
Hence
$$\frac{1}{k^2}\left[K(k)-E(k)\right] = \frac{1}{k^2} \intop_{0}^{\pi/2}  \left[\frac{1}{\sqrt{1-k^2\sin^2\left(\theta\right)}} - \sqrt{1-k^2\sin^2\left(\theta\right)}\right]{d\theta} =\frac{1}{k^2} \intop_{0}^{\pi/2}  \left[\frac{1}{\sqrt{1-k^2\sin^2\left(\theta\right)}} - \frac{1-k^2\sin^2(\theta)}{\sqrt{1-k^2\sin^2\left(\theta\right)}}\right]{d\theta} = \intop_{0}^{\pi/2} \frac{\sin^{2}\left(\theta\right)}{\sqrt{1-k^2\sin^2\left(\theta\right)}}{d\theta} $$
Then
$$ I = \intop_{0}^{\pi/2}\frac{a^{2}\cos^{2}\left(\theta\right)}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta}  = a^2\intop_{0}^{\pi/2}\frac{1-\sin^{2}\left(\theta\right)}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta} = a^2\intop_{0}^{\pi/2}\frac{1}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta} - a^2\intop_{0}^{\pi/2}\frac{\sin^{2}\left(\theta\right)}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta} = a^2K(a) -K(a)+E(a) =E(a)-a'^2K(a)$$
where $a' = \sqrt{1-a^2}$ is the complementary modulus
Hence
$$\boxed{\intop_{0}^{a}\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}dx= E(a)-a'^2K(a)}$$
Note that Wolfram use a slight different notation for Elliptic integrals.
There are some nice approximations for the complete Elliptic integrals.
$$ K(k) \approx \frac{\pi}{2} \left(\frac{16-5k^2}{16-9k^2}\right) \quad 0\leq k\leq 0.67$$
$$ E(k) \approx \frac{\pi}{2} \left(\frac{16-7k^2}{16-3k^2}\right) \quad 0\leq k \leq 0.71$$
and you can find others when $k$ is close to $1$.
You can also find series expansions for both functions.
A: Approximating the elliptic integral seems to be difficult.
What you could do is to expand the integrand around $a=1$
$$\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}=1+\frac{a-1}{1-x^2}-\frac{(a-1)^2 x^2}{2 \left(x^2-1\right)^2}-\frac{(a-1)^3 x^2}{2
   \left(x^2-1\right)^3}-\frac{(a-1)^4 \left(x^2 \left(x^2+4\right)\right)}{8
   \left(x^2-1\right)^4}+O\left((a-1)^5\right)$$ and integrate termwise to have for the integral (without simplifications)
$$a+(a-1) \tanh ^{-1}(a)+\frac{(a-1) \left(\left(a^2-1\right) \tanh ^{-1}(a)+a\right)}{4 (a+1)}-$$ $$\frac{1}{16} (a-1)^3 \left(\tanh ^{-1}(a)-\frac{a
   \left(a^2+1\right)}{\left(a^2-1\right)^2}\right)+\frac{(a-1) \left(-9 a^5+40 a^3+9 \left(a^2-1\right)^3 \tanh ^{-1}(a)+9
   a\right)}{384 (a+1)^3}$$
