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

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)
$$a+(a-1) \tanh ^{-1}(a)+\frac{(a-1) \left(\left(a^2-1\right) \tanh ^{-1}(a)+a\right)}{4 (a+1)}-$$ $$\frac{1}{16} (a-1)^3 \left(\tanh ^{-1}(a)-\frac{a
   \left(a^2+1\right)}{\left(a^2-1\right)^2}\right)+\frac{(a-1) \left(-9 a^5+40 a^3+9 \left(a^2-1\right)^3 \tanh ^{-1}(a)+9
   a\right)}{384 (a+1)^3}$$
