<s>[INST] Series expansion of Elliptic integral $F(\sin^{-1}(1-x)|-1)$ where $0I understand the leading term $F(\pi/2\mid-1)$ would be $K(-1)$. I also want the next term contributing from $x$. Actually I could not expand $\sin^{-1}(1-x)$ (when $0<x\ll1$) around 1. Any help would be highly appreciated.
Thanks a lot.     
 [/INST] For more than the first term.
We can start using $$\frac d {dx}F\left(\left.\sin ^{-1}(1-x)\right|-1\right)=-\frac{1}{\sqrt{1-(1-x)^2} \sqrt{1+(1-x)^2}}\tag 1$$ and use Taylor expansions around $x=0$ $$\sqrt{1-(1-x)^2}=\sqrt{2} \sqrt{x}-\frac{x^{3/2}}{2 \sqrt{2}}-\frac{x^{5/2}}{16
   \sqrt{2}}-\frac{x^{7/2}}{64 \sqrt{2}}-\frac{5 x^{9/2}}{1024
   \sqrt{2}}+O\left(x^{11/2}\right)$$ $$\sqrt{1+(1-x)^2}=\sqrt{2}-\frac{x}{\sqrt{2}}+\frac{x^2}{4 \sqrt{2}}+\frac{x^3}{8 \sqrt{2}}+\frac{3
   x^4}{64 \sqrt{2}}+\frac{x^5}{128 \sqrt{2}}+O\left(x^6\right)$$ which makes the denominator of $(1)$ to be $$2 \sqrt{x}-\frac{3 x^{3/2}}{2}+\frac{7 x^{5/2}}{16}+\frac{5 x^{7/2}}{64}+\frac{11
   x^{9/2}}{1024}+O\left(x^{11/2}\right)$$ So 
$$\frac d {dx}F\left(\left.\sin ^{-1}(1-x)\right|-1\right)=-\frac{1}{2 \sqrt{x}}-\frac{3 \sqrt{x}}{8}-\frac{11 x^{3/2}}{64}-\frac{7
   x^{5/2}}{256}+\frac{141 x^{7/2}}{4096}+O\left(x^{9/2}\right)$$ Integrating 
$$F\left(\left.\sin ^{-1}(1-x)\right|-1\right)=-\sqrt{x}-\frac{x^{3/2}}{4}-\frac{11 x^{5/2}}{160}-\frac{x^{7/2}}{128}+\frac{47
   x^{9/2}}{6144}+O\left(x^{11/2}\right)+C$$ and $$C=F\left(\left.\sin ^{-1}(1)\right|-1\right)=F(\frac\pi 2|-1)=K(-1)=\frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{2 \pi }}$$.
</s>