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.     
 A: We may put $1-y=\sin{\theta}$ in the integral, so $dy = -\cos{\theta} d\theta $, $\cos{\theta}=\sqrt{y(2-y)}$ on the interval in question, to find 
$$ F(\arcsin{(1-x)}\mid -1) = \int_x^1 \frac{dy}{\sqrt{y(2-y)(1+(1-y)^2)}} = \int_x^1 \frac{dy}{\sqrt{y(2-y)(2-2y+y^2)}}. $$
Of course,
$$F(\arcsin(1) \mid -1) = F(\pi/2 \mid -1) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1+\sin^2{\theta}}} = K(-1) = \frac{\Gamma(1/4)^2}{4\sqrt{2\pi}}, $$
so this is the constant term (this particular value is found using the Beta-function integral). The rest comes from
$$ F(\arcsin{(1-x)}\mid -1)-F(\arcsin{(1)}\mid -1) = -\int_0^x \frac{dy}{\sqrt{y(2-y)(2-2y+y^2)}}, $$
and since $x$ is small, we can do by expanding the integrand:
$$ \frac{1}{\sqrt{y(2-y)(2-2y+y^2)}} = \frac{1}{2} y^{-1/2} \left(1 - y \left(\frac{3}{2} +  y - \frac{1}{4} y^2 \right) \right)^{-1/2} = \frac{1}{2} y^{-1/2} + O(y^{1/2}) $$
by using the binomial expansion on the bracket. Integrating term-by-term gives
$$ F(\arcsin{(1-x)}\mid -1) = K( -1) - x^{1/2} + O(x^{3/2}) = \frac{\Gamma(1/4)^2}{4\sqrt{2\pi}} - x^{1/2} + O(x^{3/2}), $$
and you can find more of the terms by a taking more terms of the expansion of $\left(1 - y \left(\frac{3}{2} +  y - \frac{1}{4} y^2 \right) \right)^{-1/2}$.

As for $\arcsin{(1-x)}$, we can use the same trick on its integral: putting $y=1-u$, $1-u^2 = 1-(1-y)^2 = y(2-y)$
$$ \arcsin{(1-x)} = \int_0^{1-x} \frac{du}{\sqrt{1-u^2}} = \int_x^1 \frac{du}{\sqrt{y(2-y)}}, $$
and then you can expand the square root binomially and get an expansion in odd powers of $x^{1/2}$ in the same way.
A: 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 $$C=F\left(\left.\sin ^{-1}(1)\right|-1\right)=F(\frac\pi 2|-1)=K(-1)=\frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{2 \pi }}$$.
