finding the expansion of $\arcsin(z)^2$

Is there a fast and nice way to find the expansion of $\arcsin(z)^2$ without squaring expansion of $\arcsin(z)$ ?

For $|z|<1$ show that $$(\sin^{-1}(z))^2 = z^2 + \frac{2}{3}\cdot \frac{z^4}{2} + \frac{2}{3}\cdot\frac{4}{5}\cdot \frac{z^6}{3}+ \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot \frac{z^8}{4} + \dots$$

It should have something like $c_{2n} = \frac{2^{2n}n!^2}{(2n+1)!n}$ as coefficient maybe we could use Residue theorem to evaluate it.

To prove that the formula :

$$2\;\arcsin(x)^2=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2\binom{2n}{n}}$$

is right we may use the method proposed by the brothers Borwein at the end of 'Pi and the AGM'.

Observe that : $$x \frac{d}{dx}(\arcsin\ x)^2=\frac{2x \arcsin\ x}{\sqrt{1-x^2}}$$ and use the fact that both $\ \displaystyle f(x)= \frac{\arcsin\ x}{\sqrt{1-x^2}}\ $ and $\ \displaystyle F(x)=\frac{1}{2x}\sum_{m=1}^{\infty} \frac{(2x)^{2m}}{m\binom{2m}{m}}$

satisfy the differential equation : $\;(1-x^2)f'=1+xf$

From $f(0)=F(0)$ me may then integrate relatively to $x$ (under the sum sign) $\;f(x)=F(x)\;$ to get (the $2x$ at the denominator disappears and a $2m$ appears) : $$\arcsin(x)^2=\sum_{m=1}^{\infty} \frac{1}{2m}\frac{(2x)^{2m}}{m\binom{2m}{m}}$$

(this was stolen from my earlier answer with better references... robjohn's answer in that link should be helpful too (+1) !)

For alternative proofs of the expansion of $\arcsin(x)^2$ and many other functions you may consult this table (3 proofs are exposed by clicking on 'Ausklappen' at the right). The first proof for the expansion of the derivative for example uses $\,\displaystyle\int_0^{\frac {\pi}2} \sin^{2n-1}\theta\ d\theta=\frac{2^{2n-1}}{n\binom{2n}{n}}$.

(from joriki's nice answer where the coefficients of $x^{-n}\,$ in $\;e^{-\frac 1{\sin(x)}}$ were given using the expansions of $\arcsin(x)^n$ for $n=1,2,3$)

Expansions of higher powers of $\arcsin(x)^n$ may be found too in the table or in Mhenni's answer here (in terms of gamma and polygamma functions $\psi^{(n)}$ up to $n=6$).

A related problem. It seems we can go beyond this. Here is the Taylor series expansion of $ \arcsin(x)^3 $

$$\arcsin(x)^3 = \frac{3}{4\sqrt {\pi }}\sum _{m=1}^{\infty }\,{\frac { \left( {\pi }^{2}-2\,\psi'\left( m+\frac{1}{2} \right) \right) \Gamma\left( m+\frac{1}{2} \right) {x}^{2\,m+1}}{ \left( 2\,m+1 \right) \Gamma \left( m+1 \right) }}.$$

Added: Here is the requested series

$$ \frac{\arcsin(x)}{\sqrt{1-x^2}}=\frac{\sqrt{\pi}}{2}\sum _{m=0}^{\infty}{\frac {\Gamma \left( m+1 \right)\, {x}^{2\,m+1}}{\Gamma \left( m+\frac{3}{2} \right) }}.$$

Added: Here are the power series of $\arcsin(x)^4$ and $\arcsin(x)^6$ respectively

$$ \frac{\sqrt {\pi }}{4}\sum _{m=0}^{\infty}\,{\frac { \left( {\pi }^{2}-6\, \psi' \left( m+1 \right)\right) \Gamma \left( m+1\right){x}^{2\, m+2}}{ \left( m+1 \right) \Gamma \left( m+\frac{3}{2} \right) }},$$

$$\frac{3\sqrt {\pi }}{32} \sum _{m=0}^{\infty }{}\,{\frac { \left( 60\, \left( \psi' \left( m+1 \right) \right) ^{2}-20\,\psi' \left( m+1 \right) {\pi }^{2}+{\pi }^{4}+10\,\psi''' \left( m+1 \right) \right) \Gamma \left( m+1 \right) {x}^{2\,m+2}}{ \left( m+1 \right) \Gamma\left( m+\frac{3}{2} \right) }}.$$