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.

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  

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



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

