Maclaurin expansion of $\arcsin x$

If I was doing this I would start with one of the very common series like $\sin(x)$, $e^x$, etc. and then use substitution. The one that fits best here in my opinion is: $$ \sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \frac{5x^4}{128} + \dotsc $$ Through substitution, we can obtain: $$ \frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} + \frac{35x^8}{128} + \dotsc $$ Then by integration: $$ \arcsin(x) = \int^x_0 \frac{1}{\sqrt{1-t^2}}\,dt = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \frac{35x^9}{1152} + \dotsc $$

Hint: Use the following binomial series $$ (1+z)^\alpha=\sum\limits_{k=0}^{+\infty}\frac{\alpha(\alpha-1)\ldots(\alpha-k+1)}{k!}z^k $$ with $z=-t^2$, $\alpha=-1/2$. Then integrate.

Change variables $t = x u$: $$ \arcsin(x) = \int_0^1 \frac{ x}{(1- x^2 u^2)^{1/2}} \mathrm{d} u $$ Now using Taylor series expansion of the integrand: $$ \frac{ x}{(1- x^2 u^2)^{1/2}} = \sum_{n=0}^\infty \frac{1}{2^{2n}}\binom{2n}{n} x^{2n+1} u^{2n} $$ Integrating term-wise: $$ \arcsin(x) = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1} $$