Finding the power series of $\arcsin x$ I'm trying to find the power series of $\arcsin x$.
This is what I did so far:
$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$, so $\arcsin x=\int \sqrt{\sum_{n=0}^{\infty}x^{2n}}$.
(for $|x|<1$)
Any hints for what should I do further? 
Thanks a lot.
 A: The generalized binomial thorem states that
$$(1+x)^{\alpha} = \sum_{k=0}^{\infty} {{\alpha\choose {k}} x^k}$$
The definition still holds 
$$ {\alpha\choose {k}} =\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}$$
As people are suggesting, you need to see that
$$\sin^{-1} x = \int \frac{dx}{\sqrt{1-x^2}}$$
thus you can integrate the binomial series of
$$(1-x^2)^{-\frac{1}{2}} = \sum_{k=0}^{\infty} (-1)^k {{-\frac{1}{2}\choose {k}} x^{2k}}$$
Using the definition shows that
$$ {-\frac{1}{2}\choose {k}} = {( - 1)^k}\frac{{(2k - 1)!!}}{{k!{2^k}}} = {( - 1)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}$$
Putting this in the sum produces:
$${\left( {1 - {x^2}} \right)^{-\frac{1}{2}}} = \sum\limits_{k = 0}^\infty  {\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}{x^{2k}}} $$
Finally this yields:
$${\sin ^{ - 1}}x = \sum\limits_{k = 0}^\infty  {\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}\frac{{{x^{2k + 1}}}}{{2k + 1}}} $$
Remember that by definition
$$0!! = 1 $$
$$(-1)!! = 1$$
If you're not comfortable with that simply put:
$${\sin ^{ - 1}}x = x + \sum\limits_{k = 1}^\infty  {\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}\frac{{{x^{2k + 1}}}}{{2k + 1}}} $$
A: Hint: binomial series. http://en.wikipedia.org/wiki/Binomial_series
