Derivation of Series Expansions 
I have the series expansion for
  $$I=\frac{1}{\sqrt{1-x^2}}$$ 
  $$|x|\lt1$$
  which is
  $$1+\sum_{k=1}^\infty\frac{1.3.5.7......(2k-1)x^{2k}}{2^kk!}$$
  or so I am assured. Given the above what is the series expansion for
  $$\int_0^x\frac{dt}{\sqrt{1-t^2}}$$
  can I simply substitute $k+1$ for $k$ and integrate the $x$ terms in the expansion above? I hope I am not completely off-course. I have made a stab in the dark which is
  $$\sin^{-1}{x}=\sum_{k=0}^\infty{\frac{x^{1+2k}\frac{1}{2}k}{k!+2kk!}}$$

 A: Term by term integration yields
$$\begin{align}
\int_0^x\frac{1}{\sqrt{1-t^2}}dt=\int_0^x \left(1+\sum_{n=1}^{\infty}\frac{(2n-1)!!\,t^{2n}}{2^n\,n!}\right)\,dt\\\\
&=x+\sum_{n=1}^{\infty}\frac{(2n-1)!!}{2^n\,n!}\int_0^x\,t^{2n}\,dt\\\\
&=x+\sum_{n=1}^{\infty}\frac{(2n-1)!!}{2^n\,n!}\frac{x^{2n+1}}{2n+1}\\\\
\end{align}$$
Therefore, we can write the series for the arcsine for $|x|<1$ as
$$\bbox[5px,border:2px solid #C0A000]{\arcsin(x)=x+\sum_{n=1}^{\infty}\frac{(2n-1)!!\,x^{2n+1}}{2^n\,n!(2n+1)}=\sum_{n=0}^{\infty}\frac{(2n)!x^{2n+1}}{4^n\,(n!)^2(2n+1)}}$$
A: You need to use the following theorem:
A power series can be integrated term by term in the interior of region of convergence.
If we restrict to real variables then a power series of the from $$f(x) = a_{0} + a_{1}x + a_{2}x^{2} + \cdots = \sum_{n = 0}^{\infty}a_{n}x^{n}\tag{1}$$ has region of convergence as an interval of type $(-R, R)$ so that the series converges for $|x| < R$. It may also converge for $x = \pm R$ but that is not important here.
The above theorem says that we can integrate the equation $(1)$ and get $$\int_{0}^{x}f(t)\,dt = a_{0}x + a_{1}\frac{x^{2}}{2} + a_{2}\frac{x^{3}}{3} + \cdots = \sum_{n = 1}^{\infty}\frac{a_{n - 1}}{n}x^{n}\tag{2}$$ and this is valid for all $x \in (-R, R)$ (or $|x| < R$).
In your case we have $$\frac{1}{\sqrt{1 - x^{2}}} = 1 + \sum_{n = 1}^{\infty} \frac{1\cdot 3\cdot 5\cdots (2n - 1) x^{2n}}{2^{n}n!}$$ which is valid for $|x| < 1$ and we can integrate it to get the series for $\sin^{1}x$ which will also be valid for $|x| < 1$.
However there is catch. Unless you are aware of the theorem (by "awareness" I mean "a complete understanding of its proof") mentioned in beginning of this answer, you should not use the this technique. Rather there are special mechanisms for each specific series.
In your case of $\sin^{-1}x$ Hardy shows us the way in his book "A Course of Pure Mathematics" on page $339$ where he gives a reduction formula for integral $$\int_{0}^{x}\sin^{2n - 1}t\,dt$$ in the form $$1 = \cos x + \frac{1}{2}\cos x\sin^{2}x + \cdots + \frac{1\cdot 3\cdots (2n - 3)}{2\cdot 4\cdots (2n - 2)}\cos x\sin^{2n - 2}x + r_{n}\tag{3}$$ where $$r_{n} = \frac{3\cdot 5\cdots (2n - 1)}{2\cdot 4\cdots (2n - 2)}\int_{0}^{x}\sin^{2n - 1}t\,dt$$ and integrating equation $(3)$ we get $$\alpha = \sin \alpha + \frac{1}{2}\cdot\frac{\sin^{3}\alpha}{3} + \cdots + \frac{1\cdot 3\cdots (2n - 3)}{2\cdot 4\cdots (2n - 2)}\cdot\frac{\sin^{2n - 1}\alpha}{2n - 1} + R_{n}\tag{4}$$ where $$R_{n} = \int_{0}^{\alpha}r_{n}\,dx = \frac{3\cdot 5\cdots (2n - 1)}{2\cdot 4\cdots (2n - 2)}\int_{0}^{\alpha}(\alpha - x)\sin^{2n - 1}x\,dx\tag{5}$$ from which we get $$0 \leq R_{n} \leq \frac{1\cdot 3\cdots (2n - 1)}{2\cdot 4\cdots (2n)}\alpha\sin^{2n}\alpha\tag{6}$$ for $0 \leq \alpha \leq \dfrac{\pi}{2}$. Letting $n \to \infty$ in $(4)$ and noting than $R_{n} \to 0$ as $n \to \infty$ it follows that (on putting $\sin \alpha = u$) $$\sin^{-1}u = u + \frac{1}{2}\frac{u^{3}}{3} + \frac{1\cdot 3}{2\cdot 4}\frac{u^{5}}{5} + \cdots\tag{7}$$ where $0 \leq u \leq 1$. Since both sides are odd functions of $u$ the formula also holds for $-1 \leq u \leq 0$ or thus for all $|u| \leq 1$.
