Finding Taylor's series of the function: $\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}$ Show that $$\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}=1+\frac{ax}{1!}+\frac{(a^2+1^2)x^2}{2!}+\frac{a(a^2+2^2)x^3}{3!}+\frac{(a^2+1^2)(a^2+3^2)x^4}{4!}+\cdots$$
My attempt:
I integrated the function and got $\frac{e^{a \sin^{-1}x}}{a}$ then I wrote the series of $e^{a \sin^{-1}x}$ but it contained terms like $(\sin^{-1}x)^2$, $(\sin^{-1}x)^3$ and so on so I could not find the series. My idea was to find the series of the anti derivative of the function and then to derivate the obtained series. Any other way to do it?

We can use your idea of finding the antiderivative. We have:
\begin{align}
e^{a\sin^{-1}(x)}& =1+a\sin^{-1}x+\dfrac{(a\sin^{-1}x)^2}{2!}+...+\dfrac{(a\sin^{-1}x)^5}{5!}+...\\
&=1+a\left(x+\dfrac{x^3}{6}+\dfrac{3x^5}{40}+...\right)+\dfrac{a^2}{2!}\left(x+\dfrac{x^3}{6}+...\right)^2+\dfrac{a^3}{3!}\left(x+\dfrac{x^3}{6}+...\right)^3+\dfrac{a^4}{4!}x^4+\dfrac{a^5}{5!}x^5+...\\
&=1+a\left(x+\dfrac{x^3}{6}+\dfrac{3x^5}{40}+...\right)+\dfrac{a^2}{2!}\left(x^2+\dfrac{x^4}{3}+...\right)+\dfrac{a^3}{3!}\left(x^3+\dfrac{x^5}{2}+...\right)+\dfrac{a^4}{4!}x^4+\dfrac{a^5}{5!}x^5+...\\
&=1+ax+\dfrac{a^2x^2}{2!}+\left(\dfrac{ax^3}{3!}+\dfrac{a^3x^3}{3!}\right)+\left(\dfrac{a^2x^4}{3!}+\dfrac{a^4x^4}{4!}\right)+\left(\dfrac{3ax^5}{40}+\dfrac{a^3x^5}{2\cdot3!}+\dfrac{a^5x^5}{5!}\right)+...\\
&=1+ax+\dfrac{a^2x^2}{2!}+\dfrac{(a+a^3)x^3}{3!}+\dfrac{(4a^2+a^4)x^4}{4!}+\dfrac{(9a+10a^3+a^5)x^5}{5!}+...\\
\end{align}
Thus, $\dfrac{1}{a}e^{a\sin^{-1}(x)}=\dfrac{1}{a}+x+\dfrac{ax^2}{2!}+\dfrac{(1+a^2)x^3}{3!}+\dfrac{(4a+a^3)x^4}{4!}+\dfrac{(9+10a^2+a^4)x^5}{5!}+...$
Finally, take the derivative on both sides and we are done.
