<s>[INST] 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?
 [/INST] Hint
You face a problem of composition of series.
Start with
$$\sin ^{-1}(x)=x+\frac{x^3}{6}+\frac{3 x^5}{40}+O\left(x^7\right)$$
$$a\sin ^{-1}(x)=ax+a\frac {x^3}{6}+a\frac{3 x^5}{40}+O\left(x^7\right)$$ Now, use
$$e^{a \sin ^{-1}(x)}=\exp\Big[ax+a\frac {x^3}{6}+a\frac{3 x^5}{40}+O\left(x^7\right) \Big]$$ when done, work the denominator and use long division.
Edit
Your idea of using the antiderivative is very good. You could even continue using the logarithm of it and then go backward.
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