Find $f(x) = x + \frac{2}{3}x^3 + \frac{2\cdot4}{3\cdot5}x^5 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^7+\cdots$ where $|x|<1$ 
Find $$f(x) = x + \frac{2}{3}x^3 + \frac{2\cdot4}{3\cdot5}x^5 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^7+\cdots +\infty\,,\quad|x|<1$$

My solution:
$$f'(x) = 1 + 2x^2 + \frac{2}{3}\cdot4x^4 + \frac{2\cdot4}{3\cdot5}\cdot6x^6 + \cdots + \infty $$
$$ = 1+x\left(\frac{d}{dx} xf(x)\right) $$ (By observation)
$$ = 1+x^2f'(x)+xf(x)$$
$\implies (1-x^2)\frac{dy}{dx} = 1 + xy$ (where $y=f(x), \frac{dy}{dx}=f'(x)$)
This reduces to a linear first order differential equation, which along with $f(0)=0$, gives $$f(x) = \frac{\sin^{-1}(x)}{\sqrt{1-x^2}}$$
Is there any other way to solve this question? Preferably along the lines of Taylor series, infinite GP, fubini theorem, etc.? 
 A: The question asks for other ways to identity the function whose
power series expansion is
$$ f(x) := x + \frac{2}{3}x^3 + \frac{2\cdot4}{3\cdot5}x^5 +
 \frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^7+ 
 \ldots  .$$
There are many potential methods but maybe finding numerical values of
the function will help to identify it. For example, $\, f(i) \approx
.62322...i \,$ is OEIS sequence A196525 which
is given as the value of $\,\ln(1+\sqrt{2})/\sqrt{2}.\,$ This is not much
of a clue. Let us try $f(1/2) \approx .60959... \,$ which is
OEIS sequence A073010 which is given as the
value of $\,\pi/\sqrt{27}.\,$ Recall that $\,\sin(\pi/6)=1/2,\,$ and
that $\,\cos(\pi/6)=\sqrt{3}/2.\,$ A possible result now is that
$\, f(1/2) = \frac{\pi/6}{\sqrt{1-1/4}} = 
\frac{\sin^{-1}(1/2)}{\sqrt{1-(1/2)^2}}.\,$ Referring back to 
$\, f(i) = \frac{\ln(1+\sqrt{2})}{\sqrt{2}}\,$ we can find that
$\, \ln(1+\sqrt{2}) \approx .88137... \,$ which is 
OEIS sequence A091648 which is listed as
the value of $\, \sin^{-1}(i)/i.\,$ Thus, 
$\, f(i) = \frac{\sin^{-1}(i)}{\sqrt{1-i^2}}.\,$
These two numerical evaluations suggest that
$\, f(x) = \frac{\sin^{-1}(x)}{\sqrt{1-x^2}}\,$ and this can be
verified by several methods.
