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.? 

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.
