Prove that $x + \frac{2x^3}{3} + \cdots + \frac{2\cdot 4 \cdot \cdots 2nx^{2n+1}}{3\cdot 5 \cdot (2n+1)}+\cdots = \frac{\arcsin(x)}{\sqrt{1-x^2}}$ 
Prove that $x + \dfrac{2x^3}{3} + \cdots + \dfrac{2\cdot 4 \cdot \cdots 2nx^{2n+1}}{3\cdot 5 \cdot (2n+1)}+\cdots = \dfrac{\arcsin(x)}{\sqrt{1-x^2}}.$

Let $f(x) = x + \dfrac{2}3x^3 +\dfrac{8}{15}x^5 + \cdots + \dfrac{1}2n!\cdot\dfrac{n!}{(2n+1)!}(2x)^{2n+1}+\cdots.$
Then $xf(x) = x^2 + \dfrac{2}3 x^4 + \dfrac{8}{15}x^6 + \cdots + \dfrac{1}{4}n!\cdot \dfrac{n!}{(2n+1)!}(2x)^{2n+2}+\cdots.$
Also, $f'(x) = 1 + 2x^2 + \dfrac{8}3 x^4 + \cdots + n!\cdot \dfrac{n!}{(2n)!}(2x)^{2n}+\cdots$
and $(1-x^2)f'(x) =1+x^2 + \dfrac{2}3 x^4 + \cdots + [n!\cdot \dfrac{n!}{(2n)!}2^{2n}-(n-1)!\cdot \dfrac{(n-1)!}{(2n-2)!}2^{2n-2}]x^{2n}+\cdots\\
=1+x^2 + \dfrac{2}3 x^4 + \cdots + \dfrac{1}{4}\cdot(n-1)!\cdot \dfrac{(n-1)!}{(2n-1)!}\cdot(2x)^{2n}+ \cdots.$
Hence the derivative of $\sqrt{1-x^2}f(x)$ is $\sqrt{1-x^2}f'(x)-\dfrac{xf(x)}{\sqrt{1-x^2}} = \dfrac{1}{\sqrt{1-x^2}}((1-x^2)f'(x) - xf(x))=\dfrac{1}{\sqrt{1-x^2}}.$ Integrating, we see that $\sqrt{1-x^2}f(x) =\arcsin(x)+C.$ Plugging in the constant $C=0$ and dividing both sides by $\sqrt{1-x^2}$ gives the desired result.

Are there other approaches to solving this problem?

 A: In this proof i will use some proprieties of Wallis'integral (Which easy to show by simple induction),
We have :
$$W_{2n+1}=\prod_{k=1}^n\frac{2k}{2k+1}$$
So your series is equal to :
$$f(x)=\sum_{n=0}^\infty W_{2n+1}x^{2n+1}=\sum_{n=0}^\infty \biggl(\int_0^{\pi/2}\sin^{2n+1}t\ dt\biggr)x^{2n+1}$$
Then we put $t=\arcsin u$, therefore :
$$f(x)=\sum_{n=0}^\infty \biggl(\int_0^{1}\frac{u^{2n+1}}{\sqrt{1-u^2}}\ dt\biggr)x^{2n+1}=\int_0^1\biggl(\frac{ux}{\sqrt{1-u^2}}\sum_{n=0}^\infty(ux)^{2n}\biggr)du$$
Then we Know calculate the geometric series :
$$f(x)=\int_0^{1}\frac{ux\ du}{\sqrt{1-u^2}(1-(ux)^2)}$$
Let :
$$t^2=\frac{1-u^2}{1-(ux)^2}⇒u^2=\frac{1-t^2}{1-(tx)^2}⇒udu=-\frac{t(1-x^2)}{\biggl(1-t^2x^2\biggr)^2}dt$$
Therefor :
$$f(x)=\frac{1}{\sqrt{1-x^2}}\int_0^1\frac{x\ du}{\sqrt{1-(tx)^2}}=\frac{\arcsin x}{\sqrt{1-x^2}}$$
A: Like this
$$ \sum_{n=0}^{\infty} \frac{x^{2n+1}(2n)!!}{(2n+1)!!}=\sum_{n=0}^{\infty} \int_{0}^{\frac{\pi}{2}}x^{2n+1}\sin^{2n+1} t dt=\int_{0}^{\frac{\pi}{2}} \frac{x \sin t}{1-x^{2}\sin^{2} t} dt=-\int_{0}^{\frac{\pi}{2}}\frac{1}{1-x^{2}+x^{2}\cos^{2} t} d(x\cos t)=\frac{1}{\sqrt{1-x^{2}}} \arctan{\frac{x}{\sqrt{1-x^2}}}=\frac{\arcsin x}{\sqrt{1-x^2}} $$
