calculus - Closed form expression for infinite series

I was given the following function:

$$f(x) = x + \frac{2x^3}{1\cdot3} + \frac{2\cdot4x^5}{1\cdot3\cdot5} + \frac{2\cdot4\cdot6x^7}{1\cdot3\cdot5\cdot7}...$$ $$\forall x \in [0,1)$$

And then I was asked to find the value of $f(\frac{1}{\sqrt{2}})$, which obviously requires me to compute the closed form expression of the infinite series.

I tried 'Integration as a limit of sum' but I was unable to modify the expression accordingly. How do I approach the problem?

Answer

Another answer. We use formulas

$$\int_0^{\pi/2}\sin^{2n+1}sds=\frac{(2n)!!}{(2n+1)!!},$$ $$\sum_{k=0}^{\infty }{\left. {{z}^{2 k+1}}\right.}=\frac{z}{1-{{z}^{2}}},\quad |z|<1.$$ Then $$f(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}(2n)!!}{(2n+1)!!}\\=\sum_{n=0}^{\infty}x^{2n+1}\int_0^{\pi/2}\sin^{2n+1}sds\\ =\int_0^{\pi/2}\left(\sum_{n=0}^{\infty}(x\sin s)^{2n+1} \right)ds\\= \int_0^{\pi/2}\frac{x\sin s}{1-x^2\sin^2s}ds\\={\frac {1}{\sqrt {1-x^2}}\arctan \left( {\frac {x}{\sqrt {1-x^2}}} \right) }$$ We get $$f\left(\frac{1}{\sqrt2}\right)=\sqrt2\arctan1=\frac{\sqrt2\pi}{4}$$