Generating functions of Wallis integrals

simple question, how would one go about proving that

$$\sum_{n\ge 0}^{}W_{n}x^{n}=\int_{0}^{\frac{\pi}{2}}\frac{dt}{1-x\cos t}$$

from this, it results with substituing $u= \tan(t/2)$ that

$$\sum_{n\ge 0}^{}W_{n}x^{n}=\int_{0}^{1}\frac{2dt}{1-x+u^2(1+x)}$$and noting that $f(-1)=1$ then for all $x$ in the open interval $]-1;1[$ we finish with:

$$\sum_{n\ge 0}^{}W_{n}x^{n}=\frac{2}{\sqrt{1-x^2}}\arctan\left( \sqrt{\frac{1+x}{1-x}} \right)$$

I'm probably just blind, but I have no clue as to how to get the first integral relation with the generating function of the Wallis integral.