Consider the function
$$f(a) = \sum_{k=0}^{\infty} \frac{1\cdot3\cdots(2k-1)}{2\cdot4\cdots(2k)}\cdot a^{2n+2k}$$
$$|a| \leq 1$$
What you want is $$\int_{0}^{1} f(a) da$$
The $k^{th}$ coefficient can be written as
$$\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}{\sin^{2k}x}dx$$
(again Wallis's like product as in the other question).
Thus we have
$$f(a) = \frac{2a^{2n}}{\pi} \int_{0}^{\frac{\pi}{2}} \sum_{k=0}^{\infty} {(a\sin x)^{2k}} dx$$
$$ = \frac{2a^{2n}}{\pi} \int_{0}^{\frac{\pi}{2}} {\frac{1}{1-(a\sin x)^2}}dx$$
Now $$\frac{1}{1-(a\sin x)^2} = \frac{1}{\cos^2 x + (\sqrt{1-a^2}\sin x)^2} = \frac{\sec^2 x}{1+(\sqrt{1-a^2}\tan x)^2}$$ which is the derivative of
$$\frac{\arctan(\sqrt{1-a^2}\tan x)}{\sqrt{1-a^2}}$$ whose integral between $0$ and $\frac{\pi}{2}$ is $$\frac{\pi}{2\sqrt{1-a^2}}$$
Thus we get $$f(a) = \frac{a^{2n}}{\sqrt{1-a^2}}$$
(So in fact, we have shown the identity which Qiaochu uses in his answer).
Hence the required sum is $$\int_{0}^{1} \frac{a^{2n}}{\sqrt{1-a^2}}da$$ which can easily be found by the substitution $a = \sin x$ (like in Qiaochu's answer) and which gives us the result to be
$$\int_{0}^{\frac{\pi}{2}} {\sin^{2n}x}dx$$
Now this integral for positive integer $n$ can easily be seen to be what Mariano got. I am unsure of whether that formula holds for an arbitrary real number $n$.