Limit of Gamma integrals

This is nothing mysterious, it is the continuous analogue of $\Sigma 2^{-n} \binom{n}{k} = 1$ with factorials interpolated by Gamma functions that vary quite smoothly in the middle range where the sum is concentrated and has approximately Gaussian distribution, $k = n/2 + O(\sqrt{n})$.

The ratio between the discrete step function and continuous interpolation is $1+O(1/n)$ in the middle range (using the first few terms of Stirling asymptotic expansion for $\Gamma$).

The $1/\sqrt{x(1-x)}$ is irrelevant because it is effectively a constant factor $2 + O(1/n)$ in the middle range.

The middle range has width less than $n^{1/2 + \epsilon}$.

So, writing the integral in terms of $u = nx$,

$$I_n = \int_0^n \binom{n}{u} 2^{-n} du \frac {1} {\sqrt{x(1-x)}} \sim 2 \int_0^n \binom{n}{u} 2^{-n} du \sim 2$$

with an additive error of $O(n^{\epsilon - 1/2})$. The $\epsilon$ can be a small positive constant or an infinitesimal depending on how you dissect the integral into middle and tails (we have to choose a cut for each $n$, such that the entire middle range is covered as $n \to \infty$; any set of choices will prove convergence, but the rate of convergence depends on the choices).

Limit (and even asymptotic series) of your integral can be calculated using Laplace's method and Stirling's approximation. The solution can be done in 5 steps:
1. Prove, that integral over $[0,1/n]$ can be neglected. In order to do this, use Stirling's approximation for all factorials except $(nx)!$, which can be approximated by a constant.
2. Prove, that integral over $[1/n,\delta]$ can be neglected (for $\delta<1/2$). In order to do this, use Stirling's approximation for all factorials (note, that $z!/(\sqrt{2 \pi z} (z/e)^z))$ is bounded from both sides by positive constants if $z>=1$).
3. Using symmetry you know, that your limit is equal to $\lim_{n\to\infty}\int_{\delta}^{1-\delta}dx(\dots)$.
4. Use Stirling's approximation to approximate factorials. You will get the following:
$$\int_{\delta}^{1-\delta}dx\frac{n!2^{-n}n}{(xn)!(n-xn)!\sqrt{x(1-x)}}=(1+O(n^{-1}))\sqrt{\frac{n}{2\pi}}\cdot$$ $$\int_{\delta}^{1-\delta} x^{-1}(1-x)^{-1}exp\bigl(-n(x\ln x+(1-x)\ln(1-x)+\ln 2)\bigr)dx$$
5. Apply Laplace's method to the last integral and get it's asymptotic.