Limit of binomial coefficient by using Integral of cos^n

Question

I want to prove the following:

$\lim\limits_{n\to\infty} 2^{-2n} \sqrt{n} \binom{2n}{n} =\frac{1}{\sqrt{\pi}}$

I know that I can use $I_n=\int_0^\frac{\pi}{2}(cos(x))^ndx$ and I already proved $nI_n=(n-1)I_{n-2}$ and $I_{n-1}I_n=\frac{\pi}{2n}$.

Now I'm struggling to get this together and need some help.

Answer 1

Hint:
You can use Stirling's approximation to prove that, $$\binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}}.$$