Finding the value of $\lim\limits_{n\rightarrow \infty}\sqrt{n}\int^{\frac{\pi}{4}}_{0}\cos^{2n-2}(x)\mathrm dx$ 
Finding the value of $\displaystyle \lim\limits_{n\rightarrow \infty}\sqrt{n}\int^{\frac{\pi}{4}}_{0}\cos^{2n-2}(x)\mathrm dx$

What I tried
Let $\displaystyle I_{k} =\int^{\frac{\pi}{4}}_{0}\cos^{k}(x)\mathrm dx=\int^{\frac{\pi}{4}}_{0}\cos^{k-1}(x)\cdot \cos (x)\mathrm dx$
$$ I_{k}=\left.\cos^{k-1}(x)\sin x\right|^{\frac{\pi}{4}}_{0}+(k-1)\int^{\frac{\pi}{4}}_{0}\cos^{k-2}(x)\left(1-\cos^2 x\right)\mathrm dx$$
$$ k I_{k}=\bigg(\frac{1}{2}\bigg)^{\frac{k}{2}}+(k-1)I_{k-2}$$
How do I solve it? Help me please.

Result
Let
$$f(n) = \int_0^\frac{\pi}{4} \cos(x)^n\,dx$$
then
$$\lim_{n\to \infty } \, \sqrt{n} f(n)=\sqrt{\frac{\pi }{2}} \simeq 1.2533141373155001\tag{1}$$
and
$$\lim_{n\to \infty } \, \sqrt{n} f(2n)=\frac{\sqrt{\pi }}{2} \simeq 0.8862269254527579\tag{2}$$
Here $(2)$ is the limit asked for in the OP.
Derivation
Decomposing the integration region we write
$$f(n) = i_1(n) - i_2(n)$$
where
$$i_1(n) = \int_0^\frac{\pi}{2} \cos(x)^n\,dx$$
$$i_2(n) = \int_\frac{\pi}{4}^\frac{\pi}{2} \cos(x)^n\,dx$$
Because in the integration interval of $i_2$ we have $\cos(x) \le \cos(\frac{\pi}{4})=\frac{1}{\sqrt{2}}$ we see immediately that $i_2$ goes to zero exponentially for $n\to\infty$. 
Hence we focus on $i_1$.
Changing variables $x\to\arctan(t)$, $dx\to\frac{1}{1+t^2}$ gives
$$i_1(n) = \int_0^{\infty } \left(t^2+1\right)^{-\frac{n}{2}-1} \, dt$$
then $t\to \frac{1}{\sqrt{r}}$ gives
$$i_1(n) = \frac{1}{2} \int_0^{\infty } r^{-\frac{3}{2}} \left(\frac{1}{r}+1\right)^{-\frac{n}{2}-1} \, dr$$
And finally $r\to \frac{1}{z}-1$ leads to
$$i_1(n) = \frac{1}{2} \int_0^1 z^{-\frac{1}{2}} (1-z)^{\frac{n}{2}-\frac{1}{2}} \, dz \\
= \frac{1}{2} B(\frac{1}{2},\frac{n+1}{2})=\frac{1}{2}
\frac{\Gamma(\frac{1}{2}) \Gamma(\frac{1+n}{2})} {\Gamma(1+\frac{n}{2})}\tag{3}$$
The asymptotic behavour is then found using Stirling's formula to give the results announced in the beginning.
Notice that for (2) we have to let $n\to 2n$ in $(3)$ before taking the limit.
Remark
Herewith we have also found that for the more general integral we have for any positive $k$ that
$$\lim_{n\to \infty } \, \sqrt{k n}\int_0^\frac{\pi}{4} \cos(x)^{k n}\,dx = \sqrt{\frac{\pi}{2}}\tag{4}$$
