A variant on Laplace's method $\int_{-\pi}^{\pi} \cos^{2n} (x/2) \cos( m x) dx \sim \frac{1}{\sqrt{n} }$ I am trying to estimate some probability using the inversion formula of characteristics of some discrete random variable and it finally boils down to the following integration which is very similar to the situation in the Laplace's method.
I wonder that if the following holds 
$$\exists C>0\; s.t.\; \int_{-\pi}^{\pi} \cos^{2n} (x/2) \cos( m x) dx \geq \frac{C}{\sqrt{n}},\forall m\leq \sqrt n,$$
or just $$\int_{-\pi}^{\pi} \cos^{2n} (x/2) \cos( [\sqrt{n}]x) dx \sim \frac{1}{\sqrt{n}},$$
where $\sim$ means of the same order as $n$ increases to infinity.
I got stuck in using Laplace's method as $\cos( [\sqrt{n}]x)$ can vary a lot in a neighborhood of $0$ and the original Laplace's method does not work. 

Hint. As an alternate approach, one may recall that (see here p. 397 (446))
$$
\int_{0}^{\pi/2} \cos^{2n} x \: \cos( 2m x) \:dx =\frac{\pi}{2^{2n+1}}\frac{(2n)!}{(m+n)!(n-m)!},\qquad n\geq m, \tag1
$$ then, as $n \to \infty$, by the use of Stirling's formula, one gets
$$
\int_{0}^{\pi/2} \cos^{2n} x \: \cos( 2m x) \:dx \sim \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{n}},\qquad 0\leq m<n. \tag2
$$ 
We conclude by parity of the initial integrand and by an easy change of variable:

$$
\int_{-\pi}^{\pi} \cos^{2n} (x/2) \cos( m x) dx \sim \frac{2\sqrt{\pi}}{\sqrt{n}},\qquad 0\leq m<n, \,n \to \infty. \tag3
$$

