Using the theorem of dominated convergence, it's easy to prove that
\begin{align*} \lim_{n \rightarrow \infty} \int_0^{\frac{1}{2}\pi \sqrt{n}} \cos \left( \frac{t}{\sqrt{n}} \right)^{n} \, dt = \int_0^\infty e^{-t^2/2} \, dt. \end{align*} How would one go about estimating the rate of convergence of this sequence of integrals? In other words:
How can one get an explicit expression for a function $R(n) = o(1)$ such that \begin{align*} \int_0^{\frac{1}{2}\pi \sqrt{n}} \cos \left( \frac{t}{\sqrt{n}} \right)^{n} \, dt = \int_0^\infty e^{-t^2/2} \, dt + O\left( R(n) \right) \end{align*} as $n \rightarrow \infty$?
For example, the Taylor series estimate $\cos (t/\sqrt{n})^{n-2} = \exp ( -t^2/2) + O(t^4 / n)$ doesn't seem to be of any use since the remainder term integrates to a constant times $n^{3/2}$.
Edit: Just to clarify, the integrand I am considering is $\left( \cos (t/\sqrt{n}) \right)^{n}$.
Note that $$\int_0^{\frac{1}{2}\pi \sqrt{n}} \cos \left( \frac{t}{\sqrt{n}} \right)^{n} \, dt=\sqrt n \int_0^{\frac \pi 2} \cos(t)^n dt = \sqrt n W_n$$ where $W_n$ denotes the Wallis integral.
Since $\displaystyle W_n=\frac{\sqrt{\pi}}2\frac{\Gamma\left(\frac{n+1}2 \right)}{\Gamma\left(\frac n2 +1\right)}$, asymptotics of the Gamma function yield
$$\sqrt n W_n = \sqrt{2\pi} - \frac 14 \sqrt{\frac{\pi}{2}}\frac{1}{\sqrt n} + O\left(\frac{1}{n^{3/2}}\right) = \sqrt{2\pi} + O\left(\frac{1}{\sqrt n} \right)$$