Stationary Phase, approximate the leading behavior of the integral and determine the dependence on n. The given integral is:
$$\int_{-\pi/2}^{\pi/2}\cos(nt-\lambda\cos t) dt, \lambda \rightarrow \infty $$
This can be rewritten as a complex exponential so it's $$ Re \left[ \int_{-\pi/2}^{\pi/2}\ e^{i(nt-\lambda\cos t)} dt \right] \lambda \rightarrow \infty$$
If it was just $e^{-i\lambda \cos t}$, the question would be relatively straight forward as the only contribution to the leading behavior is in the neighborhood of $t=0$ as that is the maximum of the function in the exponential and then you could just taylor expand the cosine function, but I don't know how to get the leading behavior as a function of $n$.
 A: Expanding the $\cos$ and using parity properties, the given integral can be written as
\begin{align}
I_n(\lambda)&=\int_{-\pi/2}^{\pi/2}\cos(nt-\lambda\cos t) \,dt\\
&=\int_{-\pi/2}^{\pi/2}\left[\cos(nt)\cos(\lambda\cos t)+\sin(nt)\sin(\lambda\cos t)\right] \,dt\\
&=2\int_0^{\pi/2}\cos(nt)\cos(\lambda\cos t) \,dt\\
&=2\Re \int_0^{\pi/2}\cos(nt)e^{-i\lambda\cos t} \,dt
\end{align}
This form allows to use directly the method of stationary phase. 
\begin{equation}
I_n(\lambda)=\int_{0}^{\pi/2}e^{i\lambda p(t)}q(t)\,dt
\end{equation}
with
\begin{align}
p(t)&=-\cos(t)=-1+\frac{1}{2}t^2+\cdots\\
q(t)&=\cos nt=1-\frac{1}{2}n^2t^2+\cdots
\end{align}
To obtain the leading asymptotic contribution which originates from the region $t\sim 0$, we may develop
for $t\to 0$,
\begin{align}
p(t)&\sim p(0)+\sum_{s=0}^{\infty}p_{s}t^{s+\mu}\\
q(t)&\sim\sum_{s=0}^{\infty}q_{s}t^{s+\nu-1}
\end{align}
with $p(0)=-1,\mu=2,p_0=1/2,\ldots,p_{2k-1}=0,p_{2k}=\tfrac{(-1)^{k+1}}{(2k)!} $ and
$\nu=1,q_0=1,q_1=0,q_2=-n^2t^2/2\ldots$ The leading contribution is then (eq. 2.3.23)
\begin{equation}
I_n(\lambda)\sim 2\Re\left[e^{-i\lambda +i\pi/4}\Gamma\left(\frac{1}{2}%
\right)\frac{b_{0}}{\lambda^{1/2}}\right]
\end{equation} 
where $b_0=q_0/(\mu p_0^{\nu/\mu})=\sqrt{2}/2$. Finally,
\begin{equation}
I_n(\lambda)\sim \sqrt{2\pi}\cos\left(  \lambda -\frac{\pi}{4}\right)\lambda^{-1/2}
\end{equation} 
Higher order terms are easily obtained from both $t=0$ and $t=\pi/2$ contributions.
