Equivalent of $\int_0^{\pi/2}\cos^n(\sin(x))dx$ Let $\displaystyle u_n=\int_0^{\pi/2}\cos^n(\sin(x))dx$.
How can I find an equivalent of $u_n$ when $n\to\infty$ ?
 A: The answer is $u_n \sim \sqrt{\frac{\pi}{2n}}$.
To prove this, first note that for any fixed $\delta > 0$, $\int_\delta^{\pi/2} \cos^{n} (\sin x) \, dx = O(r^n)$, where $r = \cos \sin \delta$. This is $o(1/\sqrt{n}$. Therefore we can replace the $\pi/2$ with a smaller $\delta$ without changing anything.
Now make the substitution $y = \sin x$. Then we get 
$$\int_0^\delta \cos^{n} (\sin x) \, dx = \int_0^\alpha \frac{\cos^n y}{\sqrt{1-y^2}} \, dy,$$
where $\alpha = \sin \delta$ is small and fixed. By taking $\alpha$ small enough, the factor $\frac{1}{\sqrt{1-y^2}}$ can be made as close to $1$ as we like, hence for the purposes of finding an equivalent, we may treat it as being equal to $1$.
Thus our problem is to find an equivalent of $\int_0^\alpha \cos^n y \, dy$. On $[0,\alpha]$ we have an inequality 
$$1 - \frac{1}{2} y^2 \leq \cos y \leq 1 - (1/2 - \beta)y^2/2,$$
where $\beta$ is a positive constant that can be made as small as we like by selecting $\alpha$ to be small. Therefore it will be enough to find an equivalent of $\int_0^{\alpha} (1-c y^2)^n \, dy$ where $c$ is a constant close to $1/2$.
Make the substitution $v = y \sqrt{n}$ and the integral becomes
$$\frac{1}{\sqrt{n}} \int_0^{\alpha\sqrt{n}} (1-cv^2/n)^n \, dv.$$
The integrand is bounded above by $e^{-cv^2}$, whose integral converges, so we can apply the Lebesgue dominated convergence theorem to conclude that the integral tends to $\int_0^{+\infty} e^{-cv^2} \, dv = \sqrt{\frac{\pi}{4c}}$ as $n \to +\infty$. Since we can select $c$ as close as we like to $1/2$, it follows that $u_n \sim \sqrt{\frac{\pi}{2n}}$.
A: Writing
$$
\cos^n(\sin x) = \exp(n\log(\cos\sin x)) \stackrel{\text{def}}{=} e^{n\varphi(x)},
$$
one can differentiate to check that $\varphi(x)$ is decreasing on $(0,\pi/2)$ and hence has a maximum when $x=0$.  Near there we have
$$
\varphi(x) = -\frac{1}{2}x^2 + O(x^4),
$$
so by the Laplace method
$$
\begin{align}
\int_0^{\pi/2} \cos^n(\sin x)\,dx &= \int_0^{\pi/2} e^{n\varphi(x)}\,dx \\
&\sim \int_0^\infty e^{-nx^2/2}\,dx \\
&= \sqrt{\frac{\pi}{2n}}.
\end{align}
$$
