TITLE: Show that $\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \sqrt{2\pi}$
QUESTION [5 upvotes]: I wish to show
$$\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \sqrt{2\pi}$$
I've tried substitution and integration by parts to get a recursive formula for the integral, but so far it hasn't worked. Any help would be appreciated, thanks.

REPLY [3 votes]: If we write
$$
\int_0^{\pi/2} \sin^n x \cos^{n-2} x\,dx = \int_0^{\pi/2} \cos^{-2} x \exp\Bigl[n\log(\sin x \cos x)\Bigr]\,dx
$$
and note that the quantity $\log(\sin x \cos x)$ achieves a maximum over the interval of integration at $x = \pi/4$, then by calculating
$$
\log(\sin x \cos x) = -\log 2 - 2 \left(x-\frac{\pi}{4}\right)^2 + O\left(\left(x-\frac{\pi}{4}\right)^4\right)
$$
and
$$
\cos^{-2} x = 2 + O\left(x-\frac{\pi}{4}\right)
$$
as $x \to \pi/4$ we may appeal to the Laplace method to conclude that
$$
\int_0^{\pi/2} \sin^n x \cos^{n-2} x\,dx \sim \int_{-\infty}^{\infty} 2 e^{-n\log 2-2ny^2}\,dy = \frac{\sqrt{2\pi}}{2^n\sqrt{n}}
$$
as $n \to \infty$.