Evaluate $\lim_{n\to\infty} \sqrt{n}\int_{0}^{\pi/2} \sin^{n} x dx$ Question:
Compute $I_n=\lim\limits_{n\to\infty} \sqrt{n}\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx$.
Attempt:
We know the famous result that $\lim\limits_{n\to\infty}\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx=0$,
and by dividing it into two parts $\int_{0}^{\frac{\pi}{2}-n^{-\alpha}}+\int_{\frac{\pi}{2}-n^{-\alpha}}^{\frac{\pi}{2}}(0<\alpha<\frac{1}{2})$ ,
we can show that $\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx=O(\frac{1}{n^{\alpha}})$.
There is a hint which says by substitution we can rewrite the limits with $\Gamma$, and eventually get $I_n\sim c\frac{1}{\sqrt{n}},c>0$.
Is there any method without using the $\Gamma$ function? (Answers using Gamma function are also appreciated.)
 A: You may show that the main contribution comes from the neighbourhood of $\frac{\pi}{2}$. Then, by dominated convergence,
\begin{align*}
\sqrt n \int_{\pi /2 - \varepsilon }^{\pi /2} {\sin ^n x\,dx} & \approx \sqrt n \int_{\pi /2 - \varepsilon }^{\pi /2} {\left( {1 - \frac{1}{2}\left( {\frac{\pi }{2} - x} \right)^2 } \right)^n dx} = \sqrt n \int_0^\varepsilon  {\left( {1 - \frac{1}{2}t^2 } \right)^n dt} \\ & = \int_0^{\sqrt n \varepsilon } {\left( {1 - \frac{{s^2 }}{{2n}}} \right)^n ds}  \to \int_0^{ + \infty } {e^{ - s^2 /2} ds} =\sqrt\frac{\pi}{2}.
\end{align*}
I let you make this argument rigorous. You may study the Laplace method.
A: 
Answers using Gamma function are also appreciated.

I'll hold you to that: linearizing $\ln\Gamma(z)$,$$\int_0^{\pi/2}\sin^nxdx=\frac12\operatorname{B}\left(\frac12,\,\frac{n+1}{2}\right)\stackrel{\ast}{=}\frac{\sqrt{\pi}}{2}\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n+2}{2})}\sim\frac{\sqrt{\pi}}{2}\sqrt{\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n+3}{2})}}=\frac{\sqrt{\pi}}{2}\sqrt{\frac{2}{n+1}}\sim\sqrt{\frac{\pi}{2n}}$$(see here for $\stackrel{\ast}{=}$), so $\lim_{n\to\infty}\sqrt{n}\int_0^{\pi/2}\sin^nxdx=\sqrt{\frac{\pi}{2}}$, in agreement with @Gary's answer.
A: If $I_n=\int_0^{\pi/2}\sin^n x\, dx$ then we have $$I_{n}=\frac{n-1}{n}I_{n-2}\tag{1}$$ (prove this using integration by parts) and then we can note that integrand is a decreasing function of $n$ hence $$I_{n+2}<I_{n+1}<I_n$$ This implies that $$\frac{I_{n+2}}{I_n}=\frac{n+1}{n+2}<\frac{I_{n+1}}{I_n}<1$$ Taking limits as $n\to\infty $ we see that $I_{n+1}/I_n\to 1$ as $n\to\infty $.
Let $a_n=\sqrt{n} I_n$ so that $a_{n+1}/a_n\to 1$. Next we prove that $a_{2n}\to\sqrt{\pi/2}$ and by the result noted in last sentence we also have $a_{2n+1}\to\sqrt{\pi/2}$ so that $a_n$ also tends to same limit.
Using $(1)$ repeatedly we get $$I_{2n}=\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdots\frac{1}{2}\cdot\frac{\pi} {2}=\frac{\pi}{2}x_n\text{ (say)} \tag{2}$$ and $$I_{2n+1}=\frac{2n}{2n+1}\cdot \frac{2n-2}{2n-1}\cdots \frac{2}{3}=\frac{1}{(2n+1) x_n} \tag{3}$$ Dividing $(2)$ by $(3)$ we get $$\frac{I_{2n}}{I_{2n+1}}=(2n+1)x_n^2\cdot\frac{\pi}{2}$$ Taking square roots and further taking limits as $n\to\infty $ we get $$\sqrt{2n}x_n\to\sqrt{\frac{2}{\pi}}$$ Using $(2)$ we now have $$a_{2n}=\sqrt{2n}I_{2 n} =\sqrt{2n}x_n\cdot\frac{\pi}{2}\to\sqrt{\frac{\pi}{2}}$$
A: Using @J.G.' answer and Stirling approximation
$$\int_0^{\frac \pi 2}\sin^n(x)\,dx=\frac{\sqrt \pi }{2}\frac{\Gamma \left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n+2}{2}\right)}=\sqrt{\frac{\pi }{2n}}\,\,\Bigg[1-\frac{1}{4 n}+\frac{1}{32 n^2}+\frac{5}{128 n^3}-\frac{21}{2048
   n^4}+O\left(\frac{1}{n^5}\right) \Bigg]$$
A: Proceeding along your approach, without using the gamma function
Let
\begin{align*}
 J_n &:= \sqrt{n}\int_{\pi/2 - n^{-1/3}}^{\pi/2} \sin^n x \,\mathrm{d} x, \\
 K_n &:= \sqrt{n}\int_0^{\pi/2 - n^{-1/3}} \sin^n x \,\mathrm{d} x.
\end{align*}
We have
$$J_n
= \sqrt{n} \int_0^{n^{-1/3}} \cos^n y \,\mathrm{d} y
= \int_0^{n^{1/6}} \cos^n \frac{z}{\sqrt n}\, \mathrm{d} z
= \int_0^{n^{1/6}}
\mathrm{e}^{n\ln \cos \frac{z}{\sqrt n}}\mathrm{d} z.$$
Fact 1: For $0 \le u \le 1$, it holds that
$$-\frac{u^2}{2} - u^4 \le \ln \cos u \le - \frac{u^2}{2}.$$
(The proof is not difficult. Omitted.)
Using Fact 1, we have, for all $0 \le z \le n^{1/6}$,
$$-\frac{z^2}{2} - \frac{(n^{1/6})^4}{n}\le -\frac{z^2}{2} - \frac{z^4}{n} \le n\ln \cos \frac{z}{\sqrt n} \le -\frac{z^2}{2}.$$
Thus, we have
$$\mathrm{e}^{-n^{-1/3}}\int_0^{n^{1/6}}
\mathrm{e}^{-z^2/2}\mathrm{d} z \le J_n \le \int_0^{n^{1/6}}
\mathrm{e}^{-z^2/2}\mathrm{d} z.$$
Thus,
$$\lim_{n\to \infty} J_n = \int_0^\infty \mathrm{e}^{-z^2/2}\mathrm{d} z = \sqrt{\pi/2}.$$
$\phantom{2}$
On the other hand, we have
\begin{align*}
 K_n 
 &\le  \sqrt{n}\, \cdot \frac{\pi}{2} \sin^n \left(\pi/2 - n^{-1/3}\right)\\
 &= \frac{\pi}{2} \sqrt{n}\, \cos^n \frac{1}{\sqrt[3]{n}}\\
 &\le \frac{\pi}{2} \sqrt{n}\,
 \left(1 - \frac{1}{4\sqrt[3]{n^2}}\right)^n
\end{align*}
where we have used $\cos y \le 1 - \frac{y^2}{4}$
for all $y\in [0, \pi/2]$ (easy). Also,
$$\lim_{n\to \infty} \frac{\pi}{2} \sqrt{n}\,
\left(1 - \frac{1}{4\sqrt[3]{n^2}}\right)^n = 0.$$
Thus,
$$\lim_{n\to \infty} K_n  = 0.$$
$\phantom{2}$
Finally, we have
$$\lim_{n\to \infty} I_n = \lim_{n\to \infty} J_n + \lim_{n\to \infty} K_n = \sqrt{\pi/2}.$$
