How find this integral limit 
Find this limit $$\lim_{n\to\infty}\dfrac{\displaystyle\int_{0}^{\frac{\pi}{2}}x\left(\dfrac{\sin{(nx)}}{\sin{x}}\right)^4dx}{n^2}.$$

This problem is post my teacher(tian275461, he is very popular in China), and he told me this problem answer is $\ln{2}$, but I can't prove it. Thank you.

With the substitution $y = nx$, we have
\begin{align}
I_n &= \int_0^{n\pi/2} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y \\
&= \underbrace{\int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y}_{I_{n,1}} + 
\underbrace{\int_{\sqrt{n}}^{n\pi/2} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y}_{I_{n,2}}.
\end{align}
Since $u \mapsto \frac{u}{\sin u}$ is increasing on $(0, \frac{\pi}{2}]$, we have
$$1 \le \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \le \left(\frac{\frac{1}{\sqrt{n}}}{\sin \frac{1}{\sqrt{n}}}\right)^4,
\ \forall 0 < y \le \sqrt{n}.$$
Thus, we have
$$\int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3}\mathrm{d} y \le I_{n,1} \le \left(\frac{\frac{1}{\sqrt{n}}}{\sin \frac{1}{\sqrt{n}}}\right)^4 \int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3} \mathrm{d} y.$$
By the squeeze theorem, we have $\lim_{n\to \infty} I_{n,1} = \int_0^{\infty} \frac{\sin^4 y}{y^3}\mathrm{d} y = \ln 2$
(see e.g. Laplace transform:$\int_0^\infty \frac{\sin^4 x}{x^3} \, dx $).
Also, we have
$$I_{n,2} \le \int_{\sqrt{n}}^{n\pi/2} \frac{1}{\sqrt{n}^3} \left(\frac{\pi}{2}\right)^4 \mathrm{d} y$$
and thus $\lim_{n\to \infty} I_{n,2} = 0$.
Thus, $\lim_{n\to \infty} I_n = \ln 2$.
