How to improve $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$ I have proved this inequality $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$.
Using $\left|\sin(nx)\right|\leq n\left|\sin(x)\right|$ on $[0,\frac{\pi}{2n}]$ and $\frac{\left|\sin(nx)\right|}{\left|\sin(x)\right|}\leq\frac{\pi}{2x}$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$,we can have
$$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}+\frac{\pi^{2}}{8}\left(n^{2}-1\right)<\frac{\pi^{2}n^{2}}{4}.$$
But using mathematica I found this inequality can still be improved.
And after calculating some terms I found it seems that when $n\geq 2$ we can have
$$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}.$$
But I cannot prove this.So is there any method to improve my result?Any help will be thanked.

We have the elementary estimate
$$1 \le \frac{z^4}{\sin^4 z} \le 1 + z^2 \varepsilon$$
where
$$\varepsilon= \frac{\pi^2}{4} - \frac{4}{\pi^2}.$$
Let $z = (y/n)$ and multiply both sides by $\sin^4 y/y^4$. Then for $y \in [0,n \pi/2]$, 
one has:
$$ \frac{\sin^4 y}{y^4}
\le 
\left(\frac{\sin(y)/n}{\sin (y/n)}\right)^4
\le \frac{\sin^4 y}{y^4} + \frac{\sin^4 y}{y^2 n^2}  \cdot  \varepsilon
$$
Make the substitution $x = y/n$ in the integral, it becomes
$$I_n:=n^2 \int^{n \pi/2}_{0}  y \left(\frac{\sin(y)/n}{\sin(y/n)}\right)^4 dy$$
and thus
$$n^2 \int^{n \pi/2}_{0}  \frac{\sin^4 y}{y^3} dy \le I_n
\le n^2 \int^{n \pi/2}_{0}  \frac{\sin^4 y}{y^3} dy + \varepsilon \cdot \int^{n \pi/2}_{0}   \frac{\sin^4 y}{y} dy$$
The lower bound is asymptotic to
$$n^2 \int^{\infty}_{0}  \frac{\sin^4 x}{x^3}  dx = n^2 \log 2,$$
and in fact since
$$n^2 \int^{\infty}_{n \pi/2}  \frac{\sin^4 x}{x^3} 
\le n^2 \int^{\infty}_{n \pi/2}  \frac{1}{x^3}  = \frac{2}{\pi^2} $$
one even has the lower bound
$$I_n \ge n^2 \log 2 - \frac{2}{\pi^2}$$
On the other hand, an upper bound is given by
$$ n^2 \int^{\infty}_{0}  \frac{\sin^4 y}{y^3} dy  = 
 \varepsilon \cdot \int^{1}_{0} \frac{\sin^4 y}{y} +  
 \varepsilon \cdot \int^{n \pi/2}_{1}   \frac{1}{y} dy$$
$$ = n^2 \log 2 + \eta + \varepsilon \log(n \pi/2)$$
where $\eta \sim 0.160629\ldots$ and $\varepsilon \sim 2.062116\ldots$.
From this you can obtain your explicit bound for $n \ge 3$ and check $n = 2$ by hand. Of course, it gives a more precise bound for larger $n$, and it's clear that you can push this much further if you want to.
