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.
 A: 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.
A: Alternative solution:
When $n = 2, 3, 4$, the inequality is verified directly.
In the following, assume that $n\ge 5$.
Let
$$I_n = \int_0^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x.$$
We have
\begin{align}
I_n &= \underbrace{\int_0^{\pi/n} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x}_{I_{n,1}}
+ \underbrace{\int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x}_{I_{n,2}}.
\end{align}
First, we have
\begin{align}
I_{n,1} &\le \int_0^{\pi/n} \frac{x}{n^2}(\sin nx)^4 \frac{1}{x^4} \left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4\mathrm{d} x \\
&= \frac{1}{n^2}\left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4
\int_0^{\pi/n} \frac{(\sin nx)^4}{x^3} \mathrm{d} x \\
&= \left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4
\int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y\\
&\le \left(\frac{\frac{\pi}{5}}{\sin\frac{\pi}{5}}\right)^4
\int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y
\end{align}
where we have used: i) $\frac{\sin x}{x} \ge \frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}}$
on $0 \le x \le \frac{\pi}{n}$; ii) the substitution $y = nx$; iii) $\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}$ is non-increasing for $n\ge 2$.
Second, we have
\begin{align}
I_{n, 2} &= \int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x\\
&\le \int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\pi}{2x}\right)^4\mathrm{d} x \\
&= -\frac{\pi^2}{8n^2} + \frac{\pi^2}{32}\\
&\le \frac{\pi^2}{32}
\end{align}
where we have used $\sin x \ge \frac{2}{\pi}x$ for $0 \le x \le \frac{\pi}{2}$.
Thus, we have
$$I_n \le \left(\frac{\frac{\pi}{5}}{\sin\frac{\pi}{5}}\right)^4
\int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y
+ \frac{\pi^2}{32} < \frac{\pi^2}{8}.$$
We are done.
A: Let $ a_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx $
$$ a_{n}-a_{n-1}=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos(2n-2)x-\cos 2nx}{\sin^2 x} dx=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x}{\sin x} dx $$
$$ a_{n}-a_{n-1}-(a_{n-1}-a_{n-2})=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x-\sin(2n-3)x}{\sin x} dx=2\int_{0}^{\frac{\pi}{2}} \cos(2n-2)x dx=0 $$
$$ a_{n}-2a_{n-1}+a_{n-2}=0 $$
So $ a_{n} $ is an arithmetic sequence,$ a_{n}=a+bn,a_{0}=0,a_{1}=\frac{\pi}{2} $
We get
$$ a_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx=\frac{n\pi}{2} $$
Thus
$$ \int_{0}^{\frac{\pi}{2}}x \frac{\sin^4 nx}{\sin^4 x} dx=\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} \frac{\left| \sin nx \right |}{\sin x} \left| \sin nx \right | \frac{\sin^2 nx}{\sin^2 x} dx ＜ \int_{0}^{\frac{\pi}{2}} \frac{\pi}{2} \times n \times 1 \times \frac{\sin^2 nx}{\sin^2 x} dx=\frac{n^2 \pi^2}{4} $$
