Prove: $\int_0^{\frac{\pi}{2}}t(\frac{\sin nt}{\sin t})^4dt<\frac{\pi^2n^2}{4}$ I have a question about integral. Prove:
$$\int_0^{\frac{\pi}{2}}t\left(\dfrac{\sin(nt)}{\sin(t)}\right)^4dt<\dfrac{\pi^2n^2}{4}$$
I have tried several methods including $\sin(t)\geq\frac{2t}{\pi}$, but I can't work it out.

First note that
$$I:=\int_0^{\frac{\pi}{2}}x\left(\frac{\sin nx}{\sin x}\right)^4\mathrm{d}x=\int_0^{\frac{\pi}{2}}\frac{x}{\sin x}\frac{|\sin nx|}{\sin x}|\sin nx|\left(\frac{\sin nx}{\sin x}\right)^2\mathrm{d}x.$$
We can obtain the following inequalities with little effort :
$$1<\frac{x}{\sin x}<\frac{\pi}{2}(0<x<\pi),\ |\sin nx|\leqslant n|\sin x|(x\in \mathbb{R}).$$
Thus with them we get
$$I<\frac{n\pi}{2}\int_0^{\frac{\pi}{2}}\left(\frac{\sin nx}{\sin x}\right)^2\mathrm{d}x=:\frac{n\pi}{2}J_n.\tag{1}$$
To finish the proof, it's sufficient to evaluate that :
\begin{align*}
J_{n+1}-J_n& =\int_0^{\frac{\pi}{2}}\frac{\sin^2 (n+1)x-\sin^2 nx}{\sin^2  x}\mathrm{d}x\\
& =\int_0^{\frac{\pi}{2}}\frac{\sin(2n+1)x}{\sin x}\mathrm{d}x=:K_n.
\end{align*}
Since $K_{n+1}-K_n=\int_0^{\frac{\pi}{2}}2\cos 2nx\ \mathrm{d}x=0$, we claim that $J_n=\frac{n\pi}{2}$.
With (1), we get $I_n<\frac{n^2\pi^2}{4}$.
