<s>[INST] 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.
 [/INST] Assume that $n\in \mathbb{N}$.
Then the ratio of sines is polynomial in cosines:
$$
   \frac{\sin(n t)}{\sin(t)} = \cos((n-1) t) + \cos(t) \frac{\sin((n-1) t}{\sin(t)} = \ldots = \sum_{k=1}^n \cos((n-k) t) \cdot \cos^{k-1}(t) \leqslant n
$$
Since $\sin(t)$ is increasing on the interval $\left(0,\frac{\pi}{2}\right)$, and since $\frac{\sin(n t)}{\sin(t)} < n$ for $0<t<\frac{\pi}{2n}$, we have:
$$
  \int_0^{\tfrac{\pi}{2}} t \cdot \left(  \frac{\sin(n t)}{\sin(t)}  \right)^4 \mathrm{d} t = \sum_{k=1}^n \int_{\tfrac{k-1}{n} \cdot \tfrac{\pi}{2}}^{\tfrac{k}{n} \cdot \tfrac{\pi}{2}} t \cdot \left(  \frac{\sin(n t)}{\sin(t)}  \right)^4 \mathrm{d} t < \\ 
 \int_0^{\pi/(2n)} t n^4 \mathrm{d} t + \sum_{k=2}^n \int_{\tfrac{k-1}{n} \cdot \tfrac{\pi}{2}}^{\tfrac{k}{n} \cdot \tfrac{\pi}{2}} t \cdot \left(  \frac{\sin(n t)}{\sin\left( \frac{k-1}{n} \cdot \tfrac{\pi}{2}  \right)}  \right)^4 \mathrm{d} t  = \\
 \frac{\pi^2 n^2}{8}  + \sum_{k=2}^n \int_{\tfrac{k-1}{n} \cdot \tfrac{\pi}{2}}^{\tfrac{k}{n} \cdot \tfrac{\pi}{2}} t \cdot \left(  \frac{\sin(n t)}{\sin\left( \frac{k-1}{n} \cdot \tfrac{\pi}{2}  \right)}  \right)^4 \mathrm{d} t
$$
The remaining bounding integral is not hard to compute:
$$
   \int_{\tfrac{k-1}{n} \cdot \tfrac{\pi}{2}}^{\tfrac{k}{n} \cdot \tfrac{\pi}{2}} t \cdot \sin^4(n t)  \mathrm{d} t \stackrel{t = \tfrac{(k-1)\pi}{2n} + \tfrac{u}{2}}{=} \int_0^{\tfrac{\pi}{2}} \frac{2u+ \pi(k-1)}{2n^2} \left( \frac{1+(-1)^k}{2} \cos^4 u + \frac{1-(-1)^k}{2} \sin^4 u  \right) \mathrm{d} u = \frac{3 \pi^2 \cdot (2k-1) - 16 (-1)^k}{64 n^2}
$$
The upper bound then becomes:
$$
  \frac{\pi^2 n^2}{8} + \sum_{k=2}^n \frac{3 \pi^2 \cdot (2k-1) - 16 (-1)^k}{64 n^2 \cdot \sin^4\left( \frac{k-1}{n} \cdot \tfrac{\pi}{2}  \right)}
$$
It is easy to check numerically that this bound is more crude than the one you seek to establish.

Added
Notice the series expansion around $t=0$:
$$
    \left(\frac{\sin(n t)}{\sin(t)} \right)^4 = n^4 \left( 1 - \frac{2}{3} (n^2-1) t^2 + \mathcal{o}(t^2) \right)
$$
This suggests looking for a bound in the form $\exp(-(n^2-1) t^2 \alpha)$. Suppose we fix a small enough   $\alpha$, such that
$$
   \forall_{0 < t < \tfrac{\pi}{2}} \left(\frac{\sin(n t)}{\sin(t)} \right)^4 \leqslant \exp(-(n^2-1) t^2 \alpha)
$$
Then
$$
  \int_0^{\pi/2} t \cdot \left(\frac{\sin(n t)}{\sin(t)} \right)^4 \mathrm{d} t < \int_0^{\pi/2} t  \exp(-(n^2-1) t^2 \alpha) \mathrm{d} t = n^4 \frac{1 - \exp(-\alpha (n^2-1) \pi^2/4)}{\alpha (n^2-1)} = \frac{ \pi^2 n^2}{8} \cdot  \exp\left(-\frac{\pi^2 \alpha}{8} (n^2-1)\right) \cdot n^2 \operatorname{sinch}\left(\frac{\pi^2 \alpha}{8} (n^2-1)\right) 
$$
where $\operatorname{sinch}(x) = \frac{\sinh(x)}{x}$ and is an increasing function of $x$, therefore:
$$
 \frac{ \pi^2 n^2}{8} \cdot  \exp\left(-\frac{\pi^2 \alpha}{8} (n^2-1)\right) \cdot n^2 \operatorname{sinch}\left(\frac{\pi^2 \alpha}{8} (n^2-1)\right)  < \frac{ \pi^2 n^2}{8} \cdot  \exp\left(-\frac{\pi^2 \alpha}{8} (n^2-1)\right) \cdot n^2 \operatorname{sinch}\left(\frac{\pi^2 \alpha}{8} n^2 \right) <  \frac{ \pi^2 n^2}{8} \cdot  \exp\left(\frac{\pi^2 \alpha}{8}\right) 
$$
</s>