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.
 A: 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:
$$
 \frac{ \pi^2 n^2}{8} \cdot  \exp\left(-\frac{\pi^2 \alpha}{8} (n^2-1)\right) \cdot n^2 \operatorname{sinch}\left(\frac{\pi^2 \alpha}{8} (n^2-1)\right)  < \frac{ \pi^2 n^2}{8} \cdot  \exp\left(-\frac{\pi^2 \alpha}{8} (n^2-1)\right) \cdot n^2 \operatorname{sinch}\left(\frac{\pi^2 \alpha}{8} n^2 \right) <  \frac{ \pi^2 n^2}{8} \cdot  \exp\left(\frac{\pi^2 \alpha}{8}\right) 
$$
A: Hint: prove that $\dfrac{\sin n t}{\sin t}\le n$ for $0<t<\frac\pi{2n}$.
A: Write
$$
\frac{\sin(n\,t)}{\sin t}=\frac{\sin(n\,t)}{t}\cdot\frac{t}{\sin t}.
$$
From $\sin t\le t$ it follows that
$$
\frac{|\sin(n\,t)|}{t}\le\min(n,t^{-1}),\quad t>0.\tag1
$$
From convexity, it follows that
$$
\Bigl(\frac{t}{\sin t}\Bigr)^4\le1+\Bigl(\Bigl(\frac{\pi}{2}\Bigr)^3-\frac{2}{\pi}\Bigr)t,\quad0\le t\le\frac{\pi}{2}.\tag2
$$
From (1) and (2) we get that
$$\begin{align*}
\int_0^{\frac{\pi}{2}}t\Bigl(\dfrac{\sin(nt)}{\sin t}\Bigr)^4dt&\le\int_0^{\frac{\pi}{2}}t\bigl(\min(n,t^{-1})\bigr)^4\Bigl(1+\Bigl(\Bigl(\frac{\pi}{2}\Bigr)^3-\frac{2}{\pi}\Bigr)t\Bigr)dt\\
&=n^4\int_0^{\frac1n}t\,\Bigl(1+\Bigl(\Bigl(\frac{\pi}{2}\Bigr)^3-\frac{2}{\pi}\Bigr)t\Bigr)dt+\int_{\frac1n}^{\frac\pi2}t^{-3}\Bigl(1+\Bigl(\Bigl(\frac{\pi}{2}\Bigr)^3-\frac{2}{\pi}\Bigr)t\Bigr)dt\\
&=n^2+\frac{(\pi ^4-16) n}{6 \pi }-\frac{\pi ^4-8}{4 \pi ^2}\\
&<\frac{\pi^2}{4}\,n^2
\end{align*}$$
if $n>2$. The cases $n=1$ and $n=2$ can be checked by direct computation.

Note
A better estimate can be obtained using the inequality
$$
\frac{\sin t}{t}\le\min\Bigl(\frac{\pi}{2},1+(1-\frac{2}{\pi})t,\frac{6}{6-t^2}\Bigr).
$$

A: On the interval $0 < t < {\pi \over 2n}$, use the estimate ${\sin(nt) \over \sin(t)} < n$, giving that
$$\int_0^{\pi \over 2n} t({\sin(nt) \over \sin(t)})^4 \,dt < \int_0^{\pi \over 2n}n^4 t\,dt$$
$$= n^4 {({\pi \over 2n})^2 \over 2} = {\pi^2 n^2 \over 8}$$
On the interval ${\pi \over 2n} < t < {\pi \over 2}$, use the estimates $|\sin(nt)| \leq 1$
and $\sin(t) > {2t \over \pi}$, giving the estimate
$$\int_{\pi \over 2n}^{\pi \over 2} t({\sin(nt) \over \sin(t)})^4 \,dt < \int_{\pi \over 2n}^{\pi \over 2}t ({\pi \over 2t})^4\,dt$$
$$= {\pi^4 \over 16}\int_{\pi \over 2n}^{\pi \over 2} t^{-3}\,dt$$
$$< {\pi^4 \over 16}{1 \over 2}({\pi \over 2n})^{-2}$$
$$= {\pi^2 n^2 \over 8}$$
Adding this to the first part of the integral, we get the upper bound of ${\pi^2 n^2 \over 4}$ as needed.

As for the proof that ${\displaystyle {\sin(nt) \over \sin(t)} < n}$ for ${\displaystyle 0 < t < {\pi \over 2n}}$, it's equivalent to ${\displaystyle {\sin(nt) \over nt} < {\sin(t) \over t}}$. This follows from the fact that ${\displaystyle {\sin(x) \over x}}$ is decreasing on ${\displaystyle (0,{\pi \over 2}]}$; the same fact gives that ${\displaystyle {\sin(t) \over t} < {\sin({\pi \over 2}) \over {\pi \over 2}} = {2 \over \pi}}$ for ${\displaystyle 0 < t < {\pi \over 2}}$, which we also used above. 
A: 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 $I_n<\frac{n^2\pi^2}{4}$.
