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Prove (or disprove) the following conjecture:

Given $s>0$, function $f(x)$ is defined for all $x\in[-1;1]$, then:
$$\lim_{x\to\infty}\frac{1}{x^s}\int_0^x t^{s-1}f(\sin(t))\,dt=\frac{1}{2{\pi}s}\int_0^{2\pi} f(\sin(t))\,dt$$

I can only prove the conjecture for $s=2$ (using $\int_0^{(2n+1)\pi}tf(\sin(t))\,dt=\frac{(2n+1)\pi}{2}\int_0^{(2n+1)\pi}f(\sin(t))\,dt$ ($*$)) and $s=1$ (obvious). However, the substitution trick $t=(2n+1)\pi-u$ in ($*$) doesn't work for $s$ is even or $s$ is non-integer. It seems to me that the conjecture is true but all my tools for evaluating limit problems don't work here.
(Update) The answer to the problem above was done in What is $\lim_{t\rightarrow \infty }\int_{a}^{b}f(x,\sin(tx))dx$？
(I moved my generalization to another post)

Quý Nhân
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    With no loss of generality you may assume that $f\ge 0.$ It suffices to consider $x=2n\pi$ with $n\to \infty.$ Then you can split the integral over $[2(k-1)\pi,2k\pi]$. Each integral can be estimated below and above by the appropriate values of $t^{s-1}$ times the integral of $f(\sin t).$ It should work I hope. – Ryszard Szwarc Feb 07 '23 at 07:16
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    Related: https://math.stackexchange.com/questions/4566609/what-is-lim-t-rightarrow-infty-int-abfx-sintxdx/4569585#4569585 – Gerd Feb 08 '23 at 07:01
  • @Gerd Amazing proof! How unfortunate that the original post was closed. By the way, I wonder whether we can generalize the formula for general curves and n-dimensional analogy – Quý Nhân Feb 08 '23 at 14:37
  • Thanks. The proof relies on properties of sinus. For a generalization to periodic functions I think a different method is needful. – Gerd Feb 08 '23 at 14:53

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