(Dis)Proving $\lim_{x\to\infty}\frac1{x^s}\int_0^xt^{s-1}f(\sin t)dt=\frac1{2\pi s}\int_0^{2\pi}f(\sin t)dt$, for $s>0$ and $f$ defined on $[-1,1]$

Question

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)