I find the following *averaged-integra*l amusing and intriguing, to say the least. Is there any proof?



> 
> For any pair of integers $n\geq k\geq0$, we have
>  $$\frac1{\pi}\int\_0^{\pi}\frac{\sin^n(x)}{\sin^k(\frac{kx}n)\sin^{n-k}\left(\frac{(n-k)x}n\right)}dx=\binom{n}k. \tag1$$ 
> 
> 
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I also wonder if there's any reason to relate these with [an MO question that I just noticed](https://mathoverflow.net/questions/109372/interpolating-a-sum-of-binomial-coefficients-using-a-sin-function?rq=1). Perhaps by inverting?


**AN UPDATE.** I'm extending the above to a stronger conjecture shown below.



> 
> For non-negative reals with $r\geq s$, a generalization is given by
>  $$\frac1{\pi}\int\_0^{\pi}\frac{\sin^r(x)}{\sin^s(\frac{sx}r)\sin^{r-s}\left(\frac{(r-s)x}r\right)}\,dx
> =\binom{r}{s}. \tag2$$
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> 
> 



integral of a "sin-omial" coefficients=binomial
Tonight I read [here](https://mathoverflow.net/a/255031/4312) [the answer by esg to another your question] that $\frac1{2\pi}\int\_{-\pi}^\pi e^{-ik t}(1+e^{it})^ndt=\binom{n}{k}$, which is, well, obvious at least when both $n$ and $k$ are positive integers: just expand the binomial $(1+e^{it})^n$ and integrate. Denoting $\alpha=k/n$ we may rewrite this as $\frac1{2\pi}\int\_{-\pi}^\pi (f(t))^n dt=\binom{n}{\alpha n}$, where the function $f(t)=(1+e^{it})e^{-i\alpha t}$ is complex-valued. For making it real-valued, we change the path between the points $-\pi$ and $\pi$. The value of the integral does not change (since $f^n$ is analytic between two paths, for integer $n$ it is simply entire function.) On the second path $f$ takes real values. Namely, for $t\in (-\pi,\pi)$ we define $s(t)=\ln \frac{\sin (1-\alpha)t}{\sin \alpha t}$. It is straightforward (some elementary high school trigonometry) that $$f(t+is(t))=\frac{\sin t}{\sin^{\alpha} \alpha t\cdot \sin^{1-\alpha}(1-\alpha)t},$$
so we replace the path from $(-\pi,\pi)$ to $\{t+s(t)i:t\in (-\pi,\pi)\}$ (limit values of $s(t)$ at the endpoints are equal to 0) and take only the real part of the integral (this allows to replace $d(t+s(t)i)$ to $dt$ in the differential). We get
$$
\frac1{2\pi}\int\_{-\pi}^\pi \frac{\sin^n t}{\sin^{\alpha n} \alpha t\cdot \sin^{(1-\alpha)n}(1-\alpha)t}dt=\binom{n}{\alpha n}
$$
as desired.


