<s>[INST] For which $n$ is $ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx}= \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$? I have been trying to figure out for which $n$ is $$ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx} = \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$$
Using maple I got the following values below 100 
$$
n = 1
,\,2
,\,12
,\,13
,\,14
,\,24
,\,25
,\,26
,\,36
,\,37
,\,38
,\,48
,\,49
,\,50
,\,60
,\,61
,\,62
,\,73
,\,74
,\,85
,\,86
,\,98
,\,110
$$
But I am having a hard time seeing any pattern. I tried calculating the integral directly.
The integral seems to converge for all $n$, but a closed form seems hard. Any help would be appreciated =)
EDIT: I ran a few more tests and can not find any more integer values that work. Are the list above exhaustive?
 [/INST] Something is amiss. We have
$$\int_0^{\pi/2} \frac{dx}{2+\sin (nx)} = \frac{1}{n}\int_0^{n\pi/2} \frac{dy}{2+\sin y}.$$
Using
$$\int_0^{2\pi} \frac{dy}{2+\sin y} = \frac{2\pi}{\sqrt{3}},$$
writing $n = 4k + r$ with $0 \leqslant r \leqslant 3$, we find
$$\int_0^{\pi/2}\frac{dx}{2+\sin (nx)} = \frac{k}{4k+r}\cdot \frac{2\pi}{\sqrt{3}} + \frac{1}{4k+r}\int_0^{r\pi/2}\frac{dy}{2+\sin y}.$$
In particular for $r = 0$, we have
$$\int_0^{\pi/2}\frac{dx}{2+\sin (4kx)} = \frac{1}{4}\frac{2\pi}{\sqrt{3}} = \frac{\pi}{2\sqrt{3}} \neq \frac{\pi}{3\sqrt{3}}.$$
No $n$ divisible by $4$ satisfies the equation. For $n = 4k+1$, we have the value
$$\frac{k}{4k+1} \frac{2\pi}{\sqrt{3}} + \frac{1}{4k+1}\frac{\pi}{3\sqrt{3}} = \frac{6k+1}{4k+1} \frac{\pi}{3\sqrt{3}},$$
which only equals $\frac{\pi}{3\sqrt{3}}$ for $k = 0$.
For $n = 4k+2$, we have
$$\frac{k}{4k+2}\frac{2\pi}{\sqrt{3}} + \frac{2}{4k+2} \frac{\pi}{3\sqrt{3}} = \frac{6k+2}{4k+2}\frac{\pi}{3\sqrt{3}},$$
which also only equals $\frac{\pi}{3\sqrt{3}}$ for $k = 0$.
For $n = 4k+3$, we have
$$\frac{k}{4k+3}\frac{2\pi}{\sqrt{3}} + \frac{1}{4k+3}\frac{4\pi}{3\sqrt{3}} = \frac{6k+4}{4k+3}\frac{\pi}{3\sqrt{3}},$$
which never equals $\frac{\pi}{3\sqrt{3}}$. Thus $n = 1$ and $n = 2$ are the only solutions.
</s>