<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] Sub $y=n x$ and use a Weierstrass substitution $t=\tan{(y/2)}$ and get that the integral is
$$\frac1{n} \int_0^{\tan{(\pi n/4)}} \frac{dt}{1+t+t^2} $$
Evaluate this by expressing the denominator as $(t+1/2)^2+3/4$ to get an expression for the integral and the stated equation:
$$\frac{2}{\sqrt{3} n} \arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{\pi}{4} n \right )+1}} \right ] } = \frac{\pi}{3 \sqrt{3}} $$
A quick scan shows this works for $n=1$ and $n=2$.  I guess you would have to define the branch of the arctangent according to the value of $n$.  For example, add $2 \pi$ to each multiple of $8$ for $n$.  For example,
$$\arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{\pi}{4}  \right )+1}} \right ] } = \frac{\pi}{6} $$
$$\arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{9 \pi}{4}  \right )+1}} \right ] } = \frac{\pi}{6} + 2\pi$$
$$\arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{17 \pi}{4}  \right )+1}} \right ] } = \frac{\pi}{6} + 4\pi$$
and so on.  Thus, for $n=12$, we have
$$\frac{2}{12 \sqrt{3}} \arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{12 \pi}{4}  \right )+1}} \right ] } = \frac{1}{6 \sqrt{3}} 2 \pi = \frac{\pi}{3 \sqrt{3}} $$
For $n=13$, we have
$$\frac{2}{13\sqrt{3}} \arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{13 \pi}{4}  \right )+1}} \right ] } = \frac{2 }{13 \sqrt{3}} \left ( 2 \pi+\frac{\pi}{6}\right ) = \frac{\pi}{3 \sqrt{3}} $$
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