= Solution
The <tangent half-angle substitution> $t=\tan(x/2)$ gives $\sin x=2t/(1+t^2)$ and $dx=2\,dt/(1+t^2)$. The endpoints $x=0,\pi/2$ become $t=0,1$, respectively. Hence
$$
\int_0^{\pi/2}\frac{dx}{2+\sin x}=\int_0^1\frac{2\,dt}{2(1+t^2)+2t}=\int_0^1\frac{dt}{t^2+t+1}.
$$
<Completing the square>, $t^2+t+1=(t+1/2)^2+3/4$, then gives
$$
\int_0^1\frac{dt}{t^2+t+1}=\frac2{\sqrt3}\left[\arctan\frac{2t+1}{\sqrt3}\right]_0^1=\frac2{\sqrt3}\left(\frac\pi3-\frac\pi6\right)=\boxed{\frac\pi{3\sqrt3}}.
$$
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