Solution
= 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 become zero and one, so
$$
\int_0^{\pi/2}\frac{dx}{2+\sin x}
=\int_0^1\frac{dt}{t^2+t+1}.
$$
Use <completing the square>, $t^2+t+1=(t+1/2)^2+3/4$, and integrate:
$$
\frac2{\sqrt3}\left[\arctan\frac{2t+1}{\sqrt3}\right]_0^1
=\frac2{\sqrt3}\left(\frac\pi3-\frac\pi6\right).
$$
Thus both integrals equal $\boxed{\pi/(3\sqrt3)}$.