$$\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx$$
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
$$z=\alpha(x)=e^{ix}$$
$$\sin(x)=\frac{z^2-1}{2zi}$$ $$2+\sin(x)=\frac{4zi+z^2-1}{2zi}$$ $$\frac{1}{2+\sin(x)}=\frac{2zi}{4zi+z^2-1}$$
$$\int_\alpha \frac{2zi}{4zi+z^2-1} \ \ \frac{1}{zi}$$
$$4zi+z^2-1=0$$ $$z_1=\frac{-4i+2\sqrt{3}i}{2}$$ $$z_2=\frac{-4i-2\sqrt{3}i}{2}$$
$$\rvert z_1 \rvert <1$$ $$\rvert z_2 \rvert >1$$
Usando el teorema del residuo:
$$\int_\alpha \frac{2}{4zi+z^2-1}=2 \pi i \ \lim_{z\rightarrow-2i+\sqrt{3}i} \ \frac{2}{z+2i+\sqrt{3}i}=\frac{2 \pi}{\sqrt{3}}$$
¿Es correcto?
¡Gracias!