Compute the real integral:
$$\int_0^{2{\pi}} \frac{dx}{2+\sin(x)}.$$
The idea here is to convert this real integral into a complex integral, and then solve it using the Cauchy integration form.
Here is what I have thus far:
Consider that $ \sin(x)=\frac{e^{ix}-e^{-ix}}{2i} $.
Thus,
$$\int_0^{2{\pi}} \frac{dx}{2+\sin(x)}=\int_0^{2{\pi}}\frac{dx}{2+\frac{e^{ix}-e^{-ix}}{2i}}= \int_0^{2{\pi}}\frac{dx}{2+\frac{e^{ix}-e^{-ix}}{2i}}\cdot\frac{2ie^{ix}}{2ie^{ix}}=\int_0^{2{\pi}}\frac{2ie^{ix}}{(e^{ix})^2+4ie^{ix}-1}dx.$$
If we let $z=e^{ix}$, where $dz=ie^{ix}$, the integral becomes:
$$\int_c\frac{2}{z^2+4iz-1}dz$$
I found that the roots of the denominator are $z=-i(2+\sqrt{3})$, and $z=i(\sqrt{3}-2)$. We can again rewrite the integral as:
$$2 \int_c\frac{\frac{1}{(z+(2i+\sqrt{3}i)}}{(z-(-2i+\sqrt{3}i)}dz.$$
At this point I am confused as to how I would actually going about applying the cauchy integration formula... Hopefully I've done my math properly thus far. Any help is greatly appreciated.
I may have figured something out after some more reading,
I think it is the case that I should use $\vert z \vert =1 $ as my disk. Plus I may have the formula worked out, so I think everything becomes
$$2 \int_{\vert z \vert =1}\frac{\frac{1}{(z+(2i+\sqrt{3}i)}}{(z-(-2i+\sqrt{3}i)}dz=4\pi i \cdot \frac{1}{i(\sqrt 3 -2)+i(\sqrt3 + 2)} = \frac{4\pi}{\sqrt 3}$$
Still not positive on this, but seems closer.
Looks like I missed a $2$ in my denominator on the last step... should read:
$$\frac{4\pi}{2\sqrt 3}=\frac{2\pi}{\sqrt 3}$$
I answered the question in my edits of the original question. The last stipulation is that the Cauchy integration formula can be applied in this manner because $$f(z)= \frac{1}{(z+(2i+\sqrt{3}i))},$$ is holomorphic on the unit disc.