Integral $\int_{-\pi}^{\pi}\frac{\operatorname {d}\phi}{\sin\phi+4} $

$\newcommand{\d}{\mathrm{d}}$I have this integral which I consider as challenging.

\begin{equation} \int_{-\pi}^{\pi}\frac{\d\phi}{\sin\phi+4} \end{equation}

I rewrite the denominator using $\sin\phi=\frac{e^{i\phi}-e^{-i\phi}}{2i}$

\begin{equation} \int_{-\pi}^{\pi}\frac{\d\phi}{\frac{e^{i\phi}-e^{-i\phi}}{2i}+4} \end{equation}

So, at this stage it is written with the form for complex numbers. But since I want to solve this using the Residue Theorem, how does one do that on this form?

Thanks

You've got $$\int_{-\pi}^\pi {d\phi\over {(e^{i\phi}+e^{-i\phi})/2i}+4}$$ Now let $z=e^{i\phi}$ then the integral becomes $$\oint {dz\over iz[(z+1/z)/2i+4]}=\oint {dz\over z^2/2+4iz+1/2}$$ Can you proceed now?