Use complex substitution to evaluate integral Use the substitution $z = e^{i\theta}$ to evaluate
$$\int_{0}^{2\pi} \frac{d\theta}{\sin(\theta)-2}$$
Can somebody point me in the right direction? 
 A: Combine the highlights of Mary Star's and Dr. MV's answers.
Substitute $z=e^{i\theta}$:
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}\theta}{\sin(\theta)-2}
&=\oint\overbrace{\frac{2i}{z-\frac1z-4i}}^{\frac1{\sin(\theta)-2}}\overbrace{\ \ \ \ \frac{\mathrm{d}z}{iz}\ \ \ \ }^{\mathrm{d}\theta}\\
&=\oint\frac{2\,\mathrm{d}z}{z^2-4iz-1}\\
&=\oint\frac1{i\sqrt3}\left(\color{#C00000}{\frac1{z-i\left(2+\sqrt3\right)}}-\color{#00A000}{\frac1{z-i\left(2-\sqrt3\right)}}\right)\mathrm{d}z\\
&=\frac{2\pi i}{i\sqrt3}(\color{#C00000}{0}-\color{#00A000}{1})\\[3pt]
&=-\frac{2\pi}{\sqrt3}
\end{align}
$$
where $i\left(2+\sqrt3\right)$ is outside the unit circle and $\frac1{z-i\left(2-\sqrt3\right)}$ has a residue of $1$.
A: Let $z=e^{i\theta}$ so that
$$\begin{align}
\int_0^{2\pi} \frac{1}{\sin \theta -2}\,d\theta &=\oint_{|z|=1}\frac{2}{\left(z-i(2+\sqrt 3)\right)\left(z-i(2-\sqrt 3)\right)}\,dz \tag 1\\\\
&=2\pi i \left(\frac{2}{-i2\sqrt 3)}\right) \tag 2\\\\
&=-\frac{2\pi}{\sqrt 3}
\end{align}$$
where in going from $(1)$ to $(2)$, we invoked the Residue Theorem.
A: Hint: 
$$e^{i\theta}=\cos \theta +i \sin \theta$$ 
$$e^{-i\theta}=\cos \theta -i \sin \theta$$  
So $$\sin \theta=\frac{e^{i\theta}-e^{-i \theta}}{2i}\Rightarrow \sin \theta=i\frac{e^{-i\theta}-e^{i \theta}}{2}=i\frac{z^{-1}-z}{2}$$ 
