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?

  • $\begingroup$ Pointing: $\mathrm{e}^{\mathrm{i}\theta} = \cos \theta + \mathrm{i} \sin \theta$. $\endgroup$ – Eric Towers Dec 17 '15 at 18:17

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 $1$.

  • $\begingroup$ How did you compute that residue? $\endgroup$ – Dr. John A Zoidberg Dec 17 '15 at 20:59
  • $\begingroup$ the residue of $f$ at $z=\alpha$ is the coefficient of $\frac1{z-\alpha}$ in the Laurent expansion of $f$ at $\alpha$. $\endgroup$ – robjohn Dec 18 '15 at 1:09

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.

  • 1
    $\begingroup$ You seem to be missing a minus sign (the integrand is negative everywhere) $\endgroup$ – robjohn Dec 17 '15 at 19:09
  • $\begingroup$ How did you figure the domain would be $|z|=1$? $\endgroup$ – YoTengoUnLCD Dec 17 '15 at 19:26
  • $\begingroup$ @robjon Rob. Thanks for the catch. +1 when one is careless and overlooks the easy things while doind lots of arithmetic in one's head, ... well you get the point. Edited. - Mark $\endgroup$ – Mark Viola Dec 17 '15 at 20:19
  • $\begingroup$ @yotengounlcd For $z=e^{i\theta}$, $|z|=1$ and as $\theta$ goes from $0$ to $2\pi$, the entire unit circle is traced. $\endgroup$ – Mark Viola Dec 17 '15 at 20:21


$$e^{i\theta}=\cos \theta +i \sin \theta$$ $$e^{-i\theta}=\cos \theta -i \sin \theta$$

So $$\sin \theta=\frac{e^{i\theta}-e^{-i \theta}}{2i}\Rightarrow \sin \theta=i\frac{e^{-i\theta}-e^{i \theta}}{2}=i\frac{z^{-1}-z}{2}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.