Guidance or advice with $I=\int_0^{2\pi}\frac{1}{4+\cos t}dt$

Question

Let $$ \begin{align} I=\int_0^{2\pi}\frac{1}{4+\cos t}dt \end{align} $$

I would like to evaluate this integral using cauchhy's Integral formula, I understand that I have to convert this into a form like $\int_{\gamma}\frac{f(z)}{z-z_0}dz$ .

I tried using $\cos t=\frac{e^{it}+e^{-it}}{2}$, but I didn't get anywhere. I haven't seen an example like this.

Answer 1

Note that your integral can be rewritten as follows:

$$\color{blue}{ I = \int^{2\pi}_0 \frac{2}{8+ e^{it}+ e^{-it}} \, \mathrm{d}t = \int^{2\pi}_0 \frac{2 e^{it}}{e^{i 2 t} + 8 e^{it} + 1} \, \mathrm{d}t}$$

Define now $z \equiv e^{it}$, $\mathrm{d}z = ie^{it} \, \mathrm{d}t = i z \, \mathrm{d}t $, so:

$$ \color{blue}{I = \frac{2}{i} \int_\gamma \frac{\mathrm{d}z}{z^2 + 8 z + 1}}$$

What should then $\gamma$ be? Can you take it from here?

Cheers!

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Spoiler:

$$ I = \dfrac{2}{i}\times 2\pi i \times \displaystyle\lim_{z\to z_1} \frac{z-z_1}{z^2+8z+1} = \frac{2 \pi}{\sqrt{15}}, \quad z_1 = -4+\sqrt{15}$$

Answer 2

Let $z=e^{it}$, then $dz=ie^{it}\ dt$ or $dt=\dfrac{dz}{iz}$, and $\cos t=\dfrac{e^{it}+e^{-it}}{2}=\dfrac{z+z^{-1}}{2}$. \begin{align} \int_0^{2\pi}\frac{1}{4+\cos t}dt&=\oint_C\frac{1}{4+\frac{z+z^{-1}}{2}}\cdot\frac{dz}{iz}\\ &=\frac2i\oint_C\frac{1}{z^2+8z+1}dz, \end{align} where $C$ is the circle of unit radius with its center at the origin. The poles are $z_1=-4-\sqrt{15}$ and $z_2=\sqrt{15}-4$, but only $z_2$ lies inside $C$. The rest is finding the residue at $z_2$.

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{I\equiv\int_{0}^{2\pi}{1 \over 4 + \cos\pars{t}}\,\dd t :\ {\large ?}}$.

\begin{align} I&=\int_{-\pi}^{\pi}{1 \over 4 - \cos\pars{t}}\,\dd t =2\ \overbrace{\int_{0}^{\pi}{1 \over 4 - \cos\pars{t}}\,\dd t} ^{\ds{\mbox{Set}\ x \equiv \tan\pars{t \over 2}}}\ =\ \\[3mm]&=2\int_{0}^{\infty}{1 \over 4 - \pars{1 - x^{2}}/\pars{1 + x^{2}}}\, {2\,\dd x \over 1 + x^{2}} =4\int_{0}^{\infty}{1 \over 5x^{2} + 3}\,\dd x \\[5mm]&={4 \over 3}\,\root{3 \over 5} \int_{0}^{\infty}{1 \over \pars{\root{5/3}x}^{2} + 1}\,\root{5 \over 3}\dd x ={4 \over 3}\,{\root{15} \over 5}\ \overbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}^{\ds{=\ {\pi \over 2}}} =\color{#66f}{\large{2\root{15} \over 15}\,\pi} \\[3mm]&\approx {\tt 1.6223} \end{align}