Evaluate the integral $\displaystyle \int_0^{2\pi} \frac {d\theta}{5 - 3\cos(\theta)}$.

Hint: put $z = e^{i\theta}$.

Is there a way to solve this without using the Residue Theorem and $\tan(z)$? Is Cauchy's integral formula applicable?

  • $\begingroup$ It might be fun to do this by finding a geometric interpretation. It is reminiscent of problems where I've seen that done but I'm not adept at that just now. $\endgroup$ – Michael Hardy Jul 27 '17 at 2:26


with $t=\theta-\pi $, it becomes

$$I=\int_{-\pi}^\pi \frac{dt}{5+3\cos (t)}$$ $$ =2\int_0^\pi\frac {dt}{5+3\cos (t)}$$

now put $$u=\tan \left(\frac {t}{2}\right) .$$


$$\cos (t)=\frac {1-u^2}{1+u^2} $$ $$dt=\frac {2\,du}{1+u^2} $$

You will find $$I=2\int_0^{+\infty}\frac {du}{4+u^2} =\int_0^{+\infty}\frac {dv}{1+v^2}$$ $$=\Bigl [\arctan (v)\Bigr]_0^{+\infty}=\frac {\pi}{2} $$

where $u=2v $.

  • $\begingroup$ This substitution is probably due to Euler, but Stewart's calculus textbooks, and probably others as well, wrongly attribute it to Weierstrass, who was born long after Euler died. $\endgroup$ – Michael Hardy Jul 27 '17 at 0:19
  • 1
    $\begingroup$ I would call this a sketch rather than a hint. Calling a sketch of a solution a "hint" seems to by done somewhat frequently. $\endgroup$ – Michael Hardy Jul 27 '17 at 2:27

Another approach.

If $|b|<1$ then we can write:

$$\frac{1}{1-b\cos x}=\sum_{n=0}^{\infty} b^n\cos^n x$$

The constant term of the Fourier expansion of $\cos^n x$ is zero when $n$ odd and $\frac{1}{2^n}\binom{n}{n/2}$ when $n$ is even.

Now, $\frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{1-b\cos x}$ is the constant of the Fourier series expansion, and, since the series converges absolutely, you get that:

$$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{1-b\cos x}=\sum_{n=0}^{\infty}\left(\frac{b}{2}\right)^{2n}\binom{2n}{n}$$

But when $|z|<\frac{1}{4}$, $$\sum_{n=0}^{\infty} \binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}.$$

So, with $z=b^2/4$, $$\int_{0}^{2\pi}\frac{dx}{1-b\cos x}=2\pi \frac1{\sqrt{1-b^2}}$$

Setting $b=\frac{3}{5}$ we get:

we get:

$$\int_{0}^{2\pi}\frac{dx}{1-\frac{3}{5}\cos x}=\frac{5\pi}{2}$$

And thus your integral is $\frac{\pi}{2}$.

You get the general formula, when $|b|<a$ that:

$$\int_{-\pi}^{\pi}\frac{dx}{a+b\cos x}=\frac{2\pi}{\sqrt{a^2-b^2}}$$


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{2\pi}{\dd\theta \over 5 - 3\cos\pars{\theta}} = \int_{-\pi}^{\pi}{\dd\theta \over 5 + 3\cos\pars{\theta}} = 2\int_{0}^{\pi}{\dd\theta \over 5 + 3\cos\pars{\theta}} \\[5mm] = {} & 2\bracks{\int_{0}^{\pi/2}{\dd\theta \over 5 + 3\cos\pars{\theta}} + \int_{\pi/2}^{\pi}\!\!{\dd\theta \over 5 + 3\cos\pars{\theta}}} = 2\bracks{\int_{0}^{\pi/2}\!\!{\dd\theta \over 5 + 3\cos\pars{\theta}} + \int_{-\pi/2}^{0}{\dd\theta \over 5 - 3\cos\pars{\theta}}} \\[5mm] = {} & 20\int_{0}^{\pi/2}{\dd\theta \over 25 - 9\cos^{2}\pars{\theta}} = 20\int_{0}^{\pi/2}{\sec^{2}\pars{\theta} \over 25\sec^{2}\pars{\theta} - 9} \,\dd\theta = 20\int_{0}^{\pi/2}{\sec^{2}\pars{\theta} \over 25\tan^{2}\pars{\theta} + 16} \,\dd\theta \\[5mm] = {} & 20\,{1 \over 16}\,{4 \over 5}\int_{0}^{\pi/2}{% \bracks{5\sec^{2}\pars{\theta}/4}\dd\theta \over \bracks{5\tan\pars{\theta}/4}^{\,2} + 1} \,\,\,\stackrel{t\ =\ 5\tan\pars{\theta}/4}{=}\,\,\, \int_{0}^{\infty}{\dd t \over t^{2} + 1} = \bbx{\pi \over 2} \end{align}


Using your hint the integral is $2i\int_{S^1}\frac{dz}{3z^2-10z+3}=\frac{i}{4}\int_{S^1}\frac{dz}{z-3}+\frac{3i}{4}\int_{S^1}\frac{dz}{3z-1}$

Now you can use Cauchy's formula on each term.

[Although I don't know why you don't like the residue theorem. This and Cauchy;s formula are pretty much the same stuff.]

  • $\begingroup$ What would it look like by using the Residue Theorem? $\endgroup$ – Hwi Moon Jul 27 '17 at 0:24

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