Evaluate the integral $\int\limits_0^{2\pi} \frac {d\theta}{5 - 3\cos(\theta)}$. 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? 
 A: hint
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 $$u=\tan \left(\frac {t}{2}\right) .$$
using
$$\cos (t)=\frac {1-u^2}{1+u^2} $$
$$dt=\frac {2\,du}{1+u^2} $$
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 $.
A: 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:
$$\int_{-\pi}^{\pi}\frac{dx}{a+b\cos x}=\frac{2\pi}{\sqrt{a^2-b^2}}$$
A: 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.]
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
