Evaluate the integral $\int_0^{2 \pi} {\cos^2 \theta \over a + b \cos \theta}\; d\theta$ Given $a > b > 0$ what is the fastest possible way to evaluate the following integral using Residue theorem. I'm confused weather to take the imaginary part of $z^2$ or whole integral.
$$\int_0^{2 \pi} {\cos^2 \theta \over a + b \cos \theta}\; d\theta$$
 A: If $z=e^{i\theta}$, then $\cos\theta=\dfrac 1 2\left(z+\dfrac1z\right)$, and $dz=ie^{i\theta}\,d\theta/2=iz\,d\theta/2$, so $-2i\dfrac{dz}{z} = d\theta$.  Then
$$
\int_0^{2\pi} \frac{\cos^2\theta}{a+b\cos\theta} d\theta = \int\limits_\text{circle} \frac{\frac14\left(z+\frac1z\right)^2}{a+\frac b2\left(z+\frac1z\right)}(-i)\frac{dz}{z} = -i\int\limits_\text{circle} \frac{z^4+2z^2+1}{2z^2(2az+b(z^2+1))} dz.
$$
This function has a double pole at $z=0$ and simple poles at $\dfrac{-a\pm\sqrt{a^2-b^2}}{b}$.  So the question is: for which values of $a,b$ are the simple poles inside the circle?  If there's just one simple pole inside the circle at $c$, then the integral becomes
$$
\int\limits_\text{circle} \frac{g(z)}{z-c} dz = 2\pi i g(c),
$$
where $g(z)$ is whatever's left after you've factored out $1/(z-c)$.  If there's more than one simple pole, you need a sum: take values of $g$ at those points and sum them.
A: We should have 
$$\frac{\cos^{2}\theta}{a+b\cos\theta}=\frac{1}{b}\cos\theta-\frac{\frac{a}{b}\cos\theta}{a+b\cos\theta}.$$
The first part of the definite integral is easily evaluated. The second part is the same as evaluating 
$$\int \frac{\cos\theta}{a+b\cos\theta}d\theta=\int\frac{1}{b}\left(1-\frac{a}{a+b\cos\theta}\right)d\theta.$$
So we only need to evaluate 
$$\int\frac{1}{a+b\cos\theta}d\theta.$$ 
This can be done by various trignometry identities such as using $\tan(\theta/2)$. A detailed step by step proof can be found here (click show steps). 
A: Hint: put $z={\rm e}^{i\theta}$ and change the integral to the form $ \int_{|z|=1}f(z)\,dz \,,$ then use residue theorem. Exploit the identity 
$$ \cos\theta = \frac{1}{2}\left({\rm e}^{i \theta} + {\rm e}^{- i \theta}\right)\,. $$
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$\ds{\int_{0}^{2\pi}{\cos^{2}\pars{\theta} \over a + b\cos\pars{\theta}}\,
     \dd\theta:\ {\large ?}.\qquad\qquad a > b > 0}$.

\begin{align}&\color{#c00000}{%
\int_{0}^{2\pi}{\cos^{2}\pars{\theta} \over a + b\cos\pars{\theta}}\,\dd\theta}
=2\int_{0}^{\pi}{\cos^{2}\pars{\theta} \over a - b\cos\pars{\theta}}\,
\dd\theta
=-\,{2 \over b}
\int_{0}^{\pi}{\cos^{2}\pars{\theta} \over \cos\pars{\theta} - \mu}\,\dd\theta
\\[3mm]&\mbox{where}\quad\mu \equiv {a \over b}\,,\qquad  \mu > 1
\end{align}

\begin{align}&
\color{#c00000}{\int_{0}^{2\pi}{\cos^{2}\pars{\theta} \over a + b\cos\pars{\theta}}\,\dd\theta}
=-\,{2 \over b}\int_{0}^{\pi}{%
\bracks{\cos\pars{\theta} - \mu}\bracks{\cos\pars{\theta} + \mu}  + \mu^{2}
\over \cos\pars{\theta} - \mu}\,\dd\theta
\\[3mm]&=-\,{2 \over b}\int_{0}^{\pi}\bracks{\cos\pars{\theta} + \mu}\,\dd\theta
-{2 \over b}\,\mu^{2}\int_{0}^{\pi}{\dd\theta \over \cos\pars{\theta} - \mu}
\end{align}

\begin{align}&
\color{#c00000}{\int_{0}^{2\pi}{\cos^{2}\pars{\theta} \over a + b\cos\pars{\theta}}\,\dd\theta}
=-\,{2a \over b^{2}}\,\pi - {2a^{2} \over b^{3}}
\color{#00f}{\int_{0}^{\pi}{\dd\theta \over \cos\pars{\theta} - \mu}}\tag{1}
\end{align}

With the usual
Weierstrass Tangent Half-Angle substitution:
\begin{align}&\color{#00f}{\int_{0}^{\pi}{\dd\theta \over \cos\pars{\theta} - \mu}}
=\int_{0}^{\infty}{2\,\dd t/\pars{1 + t^{2}}\over
\pars{1 - t^{2}}/\pars{1 + t^{2}} - \mu}
=-2\int_{0}^{\infty}{\dd t \over \pars{\mu + 1}t^{2} + \mu - 1}
\\[3mm]&=-2\,{1 \over \mu - 1}\,\root{\mu - 1 \over \mu + 1}\
\overbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}}^{\ds{=\ {\pi \over 2}}}
=-\,{\pi \over \root{\mu^{2} - 1}}\ =\
=-\,{b \over \root{a^{2} - b^{2}}}\,\pi
\end{align}

Replacing in $\pars{1}$:
  $$
\color{#66f}{\large\int_{0}^{2\pi}{\cos^{2}\pars{\theta} \over a + b\cos\pars{\theta}}\,\dd\theta
={2a \over b^{2}}\pars{{a \over \root{a^{2} - b^{2}}} - 1}\pi}\,,\qquad a > b > 0
$$

