Question

Q: Evaluate the integral using contour integration . Full Solution please ​

Answer 1

Consider,

${\displaystyle I =\int_0^{2\pi}\dfrac{\cos(2\theta)}{5 + 4\cos(\theta)}\ d\theta = \int_0^{2\pi}\dfrac{\cos^2(\theta) - 1}{5 + 4\cos(\theta)}\ d\theta}$

${\text{Let z = e^{i\theta} so that dz = ie^{i\theta} d\theta \implies \dfrac{dz}{iz} = {d\theta} }}$

Thus we have,

${\displaystyle \int_0^{2\pi}\dfrac{\cos^2(\theta) - 1}{5 + 4\cos(\theta)}\ d\theta = \oint_C\dfrac{\left\{\frac12\left(z + \frac1z\right)\right\}^2 - 1}{5 + 4\left\{\frac12\left(z + \frac1z\right)\right\} }\ \dfrac{dz}{iz} where C: |z| = 1}$

Simplifying the above contour,

${\displaystyle\oint_C\dfrac{\left\{\frac12\left(z + \frac1z\right)\right\}^2 - 1}{5 + 4\left\{\frac12\left(z + \frac1z\right)\right\} }\ \dfrac{dz}{iz} }$

$\displaystyle=\oint_C\dfrac{\frac14\left(z^2 + \frac1{z^2} + 2\right) - 1}{5 +2\left(z + \frac1z\right) }\ \dfrac{dz}{iz}$

$\displaystyle= \boxed{\oint_C\dfrac{1 + z^4}{2i z^2(2z^2 + 5z + 2)}dz}$

Let's assume that,

$F(z) = \dfrac{1 + z^4}{2i z^2(2z^2 + 5z + 2)}$

It's poles are given by,

$2i z^2(2z^2 + 5z + 2) = 0 \implies z = 0, 0, -2 , \dfrac{-1}{2}$

It has simple poles at $z = -\dfrac12, -2$ and pole of order $2$ at $z=0$ of which $z = -\dfrac12$ and $z = 0$ lies inside $C$.

If z = a is a pole of order m, then

$\boxed{\displaystyle\text{Res _{z = m}f(x) =\dfrac{1}{(m-1)!}\lim_{z\to a}\left\{\dfrac{d^{m-1}}{dz^{m-1}} (z-a)^m \cdot f(z)\right\} }}$

For z = 0, residue is: $\dfrac{-5}{8i}$

For z = -1/2, residue is: $\dfrac{17}{24i}$

Using cauchy's theorem we have:
$\displaystyle \oint_C\dfrac{1 + z^4}{2i z^2(2z^2 + 5z + 2)}dz =2\pi i (\sum Re F(z)) = 2\pi i \left(\dfrac{17}{24i} - \dfrac{5}{8i} \right) = \dfrac{\pi}{6}$

Thus the given integral simplifies to pi/6.