**1 Answer**

written 3.8 years ago by |

Consider the integral $∫_0^{2π} \frac{dθ}{(3+2 cosθ)} $

To evaluate above integral, let $z = e^{iθ}$

∴ $dz = ie^{iθ} \ dθ$ , $dθ= \frac{dz}{iz}$ & $cosθ=\frac{(z^2+1)}{2z}$ since θ varies from 0 to 2π, z moves around a unit circle |z|=1 , putting above values in the given integral.

$I = ∮ \frac{1}{3+2(\frac{z^2+1}{2z})} \frac{dz}{iz} = \frac{1}{i} ∮ \frac{1}{(z^2+3z+1)} \ dz $, where C is the circle |z|=1

Now poles are given by $z^2+3z+1=0$ ∴ $z = \frac{-3± \sqrt{(3^2-4)}}{2} = \frac{-3±\sqrt{5}}{2}$

Let $α = \frac{-3+ \sqrt{5}}{2}$ and $β = \frac{-3- \sqrt{5}}{2}$, out of these poles α lies within unit circle & β lies outside it.

Residue of f(z) (at z= α)

$R = lim_{z→α} \ (z-α) \frac{1}{(z-α)(z-β )} \frac{1}{i} = \frac{1}{(α-β )} \frac{1}{i}$

Now, $α-β = \frac{-3+\sqrt{5}}{2} - \frac{-3- \sqrt{5}}{2} = \sqrt{5}$

∴ Residue of $f(z) (at \ z= α) = \frac{1}{\sqrt{5}} \frac{1}{i}$, by residue theorem value of the integral

∴ $I = 2πi \frac{1}{\sqrt{5}} \frac{1}{i} = \frac{2π}{\sqrt{5}}$

$∫_0^π \frac{dθ}{(3+2 cosθ)} = \frac{1}{2} ∫_0^{2π} \frac{dθ}{(3+2 cosθ)}$ $=\frac{1}{2}. \frac{2π}{\sqrt{5}} = \frac{π}{\sqrt{5}}$