**1 Answer**

written 4.3 years ago by |

Consider $ ∫_0^{2π} \frac{cos2θ}{5+4 cosθ} \ dθ =∫_0^{2π} \frac{e^{2iθ}}{5+4 cosθ} \ dθ$

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 therefore z moves around a unit circle |z|=1 , putting above values in the given integral

$I = ∮ \frac{z^2}{5+4(\frac{z^2+1}{2z})} \frac{dz}{iz} = ∮ \frac{z^2}{i(2z^2+5z+2)} \ dz$

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

Now poles are given by (z+2)(2z+1) =0 i.e. z = $\frac{-1}{2}$ & - 2, both are poles of order 1, out of these poles z = $\frac{-1}{2}$ lies inside & z = - 2 lies outside .

$R_1 = lim_{z→ (\frac{-1}{2})} \ (z+\frac{1}{2}) \frac{z^2}{i(z+2)(2z+1)} = \frac{1}{12i} $

By residue theorem value of the integral

∴ $I = 2πi \frac{1}{12i}= \frac{π}{6}$