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Using Residue theorm evaluate $\int_{0}^{2 \pi} \frac{d \theta}{5+4cos \theta}$

Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

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Let $I = \int_{0}^{2 \pi} \frac{d \theta}{5+4cos \theta}$

We put, $z = e^{iθ}$

Then, $cos\ θ = \frac{1}{2}(z+\frac{1}{z}), d\theta = \frac{dz}{iz}$

Therefore,

$I = \int_{0}^{2 \pi} \frac{1}{5+4cos \theta} dz$

$I = \int_{|z|=1} \frac{1}{5+\frac{4}{2}(z+\frac{1}{z})} . \frac{dz}{iz}$

$I = \int_{|z|=1} \frac{1}{5+2(z+\frac{1}{z})} . \frac{dz}{iz}$

$I = \frac{1}{2i}\int_{|z|=1} \frac{1}{5z+2z^2+ 2}.dz$

Now, Poles are determined by $5z+2z^2+ 2 = 0$

That implies, (2z+1) (z+2) = 0

Or, Z = -1/2, -2

So inside the contour C there is a simple pole at z = $\frac{-1}{2}.$

Residue at the simple pole $(z = \frac{-1}{2})$

i.e. $\implies lim_{-1/2}(z + \frac{-1}{2}) \frac{1}{(2z+1) (z+2)}.dz$

$\implies lim_{-1/2}(z + \frac{-1}{2}) \frac{1}{2 (z+2)} = \frac{1}{2*(3/2)} = \frac{1}{3}$

$\therefore\ Residue\ at\ \int_{0}^{2 \pi} \frac{d \theta}{5+4cos \theta} = \pi \frac{1}{3} = \frac{\pi}{3}.$