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Question: Using Residue theorm evaluate $\int_{0}^{2 \pi} \frac{d \theta}{5+3sin \theta}$
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Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 212 views
 modified 4 months ago  • written 7 months ago by
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Put z = e$^{i \theta}$

dz = i e$^{i \theta}$ d$\theta$

$d \theta = \frac{dz}{iz}$

Therefore, $sin\theta = \frac{z-1/z}{2i} = \frac{z^2-1}{2iz}$

The integration becomes,

$=\int_c \frac{1}{5+3(z^2-1)/(2iz)} \,\, \frac{dz}{iz} \\ = \int_c \frac{2iz}{10iz + 3z^2 - 3} \frac{dz}{iz} \\ = \int_c \frac{2}{3z^2 + 10iz - 3} \,\, dz$

Consider f(z) = $\frac{2}{3z^2 + 10iz - 3}$

For poles,

$3z^2 + 10iz - 3 = 0 \\ 3z^2 + 9iz + iz - 3 = 0 \\ 3z(z+3i) + i(z+3i) = 0 \\ \therefore 3z + i = 0; z+ 3i = 0$

Therefore, $z = \frac{-i}{3}; z = -3i$ are poles of which $z = \frac{-i}{3}$ is the only pole which lies inside |z| = 1

Therefore, residue of f(z) at $z = \frac{-i}{3}$ is

$lim_{z \to (\frac{-i}{3})} (z + \frac{i}{3})[\frac{2}{(3z+i)(z+3i)}] \\ = lim_{z \to (\frac{-i}{3})} (z + \frac{i}{3})[\frac{2}{3(z+i/3)(z+3i)}] \\ = \frac{2}{3[-i/3 + 3i]} = \frac{2}{3(8i/3)} = \frac{2}{8i} = \frac{1}{4i}$

By Cauchy's Residue theorem,

$\int_c \frac{2}{3z^2 + 10iz - 3} \,\, dz = 2 \pi i(\frac{1}{4 i}) = \frac{\pi}{2}$

$\therefore \int_0^{2\pi} \frac{d\theta}{5 + 3 sin\theta} = \frac{\pi}{2}$