Evaluate $\int ^{2\pi}_{0} \frac{1}{5+3sin \theta}d\theta$

Solution :

let, $z = e^{iθ}$

$dz = ie^{iθ} \ dθ$ ,

$dθ=\frac{dz}{iz}$

$sin\theta=\frac{(z^2-1)}{ 2iz}$

since θ varies from 0 to 2π,

z moves around a unit circle |z|=1 ,

Putting values in the above integral

$\begin {aligned} I &= \oint_C \frac{1}{(5+3( \frac{z^2-1}{2iz})} \frac{dz}{iz} \\ &=\oint_C \frac{2}{(3z^2+10iz-3)} \\ &=\oint_C \frac{2}{(3z+i)(z+3i)} dz \end{aligned}$

where C is the circle |z|=1

Now poles are given by $$ (3z+i)(z+3i) =0 $$

i.e. z = $\frac{-i}{3}$ & - 3i both are poles of order 1,

Pole $z = \frac{-i}{3}$ lies inside & $z = - 3i $ lies outside .

Applying residue at $z = \frac{-i}{3}$

$\begin{aligned} &= lim_{z\rightarrow \frac{-i}{3}} ⁡(z+\frac{i}{3}) \frac{2}{(3z+i)(z+3i)} \\&= lim_{z\rightarrow \frac{-i}{3}} \frac{2(3z+i)}{3(3z+i)(z+3i)} \\ &= lim_{z\rightarrow \frac{-i}{3}} \frac{2}{3(z+3i)} \\ &=\frac{1}{4i}\end{aligned}$

By residue theorem value of the integral

∴ $I = 2πi \times \frac{1}{4i} = \frac{π}{2}$