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

$I = ∮ \frac{1}{(5+3 \frac{z^2-1}{2iz})} \frac{dz}{iz} = ∮ \frac{2}{(3z^2+10iz-3)} \ dz$

=$∮ \frac{2}{(3z+i)(z+3i)} \ dz$ , 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, out of these poles z = $\frac{-i}{3}$ lies inside & z = - 3i lies outside .

$R_1 = lim_{z→ (\frac{-i}{3})} \,\ ⁡(z+\frac{i}{3}) \frac{2}{(3z+i)(z+3i)} = \frac{1}{4i}$, by residue theorem value of the integral

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