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} $