Evaluating definite integral $\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$ Question: $$\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$$ is equals to
(a) $-\frac\pi6$
(b) $-\frac{\pi}{12}$
(c) $\frac\pi{12}$
(d) $\frac\pi6$
My attempt: Denoting given integral by $I$ and letting $z=e^{iθ}$ then given integral becomes, 
\begin{align*}
I&=\int_C\frac{1}{13-5(\frac{z-\bar{z}}{2i})}\frac{\mathrm dz}{iz}\\
&=\frac{1}{i}\int_C\frac{2i}{26iz-5z^2+5|z|^2}\mathrm dz\\
&=2\int_C\frac{\mathrm dz}{-5z^2+26iz+5}\hspace{0.5in}\text{As }C: |z|=1\\
&=2\int_C \frac{\mathrm dz}{(z-5i)(z-i/5)}\\
&=2\left(\frac{5}{24i}\int_C\frac{1}{z-5i}-\frac{5}{24i}\int_C \frac{1}{z-i/5}\right)
\end{align*}
Now as point $z=5i$ lies outside $C$ so it's integral evaluates to $0$ and by Cauchy integral formula, above becomes, 
$$I=0-2\frac{5}{24i}2\pi i = -\frac{5\pi}{6}$$
But none of the given answer matches with mine. So is am i incorrect? Please help me..stuck on this from hours...

COMMENT.-An unorthodox way. The function is continuous and its minimum and maximum are taken in $\dfrac{3\pi}{2}$ and $\dfrac{\pi}{2}$ respectively. Then we have (using areas of rectangles)
$$2\pi\cdot0.0555\approx\dfrac{\pi}{10}\lt\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta\lt 2\pi\cdot\frac 18=\frac{\pi}{4} $$
Taking into account that the answer is given to select between four possibilities, the solution is clearly now d) $\dfrac{\pi}{6}$
