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...

Use periodicity to rewrite your integral
$$I=\int_{-\pi}^{\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$$
With the change of variable $t=\tan\frac\theta2$,
$$I=\int_{-\infty}^{+\infty}\frac{2\mathrm dt}{(1+t^2)\left(13-5\dfrac{2t}{1+t^2}\right)}=\int_{-\infty}^{+\infty}\frac{2\mathrm dt}{13t^2-10t+13}$$
Then
$$I=\dfrac2{13}\int_{-\infty}^{+\infty}\frac{\mathrm dt}{\left(t-\dfrac5{13}\right)^2+\left(\dfrac{12}{13}\right)^2}=\dfrac{2}{13}\left(\dfrac{13}{12}\right)^2\int_{-\infty}^{+\infty}\frac{\mathrm dt}{\left(\dfrac{t-\frac5{13}}{\frac{12}{13}}\right)^2+1}$$
Now with the change of variable $u=\dfrac{t-\frac5{13}}{\frac{12}{13}}$,
$$I=\dfrac{2}{13}\left(\dfrac{13}{12}\right)^2\dfrac{12}{13}\int_{-\infty}^{+\infty}\dfrac{\mathrm du}{1+u^2}=\dfrac{2}{13}\left(\dfrac{13}{12}\right)^2\dfrac{12}{13}\pi=\frac{\pi}{6}$$
