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...
 A: 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 $u=\dfrac{t-\frac5{13}}{\frac{12}{13}}$,
$$I=\dfrac{2}{13}\left(\dfrac{13}{12}\right)^2\dfrac{12}{13}\int_{-\infty}^{+\infty}\dfrac{\mathrm du}{1+u^2}=\dfrac{2}{13}\left(\dfrac{13}{12}\right)^2\dfrac{12}{13}\pi=\frac{\pi}{6}$$
A: 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) $\dfrac{\pi}{6}$
A: I will show you where your mistake was, so that you may use your preferred technique in evaluating the integral. You also posted your question in multiple choice form, and I will show you a way to find the answer without solving the integral exactly.

You went wrong at the 4th equality sign in your solution when you wrote 

$I = 2\int_C \frac{\mathrm dz}{(z-5i)(z-i/5)} $ (incorrect)

I believe the best thing to do in such problems involving polynomials is to factor out the coefficient on $z^2$ or whichever power of $z$ is the largest. In doing that we see that from your third equality sign, on explicitly pulling out the -5,
$$I = -\frac{2}{5}\int_C\frac{\mathrm dz}{z^2-\frac{26}{5}iz-1}\hspace{0.5in}$$
The roots of the denominator are what you found in your OP, so you were correct in your use of the quadratic formula. However, now we have the correct factor of $-\frac{1}{5}$ that saves the day, since the rest of your reasoning is sound! That factor changes your incorrect answer of $-\frac{5\pi}{6}$ to the correct answer d) of $\frac{\pi}{6}$.
I think the simplest way to avoid this mistake is to explicitly factor out the coefficient of $z^2$ from your denominator. 

As per solving the multiple choice question without explicit evaluation:
First, note that the integrand is smallest when the denominator is largest, and the denominator is largest when $\sin{\theta}$ is minimized. Since $\sin{\theta}$ is minimized with a value of -1 which makes the denominator 18, we see  $\frac{1}{13-5\sin{\theta}} \geq \frac{1}{18}$. Then that means that $ I = \int_0^{2\pi}\frac{d\theta}{13-5\sin{\theta}} > \int_0^{2\pi} \frac{1}{18}.$ That is, $I > \frac{\pi}{9} $! This is what LordSharktheUnknown said in the comment to the OP.
Thus you see that only d) can be the correct answer without having to go to the pain of integrating it explicitly! This second method will make your problem solving faster in a timed environment. Good luck with similar questions.
