Evaluate $ \int_{0}^{2\pi} \frac{1}{12+5 \cos(t)}dt$ without residues

Question

So I have to evaluate the next integral, without using the residue theorem, but writing the next integral as an integral of a complex function over the unit circle (see substitution below):

$ \int_{0}^{2\pi} \frac{1}{12+5 \cos(t)}dt$

I know what the answer is going to be: $\frac{2\pi}{\sqrt{119}}$, but I get the following:

substituting $z=e^{it}$, $dt = \frac{dz}{iz}$ and working out the denominator:

$$-2i \oint_{|z|=1} \frac{1}{5z^2+24z+5} \,dz$$

Partial fractions:

$$\frac{-10i}{2\sqrt{119}} \oint_{|z|=1} \frac{-1}{5z+\sqrt{119}+12} -\frac{1}{-5z+\sqrt{119}-12}\,dz$$

Now by the Cauchy Riemann Theorem, I get: $\frac{-10i}{2\sqrt{119}} (\frac{-2\pi i}{5} - \frac{-2\pi i}{5}) = 0$ since here (https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula) $f(z)=1$ so the integral equals $\frac{1}{5}*f(a)*2\pi i = \frac{2\pi i}{5}$.

I thought maybe I confused some - signs, but I ran over my solutions multiple times. Perhaps someone could help me out :)

Answer 1

We first replace the integral operator with $2\int_0^\pi dt$, based on the $t\mapsto 2\pi-t$ symmetry. On this integration range, $x:=\tan\frac{t}{2}$ is strictly increasing. Since $dt=\frac{2dx}{1+x^2}$ and $\cos t=\frac{1-x^2}{1+x^2}$, the integral is $4\int_0^\infty\frac{dx}{17+7x^2}=\frac{4}{7}\frac{\pi}{2}\sqrt{\frac{7}{17}}=\frac{2\pi}{\sqrt{119}}$.

Answer 2

In the Cauchy Riemann Theorem, a pole makes a contribution only when the contour wraps around it so that the pole is in the interior. So: Locate the poles in both integrands. One has the pole inside the contour, and this pole contributes to the integral value. But the other has its only pole outside the contour, leaving none inside; so it gives zero. There is only one nonzero contribution and nothing to cancel it.