How do I calculate $\int_{0}^{2\pi} \frac{2+4\cos(t)}{5+4\sin(t)} dt$? How do I calculate $$\int_{0}^{2\pi} (2+4\cos(t))/(5+4\sin(t)) dt$$
I've recently started calculating integral via the residue theorem. Somehow I'm stuck with this certain integral. I've substituted t with e^it and received two polynoms but somehow I only get funny solutions.. Could someone please help me finding the residues?
 A: It looks like you are already familiar with the substitution $z=e^{it}$ which allows us to transform this real integral into a contour integral. Note that on the unit circle, $\bar{z} = 1/z$, so we get $\cos t = (z + 1/z)/2$, $\sin t = (z - 1/z)/2i$, and $dt = dz/iz$. Substituting in, we get
$$\int_{0}^{2\pi}\frac{2+4\cos t}{5 + 4\sin t}\,dt = \oint_{|z|=1}\frac{2 + 4(\frac{z + 1/z}{2})}{5 + 4(\frac{z-1/z}{2i})}\,\frac{dz}{iz} = \oint_{|z| = 1}\frac{2(1 + z + z^2)}{2z(z+\frac{i}{2})(z+2i)}\,dz$$
The only singularities that lie within the contour are the simple poles at $z=0$ and $z=-i/2$. 
$$\textrm{Res}(0) = \frac{2}{-2} = -1$$
$$\textrm{Res}\left(-\frac{i}{2}\right) = \frac{\frac{3}{2}-i}{\frac{3}{2}} = 1 - \frac{2i}{3}$$
Taking $2\pi i \sum \textrm{Res}$, our answer is just $4\pi/3$.
A: As mentioned in the comment, the following gives you an alternative way to compute the integral without using complex analysis.
I sketch my idea and let you finish the computation yourself. Please tell me in you need more details.
First, using that $\cos(t+\pi)=-\cos(t)$ and $\sin(t+\pi)=-\sin(t)$, you get
$$ \int_{0}^{2\pi} \frac{2+4\cos(t)}{5+4\sin(t)}dt=\int_{-\pi}^{\pi} \frac{2-4\cos(t)}{5-4\sin(t)}dt.$$
Let $u=\tan(t/2)$. Then
$$ \frac{dt}{du}=\frac{2}{1+u^2}$$
and, by the parametric formulas,
$$\frac{2-4\cos(t)}{5-4\sin(t)}=\frac{2-4\frac{1-u^2}{1+u^2}}{5-4\frac{2u}{1+u^2}}=\frac{6u^2-2}{5u^2-8u+5}.$$
Therefore
$$ \int_{-\pi}^{\pi} \frac{2-4\cos(t)}{5-4\sin(t)}dt=4\int_{-\infty}^{+\infty} \frac{3u^2-1}{(5u^2-8u+5)(u^2-1)}du.$$
You probably know how to continue from here, right?
