Complex integral 1: $\int_{0}^{2\pi} \frac{\cos^2t}{13+12\cos t}dt$ $$\int_{0}^{2\pi} \dfrac{\cos^2t}{13+12\cos t}dt$$
$$\int_{0}^{2\pi} \dfrac{\cos2t+1}{13+12\cos t}dt$$
$$\int_{0}^{2\pi} \dfrac{\cos2t}{13+12\cos t}dt+\int_{0}^{2\pi} \dfrac{1}{13+12\cos t}dt$$
Of course, $\int_{0}^{2\pi} \dfrac{1}{13+12\cos t}dt$ is rather easy I suppose to evaluate. You can take $\tan t/2 = y$ and so on.
let $I = \int_{0}^{2\pi} \dfrac{\cos2t}{13+12\cos t}dt$ and $J = \int_{0}^{2\pi} \dfrac{\sin2t}{13+12\cos t}dt$
For the other one, one ideea is to take $I+iJ = \dfrac{e^{i2t}}{13+12\cos t}dt$
But what about $I-iJ$? Is this approach good? I got stuck here...
 A: HINT:
Use Partial Fraction Decomposition to write $$\dfrac{\cos^2t}{13+12\cos t}=A\cos t+B+\dfrac C{13+12\cos t}$$
Use Weierstrass Substitution for the last part
A: We have $\cos(t) = \frac{e^{it}+e^{-it}}{2}$, hence by setting $e^{it}=z$ we have
$$ \int_{0}^{2\pi}\frac{\cos^2(t)}{13+12\cos t}\,dt = \oint_{|z|=1}\frac{-i(1+z^2)^2}{4z^2(3+2z)(2+3z)}\,dz$$
and by setting $f(z)=\frac{(1+z^2)^2}{4z^2(3+2z)(2+3z)}$ we have that $f(z)$ is a meromorphic function and its poles inside the unit disk lie at $0$ and $-\frac{2}{3}$. The residue at $0$ equals $-\frac{13}{144}$ and the residue at $-\frac{2}{3}$ equals $\frac{169}{720}$, hence the original integral equals
$$ 2\pi\left(-\frac{13}{144}+\frac{169}{720}\right)=\color{red}{\frac{13}{45}\,\pi}.$$

There is an efficient real counterpart. By symmetry the original integral equals
$$ 26\int_{0}^{\pi}\frac{\cos^2(t)}{13^2-12^2\cos^2 t}\,dt = 52\int_{0}^{\pi/2}\frac{\cos^2(t)}{13^2-12^2\cos^2 t}\,dt$$
or, by substituting $t=\arctan u$,
$$ 52\int_{0}^{+\infty}\frac{du}{(1+u^2)(25+169u^2)} $$
that is straightforward to compute by partial fraction decomposition.
