Given $\int_0^{2\pi}{\frac{d\theta}{13+12cos\theta}}$

Let,

$\begin{equation} Z = e^{i\theta}\\ \text{diff. w.r.t $\theta$ } \\ dz = ie^{i\theta}d\theta \\ dz = iz d\theta \\ d\theta = \frac{dz}{iz} \\ cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2} = \frac{z+z^{-1}}{2}\\ Cos\theta = \frac{z^2+1}{2z}\\ I= \int_0^{2\pi}{\frac{\frac{dz}{iz} }{13+12(\frac{z^2+1}{2z})}} \\ I= \int_0^{2\pi}{\frac{dz}{(13z+6z^2+6)i}} \\ I= \int_0^{2\pi}{\frac{dz}{(6z^2+13z+6)i}} \\ \text{poles,}\\ 6z^2+13z+6 = 0\\ (3z+2)(2z+3)=0\\ z=-\frac23 \ and \ z=-\frac32 \\ \therefore \text{Pole $z=-\frac23$ lies inside the circle where as the pole $z=-\frac32$ lies outside the circle.} \\ \text{Residue at $z=-\frac23$} \\ \\ \lim\limits_{z \to -\frac23} (z+\frac23) \frac{1}{3(z+\frac23)(2z+3)i} \\ \lim\limits_{z \to -\frac23} \frac{1}{3i(2z+3)} \\ = \frac{1}{3i(2(-\frac23)+3))} \\ = \frac{1}{3i(-\frac43)+3)} \\ = \frac{1}{3i(\frac53))} \\ \therefore (Residue \ at \ z= -\frac23) = \frac{1}{5i}\\ \int_0^{2\pi}{\frac{d\theta}{13+12cos\theta}} = 2\pi i (Residue \ at \ z= -\frac23) \\ = 2\pi i \times \frac{1}{5i} \\ = \frac{2\pi}{5} \\ \end{equation}$

$\begin{align} \\ \int_c{\frac{sin\pi z^2+cos\pi z^2}{(z-1)^2 (z-2)}dz}& = 2\pi i (sum \ of \ Residue )\\ & =2\pi i \text{(Residue at z=2 + Residue at z=1)} \\ & =2\pi i \text{(Residue at z=2 + Residue at z=1)} \\ &=2\pi i[1+1]\\ &=4\pi i \\ \end{align}$