If $\gamma$ is unit circles $A(0,1)$ parameterization of one positive rotation and $a\in\mathbb{R}$, $0<a<1$. Show that $$ \int\limits_0^{2\pi} \frac{dt}{1+a^2-2a \cos t}=\oint\limits_{\gamma} \frac{i}{(z-a)(az-1) }dz $$ and calculate the integral.
Do I parametrize the contour integral with $z(t)=e^{it}$ and $z'(t)=ie^{it}$ when $t\in[0,2\pi]$.
Also $\cos t=1/2 (e^{it}+e^{-it}) $ and $\sin t =1/2i (e^{it}-e^{-it})$
So that.. \begin{align} \oint\limits_{\gamma} \frac{i}{(z-a)(az-1) }dz = \int_0^{2\pi}\frac{i}{(e^{it}-a)(ae^{it}-1)}dt\\ \end{align}
Is this the best to way tackle this problem?
Following your way if we substitute the following expression $$\cos(t)=\frac{e^{it}+e^{-it}}{2}$$ in the integrand will yields $$\int^{2\pi}_{0}\frac{dt}{1+a^2-2a\cos(t)}=\int^{2\pi}_{0}\frac{dt}{1+a^2-2a(\frac{e^{it}+e^{-it}}{2})}=\int^{2\pi}_{0}\frac{dt}{1+a^2-a(e^{it}+e^{-it})}$$ The above integral is equivalent to $$\int^{2\pi}_{0}\frac{dt}{1+a^2-a(e^{it}+e^{-it})}=\int^{2\pi}_{0}\frac{dt}{1+a^2-a(e^{it}+e^{-it})}=\int^{2\pi}_{0}\frac{dt}{(ae^{it}-1)(ae^{-it}-1)}$$ Now define $z=e^{it}$ then $dz=ie^{it}dt$ so the integral can be rewriten as follows $$\int^{2\pi}_{0}\frac{dt}{(ae^{it}-1)(ae^{-it}-1)}=\int^{2\pi}_{0}\frac{e^{it}dt}{(ae^{it}-1)(a-e^{it})}=\oint_{|z|=1}\frac{-idz}{(az-1)(a-z)}=\oint_{|z|=1}\frac{idz}{(az-1)(z-a)}$$ Now since $0<a<1$ then the integrand has only a simple pole at $z_0=a$ with residue $\frac{i}{a^2-1}$. Using Residue theorem due to Cauchy yields $$\oint_{|z|=1}\frac{idz}{(az-1)(z-a)}=2\pi i\cdot \frac{i}{a^2-1}=\frac{2\pi}{1-a^2} $$