Evaluation of Real-Valued Integrals (Complex Analysis) How to get calculate the integration of follwing:
$$\int_{0}^{2\pi} \frac{dt}{a + cos t} (a>1)$$
My attempt:
let, $z=e^{it}$
$\implies dt = \frac{dz}{it}$
and 
$$\cos t = \frac{z + \frac{1}{z}}{2}$$
On substituting everything in the integral I got:
$$\frac{2}{i}\int_{c} \frac{dz}{z^2+2az+1}$$
Now how do I decompose this fraction so that I can use the Residue Theorem? Or is there anyother way to solve this??
Thanks for the help.

By way of contrast, since the factorisation has been shown by others, we can proceed like this: the roots of the denominator $z^2+2az + 1$ are distinct
$$
z_{\pm} = -a \pm \sqrt{a^2-1}
$$ 
and only one lies within $C$, i.e. $z_+$; one can use the formula for the residue in a simple pole: since $(z^2+2az+1)' = 2z+2a$,
$$
\text{Res}\frac{2}{i(z^2+2az+1)}\Big|_{z=z_+} = \frac{2}{i(2z+2a)}\Big|_{z=z_+}=
\frac{1}{i(a-a+\sqrt{a^2-1})}=\frac{1}{i}\frac{1}{\sqrt{a^2-1}}.
$$
Hence finally
$$
\int_{0}^{2\pi}\frac{dt}{a+\cos t} = i2\pi \frac{1}{i}\frac{1}{\sqrt{a^2-1}} = \frac{2\pi}{\sqrt{a^2-1}}.
$$
