Contour integration 
Consider the real-valued function
  $$u(t) = \frac{1}{13-12\cos(t)}$$
  By converting it to a contour integral along the unit circle in $\mathbb{C}$, evaluate
  $$\int_0^{2\pi} u(t)\;dt$$

I have attempted so solve this and have found the residue to be $-\frac{1}{5}$ and therefore the integral to be $$\frac{2\pi}{5}$$
could someone confirm if this is correct?

It looks like your work is correct.

I get,
$$ \int_0^{2\pi} u(t)\; dt = \int_0^{2\pi} \frac{ie^{it}}{ie^{it} \left(13-12\frac{e^{it} + e^{-it}}{2}\right)}\; dt$$
$$ = \frac{1}{i} \oint_C \frac{dz}{-6z^2+13z-6},$$
where $C$ is the unit circle.  From the quadratic equation, I find that $-6z^2+13z-6 = 0$ for $z = \frac{3}{2}, \frac{2}{3}$, so by the Residue Theorem, 
$$ \int_0^{2\pi} u(t)\; dt = \frac{1}{i} 2\pi i \operatorname{Res}\left(\frac{1}{-6z^2+13z-6}, z=\frac{2}{3}\right).$$
Since $\frac{1}{-6z^2+13z-6}$ has a simple pole at $z=\frac{2}{3}$, we have,
$$\operatorname{Res}\left(\frac{1}{-6z^2+13z-6}, z=\frac{2}{3}\right) = \lim_{z\to\frac{2}{3}} \left(z-\frac{2}{3}\right) \frac{1}{-6z^2+13z-6} = \lim_{z\to\frac{2}{3}} \frac{1}{-6(z-\frac{3}{2})} = \frac{1}{5}$$
And so putting it all together,
$$ \int_0^{2\pi} u(t)\; dt = \frac{2\pi}{5}$$
as you also calculated.
