Question: 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.

Answer: 
The problem lies in his statement "go down a finite number of times". This statement is informal and does not in it self show anything. In order to use this statement and expound on it he would actually have to show that he may do this for $k=1$, $k=2$, $k=3$, etc. or if he do not wish do do an infinite algorithm, he may use induction.
After a lot of thought and back and forth in the comments, I decided that your professor is probably right.

I'm assuming pre-induction we are allowed to do any number of finite iterations:

So if:

1)True for $n =1$

2)If true for $n$ then true for $n +1$

Then we can conclude:

3) True for $k$ after $k - 1$ iterations.

But that's all. We can't conclude that it is true for all $n$ simply because we can show it is true for arbitrary $n$. arbitrary $\not \implies$ every....yet. 

If, however, we assume 2) and 1) and that it is not true for a finite $k$ then we can show it is not true for $k - 1$.

We can re-iterate this $k -1$ times and conclude it is not true for $n = 1$ which is a contradiction. So there is no $k$ that this is not true for. Hence it is true for all $n$.

The subtle but crucial difference between these two methods, is that $k$ was not an arbitrary number, and the number of iterations to get to $n = 1$ was finite, precise and determined.

And it only takes one counter example to disprove a universal statement, but it takes an infinite multitude to prove it directly. (Unless we can prove arbitrary $\implies$ every which I think we can... maybe.)