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: 
$\dots$ is not a mathematical operator. If your prof thinks his proof is correct, then he should be able to write it without $\dots$ . When $\dots$ is used correctly, it is simply tedious to replace it with symbolic logic. In the above case, it is impossible to replace it with symbolic logic so the professors proof is wrong.

Many attempts at proofs of induction can be disregarded due to a simple transform: instead of considering the natural numbers $\mathbb N$, consider the set of polynomials with natural number coefficients, $\mathbb N [x]$. $\mathbb N [x]$ shares most properties of natural numbers, such as being injective (wrt adding one) and having a zero. It is also totally well ordered. However, it is not an inductive set. So any "proof" of induction that doesn't make an assumption true of $\mathbb N$ but not $\mathbb N[x]$ is wrong.
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.)