Question: This question is about:

$$\int\_0^{2\pi} \frac{1}{1.1+\cos x}$$

The solution is $20\pi/\sqrt{21}$.

But when using the antiderivative,given [here](http://www.symbolab.com/solver/step-by-step/%20%5Cfrac%7B20%5Carctan(%5Cfrac%7B%5Ctan(%5Cfrac%7B2%5Cpi%7D%7B2%7D)%7D%7B%5Csqrt%7B21%7D%7D)%7D%7B%5Csqrt%7B21%7D%7D/?origin=button)
I get a solution of $0$.

What is going on here?

Answer: 
See [How do you integrate $\int \frac{1}{a + \cos x} dx$?](https://math.stackexchange.com/questions/134577/how-do-you-integrate-int-frac1a-cos-x-dx) and substitute 1.1 for a. 

Also, Wolfram alpha gives the right answer, and it looks very similar to what you put in Symbolab. I think you may have entered the wrong number into Symbolab ($\frac{2\pi}{2}$ instead of $\frac{\pi}{2}$).
EDIT: following answer is incorrect. Disregard, trying to fix. 

I don't know where you are getting that crazy antiderivative, but here is how I would do it.

$$\int \frac{dx}{1.1 + cos x}$$
$$x = \cos^{-1}u, \space \space dx = \frac{-1}{1-u^2}$$

$$-\int \frac{du}{(1.1 + u)(1+u)(1-u)}$$

and now we can use partial fracitonal decomposition. 

$$\frac{A}{1.1+u}+\frac{B}{1+u}+\frac{C}{1-u} = \frac{1}{(1.1 + u)(1+u)(1-u)}$$

$$ A(1+u)(1-u) + B(1.1+u)(1-u) + C(1.1+u)(1+u) = 1$$

Skipping a few steps for the interest of time, $A = 4.7619, B=-5, C=0.238095$

Our integral is equal to $-4.7619 \ln |1.1 + u| +5 \ln |1+u| -0.238095 \ln|1-u|$

Plug in u = cos x.