Integral (universal substitution) I need a little help, this integral looks like very simple, but i have problem with it
$\int_{0}^{2\pi }\frac{1}{2-\cos x}dx$ and  i want to solve it by universal substitution, e.g. $t=\tan\frac{x}{2}$
but what about limits of integration? 
What is a solution? Thanks.
 A: Break up the integral before applying the substitution: 
$$\int^{2\pi}_0 \frac{1}{2-\cos x} dx  =  \int^{\pi}_0 \frac{1}{2-\cos x} dx + \int^{2\pi}_{\pi} \frac{1}{2-\cos x} dx =\int^{\pi}_0 \frac{1}{2-\cos x} dx +  \int^{\pi}_0 \frac{1}{2+ \cos x} dx.
$$
Now you can apply the substitution to each integral and the bounds become $0$ to $\infty$ for both. 
A: Yeah if you put $t = \tan\frac{x}{2}$, then you have $dt = \frac{1}{2}\cdot \sec^{2}\frac{x}{2} \ dx$. And from here note that $dx = \frac{2 \: dt}{1+t^{2}}$.  And when $x = 0$ you have $t = \tan(0)=0$. And when $x=2\pi$ you have $t = \tan(\pi) = 0$. 
I think the value of your integral will be $0$. But in case you want to evaluate something like 
\begin{align*}
\int \frac{1}{2-\cos{x}} \ dx &= \int\frac{1}{2 - \frac{1-\tan^{2}x/2}{1+\tan^{2}x/2}} \ dx \\ &= \int \frac{1}{2 - \frac{1-t^{2}}{1+t^{2}}} \cdot \frac{2}{1+t^{2}} \ dt \\ &=\int\frac{2}{2+2t^{2}-1+t^{2}} \ dt \\ &= \int\frac{2}{3t^{2} +1} dt = \frac{2}{3} \int \frac{1}{t^{2}+\frac{1}{3}} \ dt
\end{align*}
I guess you can evaluate the integral now by putting $t =\frac{1}{\sqrt{3}}\:\tan{v}$. This is the way how one generally evaluates integrals of the form $$\int \frac{dx}{a+b\cos{x}} \qquad \text{and} \qquad \int \frac{dx}{a+b\sin{x}}$$ While evaluating integrals of the form $a+b\sin{x}$ one uses the formula $\sin{2x} = \frac{2\tan{t}}{1+\tan^{2}{t}}$ 
