Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 5 down vote accepted

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.

share|cite|improve this answer
Thanks a lot..! – Lilly Apr 12 '12 at 14:48
The splitting is very much necessary. As an aside, that some computing environments do not do these splits is why integrals like this sometimes give spurious answers in those environments. See this for instance. – J. M. Apr 14 '12 at 9:15
Yes. The substitution should be done only on an interval where $\tan(x/2)$ is monotone! – GEdgar Oct 15 '12 at 1:51

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 $\sin{2x} = \frac{2\tan{t}}{1+\tan^{2}{t}}$

share|cite|improve this answer
The function is positive, hence the value cannot be zero. – AD. Apr 12 '12 at 15:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.