I'm trying to evaluate the following integral: $$\int_{0}^{2\pi}\frac{1}{5-3\cos x} dx$$

We can evaluate indefinite one first - $\int\frac{1}{5-3\cos x}dx = \frac{1}{2}\tan^{-1}(2\tan(\frac{x}{2})) + C$. The problem is that $\frac{1}{2}\tan^{-1}(2\tan(\frac{2\pi}{2}))-\frac{1}{2}\tan^{-1}(2\tan(\frac{0}{2}))=0$ but there's a hint on this excercise that the value of this definite integral is greater than $0$. So, what went wrong? What's the trap I've fallen into during evaluation of this integral?

  • $\begingroup$ the problem is that the primitive function has discontinuities :v $\endgroup$ – cand May 11 '15 at 22:00
  • $\begingroup$ this discrepancy amounts to a "branch cut" problem $\endgroup$ – reluctant mathematician May 11 '15 at 22:01
  • $\begingroup$ Indeed, qiubit has fallen through the branch cut :) . $\endgroup$ – Count Iblis May 11 '15 at 22:04



You should do so because the primitive function is discontinues at $x=\pi$.


the graph of $\cos x$ is symmetric about $x = \pi,$ therefore $$\int_0^{2\pi}\frac{dx}{5-3\cos x} = 2 \int_0^{\pi}\frac{dx}{5-3\cos x} = \tan^{-1}\left(2\tan(x/2)\right)\Big|_0^{\pi} =\frac{\pi}2.$$

  • $\begingroup$ Question about notation - is $\tan^{-1}\left(2\tan(x/2)\right)\Big|_0^{\pi} = \tan^{-1}\left(2\tan(\pi/2)\right)-\tan^{-1}\left(2\tan(0/2)\right)$? But $\tan(\pi/2)$ is undefined. Why is it $\pi/2$ then? $\endgroup$ – qiubit May 11 '15 at 22:12
  • $\begingroup$ @qiubit, use $\pi-$ and $tan(\pi/2 -) = \infty$ $\endgroup$ – abel May 11 '15 at 22:13

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