Background: I was in the process of solving some interesting integrals from this site, only to find out I needed a lot more practice before becoming familiar with special functions.

So while doing some problems, I encountered some difficulty with one particular integral; I happened to incorrectly copy it onto a notebook. But I'm curious to know as to how exactly I can evaluate this particular integral.

Essentially, I need help in evaluating the following integral :-

$$ \int \frac{e^x}{\left({1+\cos(x)}\right)} dx$$

Question: How exactly can I evaluate this integral?

Both solutions as well as hints would be greatly appreciated.

Note: Original problem had $\cosh(x)$ instead of $\cos(x)$.


The integrand does not possess an elementary antiderivative. This can be shown using either Liouville's theorem or the Risch algorithm. However, doing so requires advanced knowledge
of abstract algebra. Alternately, expand $~\dfrac1{1+\cos x}~$ into its binomial series, then switch the
order of summation and integration to obtain an infinite series, which you might rewrite in
terms of hypergeometric functions.

  • $\begingroup$ See also Wallis' integrals. $\endgroup$ – Lucian Feb 25 '15 at 17:51
  • $\begingroup$ By elementary you mean an antiderivative without special functions right? Because the answer seems to possess the hypergeometric function. $\endgroup$ – Kugelblitz Feb 26 '15 at 1:14
  • $\begingroup$ @Kugelblitz: Yes, you can use either hypergeometric series, or incomplete beta functions. $\endgroup$ – Lucian Feb 26 '15 at 1:48

If you have $\cosh x$ instead of $\cos x$.:$$\int \frac{e^x}{\left({1+\cosh x}\right)} dx=\int \frac{2e^{2x}}{\left({e^{2x}+2e^x+1}\right)} dx=\int\dfrac{2e^x}{(1+e^x)^2}d(e^x)$$ For your question we can use the fact that, $$\cos x=\dfrac{e^{ix}+e^{-ix}}{2}=\dfrac{e^{2ix}+1}{2e^{ix}}.$$ Then it becomes $$\int \frac{e^x}{\left({1+\cos x}\right)} dx=\int \frac{2e^{(1+i)x}}{\left({e^{2ix}+2e^{ix}+1}\right)} dx=\int\dfrac{2e^{(1+i)x}}{(1+e^{ix})^2}dx$$ Which does not seems like integrable by elementary methods as Surb commented.

  • $\begingroup$ Alright. Are there 'non-elementary' methods to evaluate this? $\endgroup$ – Kugelblitz Feb 25 '15 at 12:42
  • $\begingroup$ I am not familiar with Hypergeometric Function. But most of these type of integrals can solve using that function. There are many professional mathematicians in this site who know about that function than me. $\endgroup$ – Bumblebee Feb 25 '15 at 12:47
  • $\begingroup$ Well, I thank you so much for trying; I'll try doing something with the simplification you've made. Let's wait for others to try and answer... As you said, some use ingenious methods to integrate in the website; nevertheless I've given you an upvote, for your answer is useful. Interesting question don't you think? :) $\endgroup$ – Kugelblitz Feb 25 '15 at 12:49

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