I was told that there is a closed form to the integral $$\int_{\pi/4}^{\pi/2}\frac{\cos(2\theta)e^{\cot\theta}}{\sin^3(2\theta)\left(e^{\cot\theta}-e^{\tan\theta}\right)}d\theta$$

The given answer is

$$\frac{ \pi^2 }{48}$$

and I have verified it numerically. However, I don't know the methods used.

My question is: how to obtain the answer analytically?


closed as off-topic by RRL, Lord Shark the Unknown, Saad, Cesareo, Zacky Dec 21 '18 at 10:54

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  • 1
    $\begingroup$ So, why don't you share the closed form here? $\endgroup$ – Zacky Dec 20 '18 at 22:51
  • $\begingroup$ As not to spoil. For those that are interested, it is $\pi^2/48$ $\endgroup$ – william122 Dec 20 '18 at 22:52
  • 1
    $\begingroup$ This is a recreational problem? I mean you share it for others, or you actually need help? $\endgroup$ – Zacky Dec 20 '18 at 22:55
  • $\begingroup$ I know the answer, interested in the method. But people can have fun while solving it as well! $\endgroup$ – william122 Dec 20 '18 at 22:56
  • 1
    $\begingroup$ Probably $\tan\theta=u$ may work $\endgroup$ – Tito Eliatron Dec 20 '18 at 22:56


First of all, notice that

\begin{align*} \frac{\cos(2\theta)e^{\cot(\theta)}}{\sin^{3}(2\theta)(e^{\cot(\theta)}-e^{\tan(\theta)})} = \frac{\cot(2\theta)\csc^{2}(2\theta)}{1 - e^{\tan(\theta)-\cot(\theta)}} = \frac{\cot(2\theta)\csc^{2}(2\theta)}{1 - e^{-2\cot(2\theta)}} \end{align*}

Hence, according to the substitution $w = \cot(2\theta)$, where $\mathrm{d}w = -2\csc^{2}(2\theta)\mathrm{d}\theta $, we obtain

\begin{align*} \int\frac{\cos(2\theta)e^{\cot(\theta)}}{\sin^{3}(2\theta)(e^{\cot(\theta)}-e^{\tan(\theta)})}\mathrm{d}\theta = -\frac{1}{2}\int\frac{w}{1-e^{-2w}}\mathrm{d}w \end{align*}

Can you proceed from here?

  • $\begingroup$ Yes, thank you. $\endgroup$ – william122 Dec 20 '18 at 23:12

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