# Integral $\int_{\pi/4}^{\pi/2}\frac{\cos(2\theta)e^{\cot\theta}}{\sin^3(2\theta)\left(e^{\cot\theta}-e^{\tan\theta}\right)}d\theta$ [closed]

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

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

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, ZackyDec 21 '18 at 10:54

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

HINT

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?

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