Are you kind to let me know the way? By the way, don't you have a "curiosity" tag? $$\int_{0}^{\pi/2} \text{arctanh}(\sin x) \text{arctan}(a \tan(x)) \cos(x) \ dx, \quad a>0$$
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Which problem led you to get this integral? – Mikasa May 23 '13 at 19:50
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4What would be the point of a curiosity tag? Tags have a purpose, specifically, to let people find things that interest them or which they can answer. What tickles your curiosity, on the other hand, is highly subjective. – Thomas Andrews May 23 '13 at 19:50
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I suggest this is discussed on meta if it's of further interest (: – not all wrong May 23 '13 at 20:11
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It's $\mathrm{artanh}$ as in area tangentis hyperbolici, no arc here ... – martini May 23 '13 at 21:30
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@martini are you referring to my post? In this case I think you're joking ... – Crazy_girl May 23 '13 at 22:05
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@Crazy_girl No, I'm not joking. It's $\arctan$, but $\mathrm{artanh}$. – martini May 23 '13 at 22:38
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@martini http://mathworld.wolfram.com/Arctanh.html – Crazy_girl May 23 '13 at 22:42
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It seems it is a "more proper" way of writing it. Much like you can use sine and sinus interchangeably. Although in this case, I have to agree with martini that using arc is inherently wrong. – Jon Claus May 24 '13 at 02:59
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$$\forall a>0,\quad\int_{0}^{\pi/2}\text{artanh}(\sin x)\,\arctan(a\,\tan(x))\cos(x)\ dx=\\\frac\pi2\Re\left(\ln\left(2\,a^{-1}+2\right)-\frac{\ln\left(2\sqrt{a^{-4}-a^{-2}}+2\,a^{-2}-1\right)}{2 \sqrt{1-a^2}}\right),$$ where $\Re$ denotes the real part.
OlegK
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I still need to rigorously justify some steps of the proof, so I leave this result as a conjecture for now. – OlegK May 25 '13 at 02:49