Is it possible to evaluate this integral in a closed form? $$I=\int_0^{\pi/2}\arctan^2\left(\frac{6\sin x}{3+\cos 2x}\right)\mathrm dx$$


I will refer to the following result from my previous answer:

\begin{align*} I(r, s) &= \int_{0}^{\frac{\pi}{2}} \arctan (r \sin\theta) \arctan (s \sin\theta) \, d\theta \\ &= \pi \chi_{2} \left( \frac{\sqrt{1+r^{2}} - 1}{r} \times \frac{\sqrt{1+s^{2}} - 1}{s} \right), \end{align*}

where $\chi_{2}$ is the Legendre chi function. Using the addition formula for the arctangent, it follows that

$$ \arctan\left(\frac{6\sin x}{3 + \cos 2x} \right) = \arctan \left( \frac{\frac{3}{2}\sin x}{1 - \frac{1}{2}\sin^{2} x} \right) = \arctan (\sin x) + \arctan ( \tfrac{1}{2}\sin x). $$ So it follows that

$$ \int_{0}^{\frac{\pi}{2}} \arctan^{2}\left(\frac{6\sin x}{3+\cos 2x}\right) \, dx = I(1,1) + 2I(1,\tfrac{1}{2}) + I(\tfrac{1}{2},\tfrac{1}{2}), $$

which reduces to a combination of Legendre chi functions

$$ \pi \left\{ \chi_{2}(3 - 2\sqrt{2}) + \chi_{2}(9 - 4\sqrt{5}) + 2\chi_{2}\left( (\sqrt{2} - 1)(\sqrt{5} - 2) \right) \right\}. $$

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    $\begingroup$ Your formula for the integral of a product of arctangents seems to be very wide applicable. It would be nice if it got into published tables of integrals. $\endgroup$ Nov 18 '13 at 21:47
  • $\begingroup$ @VladimirReshetnikov I didn't know that. I don't know her real name. So, I couldn't identify the reference as associated to ${\tt @sos440}$. I just delete my previous comment. Thanks. $\endgroup$ Aug 12 '14 at 16:46
  • $\begingroup$ @VladimirReshetnikov Also, I just delete my "mini-answer" in a previous post. Thanks. $\endgroup$ Aug 12 '14 at 16:51


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    $\begingroup$ I don't mean to be combative, but I am agape at the fact that anyone would upvote Cleo's answers as long as (s)he continues to post results without any clue as to their origin. This is a Mathematics community, not a Computer Output community. Context-less answers like this are not answers but rather comments. $\endgroup$
    – Ron Gordon
    Nov 17 '13 at 19:48
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    $\begingroup$ @RonGordon I do not believe any existing CAS could do this integral. $\endgroup$ Nov 17 '13 at 23:49
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    $\begingroup$ @VladimirReshetnikov: that's not really my point. My point is that the result in isolation adds little value without the thinking behind how it was obtained. $\endgroup$
    – Ron Gordon
    Nov 18 '13 at 0:23
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    $\begingroup$ I think this answer it is overly long and verbose. A better answer of the question would certainly have been "Yes." $\endgroup$ Nov 20 '13 at 3:02
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    $\begingroup$ Anyway, I agree that an answer is not really an answer without a proper explanation or at least some pointers. $\endgroup$ Dec 30 '15 at 0:03

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