# Integral $\int_0^{\pi/2}\arctan^2\left(\frac{6\sin x}{3+\cos 2x}\right)\mathrm dx$

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

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

• 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. – Piotr Shatalin Nov 18 '13 at 21:47
• @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. – Felix Marin Aug 12 '14 at 16:46
• @VladimirReshetnikov Also, I just delete my "mini-answer" in a previous post. Thanks. – Felix Marin Aug 12 '14 at 16:51

$$I=\frac\pi2\left(2\,\operatorname{Li}_2\Big(\left(1-\sqrt2\right)\left(2-\sqrt5\right)\Big)-2\,\operatorname{Li}_2\Big(\left(1-\sqrt2\right)\left(\sqrt5-2\right)\Big)\\+\operatorname{Li}_2\left(3-\sqrt8\right)-\operatorname{Li}_2\left(\sqrt8-3\right)+\operatorname{Li}_2\left(9-\sqrt{80}\right)-\operatorname{Li}_2\left(\sqrt{80}-9\right)\right)$$

• 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. – Ron Gordon Nov 17 '13 at 19:48
• @RonGordon I do not believe any existing CAS could do this integral. – Vladimir Reshetnikov Nov 17 '13 at 23:49
• @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. – Ron Gordon Nov 18 '13 at 0:23
• I think this answer it is overly long and verbose. A better answer of the question would certainly have been "Yes." – Michael Greinecker Nov 20 '13 at 3:02
• Anyway, I agree that an answer is not really an answer without a proper explanation or at least some pointers. – Jack D'Aurizio Dec 30 '15 at 0:03