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I started like this \begin{align*} \int \frac{x}{\sin^2x-3}dx & =\int \frac{x\sec^2x}{\tan^2x-3\sec^2x}dx\\ & =-\int \frac{x\sec^2x}{2\tan^2x+3}dx\\ & = -\left [ \frac {x\tan^{-1}\left (\frac {\sqrt {2}\tan x}{\sqrt {3}}\right)}{\sqrt {6}}-\int \frac {\tan^{-1}\left (\frac {\sqrt {2}\tan x}{\sqrt {3}}\right)}{\sqrt {6}}dx\right] \end{align*}

Now i am unable to solve this further.

Soham Chatterjee
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  • Wolfram Alpha shows a monster. – AryanSonwatikar Jun 12 '20 at 06:10
  • Correct me if I'm wrong, but shouldn't it be arctan(√(2/3) tan(x)), not arctan(√(2/3) x)? You do IBP with u=x and dv=sec²(x)/(3+2tan²(x)). So you should get v=1/√6 arctan(√(2/3) tan(x)). – GhostyOcean Jun 12 '20 at 06:20
  • This is not an elementary integral, are you actually interested in writing the primitive through polylogarithms or just in the value of the integral over a specific range? Over $[0,\pi]$ we may exploit symmetry through the change of variable $x\mapsto \pi-x$. – Jack D'Aurizio Jun 12 '20 at 07:17

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As said in comments, after integration by parts, you are left with $$I=\int \tan ^{-1}\left(\sqrt{\frac{2}{3}} \tan (x)\right)\,dx$$ which is a monster that I should try to avoid.

In order to compute it, I should rather consider series expansions built around $x=k \pi$ and use for $$J_k=\int_{k\pi}^{(2k+1)\frac \pi 2} \tan ^{-1}\left(\sqrt{\frac{2}{3}} \tan (x)\right)\,dx$$ the series expansion of the integrand. This is $$\tan ^{-1}\left(\sqrt{\frac{2}{3}} \tan (x)\right)=t+\frac{t^3}{6}-\frac{3 t^7}{280}-\frac{t^9}{504}+O\left(t^{11}\right)$$ where $t=\sqrt{\frac{2}{3}} (x-\pi k)$.

For illustration, this would give $$J_0=\frac{\pi ^2}{4 \sqrt{6}}\left(1+\frac{\pi ^2}{72}-\frac{\pi ^6}{80640}-\frac{\pi ^8}{3265920} \right)\approx 1.1305$$ while the monster would give as an exact solution $$\frac{1}{2 \sqrt{6}}\Phi \left(\frac{2}{3},2,\frac{1}{2}\right)+\frac{1}{2} \log \left(\frac{3}{2}\right) \tanh ^{-1}\left(\sqrt{\frac{2}{3}}\right)\approx 1.1326$$ For sure, we could have better results pushing the expansion to higher orders.

Claude Leibovici
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