An integral with arccosine

Evaluate the integral:

$$\mathcal{J} = \int_{-1/\sqrt{3}}^{1\sqrt{3}}\frac{x^4}{1-x^4}\cos^{-1}\left ( \frac{2x}{1+x^2} \right )\, {\rm d}x$$

Solution



We have successively:

$$\begin{aligned}\mathcal{J}=\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\cos^{-1}\left ( \frac{2x}{1+x^2} \right )\,{\rm d}x &\overset{u=-x}{=\! =\! =\!} \int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\cos^{-1}\left ( -\frac{2x}{1+x^2} \right )\,{\rm d}x\\
 &= \int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\left [ \pi-\cos^{-1}\left ( \frac{2x}{1+x^2} \right ) \right ]\,{\rm d}x\\
 &=\pi\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\,{\rm d}x- \mathcal{J}
\end{aligned}$$

Hence:

$$\mathcal{J}=\frac{\pi}{2}\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\,{\rm d}x$$

The latter integral can be dealt with partial fractions. The process is a long shot but a rootine one. Finally , we get that:

$$\mathcal{J}=-\frac{2\pi}{\sqrt{3}}+\frac{\pi^2}{6}+\frac{\pi}{2}\ln \left ( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right )$$

completing the exercise.