Evaluation of $\int_{0}^{\pi/6}\tan^{-1}\sqrt{2-\tan^2(x)}\mathrm dx $

Question

$$\int_{0}^{\pi/6}\tan^{-1}\sqrt{2-\tan^2(x)}\mathrm dx $$

$1+\tan^2(x)=\sec^2(x)$

$u=\sqrt{2-\tan^2(x)}, dx={udu\over \sqrt{2-u}+(\sqrt{2-u^2})^3}$

$$\int_{\sqrt{2-\sqrt{3}/3}}^{\sqrt{2}}{udu\over \sqrt{2-u}+(\sqrt{2-u^2})^3}$$

Very messy, how do we evaluate this question in a easy manner?

Answer 1

We can approximate the value using composition of Taylor series for the integrand around $x=0$.

This would give $$\tan ^{-1}\left(\sqrt{2-\tan ^2(x)}\right)=\tan ^{-1}\left(\sqrt{2}\right)-\frac{x^2}{6 \sqrt{2}}-\frac{23 x^4}{144 \sqrt{2}}-\frac{3727 x^6}{25920 \sqrt{2}}-\frac{124627 x^8}{967680 \sqrt{2}}+O\left(x^{10}\right)$$ Using the expansion to $O\left(x^{2n}\right)$ and computing the decimal representation of the result, we should get $$\left( \begin{array}{cc} n & \text{result} \\ 1 & 0.494564 \\ 2 & 0.493675 \\ 3 & 0.493518 \\ 4 & 0.493488 \\ 5 & 0.493482 \\ 6 & 0.493481 \end{array} \right)$$ quite close to the result from Wolfram Alpha.

Answer 2

Random integral, random substitution, no clue. That reminded me of something/someone. Anyway $$ \int_{0}^{\pi/6}\arctan\sqrt{2-\tan^2(x)}\,dx = \int_{0}^{1/\sqrt{3}}\frac{\arctan\sqrt{2-u^2}}{1+u^2}\,du=\int_{\sqrt{5/3}}^{\sqrt{2}}\frac{v\arctan(v)}{\sqrt{2-v^2}(3-v^2)}\,dv $$ can be tackled through Feynman's trick, since $$ \int_{\sqrt{5/3}}^{\sqrt{2}}\frac{v^2\,dv}{\sqrt{2-v^2}(3-v^2)(1+a^2 v^2)}=\int_{5/6}^{1}\frac{\sqrt{v}\,dv}{\sqrt{1-v}(3-2v)(1+ 2a^2 v)} $$ or $$ \int_{0}^{1/\sqrt{6}}\frac{2\sqrt{1-v^2}\,dv}{(1+2v^2)((1+ 2a^2)-2a^2 v^2)}$$ can be computed by partial fraction decomposition and the outcome can be integrated over $(0,1)$ with respect to $da$ through the dilogarithm machinery. The final outcome is indeed a $\frac{\pi^2}{20}$ and the original integral is pretty similar (and probably equivalent) to Ahmed's integral. See also this similar question.

The integral can be written as $$\int_{0}^{\pi/6}\arctan\sqrt{\frac{3\cos^2\theta-1}{\cos^2\theta}}\,d\theta = \frac{1}{2}\int_{0}^{\pi/3}\arctan\sqrt{\frac{1+3\cos\theta}{1+\cos\theta}}\,d\theta$$ and directly evaluated through Theorem 4.4 of Sangchul Lee. Credits to him.