Evaluation of $\int_{0}^{\pi/6}\tan^{-1}\sqrt{2-\tan^2(x)}\mathrm dx $ $$\int_{0}^{\pi/6}\tan^{-1}\sqrt{2-\tan^2(x)}\mathrm dx $$
$1+\tan^2(x)=\sec^2(x)$
$u=\sqrt{2-\tan^2(x)}, dx={udu\over \sqrt{2-u}+(\sqrt{2-u^2})^3}$
$$\int_{\sqrt{2-\sqrt{3}/3}}^{\sqrt{2}}{udu\over \sqrt{2-u}+(\sqrt{2-u^2})^3}$$
Very messy, how do we evaluate this question in a easy manner?

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.
