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Proposition. The following holds:[1]

\begin{equation*}
\tag{1}
\int_{0}^{\frac{\pi}{2}} \arctan ( r \sin\theta ) \arctan ( s \sin\theta ) \, d\theta
= \pi \chi_{2}(\alpha \beta),
\end{equation*}

where

\begin{align*}
\alpha = \frac{\sqrt{r^{2} + 1} - 1}{r}, \quad
\beta = \frac{\sqrt{s^{2} + 1} - 1}{s}
\end{align*}

and $\chi_{2}$ is the Legendre chi function.

 

Proof. See the reference [1] below.

 

Remark. This result extends my previous result. You can check this by taking the limit $s \to \infty$ to $\text{(1)}$.


Mathematica also suggests that the following holds:


Conjecture. Let

\begin{align*}
A = \{ 1, 3, 5, 7, \ldots \}^4
\end{align*}

be the set of quadruples of odd natural numbers. Then the series expantion of the integral

\begin{align*}
\int_{0}^{\frac{\pi}{2}} \arctan \left( \frac{2x \sin\theta}{1-x^{2}} \right) \arctan \left( \frac{2y \sin\theta}{1-y^{2}} \right) \arctan \left( \frac{2z \sin\theta}{1-z^{2}} \right) \arctan \left( \frac{2w \sin\theta}{1-w^{2}} \right) \, d\theta
\end{align*}

is given by

\begin{align*}
\frac{\pi}{2} \sum_{\alpha \in A} (-1)^{|\alpha|/2} d(\alpha) \frac{x^{\alpha_{1}}}{\alpha_{1}} \frac{y^{\alpha_{2}}}{\alpha_{2}} \frac{z^{\alpha_{3}}}{\alpha_{3}} \frac{w^{\alpha_{4}}}{\alpha_{4}},
\end{align*}

where $d(\alpha)$ denotes the number of choices of signatures so that

\begin{align*}
\pm \alpha_{1} \pm \alpha_{2} \pm \alpha_{3} \pm \alpha_{4} = 0.
\end{align*}

Upon reordering $\alpha$ so that $\alpha_{1} \leq \alpha_{2} \leq \alpha_{3} \leq \alpha_{4}$, this equals

\begin{align*}
(-1)^{|\alpha|/2} d(\alpha)
= \begin{cases}
6, & \alpha_{1} = \alpha_{2} = \alpha_{3} = \alpha_{4}, \\ 
4, & \alpha_{1} = \alpha_{2} < \alpha_{3} = \alpha_{4}, \\
2, & \alpha_{1} + \alpha_{4} = \alpha_{2} + \alpha_{3} \text{ and } \alpha_{1} < \alpha_{2}, \\
-2 & \alpha_{1} + \alpha_{2} + \alpha_{3} = \alpha_{4}.
\end{cases}
\end{align*}


References

  1. Zakharia Stanley, Integral $\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}dx$ - Math StackExchange

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