The exact question:
Consider the region $$ U_n = \left \{ x_1, x_2, \ldots, x_n \geqslant 0, \sum \limits_{k=1}^n x_k ^2 \leqslant 1 \right \} . $$ Define $$ B_n = \idotsint\limits_{U_n} \left( \dfrac{x_1 + x_2 + \cdots + x_n }{\sqrt{x_1 ^2 + x_2 ^2 + \cdots + x_n^2 }} \right) \ dx_1 \ dx_2 \ dx_3 \ \cdots \ dx_n $$
Prove that $B_n$ satisfies the recurrence relation, $$\begin{equation} B_n = \cases{ 1 & n =1,2 \cr \dfrac\pi{2n-2} B_{n-2} & n>2\cr } \end{equation}$$
Context/Attempt:
(It's trivial for $n=1$)
I was trying to solve a Jacobian transformation practice problem, which turns out to be the form of $B_2$. A simple transformation of $x_1 = r \cos \theta , x_2 = r\sin \theta \implies dx_1 dx_2 = r dr d\theta$ shows that $B_2 = 1$. For my own amusement, I extended it to 3-dimensions and used spherical coordinates, $$ (x_1, x_2, x_3) = (r \cos\theta \sin \phi , r \sin\theta \sin \phi , r \cos\theta) \implies dx_1 dx_2 dx_3 = r^2 \sin\phi \ dr d\phi d\theta $$ to get $B_3 = \frac\pi4$.
Then I wondered if I can generalize this, which brought me to this PDF.
By using the generalized Jacobian transformation, I get
$$ B_n= \int_0^{\pi/2} \int_0^{\pi/2} \cdots \int_0^{\pi/2} \int_0^1 \left[ \prod_{k=1}^{n-1} \sin(\theta_k) + \sum_{m=2}^{n-1} \left(\cos\theta_{m-1} \prod_{k=m}^{n-1} \sin \theta_k\right) + \cos \theta_{n-1} \right] \cdot \left[ \ r^{n-1} \prod_{k=2}^{n-1} \sin^{k-1} \theta_k \ \right ] dr d\theta_1 d\theta_2 \cdots \ d\theta_{n-1} $$
I was hoping that all the variables $(r, \theta_1, \theta_2, \ldots, \theta_{n-1} ) $ are easily separable, but so far, I'm only able to take out $ \int_0^1 r^{n-1} \ dr = \frac1n $ from the huge expression.
By testing $n=3,4,5$ on Mathematica, I get $(B_3, B_4, B_5 ) = (\pi/4, \pi/6, \pi^2 /32) $
I then test for $n=6,7,8$ and I believed I've found a pattern (as stated above), but I'm nowhere near to a proof.
Update 1 (Jan22 2023, 3.23pm GMT+0800): Looks like there's a geometric interpretation for this. OEIS A087299.