Here is a geometric proof of the identity.
Let $I$ denote the integral. Substituting $x=R\sin\theta$ and $k=-y/R \in [-1, 1]$, we get
$$ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left| \frac{\sin\theta-k}{\cos\theta} \right|^{\alpha} \, \mathrm{d}\theta. $$
In order to simplify $I$, we introduce the function
$$ F_k(t) := \operatorname{length}\left(\left\{ \theta \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right) : \left| \frac{\sin\theta-k}{\cos\theta} \right| \leq t\right\}\right) $$
and note that $I$ is recast as
$$ I = \int_{0}^{\infty} t^{\alpha} \, \mathrm{d}F_k(t) $$
by the "change of variables". Now we will make use of the following claim:
Claim. $F_k$ does not depend on $k \in [-1, 1]$.
Once this claim is proved, we may conveniently use $F_0(t) = 2\arctan t$ to find that
$$ I
= \int_{0}^{\infty} t^{\alpha} \, \mathrm{d}F_0(t)
= 2\int_{0}^{\infty} \frac{t^{\alpha}}{t^2+1} \, \mathrm{d}t
= \frac{\pi}{\cos(\pi\alpha/2)}, $$
where the last step can be verified in a routine way. $\square$
Proof of Claim. Note that $\frac{\sin\theta-k}{\cos\theta}$ is the slope of the line joining $(0, k)$ to $(\cos\theta, \sin\theta)$. Under this geometric interpretation, $F(t)$ corresponds to the length of the circular arc $AB$ shown below:

Let $l \in [-1, 1]$ and assume WLOG that $l < k$. Then $F_k(t) - F_{l}(t)$ is equal to the length of the arc $AC$ minus the length of the arc $BD$:

Now we reflect the arc $BD$ about the $y$-axis to obtain the arc $B'D'$. Then the two arcs $AC$ and $B'D'$ are congruent by the symmetry, and hence, they have the same length. This proves that $F_k(t) - F_l(t) = 0$ and therefore the claim follows. $\square$