Integral of $\frac1{|x|^3}$ on a circular segment How can I evaluate
$$\int \int _C |(x,y)|^{-3} dx dy$$
Where $C$ is the part between the chord $AB$ and the arc $AB$, and $|(x,y)| = \sqrt{x^2 + y^2}$? The radius of the circle is $R$.
I tried using polar coordinates, but I can’t find a way to represent $C$ with polar coordinates.

 A: The tricky part of doing this in polar coordinates is the chord $AB$. Triangle $OAB$ is isosceles, so its angles $\angle OAB$ and $\angle OBA$ are both $75^\circ = \frac{5 \pi}{12}$. If $P$ is a point on segment $AB$ then from triangle $OBP$ with $OB = R$ and $\angle OBP = \angle OBA$, we have
$$ \angle POB + \angle OBP + \angle BPO = \pi $$
$$ \angle BPO = \frac{7 \pi}{12} - \angle POB $$
And from the law of sines,
$$ \frac{OB}{\sin \angle BPO} = \frac{OP}{\sin \angle OBP} $$
$$ OP = R\, \frac{\sin\left(\frac{7 \pi}{12} - \angle POB\right)}{\sin \frac{7 \pi}{12}} $$
If $\theta$ is the usual counter-clockwise angle from the $+x$-axis, then $\angle POB = \theta - \frac{\pi}{6}$ and $OP$ is the needed lower bound of $r$, so the original integral becomes
$$ \int_{\pi/6}^{\pi/3} \int_{R \sin(3 \pi/4 - \theta) / \sin(7 \pi/12)}^R r^{-3} r\, dr\, d\theta $$
A: The function over which you are integrating is rotationally symmetric around the origin $(0,0)$, thus we can rotate the coordinate system in such a ways that $A$, $B$ are at $\pm15^\circ$ from an axis and, say, $x=\mathrm{const}$ for points on $\gamma$. One such rotation is such that
$$\begin{align}
A &= (\cos 15^\circ, \sin 15^\circ) = (\cos(\pi/12), \sin(\pi/12)) \\
&= \tfrac14 (\sqrt6+\sqrt2, \sqrt6-\sqrt2) =:(c,s)\\
B &= (\cos 15^\circ, -\sin 15^\circ) = (c,-s)
\end{align}$$
and all points in the chord have $x=\cos15^\circ = c$. Then
$$\begin{align}
\iint_C |(x,y)|^{-3}dxdy
&= \int_{+\sin(\pi/12)}^{-\sin(\pi/12)}  \frac{dy}{(y^2+\cos^2(\pi/12))^{3/2}} \\
&= \int_{+s}^{-s} \frac{dy}{(y^2+c^2)^{3/2}} \\
&= \frac{y}{c^2(y^2+c^2)^{1/2}} \Big|_{y=s}^{y=-s}\\
&= \frac{-2s}{c^2(s^2+c^2)^{1/2}} = -2\frac{s}{c^2} \\
&= -2 \frac{\sqrt6-\sqrt2}{2+\sqrt3}
\end{align}$$
where in the second-last line we use $s^2+c^2=1$.
