<s>[INST] 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.

 [/INST] 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 $s^2+c^2=1$.
</s>