<s>[INST] A sine inverse integral While reviewing an old book the integral
\begin{align}
J_{3} = \int_{0}^{1} \left( \frac{\sin^{-1}(x)}{x} \right)^{3} \ dx = \frac{\pi}{2}\left( 3 \ln 2 - \frac{\pi^{2}}{8} \right).
\end{align}
was asked to be shown as an exercise. Is it possible to obtain a complete derivation of this result? As an extension of this first integral what are the values of the integrals
\begin{align}
J_{1} = \int_{0}^{1} \frac{\sin^{-1}(x)}{x} \ dx
\end{align}
and
\begin{align}
J_{2} = \int_{0}^{1} \left( \frac{\sin^{-1}(x)}{x} \right)^{2} \ dx
\end{align}
 [/INST] This is for the extension of the question you asked:
\begin{align*}
  J_1 &= \int_{0}^{1} \, \frac{\arcsin{x}}{x} \, dx \\
  &= \int_{0}^{\pi/2} \, \frac{t}{\sin{t}}\, \cos{t}\, dt \\
  &= t\, \log{\sin{t}}\Big|_0^{\pi/2}-\int_{0}^{\pi/2} \, \log{\sin{t}}\, dt \\
  &= 0 + \frac{\pi}{2}\log{2}
\end{align*}
\begin{align*}
  J_2 &= \int_{0}^{1} \, \left(\frac{\arcsin{x}}{x}\right)^2 \, dx \\
&= \int_{0}^{\pi/2} \, \left(\frac{t}{\sin{t}}\right)^2\, \cos{t} \, dt \\
&= -\frac{t^2}{\sin{t}}\Big|_0^{\pi/2}+\int_{0}^{\pi/2} \,  \frac{2\, t}{\sin{t}}\, dt \\
&= -\frac{\pi^2}{4}+4\, G \approx 1.19646127643654
\end{align*}
where $G$ is the catalan's constant
Update:
Using the results in generalized integral $I_n$,
\begin{align*}
J_4 &=  \frac{1}{16} \, \pi^{4} - \frac{1}{2} \, \pi^{2} + G {\left(\pi^{2} + 8\right)} - \frac{1}{96} \, \psi^{(3)}\left( \frac{1}{4}\right) \approx 1.49222813527376\\
J_5 &= -\frac{1}{128} \, \pi^{5} - \frac{5}{48} \, \pi^{3} + \frac{5}{12} \, {\left(6 \, \pi + \pi^{3}\right)} \log\left(2\right) - \frac{15}{8} \, \pi \zeta(3) \approx 1.69763017912507\\
J_7 &= -\frac{\pi^{7}}{768} - \frac{77 \, \pi^{5}}{1920} - \frac{7 \, \pi^{3}}{48} + \frac{7 \, \pi}{8} \log\left(16\right) + \frac{7}{60} \, {\left(\pi^{5} + 25 \, \pi^{3}\right)} \log\left(2\right) + \frac{105  \, \pi}{8}\, \zeta(5) - \frac{7}{8} \, {\left(15 \, \pi + 2 \, \pi^{3}\right)} \zeta(3) \approx 2.29253050578831
\end{align*}
</s>