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}
 A: 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*}
A: EDIT: I changed the second half of my answer.
$$ \begin{align} \int_{0}^{1} \frac{\arcsin^{3} (x)}{x^{3}} \, dx &= \int_{0}^{\pi /2} \frac{t^{3}}{\sin^{3} (t)} \,  \cos (t) \, dt \\ &= t^{3} \left(-\frac{1}{2 \sin^{2}(t)} \right)\Bigg|^{\pi /2}_{0} + \frac{3}{2} \int_{0}^{\pi /2} \frac{t^{2}}{\sin^{2} (t)} \, dt \\ &=- \frac{\pi^{3}}{16} + \frac{3}{2} \int_{0}^{\pi /2} \frac{t^{2}}{\sin^{2} (t)} \, dt \\ &= - \frac{\pi^{3}}{16} + \frac{3 \pi}{2} \int_{0}^{\infty} \frac{t}{\cosh^{2}(t)} \, dt \tag{1}\\&= - \frac{\pi^{3}}{16} + \frac{3\pi}{2} \lim_{b \to \infty} \left(t \tanh (t) \Big|_{0}^{b} - \int_{0}^{b} \tanh (t) \, dt \right) \\&= - \frac{\pi^{3}}{16} + \frac{3 \pi}{2} \lim_{b \to \infty} \left(b \tanh (b) - \ln (\cosh b) \right) \\&= - \frac{\pi^{3}}{16} + \frac{3 \pi}{2} \lim_{b \to \infty} \ln \left(\frac{2 \, e^{b \tanh (b)}}{e^{b} + e^{-b}} \right) \\ &= - \frac{\pi^{3}}{16} + \frac{3 \pi}{2} \, \ln (2) \tag{2} \\ &= \frac{\pi}{2} \left(3 \ln (2) - \frac{\pi^{2}}{8} \right)\end{align}$$

$(1)$ Integrate the function $ \frac{z^{2}}{\sin^{2}(z)}$ around a rectangular contour with vertices at $0$, $\pi/2$, $\pi/2 + iR$, and $iR$. The value of the integral is purely imaginary on the left side of the rectangle. And as $R \to \infty$, the integral vanishes along the top of the rectangle since $\left|\frac{1}{\sin^{2}(z)} \right|$ decays exponentially as $\text{Im}(z) \to \infty$.
$(2)$ $\tanh (b) \to 1$ as $b \to \infty$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{J_{3}\equiv\int_{0}^{1}\bracks{\arcsin\pars{x} \over x}^{3}\,\dd x
     ={\pi \over 2}\bracks{3\ln\pars{2} - {\pi^{2} \over 8}}:\ {\large ?}}$

Set $\ds{\quad x\equiv \sin\pars{t}\quad\imp\quad t=\arcsin\pars{x}}$:
  \begin{align}
J_{3}&\equiv\int_{0}^{\pi/2}\,{t^{3} \over\sin^{3}\pars{t}}\,
\bracks{\cos\pars{t}\,\dd t}
=\left. -\,\half\,{t^{3} \over \sin^{2}\pars{t}}\right\vert_{0}^{\pi/2}
+\int_{0}^{\pi/2}{1 \over 2\sin^{2}\pars{t}}\,3t^{2}\,\dd t
\\[3mm]&=-\,\half\pars{\pi \over 2}^{3}
+3\color{#c00000}{\int_{0}^{\pi/2}{t^{2}\,\dd t \over 1 - \cos\pars{2t}}}
\tag{1}
\end{align}

\begin{align}&\color{#c00000}{%
\int_{0}^{\pi/2}{t^{2}\,\dd t \over 1 - \cos\pars{2t}}}
={1 \over 8}\int_{0}^{\pi}{t^{2}\,\dd t \over 1 - \cos\pars{t}}
={1 \over 16}\int_{-\pi}^{\pi}{t^{2}\,\dd t \over 1 - \cos\pars{t}}
\\[3mm]&={1 \over 16}
\int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ < \pi}}
{-\ln^{2}\pars{z} \over 1 - \pars{z^{2} + 1}/\pars{2z}}\,{\dd z \over \ic z}
=-\,{1 \over 8}\,\ic
\int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ < \pi}}
{\ln^{2}\pars{z}\,\dd z \over \pars{z - 1}^{2}}
\\[3mm]&={1 \over 8}\,\ic\int_{-1}^{0}
{\ln^{2}\pars{-x} + 2\pi\ic\ln\pars{-x} - \pi^{2} \over \pars{x - 1}^{2}}\,\dd x
+
{1 \over 8}\,\ic\int_{0}^{-1}
{\ln^{2}\pars{-x} - 2\pi\ic\ln\pars{-x} - \pi^{2} \over \pars{x - 1}^{2}}\,\dd x
\\[3mm]&=-\,{\pi \over 2}\int_{0}^{1}
{\ln\pars{x}\,\dd x \over \pars{x + 1}^{2}}
=-\,{\pi \over 2}\sum_{n = 1}^{\infty}\pars{-1}^{n}n
\int_{0}^{1}\ln\pars{x}x^{n - 1}\,\dd x
\\[3mm]&=-\,{\pi \over 2}\sum_{n = 1}^{\infty}\pars{-1}^{n}n
\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}x^{\mu + n - 1}\,\dd x
={\pi \over 2}\
\underbrace{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n}}_{\ds{=\ \ln\pars{2}}}\
=\ {\pi \over 2}\,\ln\pars{2}
\end{align}

By replacing this result in $\pars{1}$:
  $$\color{#44f}{\large%
J_{3}\equiv\int_{0}^{1}\bracks{\arcsin\pars{x} \over x}^{3}\,\dd x
     ={\pi \over 2}\bracks{3\ln\pars{2} - {\pi^{2} \over 8}}}
$$

