<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] 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 $b \to \infty$
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