TITLE: A sine inverse integral
QUESTION [10 upvotes]: 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}

REPLY [7 votes]: $$
\begin{align}
&\int_0^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{1}\\
&=2\int_0^{\pi/4}\big(\log(2)+\log(\sin(t))+\log(\cos(t))\big)\,\mathrm{d}t\tag{2}\\
&=\frac\pi2\log(2)+2\int_0^{\pi/4}\log(\sin(t))\,\mathrm{d}t+2\int_{\pi/4}^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{3}\\
&=\frac\pi2\log(2)+2\int_0^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{4}\\
&=-\frac\pi2\log(2)\tag{5}
\end{align}
$$
Explanation:
$(2)$: substitute $t\mapsto2t$ and $\sin(2t)=2\sin(t)\cos(t)$
$(3)$: substitute $t\mapsto\pi/2-t$ in the integral of $\log(\cos(t))$
$(4)$: combine domains of integration
$(5)$: subtract $(4)$ from $2$ times $(1)$
$$
\begin{align}
\int_0^1\left(\frac{\sin^{-1}(x)}{x}\right)^3\,\mathrm{d}x
&=\int_0^{\pi/2}\left(\frac{t}{\sin(t)}\right)^3\,\mathrm{d}\sin(t)\tag{6}\\
&=-\frac12\int_0^{\pi/2}t^3\,\mathrm{d}\frac1{\sin^2(t)}\tag{7}\\
&=-\frac12\frac{\pi^3}8+\frac32\int_0^{\pi/2}\left(\frac{t}{\sin(t)}\right)^2\,\mathrm{d}t\tag{8}\\
&=-\frac12\frac{\pi^3}8-\frac32\int_0^{\pi/2}t^2\,\mathrm{d}\cot(t)\tag{9}\\
&=-\frac12\frac{\pi^3}8+3\int_0^{\pi/2}t\cot(t)\,\mathrm{d}t\tag{10}\\
&=-\frac12\frac{\pi^3}8+3\int_0^{\pi/2}t\,\mathrm{d}\log(\sin(t))\tag{11}\\
&=-\frac12\frac{\pi^3}8-3\int_0^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{12}\\
&=\bbox[5px,border:2px solid #C0A000]{-\frac12\frac{\pi^3}8+\frac{3\pi}2\log(2)}\tag{13}
\end{align}
$$
Explanation:
$\phantom{1}(6)$: $x=\sin(t)$
$\phantom{1}(7)$: prepare for integration by parts
$\phantom{1}(8)$: integrate by parts
$\phantom{1}(9)$: prepare for integration by parts
$(10)$: integrate by parts
$(11)$: prepare for integration by parts
$(12)$: integrate by parts
$(13)$: apply $(5)$
$$
\begin{align}
\int_0^1\left(\frac{\sin^{-1}(x)}{x}\right)^2\,\mathrm{d}x
&=\int_0^{\pi/2}\left(\frac{t}{\sin(t)}\right)^2\,\mathrm{d}\sin(t)\tag{14}\\
&=-\int_0^{\pi/2}t^2\,\mathrm{d}\frac1{\sin(t)}\tag{15}\\
&=-\frac{\pi^2}4+2\int_0^{\pi/2}\frac{t}{\sin(t)}\,\mathrm{d}t\tag{16}\\
&=-\frac{\pi^2}4-2\int_0^{\pi/2}t\,\mathrm{d}\log(\csc(x)+\cot(x))\tag{17}\\
&=-\frac{\pi^2}4+2\int_0^{\pi/2}\log(\csc(x)+\cot(x))\,\mathrm{d}t\tag{18}\\
&=-\frac{\pi^2}4+2\int_0^{\pi/2}\big(\log(1+\cos(x))-\log(\sin(x))\big)\,\mathrm{d}t\tag{19}\\
&=-\frac{\pi^2}4+2\left(2\mathrm{G}-\frac\pi2\log(2)+\frac\pi2\log(2)\right)\tag{20}\\
&=\bbox[5px,border:2px solid #C0A000]{-\frac{\pi^2}4+4\mathrm{G}}\tag{21}
\end{align}
$$
Explanation:
$(14)$: $x=\sin(t)$
$(15)$: prepare for integration by parts
$(16)$: integrate by parts
$(17)$: prepare for integration by parts
$(18)$: integrate by parts
$(19)$: $\csc(x)+\cot(x)=\frac{1+\cos(x)}{\sin(x)}$
$(20)$: apply $(8)$ from this answer and $(5)$ from above
$(21)$: simplify
where $\mathrm{G}$ is Catalan's constant.
$$
\begin{align}
\int_0^1\left(\frac{\sin^{-1}(x)}{x}\right)\mathrm{d}x
&=\int_0^{\pi/2}\left(\frac{t}{\sin(t)}\right)\mathrm{d}\sin(t)\tag{22}\\
&=\int_0^{\pi/2}t\,\mathrm{d}\log(\sin(t))\tag{23}\\
&=-\int_0^{\pi/2}\log(\sin(t))\,\mathrm{d}t\tag{24}\\
&=\bbox[5px,border:2px solid #C0A000]{\frac\pi2\log(2)}\tag{25}
\end{align}
$$
Explanation:
$(22)$: $x=\sin(t)$
$(23)$: prepare for integration by parts
$(24)$: integrate by parts
$(25)$: apply $(5)$