I came across (coincidentally) two integral evaluations, which seem to agree according to numerical tests. It did not seem easy to convert one into the other.
QUESTION. Is this true?
$$\int_0^1\left(\frac{\arcsin x}x\right)^3dx
=\frac34\pi\int_0^1\left(2\,\text{arctanh}\, x +\frac{\log(1-x^2)}x\right)dx.$$
Best Answer
The proposed equality is true.
Details: To find $$l:=\int_0^1\left(\frac{\arcsin x}x\right)^3\,dx =-\frac{1}{16} \pi \left(\pi ^2-24 \ln2\right), $$ make the substitution $t=\arcsin x$ and repeatedly integrate by parts to kill the powers of $t$ and reduce this integral to $$\int_0^{\pi/2}\ln\sin t\,dt=-\frac\pi2\,\ln2, $$ by formula 4.225.3, page 531, of Gradshteyn--Ryzhik.
To find $$r_1:=\int_0^1 2\,\text{arctanh}\, x\,dx =\ln4, $$ integrate by parts to find an antiderivative of $\text{arctanh}$. Alternatively, expand $\text{arctanh}\, x$ into the Maclaurin series (using $\text{arctanh}'x=\frac1{1-u}=1+u+u^2+\dots$ with $u=x^2$) and integrate termwise, to get $$r_1/2=\sum_1^\infty\frac1{(2j-1)2j} =\sum_1^\infty\Big(\frac1{2j-1}-\frac1{2j}\Big) =\sum_1^\infty\frac{(-1)^{k-1}}k=\ln2. $$
To find $$r_2:=\int_0^1 \frac{\ln(1-x^2)}x\,dx =-\frac12\,\sum_1^\infty\frac1{j^2}=-\frac{\pi^2}{12}, $$ expand $\ln(1-x^2)$ into the Maclaurin series (by using the Maclaurin series for $\ln(1-u)$) and integrate termwise. Now one can see that $l=\frac{3\pi}4\,(r_1+r_2)$, that is, the equality holds.