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?
>  $$\int\_0^1\left(\frac{\arcsin x}x\right)^3dx
> =\frac34\pi\int\_0^1\left(2\,\text{arctanh}\, x +\frac{\log(1-x^2)}x\right)dx.$$
> 
> 
> 



A rather curious equality: is this true?
The proposed equality is true. 




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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](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&ved=2ahUKEwi9tu7I57PkAhUEOKwKHZBWDdUQFjADegQIBBAC&url=http%3A%2F%2Ffisica.ciens.ucv.ve%2F~svincenz%2FTISPISGIMR.pdf&usg=AOvVaw01xV-Pb5aQAUw6cJvOGtpR). 


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. 


