challenging integral involving $\zeta(5)$

I ran across a curious integral that seems to be rather tough that some on the site may enjoy.

Show that $$\displaystyle \int_{0}^{1}\frac{\sqrt{1-x^{2}}}{1-x^{2}\sin^{2}(x)}dx = \frac{5\sqrt[5]{{\pi}^{8}}}{32\sqrt[5]{{\zeta(5)}^{9}}}$$

How in the world can $\zeta(5)$ be incorporated into this?. I tried series and several methods, but made no real progress. Any ideas?. Thanks very much.

Solution 1:

The purported identity is false, as GEdgar already indicated in the comment, but remarkably accurate: $$ \begin{eqnarray} \int_0^1 \frac{\sqrt{1-x^2}}{1-x^2 \sin^2(x)} \mathrm{d} x &\approx& \color{red}{ 0.91392913}60302011781728596 \\ \frac{5 \pi^{8/5}}{32 \zeta(5)^{9/5}} &\approx& \color{red}{0.91392913}77247633495515212 \end{eqnarray} $$

Here is the Mathematica code used:

In[19]:= N[
 NIntegrate[Sqrt[1 - x^2]/(1 - x^2 Sin[x]^2), {x, 0, 1}, 
  WorkingPrecision -> 60], 25]

Out[19]= 0.9139291360302011781728596

In[20]:= N[(5 Pi^(8/5))/(32 Zeta[5]^(9/5)), 25]

Out[20]= 0.9139291377247633495515212

In[21]:= % - %%

Out[21]= 1.694562171378662*10^-9