From Calculus, with Love

$\sqrt{\frac{\text{dy}^2}{\text{dx}^2}+1}=\sqrt{x \log (\sin (x)) \log (\cos (x)) (x \log (\sin (x)) \log (\cos (x))+2)+1}=x \log (\sin (x)) \log (\cos (x))+1$

$\text{}\left[\int_0^{\frac{\pi }{2}} x \log (\sin (x)) \log (\cos (x)) \, dx+\frac{\pi }{2}=\frac{\pi }{2}+i\right]$

$\text{}\left[i=\int_0^{\frac{\pi }{2}} \left(\frac{\pi }{2}-x\right) \log (\sin (x)) \log (\cos (x)) \, dx\right]$

$\text{}\left[i=\frac{1}{4} \pi \int_0^{\frac{\pi }{2}} \log (\sin (x)) \log (\cos (x)) \, dx\right]$

$\text{}\left[\int_0^{\frac{\pi }{2}} \log (\sin (x)) \log (\cos (x)) \, dx=-\frac{1}{48} \pi \left(\pi ^2-6 \log ^2(4)\right)\right]$

We can find the above integral by differentiating the Beta function twice under the integral. I wonder if we can also find the above integral with contour integration. By a change of variables and the symmetry of sin and cos, it suffices to find

$\text{}\left[\int_0^{1} \log (\sin (\pi x)) \log (\cos (\pi x)) \, dx \right]$

Let

$f(z) = \log \left(1-e^{2 i \pi z}\right) \log \left(1+e^{2 i \pi z}\right)=\log \left(-2 i e^{i \pi z} \sin (z)\right) \log \left(2 e^{i \pi z} \cos (z)\right)=\left(i \pi z+\log (\sin (z))-\frac{i \pi }{2}+\log (2)\right) (i \pi z+\log (\cos (z))+\log (2)) = -\pi ^2 z^2+\frac{\pi ^2 z}{2}+2 i \pi z \log (2)+i \pi z \log (\sin (z))+\log (2) \log (\sin (z))+i \pi z \log (\cos (z))+\log (2) \log (\cos (z))-\frac{1}{2} i \pi \log (\cos (z))+\log (\sin (z)) \log (\cos (z))+\log ^2(2)-\frac{1}{2} i \pi \log (2)$

(Edit: Argument of Sin and Cos should be Pi*z not z)

(I am brushing over some details about choosing which branch of the log.)

It is well known how to compute

$\text{}\left[\int_0^1 \log (\sin (\pi x)) \, dx\right]$

$\text{}\left[\int_0^1 \log (\cos (\pi x)) \, dx\right]$

with contour integration. I think we can also compute

$\text{}\left[\int_0^1 x \log (\sin (\pi x)) \, dx\right]$

$\text{}\left[\int_0^1 x \log (\cos (\pi x)) \, dx\right]$

in the same way (not totally sure about this though).

Hence, to find the desired integral, it suffices to find

$\int_0^1 f(x) \, dx$

which can probably done with contour integration by considering the rectangle with corners (0,0), (1,0), (0,R), (1,R) and let R-> Infinity. Note due to singularities at 0 and 1 and 1/2, you may have to consider circular sectors of radius epsilon that go around (0,0) and (1,0) and (1/2,0), and then let epsilon -> 0.

Does anybody have a nice way of finding this integral??

Anyway, the final answer is:

$\frac{\pi }{2}-\frac{\pi ^4}{192}+\frac{1}{32} \pi ^2 \log ^2(4)$