This is strange. If $r$ is a $C^2$ extremal, and $F(x,r(x),r'(x))$ we'd have $F_r(x,r(x),r'(x)) - \frac{d}{dx} F_{r'}(x,r(x),r'(x)) = 0$ for $x \in (0,\pi/2)$, which then means $\int_{a}^{b} F_r(x,r(x),r'(x)) = F_{r'}(b,r(b),r'(b)) - F_{r'}(a,r(a),r'(a))$ for $0 < a < b < \pi/2$. The LHS is

$ \int_a^b \left( \frac{1}{\pi } \ln \left( {\frac {\sin \left( x \right) }{1-\cos \left( x \right) }} \right) + \frac{1}{2} \right) {\frac {r(x)}{\sqrt {{r(x)}^{2} + {r'(x)}^{2}}}} $

which, provided $r(0)=1$, goes to a finite value as $a \rightarrow 0^+$. However, $ F_{r'}(a,r(a),r'(a)) = \left( \frac{1}{\pi } \ln \left( {\frac {\sin \left( a \right) }{1-\cos \left( a \right) }} \right) + \frac{1}{2} \right) {\frac {r'(a)}{\sqrt {{r(a)}^{2} + {r'(a)}^{2}}}}$ which goes to $\pm \infty$ as $a \rightarrow 0^+$ if $r'(0) \neq 0$. Thus we are left to conclude that $r'(0) = 0$. But 'experimental' data (via discretizations of the problem), show this isn't the case! Either what I just wrote is wrong, or something is seriously wrong!

This is strange. If $r$ is a $C^2$ extremal, and $F(x,r(x),r'(x))$ we'd have $F_r(x,r(x),r'(x)) - \frac{d}{dx} F_{r'}(x,r(x),r'(x)) = 0$ for $x \in (0,\pi/2)$, which then means $\int_{a}^{b} F_r(x,r(x),r'(x)) = F_{r'}(b,r(b),r'(b)) - F_{r'}(a,r(a),r'(a))$ for $0 < a < b < \pi/2$. The LHS is

$ \int_a^b \left( \ln \left( {\frac {\sin \left( x \right) }{1-\cos \left( x \right) }} \right) + 1 \right) {\frac {r(x)}{\sqrt {{r(x)}^{2} + {r'(x)}^{2}}}} $

which, provided $r(0)=1$, goes to a finite value as $a \rightarrow 0^+$. However, $ F_{r'}(a,r(a),r'(a)) = \left( \ln \left( {\frac {\sin \left( a \right) }{1-\cos \left( a \right) }} \right) + 1 \right) {\frac {r'(a)}{\sqrt {{r(a)}^{2} + {r'(a)}^{2}}}}$ which goes to $\pm \infty$ as $a \rightarrow 0^+$ if $r'(0) \neq 0$. Thus we are left to conclude that $r'(0) = 0$. But 'experimental' data (via discretizations of the problem), show this isn't the case! Either what I just wrote is wrong, or something is seriously wrong!