The integral has the numerical value$$\int_0^{\frac{\pi}{2}}\left( \frac{1}{\log(\sin x)}+\frac{1}{1-\sin x} \right)dx=0.86995763688\dots $$I have been unsuccesfully trying to find a closed form for this, if it exists. The term $\frac{1}{\log(\sin x)}$ directed me towards Feynman's technique, but I couldn't find a suitable place to insert a variable.
In order to make the $\log$ go, one has to consider integrals of the form$$F(a)=\int_0^{\frac{\pi}{2}}\left( \frac{\sin^ax}{\log(\sin x)}+\frac{g(a)}{1-\sin x} \right)dx $$ which after differentiation evaluates to$$F'(a)=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\frac{a+1}{2}\right)}{\Gamma\left(\frac{a+2}{2}\right)}+\int_0^{\frac{\pi}{2}}\frac{g'(a)}{1-\sin x} dx$$But at this point, it seems the first term cannot be integrated back, I tried but couldn't find an antiderivative.
Another approach could be a series expansion of the integrand, and convenientely $\frac{1}{1-\sin x}$ happens to be a nice geometric series. On the other hand, $\frac{1}{\log(\sin x)}$ has revealed to be a tough term to expand as a Taylor series and as a Fourier series as well.
Any help is appreciated! Thank you.