This question is originally a simplified version from a qualifying exam problem from Archimedean Integration Bee (Cambridge University).
I tried doing angular symmetry but that gives me $$\int_{0}^{\frac{\pi}{2}}\ln(\csc(x)+\cot(x))\sec(x)dx.$$
I thought I could add both integrals and perform reverse product rule, but unfortunately
$$\frac{d}{dx}\left [ \ln(\sec(x)+\tan(x))\ln(\csc(x)+\cot(x)) \right ]=\sec(x)\ln(\csc(x)+\cot(x))-\csc(x)\ln(\sec(x)+\tan(x)).$$
According to wolframalpha, the answer is $\frac{\pi^2}{4}$. This made me think that it uses some sort of Taylor Series somewhere, or a certain identity, but what's the quick solution to this problem?
Best Answer
Hint
If you use the tangent half-angle subsitution $$\int_0^{\frac \pi 2}\csc (x) \log (\tan (x)+\sec (x))\,dx=\int_0^{1}\frac 1 t \log \left(\frac{1+t}{1-t}\right)\,dt$$ Even the antiderivative is simple.
You could even simplify using $\frac{1+t}{1-t}=u$