Question: How to evaluate $$\int_0^{\frac{\pi}{4}} \tan(x) \ln^2(\sin(4x)) \, dx?$$
My attempt
We will denote the main integral as $\Omega$. $$\Omega=\int_0^{\frac{\pi}{4}} \tan(x) \ln^2(\sin(4x)) \, dx= \int_0^{\frac{\pi}{4}} \tan(x) \left[\left(\ln 2 + \ln(\sin(2x)) + \ln(\cos(2x))\right)^2\right] dx$$
We start by using the logarithmic property for the argument $\sin(4x)$ $$ \ln^2(\sin(4x)) = \left(\ln 2 + \ln(\sin(2x)) + \ln(\cos(2x))\right)^2 $$
Expanding the square, we obtain: $$ \ln^2(\sin(4x)) = \ln^2(2) + \ln^2(\sin(2x)) + \ln^2(\cos(2x)) + 2 \ln 2 \ln(\sin(2x)) + 2 \ln 2 \ln(\cos(2x)) + 2 \ln(\sin(2x)) \ln(\cos(2x)) $$
Substituting this back into the integral, we get: $$ \Omega = \int_0^{\frac{\pi}{4}} \tan(x) \left[\ln^2(2) + \ln^2(\sin(2x)) + \ln^2(\cos(2x)) + 2 \ln 2 \ln(\sin(2x)) + 2 \ln 2 \ln(\cos(2x)) + 2 \ln(\sin(2x)) \ln(\cos(2x))\right] dx $$
Can someone help me proceed further, or suggest any better ideas? Thanks!!