Inspired by Jake Lai

References:

• Beta Function $B(x,y)$

• Digamma Function $\psi(x)$

$\begin{aligned} &\int_0^{\pi/2} 2 \sin^{2x-1}(\theta) \cos^{2y-1}(\theta) d \theta &&= B(x,y)\\ &\int_0^{\pi/2} 4 \sin^{2x-1}(\theta) \cos^{2y-1}(\theta) \ln(\sin(\theta)) d \theta &&= \frac{d}{dx} B(x,y)\\ &\int_0^{\pi/2} 8 \sin^{2x-1}(\theta) \cos^{2y-1}(\theta) \ln(\sin(\theta)) \ln(\cos(\theta)) d \theta &&= \frac{d}{dx\, dy} B(x,y)\\ &\int_0^{\pi/2} \sin(\theta) \ln(\sin(\theta)) \ln(\cos(\theta)) d \theta &&= \frac{1}{8} {\color{#D61F06}\left . \frac{d}{dx\, dy} B(x,y) \right |_{x=1,y=1/2}}\\ &\int_0^{\pi/2} \sin(\theta) \ln(\sin(\theta)) \ln(\cos(\theta)) d \theta &&= \frac{1}{8} {\color{#D61F06}(16 - 8\ln(2)-\pi^2)} = 2 - \ln(2) - \frac{\pi^2}{8}\\ \end{aligned}$

${ \color{#D61F06}\begin{aligned} &\frac{d}{dx\, dy} B(x,y) &&=B(x,y)\left [ (\psi(x)-\psi(x+y))(\psi(y)-\psi(x+y)) - \psi^{'}(x+y) \right ]\\ & \left . \frac{d}{dx\, dy} B(x,y) \right |_{x=1,y=1/2} &&=B(1,1/2)\left [ (\psi(1)-\psi(3/2))(\psi(1/2)-\psi(3/2)) - \psi^{'}(3/2) \right ]\\ & \left . \frac{d}{dx\, dy} B(x,y) \right |_{x=1,y=1/2} &&=2\left [ (- \gamma +\gamma+2 \ln(2) - 2)(\psi(1/2)-\psi(1/2) - 2) - (\frac{\pi^2}{2} - 4) \right ]\\ & \left . \frac{d}{dx\, dy} B(x,y) \right |_{x=1,y=1/2} &&=16 - 8\ln(2)-\pi^2 \end{aligned}}$

${\color{#3D99F6}\begin{aligned} &\psi(1)= - \gamma\\ &\psi(1/2)= - \gamma- 2\ln(2)\\ &\psi(x+1)= \psi(x) + \frac{1}{x}\Rightarrow \psi(3/2)=\psi(1/2+1)= \psi(1/2) + 2 \\ &\psi(1-x) - \psi(x) = \pi \cot(\pi x)\Rightarrow - \psi(1-x)^{'} - \psi(x)^{'} = - \pi^2 \csc^2(\pi x) \\ & \Rightarrow \psi(1-1/2)^{'} + \psi(1/2)^{'} = \pi^2 \csc^2(\pi/2)\Rightarrow \psi(1/2)^{'} = \frac{\pi^2}{2} \\ &\psi(x+1)= \psi(x) + \frac{1}{x}\Rightarrow \psi(x+1)^{'}= \psi(x)^{'} - \frac{1}{x^2}\\ & \Rightarrow \psi(3/2)^{'} = \psi(1/2+1)^{'}= \psi(1/2)^{'} - 4 = \frac{\pi^2}{2} - 4 \end{aligned}}$

Great question.I enjoyed solving it.I got the answer as A=2,B=1,C=2,D=8 and E=2