Logarithms, Trigonometry, Integrals, Part 2

We start with the known result from part 1 of this problem

$\displaystyle I=\int _{ 0 }^{ \pi /2 }{ \log (\cos (x))\log (\sin (x))dx }$ =For result, try the first part.

$\displaystyle \int _{ 0 }^{ \pi /2 }{ { [\log (\cos (x))+\log (\sin (x)) }]^{ 2 }dx } =2\int _{ 0 }^{ \pi /2 }{ { (\log (\sin (x))) }^{ 2 }dx } +2I\\ \displaystyle \Rightarrow 2\int _{ 0 }^{ \pi /2 }{ { (\log (\sin (x))) }^{ 2 }dx } +2I=\int _{ 0 }^{ \pi /2 }{ { (\log (\sin (2x))) }^{ 2 }dx } +\int _{ 0 }^{ \pi /2 }{ { log }^{ 2 }2\quad dx } -2log2\int _{ 0 }^{ \pi /2 }{ { \log (\sin (x)) }dx } \\ \displaystyle \Rightarrow 2\int _{ 0 }^{ \pi /2 }{ { (\log (\sin (x))) }^{ 2 }dx } +2I=\int _{ 0 }^{ \pi /2 }{ { (\log (\sin (t))) }^{ 2 }dt } +\frac { 3\pi }{ 2 } { log }^{ 2 }2\\ \displaystyle \Rightarrow \int _{ 0 }^{ \pi /2 }{ { (\log (\sin (x))) }^{ 2 }dx } =\frac { 3\pi }{ 2 } { log }^{ 2 }2-2I\\ \displaystyle \Rightarrow \int _{ 0 }^{ \pi /2 }{ { (\log (\sin (x))) }^{ 2 }dx } -I=\int _{ 0 }^{ \pi /2 }{ \log { (sin(x)) } } \log { (tan(x)) } dx=\frac { 3\pi }{ 2 } { log }^{ 2 }2-3I$

Put value of $I$ to get answer.

$Let \quad I = \int_{0}^{\pi/2}{\log(\sin{x})\log(\tan{x})dx } \\ Now \quad I = \int_{0}^{\pi/2}{\log^2(\sin{x})dx} - \int_{0}^{\pi/2}{\log(\sin{x}) \log(\cos{x})dx} \\ I = \frac{1}{2} \left(2 \int_{0}^{\pi/2}{\log^2(\sin{x})dx} - 2 \int_{0}^{\pi/2}{\log(\sin{x}) \log(\cos{x})dx} \right) \\ I = \int_{0}^{\pi/2}{\frac{\log^2(\tan{x})dx}{2}} \\ Let \quad I_1 =\int_{0}^{\pi/2}{\frac{\tan^a{x}dx}{2}} \\ \text{Using Beta Integral} \\\int_{0}^{\pi/2}{\frac{\tan^a{x}dx}{2}} = \frac{1}{4} \beta{((a/2 + 0.5),(1 - a/2 - 0.5))} \\ \text{Using reflection formula} \\ \int_{0}^{\pi/2}{\frac{\tan^a{x}dx}{2}} = \frac{\pi}{4}\csc{\left(\frac{\pi a}{2} + \frac{\pi}{2}\right)} \\ \text{Differentiating w.r.t a twice and puting a = 0 we get } \\ \int_{0}^{\pi/2}{\frac{\log^2(\tan{x})dx}{2}} = \frac{\pi^3}{16}$ Thanks to Shivang Jindal for pointing in right direction.