Verify that $\int_{0}^{\pi/2}\tan x\ln(\tan x)\ln(\cos x)\ln^3(\sin x)\mathrm dx=-\frac{3\zeta(6)}{2^8}$

Question

$$\int_{0}^{\pi/2}\tan x\ln(\tan x)\ln(\cos x)\ln^3(\sin x)\mathrm dx=-\frac{3\zeta(6)}{2^8}$$

Any hints how to simplify this integral to more manageable to be solve.

Answer 1

Use the trigonometric form of the Beta function in (14) here http://mathworld.wolfram.com/BetaFunction.html

$$I=\underbrace{\int_{0}^{\pi/2}\tan x\ln(\cos x)\ln^4(\sin x)\mathrm dx}_{\text{Beta function}}-\underbrace{\int_{0}^{\pi/2}\tan x\ln^2(\cos x)\ln^3(\sin x)\mathrm dx}_{\text{Beta function} }.$$

Answer 2

Considering $$ \begin{aligned} J(a, b, c) &=\int_0^{\frac{\pi}{2}} \tan ^a x \cos ^b x \sin ^c x d x \\ &=\int_0^{\frac{\pi}{2}} \sin ^{a+c} x \cos ^{b-a} x d x \\ &=\frac{1}{2} B\left(\frac{a+c+1}{2}, \frac{b-a+1}{2}\right), \end{aligned} $$ then $$\boxed{I=\left.\frac{1}{2} \frac{\partial^3}{\partial a \partial b \partial c^3} B\left(\frac{a+c+1}{2}, \frac{b-a+1}{2}\right)\right|_{a=1, b=0=c}}$$