Is there any closed form for the integral $\int_0^{\frac{\pi}{2}} \frac{\ln ^n(\sin x) \ln ^m(\cos x)}{\tan x} d x?$ Inspired by the question in the post, I started to generalise the integral
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{\ln^n (\sin x) \ln (\cos x)}{\tan x} d x =& \frac{1}{n+1} \int_0^{\frac{\pi}{2}} \ln (\cos x) d\left(\ln ^{n+1}(\sin x)\right) \\
=& \frac{1}{n+1}\left[\ln (\cos x) \ln ^{n+1}(\sin x)\right]_0^{\frac{\pi}{2}}+\frac{1}{n+1} \int_0^{\frac{\pi}{2}} \frac{\sin x \ln ^{n+1}\left(\sin x\right)}{\cos x} d x\\\stackrel{IBP}{=} &\frac{1}{2^{n+2}(n+1)} \int_0^{\frac{\pi}{2}} \frac{\ln ^{n+1}\left(\sin ^2 x\right)}{\cos ^2 x} d\left(\sin ^2 x\right)\\=& \frac{1}{2^{n+2}(n+1)} \int_0^{1} \frac{\ln ^{n+1} y}{1-y} d y \quad \textrm{ where }y=\sin^2 x\end{aligned}
$$
We then deal the last integral using an infinite series on the integral
$$
\begin{aligned}
\int_0^1 \frac{y^a}{1-y} d y=\sum_{k=0}^{\infty} \int_0^1 y ^a\cdot y^k d y =\sum_{k=1}^{\infty} \frac{1}{a+k}
\end{aligned}
$$
Differentiating both sides w.r.t. $a$ by $n$ times yields $$
\int_0^1 \frac{\ln^{n+1} y}{1-y} d y=\left.\frac{\partial^{n+1}}{\partial a^{n+1}} \int_0^1 \frac{y^a}{1-y} d t\right|_{a=0} =(-1)^{n+1}(n+1) ! \sum_{k=1}^{\infty} \frac{1}{k^{n+2}}= (-1)^{n+1}(n+1) !\zeta(n+2)
$$
Then we can conclude that
$$\boxed{\int_0^{\frac{\pi}{2}} \frac{\ln^n (\sin x) \ln (\cos x)}{\tan x} d x = \frac{(-1)^{n+1} n !}{2^{n+2}} \zeta(n+2)} $$

For examples,
$$
\begin{aligned}
&I_1=\frac{1}{8} \zeta(3);\quad  I_2=-\frac{1}{16} \cdot 2 \cdot \zeta(4)=-\frac{\pi^4}{720} ;\quad  I_{12}=-\frac{12 !}{2^{14} }\zeta(12)=-\frac{691 \pi^{12}}{21840}
\end{aligned}
$$

Similarly, I want to generalise the integral further as
$$I(n,m)=\int_0^{\frac{\pi}{2}} \frac{\ln ^n(\sin x) \ln ^m(\cos x)}{\tan x} d x =\frac{m}{2^{n+m}(n+1)} \int_0^1 \frac{\ln ^{n+1} y \ln ^{m-1}(1-y)}{1-y} d y $$
Noticing that
$$
\begin{aligned}
\int_0^1 y^a(1-y)^bd y =& \int_0^1 y^a(1-y)^{b} d y =B(a+1, b+1)
\end{aligned}
$$
Therefore
$$
\boxed{I(n, m)= \frac{1}{2^{n+2}(n+1)} \left.\frac{\partial^{n+1}}{\partial a^{n+1}} \frac{\partial^{m-1}}{\partial b^{m-1}}\left(\frac{\Gamma(a+1) \Gamma(b+1)}{\Gamma(a+b+2)}\right)\right|_{a=0,b=-1}}
$$
My question: Is there a closed form for the last high derivative?
Your comments and suggestions are highly appreciated.

After submitting the post, I suddenly recognise that $I(n,m) $ is actually a high derivative of a Beta function as
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{\sin ^a x \cos ^b x}{\tan x} d x 
&=\int_0^{\frac{\pi}{2}} \sin ^{a-1} x \cos ^{b+1} x d x 
=\frac{1}{2} B\left(\frac{a}{2} , \frac{b}{2} +1\right)
\end{aligned}
$$
Then we can write down
$$
\begin{aligned}\int_0^{\frac{\pi}{2}} \frac{\ln ^n(\sin x) \ln ^m(\cos x)}{\tan x} d x & =\left.\frac{1}{2} \frac{\partial^{n+m}}{\partial a^n \partial b ^m}\left(B\left(\frac{a}{2}  , \frac{b}{2} +1\right)\right)\right|_{a,b \rightarrow 0}\\&=\frac{1}{2^{n+m+1}} \lim _{a \rightarrow 0} \lim _{b \rightarrow 1}\left[\frac{\partial^{n+m}}{\partial a^n\partial b^m} B(a, b)\right]\end{aligned}
$$
