Log Trig $\int_0^{\pi/2}\log^4 \tan \frac{x}{2}dx=\frac{5\pi^5}{32}$ Hi I am trying to integrate a log trigonometric integral given by
$$
I:=\int_0^{\pi/2}\log^4 \tan \frac{x}{2}dx=\frac{5\pi^5}{32}.
$$
This is very similar to a previous integral posted except the power of the logarithm.
Note this integral is also equal to
$$
\int_0^{\pi/2}\log^4 \tan x \, dx=\frac{5\pi^5}{32}.
$$
I have wrote I as
$$
I=\int_0^{\pi/2} \left(\log \sin \frac{x}{2}-\log \cos \frac{x}{2}\right)^4 dx
$$
but got stuck here since factoring this out seems like a mess.  Having seen how David H solved a similar integral I posted, I tried another method starting with I and using $t=\tan x/2$, and obtained
$$
I=2\int_0^{1}\log^4 t \frac{dt}{1+t^2}.
$$
Following this I tried $u=-\log t$ but got stuck after this.  Thanks, it would be nice to see a solution that doesn't reduce the integral to a difficult sum to evaluate

\begin{align}
J&=\int_0^{\frac{\pi}{2}}\log^4 \left(\tan x\right) \, dx\\
&\overset{y=\tan x}=\int_0^\infty \frac{\ln^4 x}{1+x^2}\,dx\\
J_n&=\int_0^\infty \frac{\ln^n x }{1+x^2}\,dx\\
J_0&=\frac{\pi}{2}\\
K_n&=\int_0^\infty\int_0^\infty \frac{\ln^n(xy) }{(1+x^2)(1+y^2)}\,dx\,dy\\
&\overset{u(x)=xy}=\int_0^\infty \int_0^\infty \frac{\ln^n u}{(u^2+y^2)(1+y^2)}\,du\,dy\\
&=\int_0^\infty \frac{\ln^{n+1}u }{u^2-1}\,du\\
&=\int_0^1 \frac{\ln^{n+1}u }{u^2-1}\,du+\int_1^\infty \frac{\ln^{n+1}u }{u^2-1}\,du\\
&=\Big(1+(-1)^n\Big)\int_0^1 \frac{\ln^{n+1}u }{u^2-1}\,du\\
&=\Big(1+(-1)^n\Big)\left(\int_0^1 \frac{\ln^{n+1}u }{u-1}\,du-\int_0^1 \frac{u\ln^{n+1}u }{u^2-1}\,du\right)\\
&=\Big(1+(-1)^n\Big)\left(1-\frac{1}{2^{n+2}}\right)\int_0^1 \frac{\ln^{n+1}u }{u-1}\,du\\
&=\Big(1+(-1)^n\Big)\left(1-\frac{1}{2^{n+2}}\right)(-1)^{n+2}(n+1)!\zeta(n+2)\\
&=\Big(1+(-1)^n\Big)\left(1-\frac{1}{2^{n+2}}\right)(n+1)!\zeta(n+2)\\
K_n&=\sum_{k=0}^n \binom{n}{k} J_kJ_{n-k}\\
K_2&=\pi J_2\\
K_2&=\frac{45}{4}\zeta(4)\\
K_4&=\pi J_4+6J_2^2\\
K_4&=\frac{945}{4}\zeta(6)\\
\end{align}Therefore,
\begin{align}
J_4&=\frac{K_4-6{J_2}^2}{\pi}\\
&=\frac{K_4-6\left(\frac{K_2}{\pi}\right)^2}{\pi}\\
&=\frac{K_4-6{J_2}^2}{\pi}\\
&=\frac{\frac{945}{4}\zeta(6)-6\left(\frac{45}{4}\zeta(4)\times\frac{1}{\pi}\right)^2}{\pi}\\
&=\frac{945\zeta(6)}{4\pi}-\frac{6075\zeta(4)^2}{8\pi^3}\\
\end{align}
Moreover, if you assume,
\begin{align}
\zeta(4)&=\frac{\pi^4}{90}\\
\zeta(6)&=\frac{\pi^6}{945}
 \end{align}
Therefore,
\begin{align}
J&=J(4)\\
&=\frac{945}{4\pi}\times \frac{\pi^6}{945}-\frac{6075}{8\pi^3}\times \left(\frac{\pi^4}{90}\right)^2\\
&=\boxed{\frac{5}{32}\pi^5}
\end{align}
