Evaluate $\int_0^{\pi/2}x\cot{(x)}\ln^4\cot\frac{x}{2}\,\mathrm dx$ How to evaluate the following integral ?:
$$
\int_{0}^{\pi/2}x\cot\left(\, x\,\right)\ln^{4}\left[\,\cot\left(\,{x \over 2}\,\right)\,\right]\,{\rm d}x
$$
It seems that evaluate to
$$
{\pi \over 16}\left[\,
5\pi^{4}\ln\left(\, 2\,\right) - 6\pi^{2}\zeta\left(\, 3\,\right)
-{93 \over 4}\,\zeta\left(\, 5\,\right)
\,\right]
$$
Exactly ?.
 A: Let $$J = \int_0^1 {\frac{{\arctan x{{\ln }^4}x}}{x}dx} \qquad K = \int_0^1  {\frac{x{\arctan x{{\ln }^4}x}}{{1 + {x^2}}}dx}$$
Then by M.N.C.E.'s comment, $$\tag{1}I = \int_0^{\pi /2} {x\cot x{{\ln }^4}\left( {\cot \frac{x}{2}} \right)dx} = 2J - 4K$$

Here is a symmetry of the integrand that we can exploit:
$$\begin{aligned}
K &= \int_0^1 {\frac{{x\arctan x{{\ln }^4}x}}{{1 + {x^2}}}dx}  
\\&= \int_0^1 {\frac{{\arctan x{{\ln }^4}x}}{x}dx}  - \int_0^1 {\frac{{\arctan x{{\ln }^4}x}}{{x(1 + {x^2})}}dx} 
\\&= J - \int_1^\infty  {\frac{{x\left( {\frac{\pi }{2} - \arctan x} \right){{\ln }^4}x}}{{1 + {x^2}}}dx} 
\\& = J - \int_1^\infty  {\frac{1}{x}\left( {\frac{\pi }{2} - \arctan x} \right){{\ln }^4}xdx}  + \int_1^\infty  {\frac{1}{{x(1 + {x^2})}}\left( {\frac{\pi }{2} - \arctan x} \right){{\ln }^4}xdx} 
\\& = J - J + \frac{\pi }{2}\int_1^\infty  {\frac{{{{\ln }^4}x}}{{x(1 + {x^2})}}dx}  - \int_1^\infty  {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx} 
\\&= \frac{\pi }{2}\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx}  - \int_0^\infty  {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx}  + \int_0^1 {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx}
\\& = \frac{\pi }{2}\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx}  - \int_0^\infty  {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx}  + \int_0^1 {\frac{{{{\ln }^4}x\arctan x}}{x}dx}  - \int_0^1 {\frac{{x{{\ln }^4}x\arctan x}}{{1 + {x^2}}}dx}
\\& = \frac{\pi }{4}\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx}  - \frac{1}{2}\int_0^\infty  {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx}  + \frac{J}{2} \end{aligned}$$
The fact that exponent $4$ is even is paramount here.
Plugging into $(1)$, the $J$ miraculously cancelled:
$$I = 2\underbrace{\int_0^\infty  {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx}}_{L}  - \pi \underbrace{\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx}}_{45\zeta(5)/64} $$

The crux of the problem is, indeed, evaulating the remaining integral.
Note that $$ L = \int_0^{\frac{\pi }{2}} {x\cot x{{\ln }^4}(\tan x)dx} $$ and we have the following formula:

For $-2<p<1$, $$\int_0^{\frac{\pi }{2}} {x{{\tan }^p}xdx}  = \frac{\pi
 }{4}\csc \frac{{p\pi }}{2}\left[ {\psi ( - \frac{p}{2} + 1) - 2\psi (
 - p + 1) - \gamma } \right]$$

Hence the value of $L$ follows from it by differentiating four times and set $p=-1$:
$$L = -\frac{3 \pi ^3 \zeta (3)}{16}-\frac{3 \pi  \zeta (5)}{8}+\frac{5}{32} \pi ^5 \ln 2$$
Finally, we obtain $$I = 2L - \pi \frac{45\zeta(5)}{64} = \color{blue}{-\frac{3 \pi ^3 \zeta (3)}{8}-\frac{93 \pi  \zeta (5)}{64}+\frac{5}{16} \pi ^5 \ln 2}$$
A: 
How to evaluate the following integral ?

Using M.N.C.E.'s hints, rewrite his first integral in terms of $\displaystyle\int_0^1\frac{\ln^5x}{1+x^2}dx$ using integration by parts with regard to $\dfrac{dx}x=d\big(\ln x\big)$, then expand $\dfrac1{1+x^2}$ into its binomial series, and switch the order of summation and integration. This will yield a very familiar series. Similar tricks apply for the second one as well, only first rewrite $\ln^4x$ as $\bigg[\dfrac{d^4}{dn^4}x^n\bigg]_{\large n=0}\quad$ and then switch the order of integration and differentiation. Once again, you will encounter a very familiar series.
