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 ?.


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.
