On the integral $\int_0^{\frac{\pi}{2}}x^{2n+1}\cot(x)dx$ While investigating the function $$A(z)=\int_0^\frac{\pi}{2} \frac{\sin(zx)}{\sin(x)}dx$$ I stumbled upon the integral $$\int_0^{\frac{\pi}{2}}x^{2n+1}\cot(x)dx$$ when attempting to calculate the taylor series of $A(z)$ at $z=1$. As the coefficients of the even powers in the series reduce to integrating over a polynomial which is fairly trivial, the only real problem I have is in determining the the coefficients of the odd powers as I cannot seem to find a pattern between the coefficients. 
Wolfram Alpha evaluates the first couple of integrals as:
\begin{align*}
\int_0^\frac{\pi}{2} x\cot(x)dx&=\frac{\pi\ln(2)}{2}\\
\int_0^\frac{\pi}{2}x^3\cot(x)dx&=\frac{1}{16}(\pi^3\ln(4)-9\pi\zeta(3))\\
\int_0^\frac{\pi}{2}x^5\cot(x)dx&=\frac{1}{64}(-3\pi^3\zeta(3)+225\pi\zeta(5)+\pi^5\ln(4))
\end{align*}
and in general it seems that higher powers could also be calculated in terms of the zeta function, multiples of $\pi$, and $\ln(2)$. So far I have been unsuccessful in determining a pattern for these integrals but if anyone has any ideas I would be very grateful for any help on this.
 A: Denote your integral as $\mathfrak{I}(n)$ and apply IBP by choosing  $u=x^{2n+1}$ and $\mathrm dv=\cot(x)\mathrm dx$ to get
\begin{align*}
\mathfrak{I}(n)&=\int_0^{\pi/2}x^{2n+1}\cot(x)\mathrm dx =\underbrace{\left[(2n+1)\cdot x^{2n}\log(\sin x)\right]_0^{\pi/2}}_{\to0}-(2n+1)\int_0^{\pi/2}x^{2n}\log(\sin x)\mathrm dx\\
&=-(2n+1)\int_0^{\pi/2}x^{2n}\log(\sin x)\mathrm dx
\end{align*}
Now utilizing the well-known Fourier series expansion of $\log(\sin x)$, which converges within $[0,\pi]$, and switching the order of summation and integration further gives us
\begin{align*}
\mathfrak{I}(n)&=-(2n+1)\int_0^{\pi/2}x^{2n}\log(\sin x)\mathrm dx\\
&=-(2n+1)\int_0^{\pi/2}x^{2n}\left[-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}k\right]\mathrm dx\\
&=\log(2)\left(\frac\pi2\right)^{2n+1}+(2n+1)\sum_{k=1}^\infty\frac1k\underbrace{\int_0^{\pi/2}x^{2n}\cos(2kx)\mathrm dx}_{=J}
\end{align*}
The integral $J$ can be computed via IBP again which explains the connection to values of the Riemann Zeta Function hence for integer $n$ every IBP step produces another reciprocal power of $n$ which overall combines to sums that can be expressed with the help of the Riemann Zeta Function.
As one may see the values for $n=0$ and $n=1$ can be easily verfied since for $n=0$ $J$ is overall $0$ aswell whereas for $n=1$ the latter integral can be expressed using the Dirichlet Eta Function. To be precise we got
\begin{align*}
n=0:~~~\mathfrak{I}(0)&=\log(2)\left(\frac\pi2\right)^{1}+(1)\sum_{k=1}^\infty\frac1k\underbrace{\int_0^{\pi/2}\cos(2kx)\mathrm dx}_{=0}\\
&=\frac{\pi\log(2)}2
\end{align*}
\begin{align*}
n=1:~~~\mathfrak{I}(1)&=\log(2)\left(\frac\pi2\right)^{3}+(2+1)\sum_{k=1}^\infty\frac1k\int_0^{\pi/2}x^2\cos(2kx)\mathrm dx\\
&=\log(2)\left(\frac\pi2\right)^{3}+3\sum_{k=1}^\infty\frac1k\left[\frac\pi4\frac{\cos(\pi k)}{k^2}\right]_0^{\pi/2}\\
&=\log(2)\left(\frac\pi2\right)^{3}-\frac{3\pi}4\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^3}\\
&=\log(2)\left(\frac\pi2\right)^{3}-\frac{3\pi}4\eta(3)\\
&=\log(2)\left(\frac\pi2\right)^{3}-\frac{9\pi}{16}\zeta(3)\\
&=\frac1{16}(\pi^3\log(4)-9\pi\zeta(3))
\end{align*}
Note that we used the relation $\eta(s)=(1-2^{1-s})\zeta(s)$. Similiar can be done for all integer $n$. So as at least close to a closed-form I can offer the following formula

$$\therefore~\mathfrak{I}(n)~=~\log(2)\left(\frac\pi2\right)^{2n+1}+(2n+1)\sum_{k=1}^\infty\frac1k\int_0^{\pi/2}x^{2n}\cos(2kx)\mathrm dx$$

A: Continuing off of @mrtaurho's excellent answer, we may find another form for 
$$C_k(n)=\int_0^{\pi/2}x^{2n}\cos(2kx)\mathrm dx$$
for $n\in \Bbb N$. First, we note that 
$$C_k(n)=\frac1{(2k)^{2n+1}}\int_0^{k\pi}x^{2n}\cos(x)\mathrm dx$$
Then we integrate by parts with $\mathrm dv=\cos(x)\mathrm dx$:
$$C_k(n)=\frac1{(2k)^{2n+1}}x^{2n}\sin(x)\big|_0^{k\pi}-\frac{2n}{(2k)^{2n+1}}\int_0^{k\pi}x^{2n-1}\sin(x)\mathrm dx$$
$$C_k(n)=-\frac{2n}{(2k)^{2n+1}}\int_0^{k\pi}x^{2n-1}\sin(x)\mathrm dx$$
IBP once again,
$$C_k(n)=-\frac{2n}{(2k)^{2n+1}}\left[-x^{2n-1}\cos(x)\big|_0^{k\pi}+(2n-1)\int_0^{k\pi}x^{2n-2}\sin(x)\mathrm dx\right]$$
$$C_k(n)=(-1)^k\frac{2n(k\pi)^{2n-1}}{(2k)^{2n+1}}-\frac{2n(2n-1)}{(2k)^{2n+1}}C_k(n-1)$$
$$C_k(n)=(-1)^k\frac{n\pi^{2n-1}}{2^{2n}k^2}-\frac{2n(2n-1)}{(2k)^{2n+1}}C_k(n-1)$$
So we have that 
$$\mathfrak{I}(n)=\left(\frac\pi2\right)^{2n+1}\log2-\frac{n(2n+1)}{4^n}\pi^{2n-1}\eta(3)-\frac{n(2n+1)(2n-1)}{4^n}\sum_{k\geq1}\frac{C_k(n-1)}{k^{2n+2}}$$
Which doesn't seem to give any sort of recurrence relation... :(
If I think of any new approaches I'll update my answer.
