Summing moments and Riemann zeta values Let $d\mu_n(x)=\cos^{2n}x\,dx$ and consider the averages of moments
$$\alpha_n=\frac{\int_0^{\pi/2}x^4d\mu_n(x)}{\int_0^{\pi/2}d\mu_n(x)}.$$
Then, I have encountered a curious evaluation
$$\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^2(\alpha_{n-1}-\alpha_n)
=\frac{21}8\zeta(4)-\frac32\zeta(2);$$
where $\zeta(s)$ is the Riemann zet function.

Is this true?

 A: After my first failed attempt I now follow the route suggested by Nemo --- which works smoothly.
Starting from Nemo's identities
$$F(b)\equiv\int_0^{\pi/2}\cos^{2n}x\cos bx\,dx=\frac{\pi (2n)!}{2^{2n+1}\Gamma(n+1+b/2)\Gamma(n+1-b/2)}$$
$$\int_0^{\pi/2}x^4\cos^{2n}x\,dx=\lim_{b\rightarrow 0}\frac{d^4}{db^4}F(b),$$
I arrive at
$$\alpha_n=\tfrac{3}{4}\left[ \psi ^{(1)}(n+1)\right]^2-\tfrac{1}{8}\psi ^{(3)}(n+1),$$
where $\psi^{(m)}(x)$ is the Polygamma function
$$\psi^{(m)}(x)=\frac{d^m}{dx^m}\left(\frac{1}{\Gamma(x)}\frac{d}{dx}\Gamma(x)\right),$$
$$\psi^{(m)}(n)=(-1)^{m+1}m!\sum_{k=n}^\infty\frac{1}{k^{m+1}},\;\;m\geq 1,\;\;n\in\mathbb{N}.$$
(The last equation gives the connection to Harmonic numbers mentioned by Nemo.)
The recurrence relation
$$\psi^{(m)}(x+1)=\psi^{(m)}(x)+\frac{(-1)^m m!}{x^{m+1}}$$
implies that
$$\alpha_{n-1}-\alpha_n=\tfrac{3}{2} n^{-2} \psi ^{(1)}(n)-\tfrac{3}{2}n^{-4}.$$
From here we arrive at the result in the OP,
$$\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^2(\alpha_{n-1}-\alpha_n)=\frac{3}{2}\sum_{n=1}^\infty\sum_{k=n+1}^\infty\frac{1}{k^2(n+1)^2}=\frac{21}8\zeta(4)-\frac32\zeta(2).$$

This simple final expression suggests a generalization to higher powers of $x$, but that does not seem to work. If I replace $x^4\cos^{2n}x$ in the definition of $\alpha_n$ by say $x^6\cos^{2n}x$ or $x^8\cos^{2n}x$, I find that $\alpha_{n-1}-\alpha_n$ contains products of Polygamma functions, which I have not been able to sum up in closed form.
