There is a known proof of the identity $$\sum\_{k=0}^\infty \frac1{(2k+1)^2}=\frac{\pi^2}8\,\,\,(\*)$$ (equivalent to $\sum \frac1{n^2}=\frac{\pi^2}6$) by expanding $\arcsin x$ as a power series $$\arcsin x=\sum\_{k=0}^\infty \frac{(2k-1)!!}{(2k)!!(2k+1)}x^{2k+1},$$
substituting $x=\sin t$ and integrating for $t$ from 0 to $\pi/2$. The double factorials cancel by some magic and we get $(\*)$.


I wonder whether the higher values of zeta at even integer points may be obtained on this way for other functions similar to $\arcsin$ or other clever integrals.



Higher arcsin wanted
In $[1]$ Borwein and Chamberland show that for $|x|\le 2$
\begin{align\*}
\arcsin^{2N+1}(x/2) &= (2N+1)!\sum\_{k=0}^\infty \frac{G\_N(k){2k\choose k}}{2(2k+1)4^{2k}} x^{2k+1},
 & N = 0,1,\ldots \tag{1} \\
\arcsin^{2N}(x/2) &= (2N)!\sum\_{k=1}^\infty \frac{H\_N(k)}{{2k\choose k}k^2} x^{2k}
 & N = 1,2,\ldots \tag{2}
\end{align\*}
where
\begin{align\*}
G\_0(k) &= 1 \\
G\_N(k) &=
\sum\_{n\_1=0}^{k-1}\frac{1}{(2n\_1+1)^2}
\sum\_{n\_2=0}^{n\_1-1}\frac{1}{(2n\_2+1)^2}
\cdots
\sum\_{n\_{N}=0}^{n\_{N-1}-1}\frac{1}{(2n\_{N}+1)^2}
\end{align\*}
and
\begin{align\*}
H\_1(k) &= 1/4 \\
H\_N(k) &= \frac{1}{4}
\sum\_{n\_1=1}^{k-1}\frac{1}{(2n\_1)^2}
\sum\_{n\_2=1}^{n\_1-1}\frac{1}{(2n\_2)^2}
\cdots
\sum\_{n\_{N-1}=1}^{n\_{N-2}-1}\frac{1}{(2n\_{N-1})^2}.
\end{align\*}
Letting $x=2\sin t$ in $(1)$ and $(2)$ and integrating from $0$ to $\pi/2$ we find the same type of cancellation described in the question statement, with the result
\begin{align}
\sum\_{k=0}^\infty \frac{G\_N(k)}{(2k+1)^2} &= \frac{1}{(2N+2)!}\left(\frac{\pi}{2}\right)^{2N+2} \tag{3} \\
\sum\_{k=1}^\infty \frac{H\_N(k)}{k^2} &= \frac{1}{(2N+1)!}\left(\frac{\pi}{2}\right)^{2N}. \tag{4}
\end{align}
Equations $(3)$ and $(4)$ are natural generalizations of $(\*)$. 
They give the value of $\zeta(2)$ and relate the values of $\zeta(n)$ for $n=4,6,\ldots$ to other interesting sums.


Equation $(3)$ implies for $N=0,1$ that
\begin{align\*}
\sum\_{k=0}^\infty \frac{1}{(2k+1)^2} &= \frac{\pi^2}{8} \\
\sum\_{k=0}^\infty \frac{\psi^{(1)}(k+1/2)}{(2k+1)^2} &= \frac{5\pi^4}{96},
\end{align\*}
where $\psi^{(n)}(k)$ is the polygamma function.
The last result can be reexpressed as
$$\sum\_{k=1}^\infty \frac{1}{k^4}
= \frac{\pi^4}{72} - \frac{4}{15}\sum\_{k=0}^\infty\frac{\psi^{(1)}(k+3/2)}{(2k+1)^2}.$$
(Given $\zeta(4) = \pi^4/90$ this implies 
$\sum\_{k=0}^\infty{\psi^{(1)}(k+3/2)}/{(2k+1)^2} = {\pi^4}/{96}.$)


Equation $(4)$ implies for $N=1,2$ that
\begin{align\*}
\sum\_{k=1}^\infty \frac{1}{k^2} &= \frac{\pi^2}{6} \\
\sum\_{k=1}^\infty \frac{H\_{k-1}^{(2)}}{k^2} &= \frac{\pi^4}{120},
\end{align\*}
where $H\_k^{(n)}$ are the generalized harmonic numbers of order $n$.
The last result can be reexpressed as
$$\sum\_{k=1}^\infty \frac{1}{k^4} = \sum\_{k=1}^\infty\frac{H\_k^{(2)}}{k^2} - \frac{\pi^4}{120}.$$
(Given $\zeta(4)$ this implies 
$\sum\_{k=1}^\infty{H\_k^{(2)}}/{k^2} = {7\pi^4}/{360}.$)


$[1]$ Borwein, J. M. and Chamberland, M. "Integer Powers of Arcsin." *Int. J. Math. Math. Sci.*, Art. 19381, 1-10, 2007.


