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.