Experimentally I discovered the limit below that says that
$$\lim_{n\to\infty} \int_0^{\pi/2} \frac{1}{\displaystyle \cos\left(\frac{x}{2}\right)\left(\cos\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2^2}\right)\right)\cdots \left(\cos\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2^{2n+1}}\right)\right)}+\cdots$$ $$+\frac{1}{\displaystyle \cos\left(\frac{x}{2^{2n+1}}\right)\left(\cos\left(\frac{x}{2^{2n+1}}\right)-\cos\left(\frac{x}{2}\right)\right)\cdots \left(\cos\left(\frac{x}{2^{2n+1}}\right)-\cos\left(\frac{x}{2^{2n}}\right)\right)} \ dx=2 K$$
or as @robjohn suggested
$$\lim_{n\to\infty}{\Large\int}_0^{\pi/2}\operatorname*{\Large\sum}_{j=1}^{2n+1}\left[\cos\left(\frac{x}{2^j}\right)\prod_{\substack{k=1\\k\ne j}}^{2n+1}\left(\cos\left(\frac{x}{2^j}\right)-\cos\left(\frac{x}{2^k}\right)\right)\right]^{-1}\mathrm{d}x=2 K$$ Do you see any easy way of proving this result?
Denote $\displaystyle u_j=\cos \frac{x}{2^j}$, and rewrite the integrand as \begin{align}\sum_j\frac{u_j^{-1}}{\prod_{k\neq j}\left(u_j-u_k\right)}&=\frac{1}{\prod_{m}u_m} \sum_j \prod_{k\neq j}\frac{0-u_k}{u_j-u_k}=\frac{1}{\prod_{m}u_m}, \end{align} where at the last step we used the Lagrange interpolation formula (see for example the answers to this question).
The product in the denominator is telescoping: using the double sine formula, we obtain $$\sin \frac{x}{2^{2n+1}}{\prod_{m}u_m}=\frac{1}{2^{2n+1}}\sin x.$$
Therefore, the limit reduces to $$\lim_{n\to\infty}\int_0^{{\pi}/{2}}\frac{2^{2n+1}\sin \frac{x}{2^{2n+1}}}{\sin x}dx= \int_0^{{\pi}/{2}}\frac{xdx}{\sin x}=2K,$$ where the last equality is essentially the formula (40) from Wolfram page on Catalan's constant.