On the integral $\int_{-\pi/2}^{\pi/2}\sin(x/\sin(x/\sin(x/\sin\cdots)))\,dx$

This question is the final one out of the set (see I and II), I promise!

Consider $f_1(x)=\sin(x)$ and $f_2(x)=\sin\left(\frac x{f_1(x)}\right)$ such that $f_n$ satisfies the relation $$f_n(x)=\sin\left(\frac x{f_{n-1}(x)}\right).$$ To what value does $$L:=\lim_{k\to\infty}\int_{-\pi/2}^{\pi/2} f_{2k-1}(x)\,dx$$ converge, for $k=1,2,\cdots$?

Here is a very nice graph showing the likely convergence of $f_n$:

The $R^2$ value is extremely close to $1$, and the best fit curve is given by the equation $$y=\frac{0.2091}{e^x-0.5226}+2.411$$ which implies that $$L\approx2.411$$

Are there any analytic techniques to prove this?

Similar to $\int_0^\pi\sin(x+\sin(x+\sin(x+\cdots)))\,\mathrm dx=2$, we only need to concern the integral on $[0, \pi/2]$ as mentioned by @Stijn Dietz.

Let $\operatorname{Sb}(x)$ be the inverse function of $x\sin x$ on $[0, \pi/2]$ (such function exists by the injectivity). Therefore, $$t\sin t =x\implies t=\operatorname{Sb}(x).$$

Assume $f_\infty$ exists (see 2. the third integral), then \begin{align*} f_\infty &= \sin\left(\frac x{f_\infty}\right)\\ x &=\frac x{f_\infty}\sin\left(\frac x{f_\infty}\right)\\ \operatorname{Sb}(x) &=\frac x{f_\infty}\\ f_\infty(x)&=\frac x{\operatorname{Sb}(x)}. \end{align*}

Thus, using the fact that $\operatorname{Sb(0)}=0$, $\operatorname{Sb\left(\dfrac\pi2\right)}=\dfrac\pi2$ and substituting $x = y\sin y$, $\operatorname{Sb}(x)=y$ gives \begin{align*} \int_0^{\pi/2} f_\infty(x)\,\mathrm dx & = \int_0^{\pi/2} \frac x{\operatorname{Sb}(x)}\,\mathrm dx\\ & = \int_{\operatorname{Sb}(0)}^{\operatorname{Sb}(\pi/2)} \frac {y\sin y}{y}\,\mathrm d(y\sin y)\\ & = \int_0^{\pi/2} \sin y(\sin y+y\cos y)\,\mathrm dy\\ & = \frac38\pi. \end{align*}

Therefore, $$L = 2\cdot \frac38\pi = \frac34\pi.$$