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?
 A: 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, 
  $$L = 2\cdot \frac38\pi = \frac34\pi.$$

