Inspired by all

Simpler one!

$y=\sqrt{(x-sinx)+\sqrt{(x-sinx)+....\infty}}$

$\rightarrow y^2 - y = x - \sin x .........(i)$

$|\frac{dx}{dy}|^{2}_{at~x=\frac{\pi}{2}} = (2y-1)^2$

On solving euqation $(i)$ at $x = \frac{\pi}{2}$ , we get the value of $y$

Now for the Limit part, $4\displaystyle \lim_{n\to \infty} [\frac{1}{\sqrt{n^{2}-1}}+\frac{1}{\sqrt{n^{2}-2^{2}}}+....+\frac{1}{\sqrt{n^{2}-(n-1)^{2}}}]= 2\pi \rightarrow$ by using Summation of series using Definite Integral.

Put the values, and get the answer :)