Misleading Integral

Let $y = \xi(x).........(i)$

$\implies ( y - \eta(x)) = \sqrt{\mu(x)}$ $\mbox{ if } x \in[0,\infty)$ Squaring Both sides and arranging terms $\implies ( x - \lceil{x}\rceil)^2 + ( y - \lceil{|x| -1}\rceil )^2 = \lceil{\sin^2{x}}\rceil = 1 ........(ii)$

Now $\mbox{ for }x\in( j , j + 1 ) \mbox{ where }j\in\mathbb{Z} \mbox{ , } \lceil{x}\rceil = j + 1$ $\mbox{ and }\lceil{|x|-1}\rceil = j$

$\implies$ equation (ii) represents a unit circle with center at ( 1, 0 ),( 2, 1 ),...,( j+1, j ),... and so on where $j\in\mathbb{Z}$

With similar calculation check what happens if $x \in(-\infty,0) \mbox{ [You will see that }\xi(x)\mbox{ is symmetrical about the origin]}$

Now we can easily calculate the integral by finding the area bounded by $\xi(x)$

The unit circles... From the plot of $\xi(x)$ it is clear that we need to find the area under the blue curve along x-axis. Which is very is to evaluate.

And because of symmetry we consider our calculation only on the first quadrant.

$\mathcal{ I } = \displaystyle{ \int_{-\pi}^{\pi} \xi(x)dx = 2\int_{0}^{\pi} \xi(x)dx }$

Now $\displaystyle\frac{\mathcal{ I }}{2} = \mathcal{A} = \displaystyle(\frac{3}{4}×\mbox{[Area of the unit circle ]}) + \displaystyle(3×\mbox{[Area of the unit square]}) + \mathcal{C} + \mathcal{P}$

Now $\mathcal{P} = 3×(\pi - 3 )$

And $\mathcal{C} = \frac{\arccos{(4-\pi)} -\sqrt{(\pi-3)(5-\pi)}(4-\pi)}{2}\mbox{ }$

( easy geometrical calculation )

Now $\mathcal{ I } = 2\mathcal{ A } \approx {11.66}$

$\boxed{\therefore \lceil{\mathcal{ I }}\rceil = 12}$

This is thought with which I had developed the problem , I hope I will get to see some more brilliant and efficient solutions :) :) :)