If [ . ] represents the greatest integer function, then the value of


$$\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$$ is ____________.

$\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]} dx$<br><br>$$ = \int\limits_0^1 {[ - \cos x]dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {[1 - \cos x]dx} $$ <br><br>= $$\int\limits_0^1 {\left[ { - \cos x} \right]} dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {1dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {\left[ { - \cos x} \right]} dx$$ <br><br>When 0 $\le$ x $\le$ 1 then -1 $\le$ -cosx $\le$ 0 <br><br>$\therefore$ ${\left[ { - \cos x} \right]}$ = -1 <br><br>When 1 $\le$ x $\le$ ${\sqrt {{\pi \over 2}} }$ = 1.24 then -0.33 $\le$ -cosx $\le$ 0 <br><br>$\therefore$ ${\left[ { - \cos x} \right]}$ = -1 (Integer value present in the left side of -0.33) <br><br>$$ = - \int\limits_0^1 {dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {dx - \int\limits_1^{\sqrt {{\pi \over 2}} } {dx} } } $$<br><br>$= - (x)_0^1 = - 1$<br><br>$\therefore$ $$\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$$ = 1