If $$[\cdot]$$ represents the greatest integer function, then the value of $$\left|\int_0^{\sqrt{\frac{\pi}{2}}} \left[\left[x^2\right] - \cos x\right] dx\right|$$ is ________.

We evaluate $$\left|\int_0^{\sqrt{\pi/2}} \left[\left[x^2\right] - \cos x\right] dx\right|$$, where $$[\cdot]$$ denotes the greatest integer function.

The upper limit is $$\sqrt{\pi/2} \approx 1.2533$$. As $$x$$ goes from 0 to $$\sqrt{\pi/2}$$, $$x^2$$ goes from 0 to $$\pi/2 \approx 1.5708$$.

For $$0 \leq x < 1$$: $$x^2 \in [0,1)$$, so $$[x^2] = 0$$. The integrand is $$[0 - \cos x] = [-\cos x]$$. Since $$\cos x \in (\cos 1, 1] \approx (0.5403, 1]$$ on this interval, $$-\cos x \in [-1, -0.5403)$$. At $$x = 0$$, $$-\cos 0 = -1$$ and $$[-1] = -1$$. For $$x \in (0,1)$$, $$-1 < -\cos x < 0$$, so $$[-\cos x] = -1$$. Thus $$[-\cos x] = -1$$ on $$[0, 1)$$.

For $$1 \leq x \leq \sqrt{\pi/2}$$: $$x^2 \in [1, \pi/2]$$, so $$[x^2] = 1$$. The integrand is $$[1 - \cos x]$$. Here $$\cos x$$ ranges from $$\cos(\sqrt{\pi/2}) \approx 0.309$$ to $$\cos 1 \approx 0.540$$, so $$1 - \cos x \in (0.460, 0.691)$$. Since $$0 < 1 - \cos x < 1$$, we have $$[1 - \cos x] = 0$$.

Therefore $$\int_0^{\sqrt{\pi/2}} [[x^2] - \cos x]\, dx = \int_0^1 (-1)\, dx + \int_1^{\sqrt{\pi/2}} 0\, dx = -1$$.

Taking the absolute value: $$|-1| = 1$$.

The answer is 1.