$$\int_0^5 \cos\left(\pi\left(x - \left[\frac{x}{2}\right]\right)\right)dx$$, where $$[t]$$ denotes greatest integer less than or equal to $$t$$, is equal to

We need to evaluate $$\int_0^5 \cos\left(\pi\left(x - \left[\frac{x}{2}\right]\right)\right)dx$$.

First, we determine $$\left[\frac{x}{2}\right]$$ on each interval.

For $$x \in [0,2)$$: $$\frac{x}{2} \in [0,1)$$, so $$\left[\frac{x}{2}\right] = 0$$, giving the argument $$\pi x$$.

For $$x \in [2,4)$$: $$\frac{x}{2} \in [1,2)$$, so $$\left[\frac{x}{2}\right] = 1$$, giving the argument $$\pi(x-1)$$.

For $$x \in [4,5]$$: $$\frac{x}{2} \in [2,2.5]$$, so $$\left[\frac{x}{2}\right] = 2$$, giving the argument $$\pi(x-2)$$.

Next, we evaluate each integral separately.

Therefore, $$I_1 = \int_0^2 \cos(\pi x)\,dx = \left[\frac{\sin(\pi x)}{\pi}\right]_0^2 = \frac{\sin 2\pi - \sin 0}{\pi} = 0$$

Similarly, $$I_2 = \int_2^4 \cos(\pi(x-1))\,dx = \left[\frac{\sin(\pi(x-1))}{\pi}\right]_2^4 = \frac{\sin 3\pi - \sin \pi}{\pi} = 0$$

Likewise, $$I_3 = \int_4^5 \cos(\pi(x-2))\,dx = \left[\frac{\sin(\pi(x-2))}{\pi}\right]_4^5 = \frac{\sin 3\pi - \sin 2\pi}{\pi} = 0$$

Finally, adding these results gives $$I = I_1 + I_2 + I_3 = 0 + 0 + 0 = 0$$.

The answer is Option A: $$0$$.