intlimits_{0}^{5} cos left(pileft(x-left[frac | vedclass

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(D) Let $I = \int\limits_{0}^{5} \cos \left(\pi x - \pi \left[\frac{x}{2}\right]\right) d x$.We split the integral based on the intervals where $\left

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(D) Let $I = \int\limits_{0}^{5} \cos \left(\pi x - \pi \left[\frac{x}{2}\right]\right) d x$.We split the integral based on the intervals where $\left[\frac{x}{2}\right]$ is constant:For $x \in [0, 2)$,$\left[\frac{x}{2}\right] = 0$.For $x \in [2, 4)$,$\left[\frac{x}{2}\right] = 1$.For $x \in [4, 5]$,$\left[\frac{x}{2}\right] = 2$.Thus,$I = \int\limits_{0}^{2} \cos(\pi x) d x + \int\limits_{2}^{4} \cos(\pi x - \pi) d x + \int\limits_{4}^{5} \cos(\pi x - 2\pi) d x$.Evaluating each integral:$\int\limits_{0}^{2} \cos(\pi x) d x = \left[\frac{\sin(\pi x)}{\pi}\right]_{0}^{2} = \frac{\sin(2\pi) - \sin(0)}{\pi} = 0$.$\int\limits_{2}^{4} \cos(\pi x - \pi) d x = \left[\frac{\sin(\pi x - \pi)}{\pi}\right]_{2}^{4} = \frac{\sin(3\pi) - \sin(\pi)}{\pi} = 0$.$\int\limits_{4}^{5} \cos(\pi x - 2\pi) d x = \left[\frac{\sin(\pi x - 2\pi)}{\pi}\right]_{4}^{5} = \frac{\sin(3\pi) - \sin(2\pi)}{\pi} = 0$.Therefore,$I = 0 + 0 + 0 = 0$.

$\int\limits_{0}^{5} \cos \left(\pi\left(x-\left[\frac{x}{2}\right]\right)\right) d x$,where $[t]$ denotes the greatest integer less than or equal to $t$,is equal to:

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