${\int }_0^5 \cos \left(\mathrm{\pi }\left(\mathrm{x}-\left[\frac{\mathrm{x}}{2}\right]\right)\right)\mathrm{dx}$, where [t] denotes greatest integer less than or equal to t, is equal to :

$\begin{array}{l}{\int }_0^5 \cos \left(\mathrm{\pi }\left(\mathrm{x}-\left[\frac{\mathrm{x}}{2}\right]\right)\right)\mathrm{dx}\\ ={\int }_0^2 \cos \left(\mathrm{\pi }\left(\mathrm{x}-\left[\frac{\mathrm{x}}{2}\right]\right)\right)\mathrm{dx}+{\int }_2^4 \cos \left(\mathrm{\pi }\left(\mathrm{x}-\left[\frac{\mathrm{x}}{2}\right]\right)\right)\mathrm{dx}+{\int }_4^5 \cos \left(\mathrm{\pi }\left(\mathrm{x}-\left[\frac{\mathrm{x}}{2}\right]\right)\right)\mathrm{dx}\\ ={\int }_0^2 \cos (\mathrm{\pi x}-0)\mathrm{dx}+{\int }_2^4 \cos (\mathrm{\pi x}-\mathrm{\pi })\mathrm{dx}+{\int }_4^5 \cos (\mathrm{\pi x}-2\mathrm{\pi })\mathrm{dx}\\ ={\left[\frac{\sin \mathrm{\pi x}}{\mathrm{\pi }}\right]}_0^2-{\left[\frac{\sin \mathrm{\pi x}}{\mathrm{\pi }}\right]}_2^4+{\left[\frac{\sin \mathrm{\pi x}}{\mathrm{\pi }}\right]}_4^5\\ =0\end{array}$