For $\mathrm{n}\in \mathrm{N},0<\mathrm{t}<\mathrm{\pi }/2;\text{ the value of }{\int }_0^{\mathrm{n\pi }+\mathrm{t}} \left(\right|\cos \mathrm{x}|+|\sin \mathrm{x}\left|\right)\mathrm{dx}=$

$$\begin{gathered} {\int }_0^{\mathrm{n\pi }+\mathrm{t}} (\left|\cos \mathrm{x}\right|+|\sin \mathrm{x}\left|\right)\mathrm{dx} \\ ={\int }_0^{\mathrm{n\pi }} \left(\right|\cos \mathrm{x}|+|\sin \mathrm{x}\left|\right) dx+{\int }_{\mathrm{n\pi }}^{\mathrm{n\pi }+\mathrm{t}} \left(\right|\cos \mathrm{x}|+|\sin \mathrm{x}\left|\right)\mathrm{dx} \\ =\left[{\int }_0^{2\mathrm{n}(\mathrm{\pi }/2)} \left(\right|\cos \mathrm{x}|+|\sin \mathrm{x}\left|\right)\right] +{\int }_0^{\mathrm{t}} \left(\right|\cos \mathrm{x}|+|\sin \mathrm{x}\left|\right)\mathrm{dx} \\ =2n{\int }_0^{\frac{\pi }{2}} \left(\right|\cos \mathrm{x}|+|\sin \mathrm{x}\left|\right)dx +{\int }_0^{\mathrm{t}} \left(\right|\cos \mathrm{x}|+|\sin \mathrm{x}\left|\right)\mathrm{dx} \left(\because \mathrm{period} \mathrm{of} |\cos \mathrm{x}|+|\sin \mathrm{x}| \mathrm{is} \frac{\mathrm{\pi }}{2}\right) \\ =2\mathrm{n}\left[{\int }_0^{\mathrm{\pi }/2} (\cos \mathrm{x}+\sin \mathrm{x}) dx\right] +{\int }_0^{\mathrm{t}} (\cos \mathrm{x}+\sin \mathrm{x})dx \left(\because 0<t<\frac{\pi }{2}\right) \\ =2n{(\sin x-\cos x)}_0^{\frac{\pi }{2}}+{(\sin x-\cos x)}_0^t =2\mathrm{n}\left(2\right)+(\sin \mathrm{t}-\cos t-(-1\left)\right) \\ =4n+\sin t-\cos t+1 \end{gathered}$$