${\int }_0^{41\mathrm{\pi }/2} \sin \mathrm{xdx}$ is equal to

Solution:

sin x is periodic function with period  $2\mathrm{\pi }$

$\begin{array}{l}\mathrm{I}={\int }_0^{41\mathrm{\pi }/2} \sin \mathrm{xdx}={\int }_0^{\mathrm{\pi }/2} \sin \mathrm{xdx}+{\int }_{\mathrm{\pi }/2}^{\mathrm{\pi }/2+20\mathrm{\pi }} \sin \mathrm{xdx}\\ \mathrm{I}=0+{\int }_0^{\mathrm{\pi }/2} \sin \mathrm{xdx}=1\end{array}$