If $$x\, \in \left( {2n\pi ,\,2n\pi + \pi } \right)$$     then $$\int_0^x {\left[ {\sin \,x} \right]dx,} $$    where $$\left[ x \right] = $$  greatest integer less than or equal to $$x,$$ is equal to :

Solution :
$$\eqalign{ & {\text{Here }}2n\pi < x < 2n\pi + \pi \cr & \therefore \int_0^x {\left[ {\sin \,x} \right]dx = } \int_0^{2n\pi } {\left[ {\sin \,x} \right]dx} + \int_{2n\pi }^x {\left[ {\sin \,x} \right]dx} \cr & \int_0^{2n\pi } {\left[ {\sin \,x} \right]} dx = n\int_0^{2\pi } {\left[ {\sin \,x} \right]dx} \,\,\,\,\,\left( {\because \,\left[ {\sin \,\overline {x + 2\pi } } \right] = \left[ {\sin \,x} \right]} \right) \cr & = n\left\{ {\int_0^{\frac{\pi }{2}} {\left[ {\sin \,x} \right]dx} + \int_{\frac{\pi }{2}}^\pi {\left[ {\sin \,x} \right]dx} + \int_\pi ^{\frac{{3\pi }}{2}} {\left[ {\sin \,x} \right]dx} + \int_{\frac{{3\pi }}{2}}^{2\pi } {\left[ {\sin \,x} \right]dx} } \right\} \cr & = n\left\{ {\int_0^{\frac{\pi }{2}} {0\,dx} + \int {0\,dx} + \int_\pi ^{\frac{{3\pi }}{2}} { - 1\,dx} + \int_{\frac{{3\pi }}{2}}^{2\pi } { - 1\,dx} } \right\} \cr & = n\left\{ { - \left( {\frac{{3\pi }}{2} - \pi } \right) - \left( {2\pi - \frac{{3\pi }}{2}} \right)} \right\} \cr & = - n\pi \cr & \int_{2n\pi }^x {\left[ {\sin \,x} \right]dx = } \int_{2n\pi }^x {0\,dx = 0,{\text{ if }}\,2n\pi } < x < 2n\pi + \frac{\pi }{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int_{2n\pi + \frac{\pi }{2}}^x {0\,dx = 0,{\text{ if }}\,2n\pi + \frac{\pi }{2} \leqslant x < 2n\pi + \pi } \cr & \therefore I = - n\pi + 0 = - n\pi \cr} $$