Question

$\underset{0}{overset2n\pi \int }(|\sin x|-\left[\left|\frac{\sin x}{2}\right|\right])dx$ (where [] denotes the greatest integer function and $n\in I$)

Answer 1

$I={\int }_0^{2n\pi }(|\sin x|-\left[\frac{\sin x}{2}\right])dx$

$-1\le \sin x\le 1$

$0\le |\sin x|\le 1$

$0\le \left|\frac{\sin x}{2}\right|\frac{1}{2}\Rightarrow \left[\frac{\sin x}{2}\right]=0$

$I={\int }_0^{2n\pi }\left(|\sin x|\right)dx-{\int }_0^{2n\pi }\left(0\right)dx$

$I={\int }_0^{\pi }\sin xdx+{\int }_{\pi }^{2\pi }(-\sin x)dx+{\int }_{2\pi }^{3\pi }\left(\sin x\right)dx+{\int }_{3\pi }^{4\pi }-\left(\sin x\right)dx+.....+{\int }_{(2\pi -2)\pi }^{(2\pi +1)\pi }\left(\sin x\right)dx+{}^{}{\int }_{(2n-1)\pi }^{\left(2n\pi \right)}(-\sin x)dx$

$=-\cos x{]}_0^{\pi }+\cos {]}_{\pi }^{2\pi }-\cos x{]}_{2\pi }^{3\pi }+\cos x{]}_{3\pi }^{4\pi }+...-\cos x{]}_{(2\pi -2)\pi }^{2\pi -1}+\cos x{]}_{(2n-1)\pi }^{2n\pi }$

$\Rightarrow [-(-1-1)+(1-(-1)\left)\right]+[-(-1-1)+(1-(-1)\left)\right]+...{.}_{ntimes}$

$\Rightarrow [2+2]+[2+2]+....{.}_{ntimes}$

$\Rightarrow 4+4+...{.}_{ntimes}$

$\Rightarrow 4[1+1+...{.}_{n}times]$

$\Rightarrow 4n$

$\therefore {\int }_0^{2n\pi }(|\sin x|-\left[\left|\frac{\sin x}{2}\right|\right])dx=4n$