Question

The value of ${\int }_{-2}^2\frac{\sin {}^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}dx,where\left[x\right]$=the greatest integer greater than or equal to x,is

• A. $1$
• B. $0$
• C. $4-\sin 4$
• D. none of these

Answer 1

The correct option is B $0$

$I={\int }_{-2}^2\frac{{sin}^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}dx={\int }_{-2}^0\frac{{sin}^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}dx+{\int }_0^2\frac{{sin}^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}dx$

For $-2$ and for $0$, We get

$I={\int }_{-2}^0\frac{{sin}^{2}x}{-1+\frac{1}{2}}dx+{\int }_0^2\frac{{sin}^{2}x}{0+\frac{1}{2}}dx={\int }_{-2}^0\frac{{sin}^{2}x}{-\frac{1}{2}}dx+{\int }_0^2\frac{{sin}^{2}x}{\frac{1}{2}}dx=-\frac{1}{2}{\int }_{-2}^0{sin}^{2}xdx+\frac{1}{2}{\int }_0^2{sin}^{2}xdx=-\frac{1}{2}{\left[\frac{1}{2}(x-sinxcosx)\right]}_{-2}^0+\frac{1}{2}{\left[\frac{1}{2}(x-sinxcosx)\right]}_0^2=-\frac{1}{2}(1-sin2cos2)+\frac{1}{2}(1-sin2cos2)=0$