Question

${\int }_{-2}^2\frac{{\sin }^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}dx$ where $\left[x\right]$ denotes greatest integer function $\le x,$ equals

• A. $0$
• B. $1$
• C. $2-\sin 4$
• D. $4-\sin 4$

Answer 1

The correct option is A $0$

We have, $I={\int }_{-2}^2\frac{si{n}^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}dx={\int }_{-2}^0\frac{si{n}^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}dx+{\int }_0^2\frac{si{n}^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}dx$

For $-2<x<0\Rightarrow \frac{-2}{\pi }<\frac{x}{\pi }<0\Rightarrow \left[\frac{x}{\pi }\right]=-1$

And for $0<x<2\Rightarrow 0<\frac{x}{\pi }<\frac{2}{\pi }\Rightarrow \left[\frac{x}{\pi }\right]=0$

We get,
$I={\int }_{-2}^0\frac{si{n}^{2}x}{-1+\frac{1}{2}}dx+{\int }_0^2\frac{si{n}^{2}x}{0+\frac{1}{2}}dx=-2{\int }_{-2}^{0}si{n}^{2}xdx+2{\int }_0^{2}si{n}^{2}xdx$

$I=-{[x-sinxcosx]}_{-2}^0+{[x-sinxcosx]}_0^2$

$I=0-2-sin(-2)cos(-2)+2-sin2cos2+0=0$