Question

The value of the integral $\underset{-2}{\overset{2}{\int }}\frac{{\sin }^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}dx$

(where $\left[x\right]$ denotes the greatest integer less than or equal to $x$ ) is:

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

Answer 1

The correct option is B $0$

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

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

When $x\in [0,2]$
$0\le \frac{x}{\pi }<1\Rightarrow \left[\frac{x}{\pi }\right]=0$

$\Rightarrow I=\underset{-2}{\overset{0}{\int }}\frac{{\sin }^{2}x}{\frac{-1}{2}}dx+\underset{0}{\overset{2}{\int }}\frac{{\sin }^{2}x}{\frac{1}{2}}dx$

$\Rightarrow I=-2\underset{-2}{\overset{0}{\int }}{\sin }^{2}xdx+2\underset{0}{\overset{2}{\int }}{\sin }^{2}xdx$

Replacing $x$ by $-x$ in the first integral, we get
$I=-2\underset{0}{\overset{2}{\int }}{\sin }^{2}xdx+2\underset{0}{\overset{2}{\int }}{\sin }^{2}xdx$
or, $I=0$