The value of the integral $\int_{-2}^{2} \frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}} d x$…
The value of the integral $\int_{-2}^{2} \frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}} d x$
(where $[x]$ denotes the greatest integer less than or equal to x) is
0
$\sin 4$
4
$4-\sin 4$
Solution
Let $f(x)=\frac{\sin ^{2} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}}$.
$\begin{array}{l}
\text { So, } f(-x)=\frac{\sin ^{2}(-x)}{\left[\frac{-x}{\pi}\right]+\frac{1}{2}} \quad \because[-x]=-1-[x] \\
\Rightarrow f(-x)=\frac{\sin ^{2} x}{-1-\left[\frac{x}{\pi}\right]+\frac{1}{2}}=\frac{\sin ^{2} x}{-\frac{1}{2}-\left[\frac{x}{\pi}\right]}=-f(x)
\end{array}$
$\Rightarrow f(x)$ is odd function
Hence, $\int_{-2}^{2} f(x) d x=0$