Mathematics - Definite Integration Question with Solution | TestHub

Let $f(x)=\frac{{\sin }^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}$. $\begin{array}{l}\text{ So, }f(-x)=\frac{{\sin }^2(-x)}{\left[\frac{-x}{\pi }\right]+\frac{1}{2}}\because [-x]=-1-[x]\\ \Rightarrow f(-x)=\frac{{\sin }^{2}x}{-1-\left[\frac{x}{\pi }\right]+\frac{1}{2}}=\frac{{\sin }^{2}x}{-\frac{1}{2}-\left[\frac{x}{\pi }\right]}=-f(x)\end{array}$ $\Rightarrow f(x)$ is odd function Hence, ${\int }_{-2}^{2}f(x)dx=0$