JEE Main 2019 (Online) 11th January Morning ShiftQuestion 1
The value of the integral $\int_{-2}^{2}\frac{\sin^{2}x}{\left[\frac x\pi\right]+\frac12}dx$ (where $[x]$ denotes the greatest integer less than or equal to $x$) is :
Definite Integration
Solution
Concept. On $[-2,2]$, $\left[\frac x\pi\right]$ is $-1$ for $x<0$ and $0$ for $x\ge0$. Formulas used. (1) $\int_{-2}^{0}\sin^2x\,dx=\int_0^2\sin^2x\,dx$ ($\sin^2x$ is even)Step 1 - The denominator. For $-2\le x<0$: $\frac x\pi\in(-1,0)$, so the denominator is $-1+\frac12=-\frac12$. For $0\le x\le2$: it is $\frac12$. Step 2 - Split. $$I=-2\int_{-2}^{0}\sin^2x\,dx+2\int_0^2\sin^2x\,dx.$$ Step 3 - Use (1). The two parts cancel: $I=0$. Answer: (A)