JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Definite Integrals

Because $[x]$ is an integer, the integrand separates as $[[x]-\sin x]=[x]+[-\sin x]$. On $\left(0,\frac{\pi }{2}\right]$ one has $\sin x\in (0,1]$, so $[-\sin x]=-1$; on $\left(-\frac{\pi }{2},0\right)$ one has $-\sin x\in (0,1)$, so $[-\sin x]=0$. For the first part,

so the integrand equals $-2$ on $\left[-\frac{\pi }{2},-1\right)$, $-1$ on $[-1,0)$, $-1$ on $(0,1)$ and $0$ on $\left[1,\frac{\pi }{2}\right]$. Isolated points do not affect the integral, so

The integral equals $-\pi$.