JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Definite Integrals

On $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ evaluate the integrand piecewise. For $x\in \left[-\frac{\pi }{2},0\right)$, $[\sin x]=-1$ and $[\cos x]=0$, so the denominator $3+[\sin x]+[\cos x]=2$; for $x\in \left(0,\frac{\pi }{2}\right]$, $[\sin x]=0$ and $[\cos x]=0$, so the denominator is $3$. The numerator $12(3+[x])$ equals $12$ on $\left[-\frac{\pi }{2},-1\right)$ where $[x]=-2$, $24$ on $[-1,0)$ where $[x]=-1$, $36$ on $[0,1)$ where $[x]=0$, and $48$ on $\left[1,\frac{\pi }{2}\right]$ where $[x]=1$. Hence the integrand is $6$ on $\left[-\frac{\pi }{2},-1\right)$, $12$ on $[-1,0)$, $12$ on $[0,1)$, and $16$ on $\left[1,\frac{\pi }{2}\right]$, so

The value of the integral is $11\pi +2$.