Question:
The value of $\int_{-\pi / 2}^{\pi / 2} \frac{d x}{[x]+[\sin x]+4}$, where $[\mathrm{t}]$ denotes the
greatest integer less than or equal to $t$, is:
Correct Option: , 3
Solution:
$I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{d x}{[x]+[\sin x]+4}$
$=\int_{\frac{-\pi}{2}}^{-1} \frac{d x}{-2-1+4}+\int_{-1}^{0} \frac{d x}{-1-1+4}+\int_{0}^{1} \frac{d x}{0+0+4}+\int_{1}^{\frac{\pi}{2}} \frac{d x}{1+0+4}$
$=\left(-1+\frac{\pi}{2}\right)+\frac{1}{2}(0+1)+\frac{1}{4}(1-0)+\frac{1}{5}\left(\frac{\pi}{2}-1\right)$
$=\frac{3 \pi}{5}-\frac{9}{20}=\frac{3}{20}(4 \pi-3)$