The value of

$I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{d x}{[x]+[\sin x]+4}$

$=\int_{\frac{-\pi}{2}}^{-1} \frac{d x}{-2-1+4}+\int_{-1}^{0} \frac{d x}{-1-1+4}$

$+\int_{0}^{1} \frac{d x}{0+0+4}+\int_{1}^{\frac{\pi}{2}} \frac{d x}{1+0+4}$

$\int_{\frac{-\pi}{2}}^{-1} \frac{d x}{1}+\int_{-1}^{0} \frac{d x}{2}+\int_{0}^{1} \frac{d x}{4}+\int_{1}^{\frac{\pi}{2}} \frac{d x}{5}$

$\left(-1+\frac{\pi}{2}\right)+\frac{1}{2}(0+1)+\frac{1}{4}+\frac{1}{5}\left(\frac{\pi}{2}-1\right)$

$-1+\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\frac{\pi}{2}+\frac{\pi}{10}$

$\frac{-20+10+5-4}{20}+\frac{6 \pi}{10}$

$\frac{-9}{20}+\frac{3 \pi}{5}$

Option (4)