The value of $$ \int_{\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{1}{[x]+4}\right)dx $$ where [.]denotes the greatest integer function, is

For $$x \in [-\pi/2, \pi/2]$$ where $$\pi/2 \approx 1.5708$$:

- $$x \in [-\pi/2, -1)$$: $$[x] = -2$$
- $$x \in [-1, 0)$$: $$[x] = -1$$
- $$x \in [0, 1)$$: $$[x] = 0$$
- $$x \in [1, \pi/2]$$: $$[x] = 1$$

$$I = \int_{-\pi/2}^{-1} \frac{dx}{-2+4} + \int_{-1}^{0} \frac{dx}{-1+4} + \int_{0}^{1} \frac{dx}{0+4} + \int_{1}^{\pi/2} \frac{dx}{1+4}$$
$$= \int_{-\pi/2}^{-1} \frac{dx}{2} + \int_{-1}^{0} \frac{dx}{3} + \int_{0}^{1} \frac{dx}{4} + \int_{1}^{\pi/2} \frac{dx}{5}$$
$$= \frac{1}{2}\left(-1 + \frac{\pi}{2}\right) + \frac{1}{3}(0 - (-1)) + \frac{1}{4}(1 - 0) + \frac{1}{5}\left(\frac{\pi}{2} - 1\right)$$
$$= \left(\frac{\pi}{2} - 1\right) \times \frac{7}{10} + \frac{7}{12}$$
$$= \frac{7}{60}\left(\frac{60\pi}{20} - 1\right) = \frac{7}{60}(3\pi - 1)$$