To solve the integral \(\int\limits^{\pi/2}_{-\pi/2} \frac{dx}{\left[x\right]+\left[\sin x\right]+4}\), we need to evaluate how the greatest integer function affects the integrand over the given interval and compute the resultant sum.
- Start by considering the greatest integer function, denoted \([x]\) and \([\sin x]\). The expression \([x] + [\sin x]\) changes based on the values of \(x\) in the interval \([-\pi/2, \pi/2]\).
- Split the integral into segments where \(\left[x\right]\) is constant. This means focusing on integer intervals within the limit:
- For \(x \in [-\pi/2, 0)\), \(\left[x\right] = -1\).
- For \(x \in [0, \pi/2)\), \(\left[x\right] = 0\).
- Next, consider \([\sin x]\). For \(x\) in \([-\pi/2, \pi/2]\), the range of \(\sin x\) is \([-1, 1]\). Thus, \([\sin x]\) can be \(-1$, $0$, or $1\):
- For \(x\) in \([-\pi/2, 0]\), \([\sin x] = -1\), since \(\sin x \leq 0\).
- For \(x\) close to zero but non-positive, \([\sin x] = -1\).
- For \(x\) in \((0, \pi/2)\), \([\sin x] = 0\), since \(\sin x\) is positive.
- Now, substitute back into the integrand: \(\left[x\right]+\left[\sin x\right]+4\) becomes:
- \([-1] + [-1] + 4 = 2\) for \(x \in [-\pi/2, 0)\).
- \([0] + [0] + 4 = 4\) for \(x \in [0, \pi/2]\).
- Breaking the integral into two parts:
- \(\int\limits_{-\pi/2}^{0} \frac{dx}{2} = \left[ \frac{x}{2} \right]_{-\pi/2}^{0} = \frac{0 - (-\pi/2)}{2} = \frac{\pi}{4}\)
- \(\int\limits_{0}^{\pi/2} \frac{dx}{4} = \left[ \frac{x}{4} \right]_{0}^{\pi/2} = \frac{\pi/2 - 0}{4} = \frac{\pi}{8}\)
- Add the results of these two integrals: \(\frac{\pi}{4} + \frac{\pi}{8} = \frac{2\pi}{8} + \frac{\pi}{8} = \frac{3\pi}{8}\).
Comparing with the given options, the correct answer is \(\frac{3}{20} (4\pi - 3)\).