To solve the integral \( \int_{-\pi/2}^{\pi/2} [x] + \sin x \, dx \), where \([x]\) denotes the greatest integer function, we can break the integral into segments based on the behavior of the greatest integer function and the sine function.
### Step 1: Identify the intervals
The function \([x]\) changes at integer values. The critical points in the interval \([- \pi/2, \pi/2]\) are \(-1\), \(0\), and \(1\). Thus, we can break the integral into three parts:
1. From \(-\pi/2\) to \(-1\)
2. From \(-1\) to \(0\)
3. From \(0\) to \(1\)
4. From \(1\) to \(\pi/2\)
### Step 2: Evaluate the integral on each interval
#### Interval 1: \([- \pi/2, -1]\)
In this interval, \(x\) is less than \(-1\), so \([x] = -2\). Therefore, the integral becomes:
\[
\int_{-\pi/2}^{-1} (-2 + \sin x) \, dx
\]
#### Interval 2: \([-1, 0]\)
In this interval, \([-1 \leq x < 0]\), we have \([x] = -1\). Thus, the integral becomes:
\[
\int_{-1}^{0} (-1 + \sin x) \, dx
\]
#### Interval 3: \([0, 1]\)
In this interval, \(0 \leq x < 1\), we have \([x] = 0\). Therefore, the integral becomes:
\[
\int_{0}^{1} (0 + \sin x) \, dx
\]
#### Interval 4: \([1, \pi/2]\)
In this interval, \(x\) is greater than \(1\), so \([x] = 1\). Thus, the integral becomes:
\[
\int_{1}^{\pi/2} (1 + \sin x) \, dx
\]
### Step 3: Calculate each integral
1. **For \([- \pi/2, -1]\)**:
\[
\int_{-\pi/2}^{-1} (-2 + \sin x) \, dx = \left[-2x - \cos x \right]_{-\pi/2}^{-1}
\]
Evaluating this gives:
\[
= \left[-2(-1) - \cos(-1)\right] - \left[-2(-\pi/2) - \cos(-\pi/2)\right]
\]
\[
= [2 - \cos(1)] - [\pi - 0] = 2 - \cos(1) - \pi
\]
2. **For \([-1, 0]\)**:
\[
\int_{-1}^{0} (-1 + \sin x) \, dx = \left[-x - \cos x \right]_{-1}^{0}
\]
Evaluating this gives:
\[
= [0 - \cos(0)] - [-(-1) - \cos(-1)] = [0 - 1] - [1 - \cos(1)] = -1 - 1 + \cos(1) = -2 + \cos(1)
\]
3. **For \([0, 1]\)**:
\[
\int_{0}^{1} \sin x \, dx = [-\cos x]_{0}^{1} = [-\cos(1) + \cos(0)] = [-\cos(1) + 1]
\]
4. **For \([1, \pi/2]\)**:
\[
\int_{1}^{\pi/2} (1 + \sin x) \, dx = [x - \cos x]_{1}^{\pi/2}
\]
Evaluating this gives:
\[
= \left[\frac{\pi}{2} - 0\right] - [1 - \cos(1)] = \frac{\pi}{2} - 1 + \cos(1)
\]
### Step 4: Combine all parts
Now, we combine all the results:
\[
\text{Total} = (2 - \cos(1) - \pi) + (-2 + \cos(1)) + (-\cos(1) + 1) + \left(\frac{\pi}{2} - 1 + \cos(1)\right)
\]
Simplifying this:
\[
= 2 - \pi - 2 + \frac{\pi}{2} - 1 + 1 + (-\cos(1) + \cos(1)) = \frac{\pi}{2} - \pi = -\frac{\pi}{2}
\]
### Final Result
Thus, the value of the integral is:
\[
\int_{-\pi/2}^{\pi/2} ([x] + \sin x) \, dx = -\frac{\pi}{2}
\]
