The value of `int_0^(12pi) ([sint]+[-sint])dt` is equal to (where [.] denotes the greatest integer function )

Options

A. `12pi`
B. `-12pi`
C. `-10pi`
D. `-6pi`

Answer

To solve the integral \( I = \int_0^{12\pi} \left( [\sin t] + [-\sin t] \right) dt \), where \([.]\) denotes the greatest integer function (GIF), we can follow these steps:

### Step 1: Understand the properties of the greatest integer function
The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\). For the sine function:
- \(\sin t\) varies between -1 and 1.
- Therefore, \([\sin t]\) can be either -1, 0, or 1 depending on the value of \(t\).

### Step 2: Analyze the expression \([\sin t] + [-\sin t]\)
We can use the property of the greatest integer function:
- If \(x\) is an integer, then \([x] + [-x] = 0\).
- If \(x\) is not an integer, then \([x] + [-x] = -1\).

Thus, we can conclude:
- When \(\sin t\) is an integer (0 or 1), \([\sin t] + [-\sin t] = 0\).
- When \(\sin t\) is not an integer, \([\sin t] + [-\sin t] = -1\).

### Step 3: Determine when \(\sin t\) is an integer
The sine function is an integer at:
- \(t = 0\) (where \(\sin 0 = 0\))
- \(t = \frac{\pi}{2}\) (where \(\sin \frac{\pi}{2} = 1\))
- \(t = \pi\) (where \(\sin \pi = 0\))
- \(t = \frac{3\pi}{2}\) (where \(\sin \frac{3\pi}{2} = -1\))
- \(t = 2\pi\) (where \(\sin 2\pi = 0\))

### Step 4: Break the integral into intervals
Since \(\sin t\) is periodic with a period of \(2\pi\), we can break the integral from \(0\) to \(12\pi\) into \(6\) intervals of \(2\pi\):
\[ I = 6 \int_0^{2\pi} \left( [\sin t] + [-\sin t] \right) dt \]

### Step 5: Evaluate the integral from \(0\) to \(2\pi\)
We can break the interval \(0\) to \(2\pi\) into four parts:
1. \(0\) to \(\frac{\pi}{2}\)
2. \(\frac{\pi}{2}\) to \(\pi\)
3. \(\pi\) to \(\frac{3\pi}{2}\)
4. \(\frac{3\pi}{2}\) to \(2\pi\)

**For \(0\) to \(\frac{\pi}{2}\)**:
- \(\sin t\) is between \(0\) and \(1\), so \([\sin t] = 0\) and \([- \sin t] = -1\).
- Thus, \([\sin t] + [-\sin t] = -1\).

**For \(\frac{\pi}{2}\) to \(\pi\)**:
- \(\sin t\) is between \(1\) and \(0\), so \([\sin t] = 0\) and \([- \sin t] = -1\).
- Thus, \([\sin t] + [-\sin t] = -1\).

**For \(\pi\) to \(\frac{3\pi}{2}\)**:
- \(\sin t\) is between \(0\) and \(-1\), so \([\sin t] = -1\) and \([- \sin t] = 0\).
- Thus, \([\sin t] + [-\sin t] = -1\).

**For \(\frac{3\pi}{2}\) to \(2\pi\)**:
- \(\sin t\) is between \(-1\) and \(0\), so \([\sin t] = -1\) and \([- \sin t] = 0\).
- Thus, \([\sin t] + [-\sin t] = -1\).

### Step 6: Calculate the integral
Each of the four intervals contributes \(-1\) to the integral:
\[ \int_0^{2\pi} \left( [\sin t] + [-\sin t] \right) dt = \left(-1\right) \cdot 2\pi = -2\pi \]

### Step 7: Multiply by 6
Now, we multiply by \(6\):
\[ I = 6 \cdot (-2\pi) = -12\pi \]

### Final Result
Thus, the value of the integral is:
\[ \boxed{-12\pi} \]