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, we can follow these steps:

### Step 1: Understanding the periodic nature of the sine function
The sine function, \(\sin t\), is periodic with a period of \(2\pi\). Therefore, we can simplify the integral over the interval \(0\) to \(12\pi\) by breaking it into smaller intervals of \(2\pi\).

\[ I = \int_0^{12\pi} \left( [\sin t] + [-\sin t] \right) dt = 6 \int_0^{2\pi} \left( [\sin t] + [-\sin t] \right) dt \]

### Step 2: Analyzing the expression \([\sin t] + [-\sin t]\)
We need to evaluate \([\sin t] + [-\sin t]\) over the interval \(0\) to \(2\pi\).

- For \(0 \leq t < \frac{\pi}{2}\): \(\sin t\) ranges from \(0\) to \(1\). Thus, \([\sin t] = 0\) and \([- \sin t] = -1\). Therefore, \([\sin t] + [-\sin t] = 0 - 1 = -1\).

- For \(\frac{\pi}{2} \leq t < \frac{3\pi}{2}\): \(\sin t\) ranges from \(1\) to \(-1\). Thus, \([\sin t] = 0\) and \([- \sin t] = -1\) when \(t\) is in \((\frac{\pi}{2}, \frac{3\pi}{2})\). Therefore, \([\sin t] + [-\sin t] = 0 - 1 = -1\).

- For \(\frac{3\pi}{2} \leq t < 2\pi\): \(\sin t\) ranges from \(-1\) to \(0\). Thus, \([\sin t] = -1\) and \([- \sin t] = 0\). Therefore, \([\sin t] + [-\sin t] = -1 + 0 = -1\).

### Step 3: Evaluating the integral from \(0\) to \(2\pi\)
Now we can evaluate the integral:

\[ \int_0^{2\pi} \left( [\sin t] + [-\sin t] \right) dt = \int_0^{2\pi} (-1) dt = -1 \cdot (2\pi) = -2\pi \]

### Step 4: Final calculation of \(I\)
Substituting back into our expression for \(I\):

\[ I = 6 \cdot (-2\pi) = -12\pi \]

### Conclusion
The value of the integral is:

\[ \boxed{-12\pi} \]