The value of `int_(-20pi)^(20 pi) |sin x| [ sin x] dx` is (where [.] denotes greatest integer function)

Explanation

To solve the integral \( I = \int_{-20\pi}^{20\pi} |\sin x| \lfloor \sin x \rfloor \, dx \), where \( \lfloor \cdot \rfloor \) denotes the greatest integer function, we can follow these steps:

### Step 1: Analyze the function
The sine function, \( \sin x \), is periodic with a period of \( 2\pi \). Therefore, we can simplify the integral by taking advantage of this periodicity.

### Step 2: Break the integral into intervals
Given the periodic nature of the sine function, we can express the integral over one period and then multiply by the number of periods in the interval \([-20\pi, 20\pi]\):
\[ I = \int_{-20\pi}^{20\pi} |\sin x| \lfloor \sin x \rfloor \, dx = 10 \int_{0}^{2\pi} |\sin x| \lfloor \sin x \rfloor \, dx \]
This is because there are \( 20\pi / 2\pi = 10 \) complete periods of \( \sin x \) in the interval.

### Step 3: Evaluate the integral over one period
Next, we need to evaluate:
\[ \int_{0}^{2\pi} |\sin x| \lfloor \sin x \rfloor \, dx \]
In the interval \( [0, 2\pi] \), \( \sin x \) is non-negative from \( 0 \) to \( \pi \) and non-positive from \( \pi \) to \( 2\pi \). The greatest integer function \( \lfloor \sin x \rfloor \) will be:
- \( 0 \) for \( 0 \leq x < \frac{\pi}{2} \)
- \( 1 \) for \( \frac{\pi}{2} \leq x < \pi \)
- \( 0 \) for \( \pi < x < \frac{3\pi}{2} \)
- \( -1 \) for \( \frac{3\pi}{2} < x < 2\pi \)

Thus, we can break the integral into parts:
\[ \int_{0}^{2\pi} |\sin x| \lfloor \sin x \rfloor \, dx = \int_{0}^{\frac{\pi}{2}} |\sin x| \cdot 0 \, dx + \int_{\frac{\pi}{2}}^{\pi} |\sin x| \cdot 1 \, dx + \int_{\pi}^{\frac{3\pi}{2}} |\sin x| \cdot 0 \, dx + \int_{\frac{3\pi}{2}}^{2\pi} |\sin x| \cdot (-1) \, dx \]

### Step 4: Evaluate each part
1. \( \int_{0}^{\frac{\pi}{2}} |\sin x| \cdot 0 \, dx = 0 \)
2. \( \int_{\frac{\pi}{2}}^{\pi} |\sin x| \cdot 1 \, dx = \int_{\frac{\pi}{2}}^{\pi} \sin x \, dx = [-\cos x]_{\frac{\pi}{2}}^{\pi} = -(-1 - 0) = 1 \)
3. \( \int_{\pi}^{\frac{3\pi}{2}} |\sin x| \cdot 0 \, dx = 0 \)
4. \( \int_{\frac{3\pi}{2}}^{2\pi} |\sin x| \cdot (-1) \, dx = -\int_{\frac{3\pi}{2}}^{2\pi} \sin x \, dx = -[-\cos x]_{\frac{3\pi}{2}}^{2\pi} = -(-1 - 0) = -1 \)

### Step 5: Combine the results
Now, combining these results:
\[ \int_{0}^{2\pi} |\sin x| \lfloor \sin x \rfloor \, dx = 0 + 1 + 0 - 1 = 0 \]

### Step 6: Multiply by the number of periods
Thus, the total integral is:
\[ I = 10 \cdot 0 = 0 \]

### Final Answer
The value of \( \int_{-20\pi}^{20\pi} |\sin x| \lfloor \sin x \rfloor \, dx \) is \( \boxed{0} \).