The value of `int_(0)^(2pi)[sin 2x(1+cos 3x)]dx`, where [t] denotes the greatest integer function, is:

Options

A. `-pi`
B. `2pi`
C. `pi`
D. `-2pi`

Answer

To solve the integral \( I = \int_{0}^{2\pi} \sin(2x)(1 + \cos(3x)) \, dx \), we will follow these steps:

### Step 1: Write the integral
We start with the integral:
\[ I = \int_{0}^{2\pi} \sin(2x)(1 + \cos(3x)) \, dx \]

### Step 2: Split the integral
We can split the integral into two parts:
\[ I = \int_{0}^{2\pi} \sin(2x) \, dx + \int_{0}^{2\pi} \sin(2x) \cos(3x) \, dx \]

### Step 3: Evaluate the first integral
The first integral is:
\[ \int_{0}^{2\pi} \sin(2x) \, dx \]
The integral of \(\sin(2x)\) over one complete period (from \(0\) to \(2\pi\)) is zero:
\[ \int_{0}^{2\pi} \sin(2x) \, dx = 0 \]

### Step 4: Evaluate the second integral
Now we focus on the second integral:
\[ \int_{0}^{2\pi} \sin(2x) \cos(3x) \, dx \]
We can use the product-to-sum identities:
\[ \sin(A) \cos(B) = \frac{1}{2} [\sin(A + B) + \sin(A - B)] \]
Applying this gives:
\[ \sin(2x) \cos(3x) = \frac{1}{2} [\sin(5x) + \sin(-x)] = \frac{1}{2} [\sin(5x) - \sin(x)] \]
Thus, we have:
\[ \int_{0}^{2\pi} \sin(2x) \cos(3x) \, dx = \frac{1}{2} \left( \int_{0}^{2\pi} \sin(5x) \, dx - \int_{0}^{2\pi} \sin(x) \, dx \right) \]

### Step 5: Evaluate the integrals
Both integrals \(\int_{0}^{2\pi} \sin(5x) \, dx\) and \(\int_{0}^{2\pi} \sin(x) \, dx\) are also zero:
\[ \int_{0}^{2\pi} \sin(5x) \, dx = 0 \]
\[ \int_{0}^{2\pi} \sin(x) \, dx = 0 \]

### Step 6: Combine results
Thus, the second integral evaluates to:
\[ \int_{0}^{2\pi} \sin(2x) \cos(3x) \, dx = \frac{1}{2} (0 - 0) = 0 \]

### Step 7: Final result
Combining both parts, we have:
\[ I = 0 + 0 = 0 \]

### Step 8: Apply the greatest integer function
Since the value of \(I\) is \(0\), applying the greatest integer function gives:
\[ \lfloor I \rfloor = \lfloor 0 \rfloor = 0 \]

### Conclusion
The final answer is:
\[ \boxed{0} \]