The value of `int_(0)^(2pi)[sin2x(1+cos3x)]` dx, where [t] denotes

Options

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

Explanation

To solve the integral \( I = \int_{0}^{2\pi} \left\lfloor \sin(2x)(1 + \cos(3x)) \right\rfloor \, dx \), we will break it down step by step.

### Step 1: Split the Integral
We can split the integral into two parts:
\[ I = \int_{0}^{\pi} \left\lfloor \sin(2x)(1 + \cos(3x)) \right\rfloor \, dx + \int_{\pi}^{2\pi} \left\lfloor \sin(2x)(1 + \cos(3x)) \right\rfloor \, dx \]
Let:
\[ I_1 = \int_{0}^{\pi} \left\lfloor \sin(2x)(1 + \cos(3x)) \right\rfloor \, dx \]
\[ I_2 = \int_{\pi}^{2\pi} \left\lfloor \sin(2x)(1 + \cos(3x)) \right\rfloor \, dx \]

### Step 2: Change of Variables for \( I_2 \)
For \( I_2 \), we perform a change of variables. Let \( x = 2\pi - t \). Then \( dx = -dt \). The limits change as follows:
- When \( x = \pi \), \( t = \pi \)
- When \( x = 2\pi \), \( t = 0 \)

Thus,
\[ I_2 = \int_{\pi}^{0} \left\lfloor \sin(2(2\pi - t))(1 + \cos(3(2\pi - t))) \right\rfloor (-dt) \]
This simplifies to:
\[ I_2 = \int_{0}^{\pi} \left\lfloor \sin(4\pi - 2t)(1 + \cos(6\pi - 3t)) \right\rfloor dt \]
Using properties of sine and cosine:
\[ \sin(4\pi - 2t) = -\sin(2t) \quad \text{and} \quad \cos(6\pi - 3t) = -\cos(3t) \]
Thus,
\[ I_2 = \int_{0}^{\pi} \left\lfloor -\sin(2t)(1 - \cos(3t)) \right\rfloor dt \]

### Step 3: Combine \( I_1 \) and \( I_2 \)
Now, we have:
\[ I = I_1 + I_2 = \int_{0}^{\pi} \left\lfloor \sin(2x)(1 + \cos(3x)) \right\rfloor \, dx + \int_{0}^{\pi} \left\lfloor -\sin(2x)(1 - \cos(3x)) \right\rfloor \, dx \]

### Step 4: Use the Property of the Greatest Integer Function
Using the property of the greatest integer function:
\[ \left\lfloor x \right\rfloor + \left\lfloor -x \right\rfloor = -1 \quad \text{(if \( x \) is not an integer)} \]
Thus:
\[ I = \int_{0}^{\pi} \left( \left\lfloor \sin(2x)(1 + \cos(3x)) \right\rfloor + \left\lfloor -\sin(2x)(1 - \cos(3x)) \right\rfloor \right) dx \]
This simplifies to:
\[ I = \int_{0}^{\pi} -1 \, dx = -\pi \]

### Final Answer
Therefore, the value of the integral is:
\[ I = -\pi \]