Find our value of `I = int_(-pi//2)^(pi//2) sin 2x dx` :

Options

A. Zero
B. `-1`
C. 1
D. 2

Answer

To solve the integral \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x) \, dx \), we will follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x) \, dx \]

### Step 2: Find the antiderivative
The antiderivative of \( \sin(2x) \) can be found using the formula:
\[ \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C \]
For our case, \( k = 2 \):
\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C \]

### Step 3: Evaluate the definite integral
Now we can evaluate the definite integral from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \):
\[ I = \left[-\frac{1}{2} \cos(2x)\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \]

### Step 4: Substitute the limits
Now, we substitute the upper and lower limits into the antiderivative:
\[ I = -\frac{1}{2} \cos(2 \cdot \frac{\pi}{2}) - \left(-\frac{1}{2} \cos(2 \cdot -\frac{\pi}{2})\right) \]
Calculating \( \cos(2 \cdot \frac{\pi}{2}) \) and \( \cos(2 \cdot -\frac{\pi}{2}) \):
\[ \cos(\pi) = -1 \quad \text{and} \quad \cos(-\pi) = -1 \]
Thus, we have:
\[ I = -\frac{1}{2}(-1) - \left(-\frac{1}{2}(-1)\right) \]
This simplifies to:
\[ I = \frac{1}{2} + \frac{1}{2} \]

### Step 5: Final calculation
Now, adding these values gives:
\[ I = \frac{1}{2} + \frac{1}{2} = 1 \]

### Conclusion
Thus, the value of the integral \( I \) is:
\[ \boxed{0} \]