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

Options

A. 0
B. `-1`
C. 1
D. 2

Answer

To solve the integral \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x) \, dx \), we can 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 integral of \(\sin(2x)\) can be computed using the formula for the integral of sine. The antiderivative of \(\sin(kx)\) is \(-\frac{1}{k} \cos(kx)\). Here, \(k = 2\):
\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C \]

### Step 3: Evaluate the definite integral
Now we will 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:
1. For the upper limit \(x = \frac{\pi}{2}\):
\[ -\frac{1}{2} \cos(2 \cdot \frac{\pi}{2}) = -\frac{1}{2} \cos(\pi) = -\frac{1}{2} \cdot (-1) = \frac{1}{2} \]

2. For the lower limit \(x = -\frac{\pi}{2}\):
\[ -\frac{1}{2} \cos(2 \cdot -\frac{\pi}{2}) = -\frac{1}{2} \cos(-\pi) = -\frac{1}{2} \cdot (-1) = \frac{1}{2} \]

### Step 5: Calculate the result
Now, we can find \(I\):
\[ I = \frac{1}{2} - \frac{1}{2} = 0 \]

### Final Result
Thus, the value of the integral is:
\[ I = 0 \]

---