Evaluate the following :
`int_(-pi//2)^(+pi//2) " sin 2x dx "`

Answer

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

### Step 1: Identify the integral
We need to evaluate the integral:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x) \, dx \]

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

### Step 3: Apply the limits of integration
Now we will evaluate the definite integral using the antiderivative we found:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x) \, dx = \left[-\frac{1}{2} \cos(2x)\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \]

### Step 4: Substitute the limits
Now we will substitute the upper and lower limits into the antiderivative:
1. For the upper limit \( x = \frac{\pi}{2} \):
\[ -\frac{1}{2} \cos(2 \cdot \frac{\pi}{2}) = -\frac{1}{2} \cos(\pi) = -\frac{1}{2}(-1) = \frac{1}{2} \]

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

### Step 5: Calculate the result
Now we can find the value of the definite integral:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x) \, dx = \frac{1}{2} - \frac{1}{2} = 0 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x) \, dx = 0 \]

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