Let `[a]` denote the greatest integer which is less than or equal to a. Then the value of the integral `int_(-pi//2)^(pi//2)[sinxcosx]dx` is

Options

A. `(pi)/(2)`
B. `pi`
C. `-pi`
D. `-(pi)/(2)`

Answer

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

### Step 1: Simplify the integrand
We can use the double angle identity for sine:
\[ \sin x \cos x = \frac{1}{2} \sin(2x) \]
Thus, we can rewrite the integral as:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin x \cos x \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2} \sin(2x) \, dx \]

### Step 2: Factor out the constant
We can factor out the constant \( \frac{1}{2} \):
\[ I = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x) \, dx \]

### Step 3: Evaluate the integral
Now, we will evaluate the integral \( \int \sin(2x) \, dx \):
\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C \]
Now, we will evaluate this from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \):
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x) \, dx = \left[-\frac{1}{2} \cos(2x)\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \]

### Step 4: Calculate the limits
Calculating the limits:
\[ = -\frac{1}{2} \left( \cos(\pi) - \cos(-\pi) \right) \]
Since \( \cos(\pi) = -1 \) and \( \cos(-\pi) = -1 \):
\[ = -\frac{1}{2} \left( -1 - (-1) \right) = -\frac{1}{2} \cdot 0 = 0 \]

### Step 5: Substitute back into the integral
Now substituting back into our expression for \( I \):
\[ I = \frac{1}{2} \cdot 0 = 0 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{0} \]