The `int_(0)^(pi//2)sgn(sin^2x-sinx+1/2)` dx is equal to , (where , sgn (x) denotes the sigum function of x)

Options

A. 0
B. 1
C. `pi`
D. `pi/2`

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{2}} \text{sgn}(\sin^2 x - \sin x + \frac{1}{2}) \, dx \), we will follow these steps:

### Step 1: Analyze the expression inside the signum function
We start with the expression \( \sin^2 x - \sin x + \frac{1}{2} \). We need to determine when this expression is positive, negative, or zero.

### Step 2: Completing the square
To simplify \( \sin^2 x - \sin x + \frac{1}{2} \), we can complete the square. The expression can be rewritten as follows:

\[ \sin^2 x - \sin x + \frac{1}{2} = \left(\sin x - \frac{1}{2}\right)^2 + \frac{1}{4} \]

Here, we added and subtracted \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).

### Step 3: Analyze the completed square
The completed square form \( \left(\sin x - \frac{1}{2}\right)^2 + \frac{1}{4} \) is always greater than or equal to \( \frac{1}{4} \) because a square is always non-negative.

### Step 4: Determine the signum function
Since \( \left(\sin x - \frac{1}{2}\right)^2 + \frac{1}{4} > 0 \) for all \( x \), we conclude that:

\[ \sin^2 x - \sin x + \frac{1}{2} > 0 \quad \text{for all } x \in \left[0, \frac{\pi}{2}\right] \]

Thus, we have:

\[ \text{sgn}(\sin^2 x - \sin x + \frac{1}{2}) = 1 \quad \text{for all } x \in \left[0, \frac{\pi}{2}\right] \]

### Step 5: Evaluate the integral
Now we can substitute this result into the integral:

\[ \int_{0}^{\frac{\pi}{2}} \text{sgn}(\sin^2 x - \sin x + \frac{1}{2}) \, dx = \int_{0}^{\frac{\pi}{2}} 1 \, dx \]

### Step 6: Calculate the integral
The integral of 1 over the interval from 0 to \( \frac{\pi}{2} \) is simply the length of the interval:

\[ \int_{0}^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

### Final Answer
Thus, the value of the integral is:

\[ \int_{0}^{\frac{\pi}{2}} \text{sgn}(\sin^2 x - \sin x + \frac{1}{2}) \, dx = \frac{\pi}{2} \]