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 first need to analyze the expression inside the signum function.

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

### Step 2: Find the roots
To find the roots of the quadratic equation \( \sin^2 x - \sin x + \frac{1}{2} = 0 \), we can use the quadratic formula:
\[ \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -1, c = \frac{1}{2} \).

Calculating the discriminant:
\[ D = (-1)^2 - 4 \cdot 1 \cdot \frac{1}{2} = 1 - 2 = -1 \]
Since the discriminant is negative, there are no real roots. This means \( \sin^2 x - \sin x + \frac{1}{2} \) does not cross the x-axis.

### Step 3: Determine the sign of the expression
Next, we can evaluate the expression at a point in the interval \( [0, \frac{\pi}{2}] \). Let's check \( x = 0 \):
\[ \sin^2(0) - \sin(0) + \frac{1}{2} = 0 - 0 + \frac{1}{2} = \frac{1}{2} > 0 \]
Since the expression is a continuous quadratic function and does not have any real roots, it must be positive for all \( x \) in the interval \( [0, \frac{\pi}{2}] \).

### Step 4: Apply the signum function
Since \( \sin^2 x - \sin x + \frac{1}{2} > 0 \) for all \( x \in [0, \frac{\pi}{2}] \), we have:
\[ \text{sgn}(\sin^2 x - \sin x + \frac{1}{2}) = 1 \]

### Step 5: Evaluate the integral
Now we can substitute this back into the integral:
\[ \int_{0}^{\frac{\pi}{2}} \text{sgn}(\sin^2 x - \sin x + \frac{1}{2}) \, dx = \int_{0}^{\frac{\pi}{2}} 1 \, dx \]
Calculating this integral:
\[ \int_{0}^{\frac{\pi}{2}} 1 \, dx = x \bigg|_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \frac{\pi}{2} \]