If `Delta (x)=|{:(1,cos x,1-cos x),(1+sin x,cos x,1+sinx-cosx),(sinx,sinx,1):}|` then `int_(0)^(pi//2)Delta(x)` dx is equal to

Options

A. `-(1)/(2)`
B. 0
C. `(1)/(4)`
D. `(1)/(2)`

Answer

To solve the problem, we need to evaluate the integral of the determinant given by:

\[ \Delta(x) = \begin{vmatrix} 1 & \cos x & 1 - \cos x \\ 1 + \sin x & \cos x & 1 + \sin x - \cos x \\ \sin x & \sin x & 1 \end{vmatrix} \]

We will follow these steps:

### Step 1: Simplify the Determinant
We start by simplifying the determinant. We can perform column operations to make the calculations easier.

1. Replace \( C_1 \) with \( C_1 - C_3 \):
\[ C_1 \rightarrow C_1 - C_3 \]
This gives us:
\[ \Delta(x) = \begin{vmatrix} 1 & \cos x & 1 - \cos x \\ 1 + \sin x & \cos x & 1 + \sin x - \cos x \\ \sin x - 1 & \sin x & 1 \end{vmatrix} \]

### Step 2: Further Simplify the Determinant
Next, we can perform another column operation:

2. Replace \( C_1 \) with \( C_1 - C_2 \):
\[ C_1 \rightarrow C_1 - C_2 \]
This gives us:
\[ \Delta(x) = \begin{vmatrix} 1 - \cos x & \cos x & 1 - \cos x \\ (1 + \sin x) - \cos x & \cos x & 1 + \sin x - \cos x \\ \sin x - 1 - \sin x & \sin x & 1 \end{vmatrix} \]

### Step 3: Calculate the Determinant
Now we can calculate the determinant:

3. The determinant simplifies to:
\[ \Delta(x) = \begin{vmatrix} 1 - \cos x & \cos x & 1 - \cos x \\ 1 + \sin x - \cos x & \cos x & 1 + \sin x - \cos x \\ -1 & \sin x & 1 \end{vmatrix} \]

After performing the determinant calculations (using cofactor expansion or other methods), we find:
\[ \Delta(x) = -\sin x \cos x = -\frac{1}{2} \sin(2x) \]

### Step 4: Evaluate the Integral
Now we need to evaluate the integral:
\[ \int_0^{\frac{\pi}{2}} \Delta(x) \, dx = \int_0^{\frac{\pi}{2}} -\frac{1}{2} \sin(2x) \, dx \]

4. Factor out the constant:
\[ = -\frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(2x) \, dx \]

5. The integral of \(\sin(2x)\) is:
\[ = -\frac{1}{2} \left[-\frac{1}{2} \cos(2x)\right]_0^{\frac{\pi}{2}} = -\frac{1}{2} \left[-\frac{1}{2} (\cos(\pi) - \cos(0))\right] \]

6. Substitute the limits:
\[ = -\frac{1}{2} \left[-\frac{1}{2} (-1 - 1)\right] = -\frac{1}{2} \left[-\frac{1}{2} (-2)\right] = -\frac{1}{2} \cdot 1 = -\frac{1}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_0^{\frac{\pi}{2}} \Delta(x) \, dx = -\frac{1}{2} \]