Let `Delta(x)=|(3+2sin^(4)x,2cos^(4)x,sin^(2)2x),(2sin^(4)x,3+2cos^(4)x,sin^(2)2x),(2sin^(4)x,2cos^(4)x,3+sin^(2)2x)|`
then `int_(-pi//2)^(pi//2)x Delta(x)dx` equals

Options

A. `pi^(2)`
B. `pi(pi-1)`
C. `1`
D. 0

Answer

To solve the problem, we need to evaluate the integral of \( x \Delta(x) \) over the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\), where \( \Delta(x) \) is defined as follows:

\[ \Delta(x) = \begin{vmatrix} 3 + 2\sin^4 x & 2\cos^4 x & \sin^2 2x \\ 2\sin^4 x & 3 + 2\cos^4 x & \sin^2 2x \\ 2\sin^4 x & 2\cos^4 x & 3 + \sin^2 2x \end{vmatrix} \]

### Step 1: Simplifying the Determinant
First, we can add all the columns of the determinant. The first column becomes:

\[ 3 + 2\sin^4 x + 2\cos^4 x + \sin^2 2x \]

The second and third columns will have similar expressions. Therefore, we can write:

\[ \Delta(x) = \begin{vmatrix} 3 + 2\sin^4 x + 2\cos^4 x + \sin^2 2x & 2\cos^4 x & \sin^2 2x \\ 2\sin^4 x & 3 + 2\cos^4 x + \sin^2 2x & \sin^2 2x \\ 2\sin^4 x & 2\cos^4 x & 3 + \sin^2 2x \end{vmatrix} \]

### Step 2: Factoring Out Common Terms
We can factor out the common term \( 3 + 2\sin^4 x + 2\cos^4 x + \sin^2 2x \) from the first column:

\[ \Delta(x) = (3 + 2\sin^4 x + 2\cos^4 x + \sin^2 2x) \cdot \begin{vmatrix} 1 & 2\cos^4 x & \sin^2 2x \\ 2\sin^4 x & 3 + 2\cos^4 x & \sin^2 2x \\ 2\sin^4 x & 2\cos^4 x & 3 + \sin^2 2x \end{vmatrix} \]

### Step 3: Row Operations
Next, we can perform row operations to simplify the determinant. We replace the second and third rows with the difference of those rows and the first row:

\[ \Delta(x) = (3 + 2\sin^4 x + 2\cos^4 x + \sin^2 2x) \cdot \begin{vmatrix} 1 & 2\cos^4 x & \sin^2 2x \\ 0 & 2\cos^4 x - (3 + 2\cos^4 x) & 0 \\ 0 & 2\cos^4 x - 2\cos^4 x & 3 - (3 + \sin^2 2x) \end{vmatrix} \]

This simplifies to:

\[ \Delta(x) = (3 + 2\sin^4 x + 2\cos^4 x + \sin^2 2x) \cdot \begin{vmatrix} 1 & 2\cos^4 x & \sin^2 2x \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{vmatrix} \]

### Step 4: Evaluating the Determinant
The determinant simplifies to:

\[ \Delta(x) = (3 + 2\sin^4 x + 2\cos^4 x + \sin^2 2x) \cdot (1 \cdot (-1) \cdot 0 - 0) = 0 \]

### Step 5: Evaluating the Integral
Now, we need to evaluate the integral:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \Delta(x) \, dx \]

Since \( \Delta(x) = 0 \), we have:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cdot 0 \, dx = 0 \]

### Conclusion
Therefore, the value of the integral is:

\[ \boxed{0} \]