Let `Delta(x)=|(cos^(2)x,cosxsinx,-sinx),(cosxsinx,sin^(2)x,cosx),(sinx,-cosx,0)|` then `int_(0)^(pi//2){Delta(x)+Delta'(x)]dx` equals

Options

A. `pi//3`
B. `pi//2`
C. `2pi`
D. `3pi//2`

Answer

To solve the problem, we need to evaluate the integral:

\[ \int_{0}^{\frac{\pi}{2}} \left( \Delta(x) + \Delta'(x) \right) dx \]

where

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

### Step 1: Calculate the Determinant \(\Delta(x)\)

To compute \(\Delta(x)\), we will expand the determinant using the first row:

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

Calculating each of these 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} \sin^2 x & \cos x \\ -\cos x & 0 \end{vmatrix} = (0)(\sin^2 x) - (-\cos x)(\cos x) = \cos^2 x \]

2. For the second determinant:
\[ \begin{vmatrix} \cos x \sin x & \cos x \\ \sin x & 0 \end{vmatrix} = (0)(\cos x \sin x) - (\cos x)(\sin x) = -\cos x \sin x \]

3. For the third determinant:
\[ \begin{vmatrix} \cos^2 x & \cos x \sin x \\ \cos x \sin x & \sin^2 x \end{vmatrix} = (\cos^2 x)(\sin^2 x) - (\cos x \sin x)(\cos x \sin x) = \cos^2 x \sin^2 x - \cos^2 x \sin^2 x = 0 \]

Putting it all together:
\[ \Delta(x) = \cos^2 x \cdot \cos^2 x - \cos x \sin x \cdot (-\cos x \sin x) - \sin x \cdot 0 \]
\[ = \cos^4 x + \cos^2 x \sin^2 x \]

### Step 2: Simplifying \(\Delta(x)\)

We can factor \(\Delta(x)\):
\[ \Delta(x) = \cos^2 x (\cos^2 x + \sin^2 x) = \cos^2 x \cdot 1 = \cos^2 x \]

### Step 3: Calculate the Derivative \(\Delta'(x)\)

Now we find \(\Delta'(x)\):
\[ \Delta'(x) = \frac{d}{dx}(\cos^2 x) = -2 \cos x \sin x = -\sin(2x) \]

### Step 4: Evaluate the Integral

Now we substitute \(\Delta(x)\) and \(\Delta'(x)\) into the integral:
\[ \int_{0}^{\frac{\pi}{2}} \left( \cos^2 x - \sin(2x) \right) dx \]

This can be split into two integrals:
\[ \int_{0}^{\frac{\pi}{2}} \cos^2 x \, dx - \int_{0}^{\frac{\pi}{2}} \sin(2x) \, dx \]

1. The integral of \(\cos^2 x\):
\[ \int \cos^2 x \, dx = \frac{1}{2} \left( x + \frac{\sin(2x)}{2} \right) + C \]
Evaluating from \(0\) to \(\frac{\pi}{2}\):
\[ = \frac{1}{2} \left( \frac{\pi}{2} + 0 \right) - \frac{1}{2} \left( 0 + 0 \right) = \frac{\pi}{4} \]

2. The integral of \(\sin(2x)\):
\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C \]
Evaluating from \(0\) to \(\frac{\pi}{2}\):
\[ = -\frac{1}{2} \left( \cos(\pi) - \cos(0) \right) = -\frac{1}{2} \left( -1 - 1 \right) = 1 \]

### Final Calculation

Putting it all together:
\[ \int_{0}^{\frac{\pi}{2}} \left( \Delta(x) + \Delta'(x) \right) dx = \frac{\pi}{4} - 1 \]

### Conclusion

Thus, the final answer is:
\[ \int_{0}^{\frac{\pi}{2}} \left( \Delta(x) + \Delta'(x) \right) dx = \frac{\pi}{4} - 1 \]