Let `f(alpha, beta) = =|(cos (alpha + beta), -sin (alpha + beta), cos 2beta),(sin alpha, cos alpha, sin beta),(-cos alpha, sin alpha, cos beta)|`
`if i =int_(-pi//2)^(pi//2) cos ^(2)beta(f(0,beta)+f(0,(pi)/(2)-beta))dbeta` then i is

Options

A. `e^(pi//2)`
B. 3
C. `2(2e^(pi//2)-1)`
D. `None of these

Answer

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

\[ i = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \beta \left( f(0, \beta) + f\left(0, \frac{\pi}{2} - \beta\right) \right) d\beta \]

where

\[ f(\alpha, \beta) = \begin{vmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) & \cos(2\beta) \\ \sin \alpha & \cos \alpha & \sin \beta \\ -\cos \alpha & \sin \alpha & \cos \beta \end{vmatrix} \]

### Step 1: Calculate \( f(0, \beta) \)

Substituting \(\alpha = 0\):

\[ f(0, \beta) = \begin{vmatrix} \cos \beta & -\sin \beta & \cos(2\beta) \\ 0 & 1 & \sin \beta \\ -1 & 0 & \cos \beta \end{vmatrix} \]

### Step 2: Evaluate the determinant

Using the determinant formula, we can expand it:

\[ f(0, \beta) = \cos \beta \begin{vmatrix} 1 & \sin \beta \\ 0 & \cos \beta \end{vmatrix} + \sin \beta \begin{vmatrix} 0 & \sin \beta \\ -1 & \cos \beta \end{vmatrix} + \cos(2\beta) \begin{vmatrix} 0 & 1 \\ -1 & 0 \end{vmatrix} \]

Calculating these determinants:

1. The first determinant is \( \cos \beta \cdot (1 \cdot \cos \beta - 0 \cdot \sin \beta) = \cos^2 \beta \).
2. The second determinant is \( \sin \beta \cdot (0 - (-1) \cdot \sin \beta) = \sin^2 \beta \).
3. The third determinant is \( \cos(2\beta) \cdot (0 - (-1)) = \cos(2\beta) \).

Thus,

\[ f(0, \beta) = \cos^2 \beta + \sin^2 \beta + \cos(2\beta) = 1 + \cos(2\beta) \]

### Step 3: Calculate \( f(0, \frac{\pi}{2} - \beta) \)

Substituting \(\beta\) with \(\frac{\pi}{2} - \beta\):

\[ f\left(0, \frac{\pi}{2} - \beta\right) = f(0, \beta) \text{ (since } \cos\left(\frac{\pi}{2} - \beta\right) = \sin \beta \text{ and vice versa)} \]

Thus,

\[ f\left(0, \frac{\pi}{2} - \beta\right) = 1 + \cos(2(\frac{\pi}{2} - \beta)) = 1 + \cos(\pi - 2\beta) = 1 - \cos(2\beta) \]

### Step 4: Combine \( f(0, \beta) \) and \( f(0, \frac{\pi}{2} - \beta) \)

Now, we can combine the two results:

\[ f(0, \beta) + f\left(0, \frac{\pi}{2} - \beta\right) = (1 + \cos(2\beta)) + (1 - \cos(2\beta)) = 2 \]

### Step 5: Substitute back into the integral

Now substitute this back into the integral:

\[ i = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \beta \cdot 2 \, d\beta = 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \beta \, d\beta \]

### Step 6: Evaluate the integral \( \int \cos^2 \beta \, d\beta \)

Using the identity \( \cos^2 \beta = \frac{1 + \cos(2\beta)}{2} \):

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \beta \, d\beta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos(2\beta)}{2} \, d\beta = \frac{1}{2} \left[ \beta + \frac{1}{2} \sin(2\beta) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \]

Calculating the limits:

\[ \left[ \frac{\pi}{2} + 0 \right] - \left[ -\frac{\pi}{2} + 0 \right] = \pi \]

Thus,

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \beta \, d\beta = \frac{\pi}{2} \]

### Step 7: Final calculation of \( i \)

Now substituting back:

\[ i = 2 \cdot \frac{\pi}{2} = \pi \]

### Final Answer

Thus, the value of \( i \) is

\[ \boxed{\pi} \]