Let `f(alpha, beta) =|(cos (alpha + beta), -sin (alpha + beta), cos 2beta),(sin alpha, cos alpha, sin beta),(-cos alpha, sin alpha, cos beta)|`
The value of `l=int_(0)^(pi//2)e^(beta)(f(0,0)+f((pi)/(2),beta)+f((3pi)/(2),(pi)/(2)-beta))dbeta` is

Options

A. (a) `e^(pi//2)`
B. (b) 0
C. (c) `2(2e^(pi//2)-2)`
D. (d) None of these

Answer

To solve the problem, we need to evaluate the integral

\[ L = \int_{0}^{\frac{\pi}{2}} e^{\beta} \left( f(0, 0) + f\left(\frac{\pi}{2}, \beta\right) + f\left(\frac{3\pi}{2}, \frac{\pi}{2} - \beta\right) \right) d\beta \]

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 \( f(0, 0) \)

Substituting \( \alpha = 0 \) and \( \beta = 0 \):

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

This simplifies to:

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

Calculating the determinant:

\[ = 1 \cdot \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} - 0 + 1 \cdot \begin{vmatrix} 0 & 1 \\ -1 & 0 \end{vmatrix} = 1 \cdot 1 + 1 \cdot 1 = 2 \]

Thus, \( f(0, 0) = 2 \).

### Step 2: Calculate \( f\left(\frac{\pi}{2}, \beta\right) \)

Substituting \( \alpha = \frac{\pi}{2} \) and \( \beta \):

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

This simplifies to:

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

Calculating the determinant:

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

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

Thus, \( f\left(\frac{\pi}{2}, \beta\right) = 1 \).

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

Substituting \( \alpha = \frac{3\pi}{2} \) and \( \beta = \frac{\pi}{2} - \beta \):

\[ f\left(\frac{3\pi}{2}, \frac{\pi}{2} - \beta\right) = \begin{vmatrix} \cos\left(2\pi - \beta\right) & -\sin\left(2\pi - \beta\right) & \cos(\pi - 2\beta) \\ -1 & 0 & \sin\left(\frac{\pi}{2} - \beta\right) \\ 0 & -1 & \cos\left(\frac{\pi}{2} - \beta\right) \end{vmatrix} \]

This simplifies to:

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

Calculating the determinant gives:

\[ = \cos(\beta)(0 - \cos(\beta)) - \sin(\beta)(-1 \cdot \sin(\beta)) = -\cos^2(\beta) + \sin^2(\beta) = \sin^2(\beta) - \cos^2(\beta) = -\cos(2\beta) \]

Thus, \( f\left(\frac{3\pi}{2}, \frac{\pi}{2} - \beta\right) = -\cos(2\beta) \).

### Step 4: Substitute back into the integral

Now we substitute back into the integral:

\[ L = \int_{0}^{\frac{\pi}{2}} e^{\beta} \left( 2 + 1 - \cos(2\beta) \right) d\beta \]

This simplifies to:

\[ L = \int_{0}^{\frac{\pi}{2}} e^{\beta} \left( 3 - \cos(2\beta) \right) d\beta \]

### Step 5: Evaluate the integral

We can separate the integral:

\[ L = 3 \int_{0}^{\frac{\pi}{2}} e^{\beta} d\beta - \int_{0}^{\frac{\pi}{2}} e^{\beta} \cos(2\beta) d\beta \]

The first integral evaluates to:

\[ 3 \left[ e^{\beta} \right]_{0}^{\frac{\pi}{2}} = 3 \left( e^{\frac{\pi}{2}} - 1 \right) \]

For the second integral, we can use integration by parts or a known result:

\[ \int e^{\beta} \cos(2\beta) d\beta = \frac{e^{\beta}}{5} (2 \cos(2\beta) + \sin(2\beta)) \]

Evaluating this from \( 0 \) to \( \frac{\pi}{2} \):

\[ = \left[ \frac{e^{\beta}}{5} (2 \cos(2\beta) + \sin(2\beta)) \right]_{0}^{\frac{\pi}{2}} = \frac{e^{\frac{\pi}{2}}}{5} (2 \cdot 0 + 1) - \frac{1}{5} (2 + 0) = \frac{e^{\frac{\pi}{2}}}{5} - \frac{2}{5} \]

### Final Result

Combining these results gives:

\[ L = 3 \left( e^{\frac{\pi}{2}} - 1 \right) - \left( \frac{e^{\frac{\pi}{2}}}{5} - \frac{2}{5} \right) \]

Simplifying:

\[ L = 3e^{\frac{\pi}{2}} - 3 - \frac{e^{\frac{\pi}{2}}}{5} + \frac{2}{5} \]

Combining terms:

\[ L = \left( 3 - \frac{1}{5} \right)e^{\frac{\pi}{2}} - 3 + \frac{2}{5} = \frac{14}{5} e^{\frac{\pi}{2}} - \frac{13}{5} \]

Thus, the final answer is:

\[ L = \frac{14}{5} e^{\frac{\pi}{2}} - \frac{13}{5} \]