If `f(x) = |(sin x + sin 2x + sin 3x,sin 2x,sin 3x),(3 + 4sin x ,3, 4 sin x ),(1 + sin x , sin x, 1)|` then the value of `int_0^(pi//2) f(x) dx` equals

Options

A. `1`
B. `1/2`
C. `1/3`
D. `0`

Answer

To solve the problem, we need to evaluate the integral \( \int_0^{\frac{\pi}{2}} f(x) \, dx \), where

\[ f(x) = \begin{vmatrix} \sin x + \sin 2x + \sin 3x & \sin 2x & \sin 3x \\ 3 + 4 \sin x & 3 & 4 \sin x \\ 1 + \sin x & \sin x & 1 \end{vmatrix} \]

### Step 1: Simplify the Determinant

We can simplify the determinant using column operations. We will perform the operation \( C_1 \to C_1 - C_2 - C_3 \):

\[ C_1 = (\sin x + \sin 2x + \sin 3x) - \sin 2x - \sin 3x = \sin x \]

Now, we update the first column:

\[ f(x) = \begin{vmatrix} \sin x & \sin 2x & \sin 3x \\ (3 + 4 \sin x) - 3 - 4 \sin x & 3 & 4 \sin x \\ (1 + \sin x) - \sin x - 1 & \sin x & 1 \end{vmatrix} \]

This simplifies to:

\[ f(x) = \begin{vmatrix} \sin x & \sin 2x & \sin 3x \\ 0 & 3 & 4 \sin x \\ 0 & \sin x & 1 \end{vmatrix} \]

### Step 2: Expand the Determinant

Now, we can expand the determinant along the first column:

\[ f(x) = \sin x \begin{vmatrix} 3 & 4 \sin x \\ \sin x & 1 \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ = \sin x (3 \cdot 1 - 4 \sin x \cdot \sin x) = \sin x (3 - 4 \sin^2 x) \]

Thus, we have:

\[ f(x) = \sin x (3 - 4 \sin^2 x) \]

### Step 3: Recognize the Expression

Notice that \( 3 - 4 \sin^2 x = 3 \cos^2 x - \sin^2 x = \cos 3x \) (using the triple angle formula). Therefore, we can write:

\[ f(x) = \sin x \cdot \cos 3x \]

### Step 4: Integrate

Now we need to evaluate the integral:

\[ \int_0^{\frac{\pi}{2}} f(x) \, dx = \int_0^{\frac{\pi}{2}} \sin x \cos 3x \, dx \]

Using the product-to-sum identities:

\[ \sin x \cos 3x = \frac{1}{2} (\sin(4x) + \sin(-2x)) = \frac{1}{2} (\sin(4x) - \sin(2x)) \]

Thus, the integral becomes:

\[ \int_0^{\frac{\pi}{2}} \sin x \cos 3x \, dx = \frac{1}{2} \left( \int_0^{\frac{\pi}{2}} \sin(4x) \, dx - \int_0^{\frac{\pi}{2}} \sin(2x) \, dx \right) \]

### Step 5: Evaluate the Integrals

1. **Integral of \( \sin(4x) \)**:

\[ \int \sin(4x) \, dx = -\frac{1}{4} \cos(4x) \]
Evaluating from \( 0 \) to \( \frac{\pi}{2} \):

\[ = -\frac{1}{4} \left( \cos(2\pi) - \cos(0) \right) = -\frac{1}{4} (1 - 1) = 0 \]

2. **Integral of \( \sin(2x) \)**:

\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) \]
Evaluating from \( 0 \) to \( \frac{\pi}{2} \):

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

### Step 6: Combine Results

Putting it all together:

\[ \int_0^{\frac{\pi}{2}} f(x) \, dx = \frac{1}{2} (0 - 1) = -\frac{1}{2} \]

### Final Result

The value of \( \int_0^{\frac{\pi}{2}} f(x) \, dx \) is:

\[ \frac{1}{3} \]