If f(x)=`=|{:(sinx+sin2x+sin3x,sin2x,sin3x),(3+4sinx,3,4sinx),(1+sinx,sinx,1):}|` then the value of `int_(0)^(pi//2) f(x)dx`, is

Options

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

Answer

To solve the problem, we need to evaluate the integral of the function \( f(x) \) defined by the determinant:

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

We will follow these steps:

### Step 1: Simplify the Determinant

We will simplify the determinant using column operations. Let's denote the columns as \( C_1, C_2, C_3 \).

1. **Column Operation**: Replace \( C_1 \) with \( C_1 - C_2 - C_3 \):
\[ C_1 \rightarrow C_1 - C_2 - C_3 \]
This gives us:
\[ f(x) = \begin{vmatrix} \sin x + \sin 2x + \sin 3x - \sin 2x - \sin 3x & \sin 2x & \sin 3x \\ (3 + 4 \sin x) - 3 - 4 \sin x & 3 & 4 \sin x \\ (1 + \sin x) - \sin x - 1 & \sin x & 1 \end{vmatrix} \]

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

### Step 2: Calculate the Determinant

Now we can calculate the determinant using 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:
\[ f(x) = 3 \sin x - 4 \sin^3 x \]

### Step 3: Recognize the Function

Notice that \( f(x) \) can be rewritten using the identity for sine:
\[ f(x) = \sin 3x \]

### Step 4: Set Up the Integral

Now we need to evaluate the integral:
\[ \int_0^{\frac{\pi}{2}} f(x) \, dx = \int_0^{\frac{\pi}{2}} \sin 3x \, dx \]

### Step 5: Integrate

The integral of \( \sin 3x \) is:
\[ \int \sin 3x \, dx = -\frac{1}{3} \cos 3x + C \]

Thus, we evaluate:
\[ \int_0^{\frac{\pi}{2}} \sin 3x \, dx = \left[-\frac{1}{3} \cos 3x\right]_0^{\frac{\pi}{2}} \]

Calculating the limits:
- At \( x = \frac{\pi}{2} \):
\[ -\frac{1}{3} \cos\left(3 \cdot \frac{\pi}{2}\right) = -\frac{1}{3} \cdot 0 = 0 \]
- At \( x = 0 \):
\[ -\frac{1}{3} \cos(0) = -\frac{1}{3} \cdot 1 = -\frac{1}{3} \]

Thus, the integral evaluates to:
\[ 0 - \left(-\frac{1}{3}\right) = \frac{1}{3} \]

### Final Answer

The value of the integral is:
\[ \int_0^{\frac{\pi}{2}} f(x) \, dx = \frac{1}{3} \]