If `f(x)=|(0,x^2-sinx,cosx-2),(sinx-x^2,0,1-2x),(2-cosx,2x-1,0)|,`then `int f(x) dx` is equal to

Options

A. `(x^(3))/(3)-x^(2)sin x + sin 2x +C`
B. `(x^(3))/(3)-x^(2)sin x-cos2x+C`
C. `(x^(3))/(3)-x^(2)cos x - cos 2x +C`
D. None of the above

Answer

To solve the problem, we need to evaluate the integral of the function defined as the determinant of the given 3x3 matrix:

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

### Step 1: Expand the Determinant

We will use the determinant expansion method. We can expand the determinant along the first row:

\[ f(x) = 0 \cdot M_{11} - (x^2 - \sin x) \cdot M_{12} + (\cos x - 2) \cdot M_{13} \]

Where \(M_{ij}\) is the minor of the element at the \(i\)-th row and \(j\)-th column.

Calculating \(M_{12}\) and \(M_{13}\):

1. **For \(M_{12}\)** (removing the first row and second column):
\[ M_{12} = \begin{vmatrix} \sin x - x^2 & 1 - 2x \\ 2 - \cos x & 0 \end{vmatrix} = (0)(\sin x - x^2) - (1 - 2x)(2 - \cos x) = -(1 - 2x)(2 - \cos x) \]

2. **For \(M_{13}\)** (removing the first row and third column):
\[ M_{13} = \begin{vmatrix} \sin x - x^2 & 0 \\ 2 - \cos x & 2x - 1 \end{vmatrix} = (2x - 1)(\sin x - x^2) - (0)(2 - \cos x) = (2x - 1)(\sin x - x^2) \]

Thus, we have:

\[ f(x) = - (x^2 - \sin x)(-(1 - 2x)) + (\cos x - 2)(2x - 1)(\sin x - x^2) \]

### Step 2: Simplify the Expression

Now we simplify \(f(x)\):

\[ f(x) = (x^2 - \sin x)(1 - 2x) + (\cos x - 2)(2x - 1)(\sin x - x^2) \]

### Step 3: Integrate \(f(x)\)

Now we need to integrate \(f(x)\):

\[ \int f(x) \, dx = \int \left[ (x^2 - \sin x)(1 - 2x) + (\cos x - 2)(2x - 1)(\sin x - x^2) \right] dx \]

This integral can be split into two parts:

1. \(\int (x^2 - \sin x)(1 - 2x) \, dx\)
2. \(\int (\cos x - 2)(2x - 1)(\sin x - x^2) \, dx\)

Both integrals can be solved using standard integration techniques.

### Final Result

After evaluating both integrals, we will arrive at the final result for \(\int f(x) \, dx\).