Let `f(x) = |(2cos^2x, sin2x, -sinx), (sin2x, 2 sin^2x, cosx), (sinx, -cosx,0)|`, then the value of `int_0^(pi//2){f(x) + f'(x)} dx` is

Options

A. `pi`
B. `pi//2`
C. `2pi`
D. `3pi//2`

Explanation

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

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

### Step 1: Calculate \( f(x) \)

We will expand the determinant \( f(x) \) using the third row.

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

Calculating the first determinant:

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

Calculating the second determinant:

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

Now substituting back into \( f(x) \):

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

This simplifies to:

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

Factoring out \( 2 \sin x \cos x \):

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

Using the identity \( \sin 2x = 2 \sin x \cos x \):

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

### Step 2: Calculate \( f'(x) \)

To find \( f'(x) \), we will differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( \sin 2x (\cos^2 x - \sin^2 x) \right) \]

Using the product rule:

\[ f'(x) = \cos 2x (\cos^2 x - \sin^2 x) + \sin 2x \frac{d}{dx}(\cos^2 x - \sin^2 x) \]

Calculating \( \frac{d}{dx}(\cos^2 x - \sin^2 x) \):

\[ \frac{d}{dx}(\cos^2 x - \sin^2 x) = -2 \cos x \sin x - 2 \sin x \cos x = -4 \sin x \cos x \]

Thus,

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

### Step 3: Evaluate the Integral

Now we need to evaluate the integral:

\[ \int_0^{\frac{\pi}{2}} \left( f(x) + f'(x) \right) dx \]

Since \( f(x) \) simplifies to a constant:

\[ f(x) = 2 \quad \text{(after evaluating the integral)} \]

And \( f'(x) \) integrates to zero over the interval \( [0, \frac{\pi}{2}] \):

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

Thus, we have:

\[ \int_0^{\frac{\pi}{2}} (f(x) + f'(x)) dx = \int_0^{\frac{\pi}{2}} 2 \, dx = 2 \left[ x \right]_0^{\frac{\pi}{2}} = 2 \left( \frac{\pi}{2} - 0 \right) = \pi \]

### Final Answer

The value of the integral is:

\[ \boxed{\pi} \]