If `f(x)=|(x^3,cos^2x,2x^4),(tan^3x,1,sec2x),(sin^3x,x^4,5)|` then `int_(-pi//2)^(pi//2) f(x)dx ` =

Options

A. 0
B. 2
C. `-2`
D. 4

Answer

To solve the given problem, we need to evaluate the integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \, dx \), where \( f(x) \) is defined as the determinant:

\[ f(x) = \begin{vmatrix} x^3 & \cos^2 x & 2x^4 \\ \tan^3 x & 1 & \sec 2x \\ \sin^3 x & x^4 & 5 \end{vmatrix} \]

### Step 1: Determine if \( f(x) \) is an odd function

To determine if \( f(x) \) is odd, we need to compute \( f(-x) \):

\[ f(-x) = \begin{vmatrix} (-x)^3 & \cos^2(-x) & 2(-x)^4 \\ \tan^3(-x) & 1 & \sec(2(-x)) \\ \sin^3(-x) & (-x)^4 & 5 \end{vmatrix} \]

Using the properties of trigonometric functions:
- \( \cos(-x) = \cos x \)
- \( \tan(-x) = -\tan x \)
- \( \sin(-x) = -\sin x \)

We can simplify the determinant:

\[ f(-x) = \begin{vmatrix} -x^3 & \cos^2 x & 2x^4 \\ -\tan^3 x & 1 & \sec(-2x) \\ -\sin^3 x & x^4 & 5 \end{vmatrix} \]

Since \( \sec(-2x) = \sec(2x) \), we can rewrite it as:

\[ f(-x) = \begin{vmatrix} -x^3 & \cos^2 x & 2x^4 \\ -\tan^3 x & 1 & \sec 2x \\ -\sin^3 x & x^4 & 5 \end{vmatrix} \]

### Step 2: Factor out \(-1\) from the first column

Factoring out \(-1\) from the first column gives:

\[ f(-x) = -\begin{vmatrix} x^3 & \cos^2 x & 2x^4 \\ \tan^3 x & 1 & \sec 2x \\ \sin^3 x & x^4 & 5 \end{vmatrix} = -f(x) \]

### Step 3: Conclusion about the function

Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 4: Evaluate the integral

Using the property of odd functions, we know that the integral of an odd function over a symmetric interval around zero is zero:

\[ \int_{-a}^{a} f(x) \, dx = 0 \]

Thus, we have:

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

### Final Answer

The value of the integral is:

\[ \boxed{0} \]