Answer is: option3

\(\frac{1}{8} x^4\)

Solution:

We are given the equation:

\[ \int f(x) \cos(x) \, dx = f(x) \sin(x) - \int \frac{1}{2} x^3 \sin(x) \, dx \]

We need to determine which of the given functions could be \( f(x) \).

Step 1: Differentiate Both Sides

To find \( f(x) \), we start by differentiating both sides of the equation with respect to \( x \).

The left-hand side becomes:

\[ \frac{d}{dx} \left( \int f(x) \cos(x) \, dx \right) = f(x) \cos(x) \]

Now, differentiate the right-hand side. First, we differentiate \( f(x) \sin(x) \):

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

The derivative of the integral \( \int \frac{1}{2} x^3 \sin(x) \, dx \) is simply the integrand:

\[ \frac{d}{dx} \left( \int \frac{1}{2} x^3 \sin(x) \, dx \right) = \frac{1}{2} x^3 \sin(x) \]

Step 2: Simplify the Equation

After differentiating both sides, we get:

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

Canceling \( f(x) \cos(x) \) from both sides simplifies the equation to:

\[ 0 = f'(x) \sin(x) - \frac{1}{2} x^3 \sin(x) \]

Factoring out \( \sin(x) \), we have:

\[ 0 = \sin(x) \left( f'(x) - \frac{1}{2} x^3 \right) \]

Since \( \sin(x) \neq 0 \) for most values of \( x \), we can divide both sides by \( \sin(x) \):

\[ f'(x) = \frac{1}{2} x^3 \]

Step 3: Integrate to Find \( f(x) \)

Now, integrate both sides with respect to \( x \):

\[ f(x) = \int \frac{1}{2} x^3 \, dx = \frac{1}{2} \cdot \frac{x^4}{4} = \frac{1}{8} x^4 \]

Final Answer

Thus, we find that:

\[ f(x) = \frac{1}{8} x^4 \]

option C is the correct answer.