If `intf(x)cos x dx = 1/2 f^(2)(x)+C`, then `f(x)` can be

Options

A. x
B. 1
C. cos x
D. sin x

Answer

To solve the problem, we start with the given equation:

\[ \int f(x) \cos x \, dx = \frac{1}{2} f^2(x) + C \]

### Step 1: Differentiate both sides with respect to \( x \)

We will apply the Fundamental Theorem of Calculus on the left side and the chain rule on the right side.

\[ \frac{d}{dx} \left( \int f(x) \cos x \, dx \right) = \frac{d}{dx} \left( \frac{1}{2} f^2(x) + C \right) \]

### Step 2: Apply the Fundamental Theorem of Calculus

The left side simplifies to:

\[ f(x) \cos x \]

### Step 3: Differentiate the right side

Using the chain rule, we differentiate \( \frac{1}{2} f^2(x) \):

\[ \frac{d}{dx} \left( \frac{1}{2} f^2(x) \right) = f(x) f'(x) \]

Thus, the right side becomes:

\[ f(x) f'(x) \]

### Step 4: Set the derivatives equal to each other

Now we have:

\[ f(x) \cos x = f(x) f'(x) \]

### Step 5: Simplify the equation

Assuming \( f(x) \neq 0 \), we can divide both sides by \( f(x) \):

\[ \cos x = f'(x) \]

### Step 6: Integrate both sides

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

\[ \int f'(x) \, dx = \int \cos x \, dx \]

This gives us:

\[ f(x) = \sin x + C \]

### Conclusion

Thus, the function \( f(x) \) can be expressed as:

\[ f(x) = \sin x + C \]