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

Options

A. x
B. 1
C. cosx
D. sinx

Answer

To solve the given problem step by step, we start with the equation provided:

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

### Step 1: Differentiate both sides
To eliminate the integral, we differentiate both sides with respect to \(x\).

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

Using the Fundamental Theorem of Calculus on the left side, we have:

\[ f(x) \cos x \]

For the right side, we use the chain rule:

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

Thus, we have:

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

### Step 2: Simplify the equation
Assuming \(f(x) \neq 0\), we can divide both sides by \(f(x)\):

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

### Step 3: Integrate to find \(f(x)\)
Now we need to find \(f(x)\) by integrating \(f'(x)\):

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

The integral of \(\cos x\) is:

\[ f(x) = \sin x + C_1 \]

where \(C_1\) is a constant of integration.

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

\[ f(x) = \sin x + C_1 \]