A function `f` is continuous for all `x` (and not everywhere zero) such that `f^(2)(x)=int_(0)^(x)f(t)(cost)/(2+sint)dt`. Then `f(x)` is

Options

A. `1/2 In((x+cosx)/2)`
B. `1/2 In(3/(2+cosx))`
C. `1/2In((2+sinx)/2)`
D. `(cosx+sinx)/(2+sinx)`

Explanation

To solve the problem, we need to find the function \( f(x) \) given that:

\[ f^2(x) = \int_0^x \frac{f(t) \cos t}{2 + \sin t} dt \]

### Step 1: Differentiate both sides
We will differentiate both sides of the equation with respect to \( x \). Using the Leibniz rule for differentiation under the integral sign, we have:

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

For the right-hand side, applying Leibniz's rule gives:

\[ \frac{d}{dx} \left( \int_0^x \frac{f(t) \cos t}{2 + \sin t} dt \right) = \frac{f(x) \cos x}{2 + \sin x} \]

So we equate both sides:

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

### Step 2: Simplify the equation
Assuming \( f(x) \neq 0 \) (since \( f \) is not everywhere zero), we can divide both sides by \( f(x) \):

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

### Step 3: Solve for \( f'(x) \)
Now we can express \( f'(x) \):

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

### Step 4: Integrate to find \( f(x) \)
Next, we integrate \( f'(x) \):

\[ f(x) = \int \frac{\cos x}{2(2 + \sin x)} dx \]

To solve this integral, we can use substitution. Let \( u = 2 + \sin x \), then \( du = \cos x \, dx \).

Thus, the integral becomes:

\[ f(x) = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |2 + \sin x| + C \]

### Step 5: Determine the constant of integration
To find the constant \( C \), we can evaluate \( f(0) \). From the original equation, substituting \( x = 0 \):

\[ f^2(0) = \int_0^0 \frac{f(t) \cos t}{2 + \sin t} dt = 0 \]

Thus, \( f(0) = 0 \). Substituting \( x = 0 \) into our expression for \( f(x) \):

\[ f(0) = \frac{1}{2} \ln(2 + \sin(0)) + C = \frac{1}{2} \ln(2) + C = 0 \]

This gives:

\[ C = -\frac{1}{2} \ln(2) \]

### Step 6: Final expression for \( f(x) \)
Substituting \( C \) back into our expression for \( f(x) \):

\[ f(x) = \frac{1}{2} \ln(2 + \sin x) - \frac{1}{2} \ln(2) = \frac{1}{2} \ln\left(\frac{2 + \sin x}{2}\right) \]

### Conclusion
Thus, the function \( f(x) \) is:

\[ f(x) = \frac{1}{2} \ln\left(\frac{2 + \sin x}{2}\right) \]