If `f'` is a differentiable function satisfying `f(x)=int_(0)^(x)sqrt(1-f^(2)(t))dt+1/2` then the value of `f(pi)` is equal to

Options

A. `-(sqrt(3))/2`
B. `-1/2`
C. `(sqrt(3))/2`
D. `1/2`

Explanation

To solve the problem, we start with the given equation for the function \( f(x) \):

\[ f(x) = \int_{0}^{x} \sqrt{1 - f^2(t)} \, dt + \frac{1}{2} \]

We need to find the value of \( f(\pi) \).

### Step 1: Differentiate Both Sides
We differentiate both sides of the equation with respect to \( x \):

\[ f'(x) = \frac{d}{dx} \left( \int_{0}^{x} \sqrt{1 - f^2(t)} \, dt + \frac{1}{2} \right) \]

Using the Fundamental Theorem of Calculus, we get:

\[ f'(x) = \sqrt{1 - f^2(x)} \]

### Step 2: Rearranging the Equation
We can rearrange the equation obtained from differentiation:

\[ f'(x) = \sqrt{1 - f^2(x)} \]

Squaring both sides gives:

\[ (f'(x))^2 = 1 - f^2(x) \]

### Step 3: Express \( f^2(x) \)
Rearranging the equation, we find:

\[ f^2(x) + (f'(x))^2 = 1 \]

This resembles the Pythagorean identity, suggesting that \( f(x) \) could be expressed in terms of sine or cosine functions.

### Step 4: Assume a Form for \( f(x) \)
Let’s assume:

\[ f(x) = \sin(g(x)) \]

Then, differentiating \( f(x) \):

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

Substituting into our equation:

\[ \sin^2(g(x)) + \cos^2(g(x)) (g'(x))^2 = 1 \]

This simplifies to:

\[ \sin^2(g(x)) + \cos^2(g(x)) (g'(x))^2 = 1 \]

### Step 5: Solve for \( g'(x) \)
Since \( \sin^2(g(x)) + \cos^2(g(x)) = 1 \), we can conclude that:

\[ g'(x) = 1 \]

Thus, \( g(x) = x + C \) for some constant \( C \).

### Step 6: Find \( f(x) \)
Substituting back, we have:

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

### Step 7: Determine the Constant \( C \)
To find \( C \), we can use the initial condition from \( f(0) \):

\[ f(0) = \int_{0}^{0} \sqrt{1 - f^2(t)} \, dt + \frac{1}{2} = \frac{1}{2} \]

Thus:

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

This implies:

\[ C = \frac{\pi}{6} \quad \text{(since } \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \text{)} \]

### Step 8: Final Form of \( f(x) \)
Now we can express \( f(x) \):

\[ f(x) = \sin\left(x + \frac{\pi}{6}\right) \]

### Step 9: Calculate \( f(\pi) \)
Finally, we compute \( f(\pi) \):

\[ f(\pi) = \sin\left(\pi + \frac{\pi}{6}\right) = \sin\left(\frac{7\pi}{6}\right) \]

Since \( \frac{7\pi}{6} \) is in the third quadrant, we have:

\[ \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2} \]

Thus, the value of \( f(\pi) \) is:

\[ \boxed{-\frac{1}{2}} \]