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 need to find the value of \( f(\pi) \) given the function defined by the equation:

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

### Step 1: Differentiate the equation
We start by differentiating 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 Leibniz rule for differentiation under the integral sign, we have:

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

### Step 2: Rearranging the equation
We can rearrange this equation to isolate \( f'(x) \):

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

### Step 3: Separate variables
Now, we separate the variables by dividing both sides by \( \sqrt{1 - f^2(x)} \):

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

### Step 4: Integrate both sides
Next, we integrate both sides with respect to \( x \):

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

Using the substitution \( f(x) = t \), we have \( f'(x) \, dx = dt \):

\[ \int \frac{1}{\sqrt{1 - t^2}} \, dt = x + C \]

The integral on the left side is known to be \( \sin^{-1}(t) \):

\[ \sin^{-1}(f(x)) = x + C \]

### Step 5: Solve for \( f(x) \)
Now, we solve for \( f(x) \):

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

### Step 6: Determine the constant \( C \)
To find \( C \), we use the initial condition. We substitute \( x = 0 \) into the original equation:

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

Now substituting \( x = 0 \) into our expression for \( f(x) \):

\[ f(0) = \sin(0 + C) = \sin(C) \]

Setting this equal to \( \frac{1}{2} \):

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

This gives us \( C = \frac{\pi}{6} \) (since \( \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} \)).

### Step 7: Final expression for \( f(x) \)
Now we have:

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

### Step 8: Calculate \( f(\pi) \)
Finally, we calculate \( f(\pi) \):

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

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

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

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