Let `F(x)=int_(sinx)^(cosx)e^((1+sin^(-1)(t))dt` on `[0,(pi)/(2)]`, then

Options

A. `F''(c)=0" for all" c in (0,(pi)/(2))`
B. `F''(c)=0" for some" c in (0, (pi)/(2))`
C. `F''(c) = 0" for no value of " c in (0,(pi)/(2))`
D. `F(c) ne 0" for all "c in (0,(pi)/(2))`

Answer

To solve the problem, we need to find the derivative \( F'(x) \) of the function defined by the integral:

\[ F(x) = \int_{\sin x}^{\cos x} e^{(1 + \sin^{-1}(t))} \, dt \]

We will apply the Leibniz rule for differentiation under the integral sign.

### Step 1: Apply the Leibniz Rule
The Leibniz rule states that if \( F(x) = \int_{g(x)}^{h(x)} f(t) \, dt \), then:

\[ F'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x) \]

In our case:
- \( g(x) = \sin x \)
- \( h(x) = \cos x \)
- \( f(t) = e^{(1 + \sin^{-1}(t))} \)

### Step 2: Compute the Derivatives
Now we need to compute \( h'(x) \) and \( g'(x) \):

\[ h'(x) = -\sin x \]
\[ g'(x) = \cos x \]

### Step 3: Evaluate \( f(h(x)) \) and \( f(g(x)) \)
Next, we evaluate \( f(h(x)) \) and \( f(g(x)) \):

1. \( f(h(x)) = f(\cos x) = e^{(1 + \sin^{-1}(\cos x))} \)
2. \( f(g(x)) = f(\sin x) = e^{(1 + \sin^{-1}(\sin x))} = e^{(1 + x)} \)

### Step 4: Substitute into the Leibniz Rule
Now we substitute everything back into the Leibniz rule:

\[ F'(x) = e^{(1 + \sin^{-1}(\cos x))} \cdot (-\sin x) - e^{(1 + x)} \cdot \cos x \]

### Step 5: Simplify the Expression
Thus, we can write:

\[ F'(x) = -\sin x \cdot e^{(1 + \sin^{-1}(\cos x))} - \cos x \cdot e^{(1 + x)} \]

### Step 6: Find Critical Points
To find where \( F'(x) = 0 \):

\[ -\sin x \cdot e^{(1 + \sin^{-1}(\cos x))} - \cos x \cdot e^{(1 + x)} = 0 \]

This implies:

\[ \sin x \cdot e^{(1 + \sin^{-1}(\cos x))} = -\cos x \cdot e^{(1 + x)} \]

### Step 7: Analyze the Equation
Since \( e^{(1 + \sin^{-1}(\cos x))} \) and \( e^{(1 + x)} \) are always positive, we can conclude that the equation can only hold true when:

\[ \sin x = \cos x \]

This occurs at:

\[ x = \frac{\pi}{4} \]

### Conclusion
Thus, \( F'(x) = 0 \) at \( x = \frac{\pi}{4} \), which implies that \( F(x) \) has a critical point at this value.