If `f(x) = (1)/(cos xsqrt(1-tan x))` then its anti-derivative F(x) satisfying F(0) = 4 is:

Options

A. `sqrt(1+tan x)+ 4`
B. `(2)/(3)(1+ tan x)^(3//2)`
C. `2(sqrt(1+tanx)+ 1)`
D. None of these

Answer

To find the anti-derivative \( F(x) \) of the function \( f(x) = \frac{1}{\cos^2 x \sqrt{1 - \tan x}} \) that satisfies the condition \( F(0) = 4 \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \frac{1}{\cos^2 x \sqrt{1 - \tan x}} \]

### Step 2: Substitution
Let us make the substitution:
\[ t = \sqrt{1 - \tan x} \]
Then, we need to express \( dx \) in terms of \( dt \).

### Step 3: Differentiate the substitution
Differentiating both sides with respect to \( x \):
\[ t^2 = 1 - \tan x \implies 2t \frac{dt}{dx} = -\sec^2 x \]
Thus,
\[ \frac{dt}{dx} = -\frac{\sec^2 x}{2t} \]
This gives us:
\[ dx = -\frac{2t}{\sec^2 x} dt \]

### Step 4: Rewrite \( \sec^2 x \)
Recall that \( \sec^2 x = \frac{1}{\cos^2 x} \), so:
\[ dx = -2t \cos^2 x \, dt \]

### Step 5: Substitute into the integral
Now we can substitute \( dx \) and \( t \) into the integral:
\[ \int f(x) \, dx = \int \frac{1}{\cos^2 x \sqrt{1 - \tan x}} \left(-2t \cos^2 x \, dt\right) \]
This simplifies to:
\[ \int -2t \, dt \]

### Step 6: Integrate
Now we integrate:
\[ \int -2t \, dt = -t^2 + C \]

### Step 7: Substitute back for \( t \)
Recall that \( t = \sqrt{1 - \tan x} \):
\[ F(x) = -\left(\sqrt{1 - \tan x}\right)^2 + C = - (1 - \tan x) + C = \tan x - 1 + C \]

### Step 8: Apply the initial condition
We need to satisfy the condition \( F(0) = 4 \):
\[ F(0) = \tan(0) - 1 + C = 0 - 1 + C = C - 1 \]
Setting this equal to 4:
\[ C - 1 = 4 \implies C = 5 \]

### Step 9: Write the final result
Thus, the anti-derivative is:
\[ F(x) = \tan x - 1 + 5 = \tan x + 4 \]

### Conclusion
The final answer is:
\[ F(x) = \tan x + 4 \]