Consider the function `f''(x) = sec^(4) x + 4 " with " f(0) = 0 and f '(0) = 0` What is `f(x)` equal to ?

Options

A. `(2 ln sec x)/(3) + (tan^(2)x)/(6) + 2x^(2)`
B. `(3ln sec x)/(2) + (cot^(2)x)/(6) + 2x^(2)`
C. `(4ln sec x)/(3) + (sec^(2)x)/(6) + 2x^(2)`
D. `ln sec x + (tan^(4)x)/(12) + 2x^(2)`

Explanation

To find the function \( f(x) \) given that \( f''(x) = \sec^4 x + 4 \) with initial conditions \( f(0) = 0 \) and \( f'(0) = 0 \), we will follow these steps:

### Step 1: Integrate \( f''(x) \) to find \( f'(x) \)

We start with the second derivative:
\[ f''(x) = \sec^4 x + 4 \]

Integrating both sides with respect to \( x \):
\[ f'(x) = \int (\sec^4 x + 4) \, dx \]

This can be split into two parts:
\[ f'(x) = \int \sec^4 x \, dx + \int 4 \, dx \]

### Step 2: Calculate the integral of \( \sec^4 x \)

The integral of \( \sec^4 x \) is known:
\[ \int \sec^4 x \, dx = \frac{1}{3} \tan^3 x + \tan x + C_1 \]

The integral of \( 4 \) is straightforward:
\[ \int 4 \, dx = 4x + C_2 \]

Combining these results, we have:
\[ f'(x) = \frac{1}{3} \tan^3 x + \tan x + 4x + C \]

### Step 3: Apply the initial condition for \( f'(0) \)

Using the initial condition \( f'(0) = 0 \):
\[ f'(0) = \frac{1}{3} \tan^3(0) + \tan(0) + 4(0) + C = 0 \]
\[ 0 + 0 + 0 + C = 0 \implies C = 0 \]

Thus, we have:
\[ f'(x) = \frac{1}{3} \tan^3 x + \tan x + 4x \]

### Step 4: Integrate \( f'(x) \) to find \( f(x) \)

Now we integrate \( f'(x) \):
\[ f(x) = \int \left( \frac{1}{3} \tan^3 x + \tan x + 4x \right) \, dx \]

This can be split into three parts:
\[ f(x) = \frac{1}{3} \int \tan^3 x \, dx + \int \tan x \, dx + \int 4x \, dx \]

### Step 5: Calculate the integrals

1. The integral of \( \tan x \) is:
\[ \int \tan x \, dx = -\ln |\cos x| + C_3 \]

2. The integral of \( 4x \) is:
\[ \int 4x \, dx = 2x^2 + C_4 \]

3. The integral of \( \tan^3 x \) can be computed using the identity \( \tan^3 x = \tan x \cdot \sec^2 x - \tan x \):
\[ \int \tan^3 x \, dx = \frac{1}{3} \tan^3 x - \ln |\cos x| + C_5 \]

Combining these results:
\[ f(x) = \frac{1}{3} \left( \frac{1}{3} \tan^3 x - \ln |\cos x| \right) + (-\ln |\cos x|) + 2x^2 + C \]

### Step 6: Apply the initial condition for \( f(0) \)

Using the initial condition \( f(0) = 0 \):
\[ f(0) = \frac{1}{3} \left( 0 \right) - \ln |\cos(0)| + 2(0)^2 + C = 0 \]
\[ 0 - 0 + 0 + C = 0 \implies C = 0 \]

### Final Function

Thus, the function \( f(x) \) is:
\[ f(x) = \frac{1}{9} \tan^3 x - \ln |\cos x| + 2x^2 \]