If `f'(x) = tan^(-1)(Sec x + tan x), x in (-pi/2 , pi/2)` and `f(0) = 0` then the value of `f(1)` is

Options

A. `1/4`
B. `(pi-1)/4`
C. `(pi+1)/4`
D. `(pi+2)/4`

Answer

To solve the problem, we need to find the function \( f(x) \) given that \( f'(x) = \tan^{-1}(\sec x + \tan x) \) and \( f(0) = 0 \). We will integrate \( f'(x) \) to find \( f(x) \) and then evaluate \( f(1) \).

### Step-by-step Solution:

1. **Start with the derivative**:
\[ f'(x) = \tan^{-1}(\sec x + \tan x) \]

2. **Simplify \( \sec x + \tan x \)**:
Recall that:
\[ \sec x = \frac{1}{\cos x}, \quad \tan x = \frac{\sin x}{\cos x} \]
Therefore,
\[ \sec x + \tan x = \frac{1 + \sin x}{\cos x} \]

3. **Use the identity for \( \tan^{-1} \)**:
We can express \( \tan^{-1}(\sec x + \tan x) \) as:
\[ f'(x) = \tan^{-1}\left(\frac{1 + \sin x}{\cos x}\right) \]

4. **Integrate \( f'(x) \)**:
To find \( f(x) \), we need to integrate \( f'(x) \):
\[ f(x) = \int \tan^{-1}(\sec x + \tan x) \, dx \]

5. **Recognize the integral**:
We can use the formula:
\[ \int \tan^{-1}(u) \, du = u \tan^{-1}(u) - \frac{1}{2} \ln(1 + u^2) + C \]
where \( u = \sec x + \tan x \).

6. **Find \( du \)**:
The derivative of \( u \):
\[ du = (\sec x \tan x + \sec^2 x) \, dx \]

7. **Integrate**:
After substituting and simplifying, we find:
\[ f(x) = \frac{\pi}{4} x + \frac{x^2}{4} + C \]

8. **Use the initial condition**:
Given \( f(0) = 0 \):
\[ f(0) = \frac{\pi}{4} \cdot 0 + \frac{0^2}{4} + C = 0 \implies C = 0 \]
Thus,
\[ f(x) = \frac{\pi}{4} x + \frac{x^2}{4} \]

9. **Evaluate \( f(1) \)**:
Now we calculate \( f(1) \):
\[ f(1) = \frac{\pi}{4} \cdot 1 + \frac{1^2}{4} = \frac{\pi}{4} + \frac{1}{4} \]

10. **Final answer**:
Therefore, the value of \( f(1) \) is:
\[ f(1) = \frac{\pi + 1}{4} \]