`int sqrt(1+secx) dx` is equal to:

Options

A. `2 tan ^(-1) sqrt(frac{1-cosx}{cosx)}+C`
B. `tan ^(-1) sqrt(frac{1-cosx}{cosx)}+C`
C. `3tan ^(-1) sqrt(frac{1+cosx}{cosx)}+C`
D. None of these

Answer

To solve the integral \( I = \int \sqrt{1 + \sec x} \, dx \), we will follow these steps:

### Step 1: Substitute \( 1 + \sec x \)
Let \( 1 + \sec x = t^2 \). Then, we differentiate both sides:
\[ \frac{d}{dx}(1 + \sec x) = \sec x \tan x \, dx = 2t \, dt \]
This gives us:
\[ \sec x \tan x \, dx = 2t \, dt \quad \Rightarrow \quad dx = \frac{2t \, dt}{\sec x \tan x} \]

### Step 2: Express \( \sec x \) and \( \tan x \) in terms of \( t \)
From our substitution, we have:
\[ \sec x = t^2 - 1 \]
Using the identity \( \sec^2 x = 1 + \tan^2 x \), we can express \( \tan x \) as:
\[ \tan^2 x = \sec^2 x - 1 = (t^2 - 1)^2 - 1 = t^4 - 2t^2 \]
Thus,
\[ \tan x = \sqrt{t^4 - 2t^2} \]

### Step 3: Substitute into the integral
Now substituting \( \sec x \) and \( \tan x \) into the integral:
\[ I = \int \sqrt{t^2} \cdot \frac{2t \, dt}{(t^2 - 1) \cdot \sqrt{t^4 - 2t^2}} \]
This simplifies to:
\[ I = \int \frac{2t^2 \, dt}{(t^2 - 1) \cdot \sqrt{t^4 - 2t^2}} \]

### Step 4: Simplify the integral
We can simplify the expression further. Notice that:
\[ \sqrt{t^4 - 2t^2} = \sqrt{t^2(t^2 - 2)} = t \sqrt{t^2 - 2} \]
Thus, the integral becomes:
\[ I = \int \frac{2t \, dt}{t^2 - 1} \cdot \frac{1}{\sqrt{t^2 - 2}} \]

### Step 5: Make another substitution
Let \( t^2 - 2 = z^2 \), then \( t^2 = z^2 + 2 \) and differentiate:
\[ 2t \, dt = 2z \, dz \quad \Rightarrow \quad t \, dt = z \, dz \]
Now substituting this into the integral:
\[ I = \int \frac{2z \, dz}{(z^2 + 2 - 1) \cdot z} = \int \frac{2 \, dz}{z^2 + 1} \]

### Step 6: Integrate
The integral \( \int \frac{2 \, dz}{z^2 + 1} \) is a standard integral:
\[ I = 2 \tan^{-1}(z) + C \]

### Step 7: Back substitute
Recall that \( z = \sqrt{t^2 - 2} \) and \( t^2 = 1 + \sec x \):
\[ z = \sqrt{(1 + \sec x) - 2} = \sqrt{\sec x - 1} \]
Thus, we have:
\[ I = 2 \tan^{-1}(\sqrt{\sec x - 1}) + C \]

### Final Answer
The integral \( \int \sqrt{1 + \sec x} \, dx \) is equal to:
\[ I = 2 \tan^{-1}(\sqrt{\sec x - 1}) + C \]