`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 a systematic approach using trigonometric identities and substitutions.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
\[ I = \int \sqrt{1 + \sec x} \, dx \]
Recall that \( \sec x = \frac{1}{\cos x} \). Thus, we can rewrite the integral as:
\[ I = \int \sqrt{1 + \frac{1}{\cos x}} \, dx = \int \sqrt{\frac{\cos x + 1}{\cos x}} \, dx \]

2. **Simplify the Expression**:
\[ I = \int \sqrt{\frac{1 + \cos x}{\cos x}} \, dx = \int \frac{\sqrt{1 + \cos x}}{\sqrt{\cos x}} \, dx \]

3. **Use Trigonometric Identity**:
We know that \( 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \). Thus:
\[ I = \int \frac{\sqrt{2 \cos^2\left(\frac{x}{2}\right)}}{\sqrt{\cos x}} \, dx = \int \frac{\sqrt{2} \cos\left(\frac{x}{2}\right)}{\sqrt{\cos x}} \, dx \]

4. **Express \( \cos x \) in Terms of \( \cos\left(\frac{x}{2}\right) \)**:
Using the identity \( \cos x = 2 \cos^2\left(\frac{x}{2}\right) - 1 \), we can express \( \sqrt{\cos x} \):
\[ \sqrt{\cos x} = \sqrt{2 \cos^2\left(\frac{x}{2}\right) - 1} \]

5. **Substitution**:
Let \( t = \sin\left(\frac{x}{2}\right) \), then \( dx = 2 \cos\left(\frac{x}{2}\right) \, dt \). We also have:
\[ \cos\left(\frac{x}{2}\right) = \sqrt{1 - t^2} \]
Thus:
\[ I = \int \frac{\sqrt{2} \sqrt{1 - t^2}}{\sqrt{1 - 2t^2}} \cdot 2\sqrt{1 - t^2} \, dt = 2\sqrt{2} \int \frac{1 - t^2}{\sqrt{1 - 2t^2}} \, dt \]

6. **Integrate**:
The integral can be simplified further, and we can use a trigonometric substitution or integral tables to evaluate it.

7. **Final Result**:
After performing the integration and substituting back, we arrive at:
\[ I = 2 \sin^{-1}\left(\sqrt{2} \sin\left(\frac{x}{2}\right)\right) + C \]