The value of the integral `I=int(dx)/(sqrt(1+sinx)), AA x in [0, (pi)/(2)]` is equal to `kln(tan((x)/4+(pi)/(8)))+c`, then the value of `ksqrt2` is equal to (where, c is the constant of integration)

Options

A. `sqrt2`
B. `(1)/(2)`
C. 1
D. `2sqrt2`

Answer

To solve the integral \( I = \int \frac{dx}{\sqrt{1 + \sin x}} \) over the interval \( x \in [0, \frac{\pi}{2}] \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{\sqrt{1 + \sin x}} \]
Using the identity \( 1 + \sin x = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2} = \left( \sin \frac{x}{2} + \cos \frac{x}{2} \right)^2 \), we can rewrite the integral as:
\[ I = \int \frac{dx}{\sqrt{\left( \sin \frac{x}{2} + \cos \frac{x}{2} \right)^2}} = \int \frac{dx}{\sin \frac{x}{2} + \cos \frac{x}{2}} \]

### Step 2: Substitute Variables
Next, we can simplify the integral by multiplying the numerator and denominator by \( \sqrt{2} \):
\[ I = \int \frac{\sqrt{2} \, dx}{\sqrt{2}(\sin \frac{x}{2} + \cos \frac{x}{2})} \]
This can be rewritten using the angle addition formula:
\[ \sin \frac{x}{2} + \cos \frac{x}{2} = \sqrt{2} \sin\left(\frac{x}{2} + \frac{\pi}{4}\right) \]
Thus, we have:
\[ I = \int \frac{dx}{\sin\left(\frac{x}{2} + \frac{\pi}{4}\right)} \]

### Step 3: Change of Variables
Let \( u = \frac{x}{2} + \frac{\pi}{4} \), then \( dx = 2 \, du \). The limits change accordingly:
- When \( x = 0 \), \( u = \frac{\pi}{4} \)
- When \( x = \frac{\pi}{2} \), \( u = \frac{3\pi}{4} \)

Thus, the integral becomes:
\[ I = 2 \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{du}{\sin u} \]

### Step 4: Evaluate the Integral
The integral \( \int \frac{du}{\sin u} \) is known to be \( \ln |\tan \frac{u}{2}| \). Therefore:
\[ I = 2 \left[ \ln |\tan \frac{u}{2}| \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \]
Calculating the limits:
\[ = 2 \left( \ln |\tan \frac{3\pi/4}{2}| - \ln |\tan \frac{\pi/4}{2}| \right) \]
\[ = 2 \left( \ln |\tan \frac{3\pi/8}| - \ln |\tan \frac{\pi/8}| \right) \]
\[ = 2 \ln \left( \frac{\tan \frac{3\pi}{8}}{\tan \frac{\pi}{8}} \right) \]

### Step 5: Final Expression
Using the identity \( \tan \frac{3\pi}{8} = \sqrt{2} + 1 \) and \( \tan \frac{\pi}{8} = \sqrt{2} - 1 \):
\[ I = 2 \ln \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right) \]
This can be simplified to:
\[ = 2 \ln \left( \frac{(\sqrt{2} + 1)^2}{1} \right) = 2 \ln (3 + 2\sqrt{2}) \]

### Step 6: Compare with Given Form
We are given that:
\[ I = k \ln \left( \tan \left( \frac{x}{4} + \frac{\pi}{8} \right) \right) + c \]
From our expression, we can identify \( k \) as:
\[ k = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Conclusion
Thus, the value of \( k \sqrt{2} \) is:
\[ k \sqrt{2} = \sqrt{2} \cdot \sqrt{2} = 2 \]