value of `int_0^1cot^-1((sqrt(1+sinx)+(sqrt(1-sinx)))/((sqrt(1+sinx)-(sqrt(1-sinx)))))dx`

Options

A. `1/4`
B. `1/2`
C. 0
D. `1/8`

Answer

To solve the integral

\[ I = \int_0^1 \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) dx, \]

we will simplify the expression inside the integral step by step.

### Step 1: Simplifying the Argument of Cotangent Inverse

We start with the expression inside the cotangent inverse:

\[ \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}. \]

Using the half-angle identities, we can express \( \sin x \) in terms of \( \sin \frac{x}{2} \) and \( \cos \frac{x}{2} \):

\[ \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}. \]

Thus, we can rewrite \( 1 + \sin x \) and \( 1 - \sin x \):

\[ 1 + \sin x = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2} = \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2, \]
\[ 1 - \sin x = 1 - 2 \sin \frac{x}{2} \cos \frac{x}{2} = \left(\cos \frac{x}{2} - \sin \frac{x}{2}\right)^2. \]

### Step 2: Substitute and Simplify

Substituting these back into the expression gives:

\[ \frac{\sqrt{\left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2} + \sqrt{\left(\cos \frac{x}{2} - \sin \frac{x}{2}\right)^2}}}{\sqrt{\left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2} - \sqrt{\left(\cos \frac{x}{2} - \sin \frac{x}{2}\right)^2}}. \]

This simplifies to:

\[ \frac{\sin \frac{x}{2} + \cos \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2}}{\sin \frac{x}{2} + \cos \frac{x}{2} - (\cos \frac{x}{2} - \sin \frac{x}{2})}. \]

### Step 3: Further Simplification

The numerator simplifies to:

\[ 2 \cos \frac{x}{2}, \]

and the denominator simplifies to:

\[ 2 \sin \frac{x}{2}. \]

Thus, we have:

\[ \frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}} = \cot \frac{x}{2}. \]

### Step 4: Integral of Cotangent Inverse

Now we can rewrite the integral:

\[ I = \int_0^1 \cot^{-1}(\cot \frac{x}{2}) \, dx. \]

Since \( \cot^{-1}(\cot \theta) = \theta \) for \( \theta \) in the range \( (0, \pi) \), we have:

\[ I = \int_0^1 \frac{x}{2} \, dx. \]

### Step 5: Evaluate the Integral

Calculating the integral:

\[ I = \frac{1}{2} \int_0^1 x \, dx = \frac{1}{2} \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}. \]

### Final Answer

Thus, the value of the integral is:

\[ \boxed{\frac{1}{4}}. \]