Evaluate : ` int_(0)^((pi)/2)) (sqrt(cot x))/(sqrt(cot x)+ sqrt(tan x)) dx `

Answer

To evaluate the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx, \]

we can use the property of definite integrals which states that

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx. \]

Here, our upper limit is \(\frac{\pi}{2}\), so we will replace \(x\) with \(\frac{\pi}{2} - x\).

### Step 1: Substitute \(x\) with \(\frac{\pi}{2} - x\)

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cot\left(\frac{\pi}{2} - x\right)}}{\sqrt{\cot\left(\frac{\pi}{2} - x\right)} + \sqrt{\tan\left(\frac{\pi}{2} - x\right)}} \, dx. \]

### Step 2: Simplify using trigonometric identities

Using the identities \(\cot\left(\frac{\pi}{2} - x\right) = \tan x\) and \(\tan\left(\frac{\pi}{2} - x\right) = \cot x\), we can rewrite the integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} \, dx. \]

### Step 3: Combine the two integrals

Now we have two expressions for \(I\):

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx, \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} \, dx. \]

Adding these two equations:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} + \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} \right) \, dx. \]

### Step 4: Simplify the expression inside the integral

Notice that:

\[ \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} + \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} = 1. \]

Thus, we have:

\[ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx. \]

### Step 5: Evaluate the integral

The integral of 1 from 0 to \(\frac{\pi}{2}\) is simply:

\[ \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}. \]

### Step 6: Solve for \(I\)

So we have:

\[ 2I = \frac{\pi}{2} \implies I = \frac{\pi}{4}. \]

### Final Answer

Thus, the value of the integral is

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