`int_(0)^(pi/2)(dx)/(1+sqrt(cotx))`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sqrt{\cot x}} \), we can follow these steps:

### Step 1: Rewrite the Integral
Start by rewriting the integral in terms of sine and cosine:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sqrt{\cot x}} = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sqrt{\frac{\cos x}{\sin x}}} \]
This can be rewritten as:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}} = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \]

### Step 2: Use a Symmetry Property
Now, we can use the property of definite integrals:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, let \( a = \frac{\pi}{2} \):
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin\left(\frac{\pi}{2} - x\right)}}{\sqrt{\sin\left(\frac{\pi}{2} - x\right)} + \sqrt{\cos\left(\frac{\pi}{2} - x\right)}} \, dx \]
Since \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \), we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx \]

### Step 3: Add the Two Integrals
Now we can add the two expressions for \( I \):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \right) \, dx \]
The denominators are the same, so we can combine the fractions:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx = \int_{0}^{\frac{\pi}{2}} 1 \, dx \]

### Step 4: Evaluate the Integral
Now, we can evaluate the integral:
\[ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]
Thus, we have:
\[ 2I = \frac{\pi}{2} \implies I = \frac{\pi}{4} \]

### Final Answer
The value of the integral is:
\[ \boxed{\frac{\pi}{4}} \]