Evaluate : `intsqrt(1+2cotx("cosec"x+cotx))dx`

Answer

To evaluate the integral \( \int \sqrt{1 + 2 \cot x (\csc x + \cot x)} \, dx \), we will follow these steps:

### Step 1: Simplify the Expression Inside the Square Root
We start with the expression inside the square root:
\[ 1 + 2 \cot x (\csc x + \cot x) \]
We know that:
\[ \csc x = \frac{1}{\sin x} \quad \text{and} \quad \cot x = \frac{\cos x}{\sin x} \]
Substituting these identities, we get:
\[ 1 + 2 \cot x \left(\frac{1}{\sin x} + \frac{\cos x}{\sin x}\right) = 1 + 2 \cot x \cdot \frac{1 + \cos x}{\sin x} \]
This simplifies to:
\[ 1 + 2 \frac{\cos x}{\sin x} \cdot \frac{1 + \cos x}{\sin x} = 1 + 2 \frac{\cos x (1 + \cos x)}{\sin^2 x} \]

### Step 2: Use the Identity \( \csc^2 x - \cot^2 x = 1 \)
We can express \( \sin^2 x \) in terms of \( \csc^2 x \) and \( \cot^2 x \):
\[ \sin^2 x = \frac{1}{\csc^2 x} \]
Thus:
\[ 1 + 2 \cot x \cdot \frac{1 + \cos x}{\sin x} = 1 + 2 \cot x \cdot \csc x \]

### Step 3: Recognize a Perfect Square
Notice that:
\[ 1 + 2 \cot x \cdot \csc x = (\csc x + \cot x)^2 \]
Thus, we can rewrite the integral:
\[ \int \sqrt{(\csc x + \cot x)^2} \, dx = \int |\csc x + \cot x| \, dx \]
Assuming \( \csc x + \cot x \) is positive in the interval of integration, we have:
\[ \int (\csc x + \cot x) \, dx \]

### Step 4: Integrate
The integral of \( \csc x + \cot x \) is:
\[ \int (\csc x + \cot x) \, dx = -\ln |\csc x + \cot x| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ -\ln |\csc x + \cot x| + C \]