Integral of `sqrt(1+2cotx(cot x + cosec x))` w.r.t. x, is

Options

A. `2 ln cos . (x)/(2) +C`
B. `2 ln sin. (x)/(2)+C`
C. `1/2 ln cos . (x)/(2)+C`
D. `ln sin x - ln (cosec x- cot x)+C`

Answer

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

### Step 1: Simplify the Expression Inside the Square Root
We start with the expression inside the square root:
\[ 1 + 2 \cot x (\cot x + \csc x) \]
Expanding this gives:
\[ 1 + 2 \cot^2 x + 2 \cot x \csc x \]

### Step 2: Use the Identity
Recall the identity:
\[ \csc^2 x - \cot^2 x = 1 \]
We can rearrange this to express \( \csc^2 x \):
\[ \csc^2 x = 1 + \cot^2 x \]
Substituting this into our expression, we have:
\[ 1 + 2 \cot^2 x + 2 \cot x \csc x = \csc^2 x + \cot^2 x + 2 \cot x \csc x \]

### Step 3: Factor the Expression
Notice that:
\[ \csc^2 x + 2 \cot x \csc x + \cot^2 x = (\csc x + \cot x)^2 \]
Thus, we can rewrite our integral as:
\[ \int \sqrt{(\csc x + \cot x)^2} \, dx \]

### Step 4: Simplify the Integral
Since the square root and the square cancel each other out, we have:
\[ \int |\csc x + \cot x| \, dx \]
For \( x \) in the domain where \( \csc x + \cot x \) is positive, we can drop the absolute value:
\[ \int (\csc x + \cot x) \, dx \]

### Step 5: Integrate
The integral of \( \csc x + \cot x \) is a standard result:
\[ \int (\csc x + \cot x) \, dx = -\csc x + C \]

### Final Answer
Thus, the integral is:
\[ -\csc x + C \]

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