`int"cosec"xdx=?`

Options

A. `log|"cosec"x-cotx|+C`
B. `-log|"cosec"x-cotx|+C`
C. `log|"cosec "x+cotx|+C`
D. none of these

Explanation

To solve the integral \(\int \csc x \, dx\), we will use a technique that involves multiplying and dividing by a specific expression. Here’s the step-by-step solution:

### Step 1: Multiply and Divide by \(\csc x + \cot x\)
We start with the integral:
\[ \int \csc x \, dx \]
To facilitate the integration, we multiply and divide by \(\csc x + \cot x\):
\[ \int \csc x \, dx = \int \csc x \cdot \frac{\csc x + \cot x}{\csc x + \cot x} \, dx \]
This gives us:
\[ \int \frac{\csc^2 x + \csc x \cot x}{\csc x + \cot x} \, dx \]

### Step 2: Let \(t = \csc x + \cot x\)
Now, we will set:
\[ t = \csc x + \cot x \]
Next, we need to find \(dt\). The derivatives are:
\[ \frac{d}{dx}(\csc x) = -\csc x \cot x \quad \text{and} \quad \frac{d}{dx}(\cot x) = -\csc^2 x \]
Thus, we have:
\[ dt = (-\csc x \cot x - \csc^2 x) \, dx = -(\csc x \cot x + \csc^2 x) \, dx \]
This implies:
\[ dx = \frac{dt}{-(\csc x \cot x + \csc^2 x)} \]

### Step 3: Substitute into the Integral
Substituting \(dt\) into the integral, we get:
\[ \int \frac{\csc^2 x + \csc x \cot x}{\csc x + \cot x} \cdot \frac{dt}{-(\csc x \cot x + \csc^2 x)} \]
The numerator simplifies to \(dt\):
\[ = -\int \frac{dt}{t} \]

### Step 4: Integrate
The integral \(-\int \frac{dt}{t}\) is:
\[ -\log |t| + C \]
Substituting back \(t = \csc x + \cot x\), we have:
\[ -\log |\csc x + \cot x| + C \]

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