`intx"cosec"^(2)xdx=?`

Options

A. `xcotx-log|sinx|+C`
B. `-xcotx+log|sinx|+C`
C. `xtanx-log|secx|+C`
D. none of these

Explanation

To evaluate the integral \( \int x \csc^2 x \, dx \), we will use the method of integration by parts. Let's go through the steps systematically.

### Step 1: Set Up Integration by Parts
We start by letting:
- \( u = x \) (first function)
- \( dv = \csc^2 x \, dx \) (second function)

### Step 2: Differentiate and Integrate
Next, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = dx \]
- Integrate \( dv \):
\[ v = \int \csc^2 x \, dx = -\cot x \]

### Step 3: Apply Integration by Parts Formula
The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
Substituting the values we have:
\[ \int x \csc^2 x \, dx = x(-\cot x) - \int (-\cot x) \, dx \]
This simplifies to:
\[ = -x \cot x + \int \cot x \, dx \]

### Step 4: Integrate \( \cot x \)
Now we need to evaluate \( \int \cot x \, dx \):
\[ \int \cot x \, dx = \int \frac{\cos x}{\sin x} \, dx \]
To solve this, we can use the substitution:
- Let \( t = \sin x \), then \( dt = \cos x \, dx \).

Substituting gives:
\[ \int \cot x \, dx = \int \frac{1}{t} \, dt = \ln |t| + C = \ln |\sin x| + C \]

### Step 5: Combine Results
Now, substituting back into our expression:
\[ \int x \csc^2 x \, dx = -x \cot x + \ln |\sin x| + C \]

### Final Answer
Thus, the final result is:
\[ \int x \csc^2 x \, dx = -x \cot x + \ln |\sin x| + C \]

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