`int(x)/(1-cosx)dx=`

Options

A. `xcot((x)/(2))+2log|sin((x)/(2))|+c`
B. `xcot((x)/(2))-2log|sin((x)/(2))|+c`
C. `-xcot((x)/(2))+2log|sin((x)/(2))|+c`
D. `-xcot((x)/(2))-2log|sin((x)/(2))|+c`

Answer

To solve the integral \( \int \frac{x}{1 - \cos x} \, dx \), we will use integration by parts. Here are the steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{x}{1 - \cos x} \, dx \]
We can use the identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \) to rewrite the integral:
\[ I = \int \frac{x}{2 \sin^2\left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int \frac{x}{\sin^2\left(\frac{x}{2}\right)} \, dx \]

### Step 2: Apply Integration by Parts
Let \( u = x \) and \( dv = \frac{dx}{\sin^2\left(\frac{x}{2}\right)} \). Then, we differentiate \( u \) and integrate \( dv \):
\[ du = dx \]
To find \( v \), we need to integrate \( \frac{1}{\sin^2\left(\frac{x}{2}\right)} \):
\[ v = -\cot\left(\frac{x}{2}\right) \]

### Step 3: Use the Integration by Parts Formula
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we substitute our values:
\[ I = \frac{1}{2} \left( x \cdot \left(-\cot\left(\frac{x}{2}\right)\right) - \int -\cot\left(\frac{x}{2}\right) \, dx \right) \]
This simplifies to:
\[ I = -\frac{x}{2} \cot\left(\frac{x}{2}\right) + \frac{1}{2} \int \cot\left(\frac{x}{2}\right) \, dx \]

### Step 4: Integrate \( \cot\left(\frac{x}{2}\right) \)
The integral of \( \cot\left(\frac{x}{2}\right) \) is:
\[ \int \cot\left(\frac{x}{2}\right) \, dx = -\ln\left|\sin\left(\frac{x}{2}\right)\right| + C \]

### Step 5: Substitute Back
Substituting back into our equation for \( I \):
\[ I = -\frac{x}{2} \cot\left(\frac{x}{2}\right) - \frac{1}{2} \ln\left|\sin\left(\frac{x}{2}\right)\right| + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{x}{1 - \cos x} \, dx = -\frac{x}{2} \cot\left(\frac{x}{2}\right) - \frac{1}{2} \ln\left|\sin\left(\frac{x}{2}\right)\right| + C \]

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