`int (1-cos x)/(1+cos x) dx=?`

Options

A. `2 tan ((x)/(2))+ x+c`
B. ` tan ((x)/(2)) + x+c`
C. `tan ((x)/(2)) + 2x + c`
D. none of these

Answer

To solve the integral \(\int \frac{1 - \cos x}{1 + \cos x} \, dx\), we can follow these steps:

### Step 1: Simplify the integrand
We start by using the trigonometric identities for \(1 - \cos x\) and \(1 + \cos x\):
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]
\[ 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite the integrand:
\[ \frac{1 - \cos x}{1 + \cos x} = \frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)} = \frac{\sin^2\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)} = \tan^2\left(\frac{x}{2}\right) \]

### Step 2: Rewrite the integral
Now we can rewrite the integral:
\[ \int \frac{1 - \cos x}{1 + \cos x} \, dx = \int \tan^2\left(\frac{x}{2}\right) \, dx \]

### Step 3: Use the identity for \(\tan^2\)
We know that:
\[ \tan^2\theta = \sec^2\theta - 1 \]
So, we can rewrite the integral:
\[ \int \tan^2\left(\frac{x}{2}\right) \, dx = \int \left(\sec^2\left(\frac{x}{2}\right) - 1\right) \, dx \]

### Step 4: Split the integral
Now we can split the integral into two parts:
\[ \int \tan^2\left(\frac{x}{2}\right) \, dx = \int \sec^2\left(\frac{x}{2}\right) \, dx - \int 1 \, dx \]

### Step 5: Integrate each part
1. The integral of \(\sec^2\left(\frac{x}{2}\right)\):
\[ \int \sec^2\left(\frac{x}{2}\right) \, dx = 2 \tan\left(\frac{x}{2}\right) + C_1 \]
(using the substitution \(u = \frac{x}{2}\), \(du = \frac{1}{2}dx\))

2. The integral of \(1\):
\[ \int 1 \, dx = x + C_2 \]

### Step 6: Combine the results
Combining both results, we have:
\[ \int \tan^2\left(\frac{x}{2}\right) \, dx = 2 \tan\left(\frac{x}{2}\right) - x + C \]

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

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