`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: Rewrite the integral using trigonometric identities
We know that:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]
and
\[ 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \]
Therefore, we can rewrite the integral as:
\[ \int \frac{1 - \cos x}{1 + \cos x} \, dx = \int \frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)} \, dx \]
The 2s in the numerator and denominator cancel out:
\[ = \int \frac{\sin^2\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)} \, dx \]

### Step 2: Simplify the integral
Using the identity \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \), we can rewrite the integral as:
\[ = \int \tan^2\left(\frac{x}{2}\right) \, dx \]

### Step 3: Use the identity for \( \tan^2 \)
Recall that:
\[ \tan^2 \theta = \sec^2 \theta - 1 \]
Thus, we can express the integral as:
\[ = \int \left(\sec^2\left(\frac{x}{2}\right) - 1\right) \, dx \]
This can be separated into two integrals:
\[ = \int \sec^2\left(\frac{x}{2}\right) \, dx - \int 1 \, dx \]

### Step 4: Solve the integrals
For the first integral, we can use a substitution. Let \( t = \frac{x}{2} \), then \( dx = 2 \, dt \):
\[ \int \sec^2\left(\frac{x}{2}\right) \, dx = \int \sec^2(t) \cdot 2 \, dt = 2 \int \sec^2(t) \, dt \]
The integral of \( \sec^2(t) \) is \( \tan(t) \):
\[ = 2 \tan(t) + C = 2 \tan\left(\frac{x}{2}\right) + C \]
For the second integral:
\[ \int 1 \, dx = x \]

### Step 5: Combine the results
Putting it all together, we have:
\[ \int \tan^2\left(\frac{x}{2}\right) \, dx = 2 \tan\left(\frac{x}{2}\right) - x + C \]

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