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

Options

A. `2tan""x/2-x+c`
B. `1/2tan""x/2-x+c`
C. `x-1/2tan""x/2+c`
D. `x-2tan""x/2+c`

Answer

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

### Step 1: Rewrite the integrand
We can rewrite the integrand using trigonometric identities. Recall that:
- \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \)
- \( 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \)

Thus, we can express \( \frac{\cos x - 1}{\cos x + 1} \) as:
\[ \frac{\cos x - 1}{\cos x + 1} = \frac{- (1 - \cos x)}{1 + \cos x} = -\frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)} = -\tan^2\left(\frac{x}{2}\right) \]

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

### Step 3: Use the identity for \(\tan^2\)
Recall the identity:
\[ \tan^2 x = \sec^2 x - 1 \]
Thus,
\[ -\int \tan^2\left(\frac{x}{2}\right) \, dx = -\int \left(\sec^2\left(\frac{x}{2}\right) - 1\right) \, dx \]

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

### Step 5: Integrate each part
1. The integral of \(1\) is simply \(x\).
2. For the integral of \(\sec^2\left(\frac{x}{2}\right)\), we use the substitution \(u = \frac{x}{2}\), hence \(dx = 2 \, du\):
\[ -\int \sec^2\left(u\right) \cdot 2 \, du = -2 \tan\left(u\right) + C = -2 \tan\left(\frac{x}{2}\right) + C \]

### Step 6: Combine the results
Combining both parts, we have:
\[ \int \frac{\cos x - 1}{\cos x + 1} \, dx = -2 \tan\left(\frac{x}{2}\right) + x + C \]

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