Evaluate the following integrals:
`int frac{dx}{1+cosx}`

Answer

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

### Step 1: Rewrite the denominator using a trigonometric identity
We know that:
\[ 1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right) \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{dx}{2 \cos^2 \left(\frac{x}{2}\right)} \]

### Step 2: Factor out the constant
We can factor out the constant \( \frac{1}{2} \) from the integral:
\[ I = \frac{1}{2} \int \frac{dx}{\cos^2 \left(\frac{x}{2}\right)} \]

### Step 3: Use the secant function
Recall that \( \frac{1}{\cos^2 \theta} = \sec^2 \theta \). Therefore, we can write:
\[ I = \frac{1}{2} \int \sec^2 \left(\frac{x}{2}\right) dx \]

### Step 4: Integrate using the known integral of secant squared
The integral of \( \sec^2 \theta \) is \( \tan \theta \). Thus, we have:
\[ I = \frac{1}{2} \cdot 2 \tan \left(\frac{x}{2}\right) + C \]
where \( C \) is the constant of integration.

### Step 5: Simplify the expression
The \( 2 \) in the numerator and the \( \frac{1}{2} \) in front will cancel out:
\[ I = \tan \left(\frac{x}{2}\right) + C \]

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