`int(cos 2x)/(cosx) dx` is equal to

Options

A. `2 sin x + log (sec x - tan x) + c`
B. `2 sin x - log(sec x - tan x) + c`
C. `2 sin x + log( sec x + tan x) + c`
D. `2 sin x - log(sec x + tan x) + c`

Answer

To solve the integral \(\int \frac{\cos 2x}{\cos x} \, dx\), we can follow these steps:

### Step 1: Rewrite \(\cos 2x\)
We start by using the double angle formula for cosine:
\[ \cos 2x = 2\cos^2 x - 1 \]
Thus, we can rewrite the integral as:
\[ \int \frac{\cos 2x}{\cos x} \, dx = \int \frac{2\cos^2 x - 1}{\cos x} \, dx \]

### Step 2: Separate the Integral
Now, we can separate the integral into two parts:
\[ \int \frac{2\cos^2 x - 1}{\cos x} \, dx = \int \frac{2\cos^2 x}{\cos x} \, dx - \int \frac{1}{\cos x} \, dx \]
This simplifies to:
\[ \int 2\cos x \, dx - \int \sec x \, dx \]

### Step 3: Integrate Each Term
Now, we can integrate each term separately.

1. For \(\int 2\cos x \, dx\):
\[ \int 2\cos x \, dx = 2\sin x \]

2. For \(\int \sec x \, dx\):
The integral of \(\sec x\) is:
\[ \int \sec x \, dx = \ln |\sec x + \tan x| + C \]

### Step 4: Combine the Results
Now, we can combine the results from the two integrals:
\[ \int \frac{\cos 2x}{\cos x} \, dx = 2\sin x - \ln |\sec x + \tan x| + C \]

### Final Answer
Thus, the final answer is:
\[ 2\sin x - \ln |\sec x + \tan x| + C \]