The value of `int(cos2x)/(cosx)` dx is equal to

Options

A. `2sinx-ln|secx+tanx|+C`
B. `2sinx-ln|secx-tanx|+C`
C. `2sinx+ln|secx+tanx|+C`
D. `sinx-ln|secx-tanx|+C`

Answer

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

### Step 1: Rewrite the integrand using the double angle formula
We know that:
\[ \cos 2x = 2 \cos^2 x - 1 \]
Thus, we can rewrite the integral:
\[ I = \int \frac{2 \cos^2 x - 1}{\cos x} \, dx \]

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

### Step 3: Integrate each part
1. The integral of \( 2 \cos x \):
\[ \int 2 \cos x \, dx = 2 \sin x \]

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

### Step 4: Combine the results
Putting it all together, we have:
\[ I = 2 \sin x - \ln |\sec x + \tan x| + C \]

### Final Answer
Thus, the value of the integral is:
\[ I = 2 \sin x - \ln |\sec x + \tan x| + C \]

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