`inte^(-x)(1-tanx)secx dx` is equal to

Options

A. `e^(-x)sec x+C`
B. `e^(-x)tan x+C`
C. `-e^(-x)tanx+C`
D. none of these

Explanation

To solve the integral \( \int e^{-x} (1 - \tan x) \sec x \, dx \), we can break it down into simpler parts and apply integration techniques.

### Step-by-Step Solution:

1. **Rewrite the Integral:**
\[ \int e^{-x} (1 - \tan x) \sec x \, dx = \int e^{-x} \sec x \, dx - \int e^{-x} \tan x \sec x \, dx \]

2. **Identify the First Integral:**
Let \( I_1 = \int e^{-x} \sec x \, dx \).

3. **Identify the Second Integral:**
Let \( I_2 = \int e^{-x} \tan x \sec x \, dx \).

4. **Use Integration by Parts for \( I_1 \):**
We can use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Here, let \( u = \sec x \) and \( dv = e^{-x} \, dx \). Then, we have:
\[ du = \sec x \tan x \, dx \quad \text{and} \quad v = -e^{-x} \]

5. **Apply Integration by Parts:**
\[ I_1 = -e^{-x} \sec x + \int e^{-x} \sec x \tan x \, dx \]

6. **Combine the Integrals:**
Now we can substitute back into the original integral:
\[ \int e^{-x} (1 - \tan x) \sec x \, dx = -e^{-x} \sec x + \int e^{-x} \sec x \tan x \, dx - \int e^{-x} \tan x \sec x \, dx \]

Notice that \( \int e^{-x} \sec x \tan x \, dx - \int e^{-x} \tan x \sec x \, dx = 0 \).

7. **Final Result:**
Therefore, we have:
\[ \int e^{-x} (1 - \tan x) \sec x \, dx = -e^{-x} \sec x + C \]

### Answer:
\[ \int e^{-x} (1 - \tan x) \sec x \, dx = -e^{-x} \sec x + C \]