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

Options

A. `e^(tanx)cosx +C`
B. `e^(tanx)sinx +C`
C. `-e^(tanx)cosx +C`
D. `e^(tanx)sec x +C`

Explanation

To solve the integral \( \int e^{\tan x} (\sec x - \sin x) \, dx \), we can break it down into two separate integrals:

\[ \int e^{\tan x} \sec x \, dx - \int e^{\tan x} \sin x \, dx \]

### Step 1: Solve the first integral \( \int e^{\tan x} \sec x \, dx \)

We will denote this integral as \( I_1 = \int e^{\tan x} \sec x \, dx \).

### Step 2: Solve the second integral \( \int e^{\tan x} \sin x \, dx \)

We will denote this integral as \( I_2 = \int e^{\tan x} \sin x \, dx \).

### Step 3: Apply Integration by Parts to \( I_2 \)

For the integral \( I_2 \), we will use integration by parts, where we let:
- \( u = \sin x \) and \( dv = e^{\tan x} \, dx \)

Now we need to find \( du \) and \( v \):
- \( du = \cos x \, dx \)
- To find \( v \), we need to integrate \( dv \):

Using the substitution \( w = \tan x \), we have \( dw = \sec^2 x \, dx \), so:
\[ v = \int e^{\tan x} \, dx = e^{\tan x} \]

### Step 4: Apply the Integration by Parts Formula

Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we get:
\[ I_2 = \sin x \cdot e^{\tan x} - \int e^{\tan x} \cos x \, dx \]

### Step 5: Substitute back into the original integral

Now substituting \( I_2 \) back into the original integral:
\[ \int e^{\tan x} (\sec x - \sin x) \, dx = I_1 - \left( \sin x \cdot e^{\tan x} - \int e^{\tan x} \cos x \, dx \right) \]

### Step 6: Combine the integrals

Now we have:
\[ \int e^{\tan x} \sec x \, dx - \sin x \cdot e^{\tan x} + \int e^{\tan x} \cos x \, dx \]

### Step 7: Simplify

Notice that \( I_1 \) and \( \int e^{\tan x} \sec x \, dx \) will cancel each other out:
\[ I_1 - I_1 + \int e^{\tan x} \cos x \, dx - \sin x \cdot e^{\tan x} = -\sin x \cdot e^{\tan x} + C \]

### Final Answer

Thus, the final answer is:
\[ \int e^{\tan x} (\sec x - \sin x) \, dx = e^{\tan x} \cos x + C \]