`(i) int e^(x). "[log (sec x+tan x) + sec x dx "`
`(ii) int (e^(-x)(cos x-sin x))/(cos^(2) x) dx`

Answer

Let's solve the given integrals step by step.

### Part (i):
We need to evaluate the integral:
\[ I_1 = \int e^x \left( \log(\sec x + \tan x) + \sec x \right) dx \]

**Step 1: Identify \(f(x)\) and \(f'(x)\)**
Let:
\[ f(x) = \log(\sec x + \tan x) \]
Then, we need to find \(f'(x)\):
\[ f'(x) = \frac{d}{dx} \log(\sec x + \tan x) = \frac{1}{\sec x + \tan x} \cdot \left(\sec x \tan x + \sec^2 x\right) \]
This simplifies to:
\[ f'(x) = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} = \sec x \]

**Step 2: Apply the standard result**
The standard result states:
\[ \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \]
In our case:
\[ \int e^x \left( \log(\sec x + \tan x) + \sec x \right) dx = e^x \log(\sec x + \tan x) + C \]

Thus, the solution to the first part is:
\[ I_1 = e^x \log(\sec x + \tan x) + C \]

### Part (ii):
We need to evaluate the integral:
\[ I_2 = \int \frac{e^{-x} (\cos x - \sin x)}{\cos^2 x} dx \]

**Step 1: Substitute**
Let:
\[ t = -x \implies dx = -dt \]
Then, substituting into the integral:
\[ I_2 = \int \frac{e^t (\cos(-t) - \sin(-t))}{\cos^2(-t)} (-dt) = -\int \frac{e^t (\cos t + \sin t)}{\cos^2 t} dt \]
This can be rewritten as:
\[ I_2 = \int \frac{e^t (\cos t + \sin t)}{\cos^2 t} dt \]

**Step 2: Simplify the integral**
We can split the integral:
\[ I_2 = \int e^t \sec^2 t dt + \int e^t \tan t \sec t dt \]

**Step 3: Apply the standard result**
Let:
\[ f(t) = \sec t \quad \text{and} \quad f'(t) = \sec t \tan t \]
Using the standard result:
\[ \int e^t [f(t) + f'(t)] dt = e^t f(t) + C \]
Thus:
\[ I_2 = e^t \sec t + C \]

**Step 4: Substitute back for \(t\)**
Substituting back \(t = -x\):
\[ I_2 = e^{-x} \sec(-x) + C = e^{-x} \sec x + C \]

### Final Answers:
1. \(I_1 = e^x \log(\sec x + \tan x) + C\)
2. \(I_2 = e^{-x} \sec x + C\)