`(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):
Evaluate the integral
\[ \int e^x \left( \log(\sec x + \tan x) + \sec x \right) dx \]

**Step 1: Identify \( f(x) \) and \( f'(x) \)**

Let
\[ f(x) = \log(\sec x + \tan x) \]
Then, we need to find \( f'(x) \):
\[ f'(x) = \frac{d}{dx} \log(\sec x + \tan x) = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx}(\sec x + \tan x) \]
Using the derivatives:
\[ \frac{d}{dx}(\sec x) = \sec x \tan x \quad \text{and} \quad \frac{d}{dx}(\tan x) = \sec^2 x \]
Thus,
\[ f'(x) = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} \]

**Step 2: Verify that \( \sec x = f'(x) \)**

Notice that:
\[ \sec x = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} \]
This confirms that \( f'(x) = \sec x \).

**Step 3: Apply the integration formula**

Using the integration formula:
\[ \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \]
we can write:
\[ \int e^x \left( \log(\sec x + \tan x) + \sec x \right) dx = e^x \log(\sec x + \tan x) + C \]

### Final Answer for Part (i):
\[ \int e^x \left( \log(\sec x + \tan x) + \sec x \right) dx = e^x \log(\sec x + \tan x) + C \]

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### Part (ii):
Evaluate the integral
\[ \int \frac{e^{-x} (\cos x - \sin x)}{\cos^2 x} dx \]

**Step 1: Rewrite the integral**

We can separate the terms:
\[ \int e^{-x} \left( \frac{\cos x}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right) dx = \int e^{-x} \left( \sec x - \tan x \sec x \right) dx \]

**Step 2: Identify \( f(x) \) and \( f'(x) \)**

Let
\[ f(x) = \sec x \]
Then,
\[ f'(x) = \sec x \tan x \]
Thus, we have:
\[ \sec x - \tan x \sec x = f(x) - f'(x) \]

**Step 3: Apply the integration formula**

Using the integration formula:
\[ \int e^{-x} (f(x) - f'(x)) dx = -e^{-x} f(x) + C \]
we can write:
\[ \int e^{-x} \left( \sec x - \tan x \sec x \right) dx = -e^{-x} \sec x + C \]

### Final Answer for Part (ii):
\[ \int \frac{e^{-x} (\cos x - \sin x)}{\cos^2 x} dx = -e^{-x} \sec x + C \]

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