(i) `int x sec^2 x dx` (ii) `int x cos^2 x dx`

Answer

Let's solve the integrals step by step.

### Part (i): \( \int x \sec^2 x \, dx \)

1. **Identify the method**: We will use integration by parts. Recall the formula:
\[ \int u \, dv = uv - \int v \, du \]
Here, we can choose:
- \( u = x \) (which means \( du = dx \))
- \( dv = \sec^2 x \, dx \) (which means \( v = \tan x \))

2. **Apply integration by parts**:
\[ \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx \]

3. **Integrate \( \tan x \)**:
The integral of \( \tan x \) is:
\[ \int \tan x \, dx = -\log |\cos x| + C = \log |\sec x| + C \]

4. **Combine the results**:
\[ \int x \sec^2 x \, dx = x \tan x - \log |\sec x| + C \]

### Final Result for Part (i):
\[ \int x \sec^2 x \, dx = x \tan x - \log |\sec x| + C \]

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### Part (ii): \( \int x \cos^2 x \, dx \)

1. **Rewrite \( \cos^2 x \)**: Use the identity:
\[ \cos^2 x = \frac{1 + \cos(2x)}{2} \]
Thus, we can rewrite the integral:
\[ \int x \cos^2 x \, dx = \int x \left(\frac{1 + \cos(2x)}{2}\right) \, dx = \frac{1}{2} \int x \, dx + \frac{1}{2} \int x \cos(2x) \, dx \]

2. **Integrate \( \frac{1}{2} \int x \, dx \)**:
\[ \frac{1}{2} \int x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4} \]

3. **Integrate \( \frac{1}{2} \int x \cos(2x) \, dx \)**: Again, use integration by parts:
- Let \( u = x \) (thus \( du = dx \))
- Let \( dv = \cos(2x) \, dx \) (thus \( v = \frac{1}{2} \sin(2x) \))

Applying integration by parts:
\[ \int x \cos(2x) \, dx = x \cdot \frac{1}{2} \sin(2x) - \int \frac{1}{2} \sin(2x) \, dx \]

4. **Integrate \( \int \sin(2x) \, dx \)**:
\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C \]

5. **Combine results**:
\[ \int x \cos(2x) \, dx = \frac{x}{2} \sin(2x) + \frac{1}{4} \cos(2x) + C \]

6. **Final combination**:
\[ \int x \cos^2 x \, dx = \frac{x^2}{4} + \frac{1}{2} \left( \frac{x}{2} \sin(2x) + \frac{1}{4} \cos(2x) \right) + C \]

### Final Result for Part (ii):
\[ \int x \cos^2 x \, dx = \frac{x^2}{4} + \frac{x}{4} \sin(2x) + \frac{1}{8} \cos(2x) + C \]

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