`(i) int " x sec"^(2) " 2x dx "" "(ii) int " x sin"^(3) " x dx "`

Answer

Let's solve the given integrals step by step.

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

1. **Identify the parts for integration by parts**:
We will use integration by parts, where we let:
- \( u = x \) (first part)
- \( dv = \sec^2(2x) \, dx \) (second part)

2. **Differentiate and integrate**:
- Differentiate \( u \):
\[ du = dx \]
- Integrate \( dv \):
\[ v = \int \sec^2(2x) \, dx = \frac{1}{2} \tan(2x) \]

3. **Apply the integration by parts formula**:
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
Substituting the values we found:
\[ \int x \sec^2(2x) \, dx = x \cdot \frac{1}{2} \tan(2x) - \int \frac{1}{2} \tan(2x) \, dx \]

4. **Integrate \(\tan(2x)\)**:
The integral of \(\tan(2x)\) is:
\[ \int \tan(2x) \, dx = -\frac{1}{2} \log|\cos(2x)| + C \]
Thus,
\[ \int \tan(2x) \, dx = -\frac{1}{2} \log|\cos(2x)| + C \]

5. **Combine everything**:
Now substituting back:
\[ \int x \sec^2(2x) \, dx = \frac{1}{2} x \tan(2x) + \frac{1}{4} \log|\cos(2x)| + C \]

### Final Answer for Part (i):
\[ \int x \sec^2(2x) \, dx = \frac{1}{2} x \tan(2x) - \frac{1}{4} \log|\cos(2x)| + C \]

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### Part (ii): \(\int x \sin^3(x) \, dx\)

1. **Rewrite \(\sin^3(x)\)**:
We can use the identity:
\[ \sin^3(x) = \frac{3 \sin(x) - \sin(3x)}{4} \]
Thus, we rewrite the integral:
\[ \int x \sin^3(x) \, dx = \frac{1}{4} \int x (3 \sin(x) - \sin(3x)) \, dx \]

2. **Distribute the integral**:
\[ = \frac{3}{4} \int x \sin(x) \, dx - \frac{1}{4} \int x \sin(3x) \, dx \]

3. **Apply integration by parts for \(\int x \sin(x) \, dx\)**:
Let:
- \( u = x \), \( dv = \sin(x) \, dx \)
- Then, \( du = dx \), \( v = -\cos(x) \)
\[ \int x \sin(x) \, dx = -x \cos(x) + \int \cos(x) \, dx = -x \cos(x) + \sin(x) \]

4. **Apply integration by parts for \(\int x \sin(3x) \, dx\)**:
Let:
- \( u = x \), \( dv = \sin(3x) \, dx \)
- Then, \( du = dx \), \( v = -\frac{1}{3} \cos(3x) \)
\[ \int x \sin(3x) \, dx = -\frac{1}{3} x \cos(3x) + \frac{1}{3} \int \cos(3x) \, dx = -\frac{1}{3} x \cos(3x) + \frac{1}{9} \sin(3x) \]

5. **Combine everything**:
Putting it all together:
\[ \int x \sin^3(x) \, dx = \frac{3}{4} \left(-x \cos(x) + \sin(x)\right) - \frac{1}{4} \left(-\frac{1}{3} x \cos(3x) + \frac{1}{9} \sin(3x)\right) \]

### Final Answer for Part (ii):
\[ \int x \sin^3(x) \, dx = -\frac{3}{4} x \cos(x) + \frac{3}{4} \sin(x) + \frac{1}{12} x \cos(3x) - \frac{1}{36} \sin(3x) + C \]

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