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Let's solve the integrals one by one using the method of substitution.

### Problem 17: \(\int \cos(3x + 2) \, dx\)

1. **Substitution:**
   Let \( u = 3x + 2 \). Then, differentiate \( u \) with respect to \( x \):
   \[
   \frac{du}{dx} = 3 \quad \text{or} \quad du = 3 \, dx
   \]
   Therefore, \( dx = \frac{du}{3} \).

2. **Rewrite the integral:**
   Substitute \( u = 3x + 2 \) into the integral:
   \[
   \int \cos(3x + 2) \, dx = \int \cos(u) \cdot \frac{du}{3}
   \]

3. **Integrate:**
   The integral of \( \cos(u) \) is \( \sin(u) \):
   \[
   \int \cos(u) \, du = \sin(u)
   \]
   So, the integral becomes:
   \[
   \frac{1}{3} \sin(u) + C
   \]

4. **Substitute back:**
   Replace \( u \) with \( 3x + 2 \):
   \[
   \int \cos(3x + 2) \, dx = \frac{1}{3} \sin(3x + 2) + C
   \]

### Problem 18: \(\int \sin(2x - 4) \, dx\)

1. **Substitution:**
   Let \( u = 2x - 4 \). Then, differentiate \( u \) with respect to \( x \):
   \[
   \frac{du}{dx} = 2 \quad \text{or} \quad du = 2 \, dx
   \]
   Therefore, \( dx = \frac{du}{2} \).

2. **Rewrite the integral:**
   Substitute \( u = 2x - 4 \) into the integral:
   \[
   \int \sin(2x - 4) \, dx = \int \sin(u) \cdot \frac{du}{2}
   \]

3. **Integrate:**
   The integral of \( \sin(u) \) is \( -\cos(u) \):
   \[
   \int \sin(u) \, du = -\cos(u)
   \]
   So, the integral becomes:
   \[
   -\frac{1}{2} \cos(u) + C
   \]

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

### Problem 20: \(\int \cos(\pi v - \sqrt{7}) \, dv\)

1. **Substitution:**
   Let \( u = \pi v - \sqrt{7} \). Then, differentiate \( u \) with respect to \( v \):
   \[
   \frac{du}{dv} = \pi \quad \text{or} \quad du = \pi \, dv
   \]
   Therefore, \( dv = \frac{du}{\pi} \).

2. **Rewrite the integral:**
   Substitute \( u = \pi v - \sqrt{7} \) into the integral:
   \[
   \int \cos(\pi v - \sqrt{7}) \, dv = \int \cos(u) \cdot \frac{du}{\pi}
   \]

3. **Integrate:**
   The integral of \( \cos(u) \) is \( \sin(u) \):
   \[
   \int \cos(u) \, du = \sin(u)
   \]
   So, the integral becomes:
   \[
   \frac{1}{\pi} \sin(u) + C
   \]

4. **Substitute back:**
   Replace \( u \) with \( \pi v - \sqrt{7} \):
   \[
   \int \cos(\pi v - \sqrt{7}) \, dv = \frac{1}{\pi} \sin(\pi v - \sqrt{7}) + C
   \]

### Summary of Solutions
1. \(\int \cos(3x + 2) \, dx = \frac{1}{3} \sin(3x + 2) + C\)
2. \(\int \sin(2x - 4) \, dx = -\frac{1}{2} \cos(2x - 4) + C\)
3. \(\int \cos(\pi v - \sqrt{7}) \, dv = \frac{1}{\pi} \sin(\pi v - \sqrt{7}) + C\)

Would you like more details or have any other questions?

### Related Questions
1. How do you handle substitution for integrals with higher-degree polynomials?
2. What if the integral involves an exponential function, e.g., \(\int e^{2x+3} \, dx\)?
3. Can you apply substitution for definite integrals? How does it change?
4. What is the antiderivative of a product of functions like \(\cos(x) \cdot e^{2x}\)?
5. How do you verify your integration result?

### Tip:
Always check your substitution by differentiating your result to ensure you return to the original integrand.

Explore detailed solutions for integrals involving trigonometric functions using the substitution method. Learn step-by-step how to integrate expressions like \(\cos(3x + 2)\), \(\sin(2x - 4)\), and \(\cos(\pi v - \sqrt{7})\). Suitable for students and professionals studying advanced mathematics.

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