05-04 - Integration by Substitution
Phase: 5 | Subject: 05-04 Prerequisites: 05-03-the-fundamental-theorem-of-calculus.md (antiderivatives), 04-04-differentiation-rules.md (chain rule) Next subject: 05-05-integration-by-parts.md
Learning Objectives
By the end of this subject, you will be able to:
- Recognise when substitution is appropriate
- Apply u-substitution to indefinite integrals
- Apply substitution in definite integrals (with limit changes)
- Use trigonometric substitutions
- Spot common substitution patterns
Core Content
The Substitution Rule
Integration by substitution reverses the chain rule.
Indefinite form:
$∫f(g(x))·g'(x)dx = ∫f(u)du = F(u) + C = F(g(x)) + C $
where u = g(x) and du = g'(x)dx.
Definite form:
$∫[a to b] f(g(x))·g'(x)dx = ∫[g(a) to g(b)] f(u)du $
Choosing u
Strategy: Look for a function whose derivative also appears (up to a constant).
Common patterns: 1. ∫f'(x)·f(x)^n dx → u = f(x) 2. ∫f(ax + b)dx → u = ax + b 3. ∫xⁿ·e^(kx)dx → u = kx or xⁿ 4. ∫sin(ax + b)dx → u = ax + b 5. ∫(ax + b)ⁿ dx → u = ax + b
Key Terms
- 05 04 Integration By Substitution
- Choosing u
- Correct: A)
- Correct: B)
- Correct: C)
- Example 1: Basic substitution
- Example 2: Definite integral with substitution
- Example 3: Trigonometric substitution
- The Substitution Rule
Worked Examples
Example 1: Basic substitution
∫2x·cos(x²)dx u = x², du = 2x dx = ∫cos(u)du = sin(u) + C = sin(x²) + C
Check: d/dx[sin(x²)] = cos(x²) · 2x. ✓
Example 2: Definite integral with substitution
∫[0 to 1] x·√(x² + 1)dx u = x² + 1, du = 2x dx, so x dx = du/2 Limits: x = 0 → u = 1, x = 1 → u = 2 = ∫[1 to 2] √u · (du/2) = (1/2)∫[1 to 2] u^(1/2)du = (1/2) · [2u^(3/2)/3][1 to 2] = (1/3)[2^(3/2) - 1] = (1/3)[2√2 - 1] ≈ 0.609
Example 3: Trigonometric substitution
∫sin(3x)dx u = 3x, du = 3dx, dx = du/3 = ∫sin(u)(du/3) = -(1/3)cos(u) + C = -(1/3)cos(3x) + C
Practice Problems
Problem 1: ∫x·e^(x²)dx
Answer
u = x², du = 2x dx. (1/2)∫e^u du = (1/2)e^(x²) + CProblem 2: ∫[0 to π] sin(x)·cos(x)dx
Answer
u = sin(x), du = cos(x)dx. Limits: 0 to 0. ∫[0 to 0] u du = 0.Problem 3: ∫(2x + 3)⁵dx
Answer
u = 2x + 3, du = 2dx. (1/2)∫u⁵ du = u⁶/12 + C = (2x + 3)⁶/12 + CProblem 4: ∫[1 to 2] (x - 1)²dx
Answer
u = x - 1, du = dx. Limits: 0 to 1. ∫[0 to 1] u² du = [u³/3][0 to 1] = 1/3.Problem 5: ∫cos²(x)·sin(x)dx
Answer
u = cos(x), du = -sin(x)dx. -∫u² du = -u³/3 + C = -cos³(x)/3 + C.Summary
Key takeaways:
- Substitution reverses the chain rule
- Choose u whose derivative appears in the integral
- For definite integrals: change limits to u-values
- Always substitute back for indefinite integrals
- du replaces g'(x)dx
Pitfalls
- Forgetting to change limits in definite integrals. When using u-substitution on ∫[a to b] f(g(x))·g'(x)dx, the limits must become g(a) and g(b). Keeping the old x-limits after switching to u gives wrong results.
- Choosing u poorly. The substitution should simplify the integral, not complicate it. If du introduces factors that don't cancel with what's already in the integrand, the substitution wasn't a good choice.
- Mishandling the dx-to-du conversion. After setting u = g(x) and du = g'(x)dx, you must solve for dx = du/g'(x). Forgetting to divide by g'(x) when substituting dx is a frequent algebraic error.
- Not substituting back for indefinite integrals. The final answer must be in terms of the original variable x. After integrating with respect to u, replace u with g(x). An answer written in u is incomplete.
- Forcing substitution where no derivative pattern exists. Substitution only works when the integrand contains both f(g(x)) and (a constant multiple of) g'(x). Trying u-sub on ∫e^(x²)dx (which has no x factor matching 2x from du) is futile — that integral has no elementary antiderivative.
Quiz
Q1: ∫2x·e^(x²)dx equals:
A) e^(x²) + C B) 2e^(x²) + C C) e^(2x) + C D) xe^(x²) + C
Answer and Explanations
**Correct: A)** - If you chose A: u = x², du = 2x dx. ∫e^u du = e^u + C = e^(x²) + C. Correct! - If you chose B: You may have kept the 2 outside: 2∫e^(x²)dx, which doesn't simplify. - If you chose C: That's the integral of 2x·e^(2x), not 2x·e^(x²). - If you chose D: You multiplied by x instead of substituting.Q2: For ∫x·cos(x²)dx, the best substitution is:
A) u = x B) u = cos(x) C) u = x² D) u = cos(x²)
Answer and Explanations
**Correct: C)** - If you chose C: u = x², du = 2x dx. We have x dx, so x dx = du/2. Perfect match. Correct! - If you chose A: du = dx. We'd have u·cos(u²)du, which is harder. - If you chose B: du = -sin(x)dx. Doesn't match the x dx part. - If you chose D: du = -2x·sin(x²)dx. Doesn't match.Q3: ∫[0 to 1] 2x(x² + 1)³ dx with u = x² + 1 becomes:
A) ∫[1 to 2] u³ du B) ∫[0 to 2] u³ du C) ∫[1 to 2] u³ (du/2) D) ∫[0 to 1] u³ du
Answer and Explanations
**Correct: A)** - If you chose A: u = x² + 1, du = 2x dx. Limits: x=0 → u=1, x=1 → u=2. ∫[1 to 2] u³ du. Correct! - If you chose B: The lower limit is wrong. x=0 gives u=1, not 0. - If you chose C: You have an extra 1/2 factor. du = 2x dx, so 2x dx = du, not du/2. - If you chose D: Limits unchanged. For definite integrals with substitution, limits must change.Q4: ∫(3x + 2)⁴ dx equals:
A) (3x + 2)⁵/5 + C B) (3x + 2)⁵/15 + C C) 12(3x + 2)³ + C D) (3x + 2)⁵ + C
Answer and Explanations
**Correct: B)** - If you chose B: u = 3x + 2, du = 3dx, dx = du/3. ∫u⁴(du/3) = u⁵/15 + C = (3x+2)⁵/15 + C. Correct! - If you chose A: You forgot that du = 3dx, so dx = du/3. You used dx = du. - If you chose C: You used the power rule on (3x+2)⁴ without substitution: 4(3x+2)³ × 3 = 12(3x+2)³. That's the DERIVATIVE, not the integral. - If you chose D: You forgot both the chain factor and the division by the new power.Q5: ∫sin(5x)dx equals:
A) cos(5x) + C B) -5cos(5x) + C C) -(1/5)cos(5x) + C D) 5sin(5x) + C
Answer and Explanations
**Correct: C)** - If you chose C: u = 5x, du = 5dx, dx = du/5. ∫sin(u)(du/5) = -(1/5)cos(u) + C = -(1/5)cos(5x) + C. Correct! - If you chose A: That's the derivative of sin(5x) without the chain factor. - If you chose B: You multiplied by 5 and kept the wrong sign for sin integral. - If you chose D: You multiplied by 5 and kept sin instead of cos.Next Steps
Next up: 05-05-integration-by-parts.md