05-05 - Integration by Parts
Phase: 5 | Subject: 05-05 Prerequisites: 05-04-integration-by-substitution.md Next subject: 05-06-trigonometric-integrals.md
Learning Objectives
By the end of this subject, you will be able to:
- Derive the integration by parts formula
- Apply LIATE/ILATE rule to choose u and dv
- Use the tabular method for repeated integration by parts
- Derive reduction formulas
Core Content
Derivation from Product Rule
Recall: d/dx[u·v] = u'·v + u·v'
Integrate both sides: ∫(u'·v + u·v')dx = u·v
Separate: ∫u'·vdx + ∫u·v'dx = u·v
Rearrange: ∫u·v'dx = u·v - ∫u'·vdx
Integration by Parts Formula:
$∫u dv = uv - ∫v du $
Choosing u and dv
LIATE rule (priority order): 1. Logarithmic (ln, log) 2. Inverse trig (arctan, arcsin) 3. Algebraic (polynomials, xⁿ) 4. Trigonometric (sin, cos) 5. Exponential (eˣ, aˣ)
Pick u from the highest priority category. The rest goes to dv.
Example: ∫x·eˣdx u = x (algebraic), dv = eˣdx du = dx, v = eˣ = x·eˣ - ∫eˣdx = x·eˣ - eˣ + C = eˣ(x - 1) + C
Tabular Method
For repeated integration by parts (polynomial × trig/exponential):
| D | I |
|---|---|
| x² | eˣ |
| 2x | eˣ |
| 2 | eˣ |
| 0 | eˣ |
Diagonal products with alternating signs: +x²·eˣ - 2x·eˣ + 2·eˣ + C
Key Terms
- LIATE rule
Worked Examples
Example 1: Basic
∫x·ln(x)dx u = ln(x) (L), dv = x dx du = (1/x)dx, v = x²/2 = (x²/2)·ln(x) - ∫(x²/2)(1/x)dx = (x²/2)ln(x) - (1/2)∫x dx = (x²/2)ln(x) - x²/4 + C
Example 2: Definite integral
∫[0 to 1] x·eˣdx = [eˣ(x - 1)][0 to 1] = e(0) - (-1) = 1
Example 3: Tabular
∫x³·sin(x)dx
| D | I |
|---|---|
| x³ | sin(x) |
| 3x² | -cos(x) |
| 6x | -sin(x) |
| 6 | cos(x) |
| 0 | sin(x) |
Result: -x³cos(x) + 3x²sin(x) + 6xcos(x) - 6sin(x) + C
Practice Problems
Problem 1: ∫x·cos(x)dx
Answer
x·sin(x) + cos(x) + CProblem 2: ∫x²·eˣdx
Answer
eˣ(x² - 2x + 2) + C (tabular)Problem 3: ∫ln(x)dx
Answer
x·ln(x) - x + CProblem 4: ∫[0 to π] x·sin(x)dx
Answer
Integration by parts: u = x, dv = sin(x)dx → du = dx, v = -cos(x). ∫x·sin(x)dx = -x·cos(x) + ∫cos(x)dx = -x·cos(x) + sin(x) + C. Evaluate: [-x·cos(x) + sin(x)][0 to π] = (-π·(-1) + 0) - (0 + 0) = π.Problem 5: ∫eˣ·sin(x)dx
Answer
Apply IBP twice. Result: (eˣ/2)(sin(x) - cos(x)) + C.Summary
Key takeaways:
- ∫u dv = uv - ∫v du (reverses product rule)
- LIATE: choose u from highest priority
- Tabular method for repeated IBP
- Sometimes need to apply IBP twice and solve for the integral
Pitfalls
- Choosing u and dv in the wrong order. LIATE is a guideline for priority — ignoring it often leads to an integral that is harder than the original. For ∫x·ln(x)dx, choosing u = x and dv = ln(x)dx makes the problem intractable while u = ln(x) works cleanly.
- Forgetting the minus sign. The formula is ∫u dv = uv − ∫v du. Dropping the minus sign and writing uv + ∫v du is a common sign error that produces completely wrong results.
- Circular integration without solving. For integrals like ∫eˣ·sin(x)dx, applying integration by parts twice gives back the original integral. You must set up an algebraic equation and solve for the integral — not keep applying IBP indefinitely.
- Mishandling du and v. After choosing u and dv, computing du = u' dx is straightforward, but finding v = ∫dv requires actual integration. Getting v wrong (wrong sign, forgetting constants) propagates through the whole calculation.
- Tabular method sign errors. The alternating signs (+ − + − ...) in the tabular method must be applied carefully to the diagonal products. A single sign error, especially in the first term, ruins the entire result.
Quiz
Q1: ∫x·eˣdx equals:
A) xeˣ + C B) xeˣ - eˣ + C C) eˣ + C D) (1/2)xeˣ + C
Answer and Explanations
**Correct: B)** - If you chose B: u = x, dv = eˣdx. du = dx, v = eˣ. x·eˣ - ∫eˣdx = xeˣ - eˣ + C. Correct! - If you chose A: You forgot to subtract ∫eˣdx. - If you chose C: That's just ∫eˣdx, ignoring the x. - If you chose D: That's not a standard form.Q2: For ∫x²·ln(x)dx, using LIATE, u should be:
A) x² B) ln(x) C) x²·ln(x) D) 1
Answer and Explanations
**Correct: B)** - If you chose B: Logarithmic (ln) has highest priority in LIATE. Correct! - If you chose A: Algebraic has lower priority than logarithmic. Choose ln(x) as u. - If you chose C: That's the entire integrand. We split it into u and dv. - If you chose D: 1 isn't one of the options in the integrand.Q3: ∫ln(x)dx equals:
A) (1/x) + C B) x·ln(x) - x + C C) ln(x) + C D) x·ln(x) + C
Answer and Explanations
**Correct: B)** - If you chose B: u = ln(x), dv = dx. du = dx/x, v = x. x·ln(x) - ∫1dx = x·ln(x) - x + C. Correct! - If you chose A: That's the derivative of ln(x). - If you chose C: That's just the original function. - If you chose D: You forgot the -x term from ∫1dx.Q4: The tabular method is useful for:
A) Any integral B) Integrals with a polynomial times exponential or trig function C) Trigonometric integrals only D) Substitution problems
Answer and Explanations
**Correct: B)** - If you chose B: Tabular method efficiently handles repeated IBP when one factor is a polynomial and the other is exp or trig. Correct! - If you chose A: Tabular only helps for specific forms. Substitution is often better for other integrals. - If you chose C: Tabular needs a polynomial factor. Pure trig integrals use identities. - If you chose D: Substitution doesn't use tabular method.Q5: To integrate ∫x²·sin(x)dx, after two applications of IBP, you get:
A) Back to the original integral B) An integral of x²·cos(x) C) An integral of sin(x) D) A polynomial
Answer and Explanations
**Correct: A)** - If you chose A: After two IBP applications on x²·sin(x), you get back to an integral of x²·sin(x) (or equivalent), allowing you to solve for it algebraically. Correct! - If you chose B: The derivative cycles: sin → cos → -sin → -cos. After two steps you're back to sin (up to sign). - If you chose C: After TWO IBP applications, you get -∫x²sin(x)dx, same form as original. - If you chose D: The polynomial degree decreases each time, eventually becoming a constant.Next Steps
Next up: 05-06-trigonometric-integrals.md