Lesson 3: Integration by Parts
Estimated time: 45-55 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Derive the integration by parts formula from the product rule
- Use the LIATE rule to choose u and dv
- Apply the tabular method for repeated integration by parts
- Handle integrals that require IBP twice, including "boomerang" integrals
Deriving the Formula
Integration by parts comes from reversing the product rule. If u and v are functions of x:
d(uv) = u dv + v du, so u dv = d(uv) − v du.
Integrating both sides:
Integration by Parts Formula
∫ u dv = uv − ∫ v du
For definite integrals: ∫ab u dv = [uv]ab − ∫ab v du.
The LIATE Rule for Choosing u
The hardest part of IBP is choosing what to call u and what to call dv. The LIATE rule is a guideline: choose u to be whichever type of function appears first in this list:
LIATE Priority
- Logarithmic: ln x, log x
- Inverse trig: arctan x, arcsin x
- Algebraic: x, x², polynomials
- Trigonometric: sin x, cos x
- Exponential: ex, 2x
Functions higher on the list become u (they simplify when differentiated).
Example 1: Classic IBP
Evaluate ∫ x ex dx.
LIATE: x is Algebraic, ex is Exponential. A before E, so u = x.
u = x, du = dx. dv = ex dx, v = ex.
∫ x ex dx = x ex − ∫ ex dx = x ex − ex + C = ex(x − 1) + C.
Example 2: Logarithmic IBP
Evaluate ∫ ln x dx.
LIATE: ln x is Logarithmic (highest priority). Let u = ln x, dv = dx.
du = (1/x) dx, v = x.
∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − x + C = x(ln x − 1) + C.
Example 3: Trig and Polynomial
Evaluate ∫ x sin x dx.
u = x (Algebraic), dv = sin x dx. du = dx, v = −cos x.
= −x cos x − ∫ (−cos x) dx = −x cos x + sin x + C.
Repeated Integration by Parts
Sometimes the new integral ∫ v du still requires IBP. You may need to apply the formula two or more times.
Example 4: IBP Twice
Evaluate ∫ x² ex dx.
First IBP: u = x², dv = ex dx. du = 2x dx, v = ex.
= x² ex − 2 ∫ x ex dx.
Second IBP: (from Example 1) ∫ x ex dx = ex(x − 1).
= x² ex − 2ex(x − 1) + C = ex(x² − 2x + 2) + C.
The Tabular Method
When you need IBP multiple times with a polynomial times an exponential/trig, the tabular method speeds things up dramatically.
Tabular Method Procedure
- List successive derivatives of u in one column (until you reach 0).
- List successive antiderivatives of dv in another column.
- Alternate signs: +, −, +, −, ...
- Multiply diagonally and add the terms.
Example 5: Tabular Method
Evaluate ∫ x³ e2x dx.
| Sign | Derivatives of x³ | Antiderivatives of e2x |
|---|---|---|
| + | x³ | (1/2)e2x |
| − | 3x² | (1/4)e2x |
| + | 6x | (1/8)e2x |
| − | 6 | (1/16)e2x |
| + | 0 |
Result: (1/2)x³e2x − (3/4)x²e2x + (6/8)xe2x − (6/16)e2x + C
= e2x[(1/2)x³ − (3/4)x² + (3/4)x − 3/8] + C.
Boomerang Integrals
Sometimes after two rounds of IBP, the original integral reappears on the right side. You can then solve for it algebraically.
Example 6: Boomerang — ∫ ex sin x dx
First IBP: u = sin x, dv = ex dx. du = cos x dx, v = ex.
I = ex sin x − ∫ ex cos x dx.
Second IBP: u = cos x, dv = ex dx. du = −sin x dx, v = ex.
I = ex sin x − [ex cos x + ∫ ex sin x dx]
I = ex sin x − ex cos x − I.
Solve: 2I = ex(sin x − cos x), so I = (ex/2)(sin x − cos x) + C.
Definite Integrals with IBP
Example 7: Definite IBP
Evaluate ∫1e ln x dx.
From Example 2: ∫ ln x dx = x(ln x − 1) + C.
[x(ln x − 1)]1e = e(1 − 1) − 1(0 − 1) = 0 + 1 = 1.
Key Takeaways
- ∫ u dv = uv − ∫ v du. Choose u using LIATE.
- The goal is for ∫ v du to be simpler than the original integral.
- Tabular method streamlines repeated IBP (polynomial times ex or trig).
- Boomerang integrals: if the original integral reappears, solve algebraically.
Check Your Understanding
1. Evaluate ∫ x cos x dx.
2. Evaluate ∫ x² sin x dx.
3. Evaluate ∫01 x e−x dx.
4. Evaluate ∫ arctan x dx.