Lesson 1: u-Substitution
Estimated time: 45-55 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Recognize integrals suitable for u-substitution
- Choose an effective u and compute du
- Apply u-substitution to indefinite and definite integrals
- Change the limits of integration when working with definite integrals
The Idea Behind u-Substitution
u-Substitution is the integration counterpart of the chain rule. Recall: d/dx [F(g(x))] = F'(g(x)) · g'(x). Running this in reverse:
u-Substitution Rule
If u = g(x) and du = g'(x) dx, then:
∫ f(g(x)) · g'(x) dx = ∫ f(u) du
The key is to identify a function and its derivative within the integrand, then substitute to simplify.
How to Choose u
General guidelines for picking u:
- Inner function: If you see a composite function, let u be the inner function.
- Complicated part: Let u be the part that makes the integral hard; its derivative should appear (or nearly appear) elsewhere in the integrand.
- Common patterns: ∫ f(ax + b) dx, ∫ x · f(x²) dx, ∫ f(sin x) cos x dx, etc.
Example 1: Basic u-Substitution
Evaluate ∫ 2x(x² + 1)5 dx.
Step 1: Let u = x² + 1. Then du = 2x dx.
Step 2: Substitute: ∫ u5 du = u6/6 + C.
Step 3: Back-substitute: (x² + 1)6/6 + C.
Example 2: Adjusting for a Constant
Evaluate ∫ x ex² dx.
Step 1: Let u = x². Then du = 2x dx, so x dx = du/2.
Step 2: ∫ eu (du/2) = (1/2)eu + C = (1/2)ex² + C.
Common u-Substitution Patterns
Pattern Recognition
| Integral Form | Let u = | Result |
|---|---|---|
| ∫ f(ax + b) dx | ax + b | (1/a) F(ax + b) + C |
| ∫ xn−1 f(xn) dx | xn | (1/n) F(u) + C |
| ∫ f(sin x) cos x dx | sin x | F(sin x) + C |
| ∫ f(ex) ex dx | ex | F(ex) + C |
| ∫ f(ln x) / x dx | ln x | F(ln x) + C |
Example 3: Trig Pattern
Evaluate ∫ sin³x cos x dx.
Step 1: Let u = sin x, du = cos x dx.
Step 2: ∫ u³ du = u4/4 + C = sin4x / 4 + C.
Definite Integrals: Changing the Bounds
For definite integrals, you have two options: (1) back-substitute and use the original bounds, or (2) change the bounds to u-values and evaluate directly in terms of u.
Changing Bounds
If u = g(x), then when x = a, u = g(a); when x = b, u = g(b):
∫ab f(g(x)) g'(x) dx = ∫g(a)g(b) f(u) du
Example 4: Definite Integral with Changed Bounds
Evaluate ∫01 x(x² + 1)³ dx.
Step 1: u = x² + 1, du = 2x dx, x dx = du/2.
Step 2: Change bounds: x = 0 → u = 1; x = 1 → u = 2.
Step 3: (1/2) ∫12 u³ du = (1/2)[u4/4]12 = (1/2)(16/4 − 1/4) = (1/2)(15/4) = 15/8.
Example 5: Exponential with Changed Bounds
Evaluate ∫0ln 3 ex/(ex + 1) dx.
Step 1: u = ex + 1, du = ex dx.
Step 2: x = 0 → u = 2; x = ln 3 → u = 4.
Step 3: ∫24 (1/u) du = [ln u]24 = ln 4 − ln 2 = ln 2.
More Challenging Substitutions
Example 6: Multiplying and Dividing to Create du
Evaluate ∫ tan x dx.
Solution: ∫ (sin x / cos x) dx. Let u = cos x, du = −sin x dx.
= ∫ (−1/u) du = −ln|u| + C = −ln|cos x| + C = ln|sec x| + C.
Example 7: Algebraic Manipulation
Evaluate ∫ x/√(1 − x²) dx.
Step 1: Let u = 1 − x², du = −2x dx, so x dx = −du/2.
Step 2: ∫ (−1/2) u−1/2 du = −u1/2 + C = −√(1 − x²) + C.
When u-Substitution Does Not Work
u-Substitution requires that the derivative of u appears (up to a constant) in the integrand. If it does not, you need a different technique:
- ∫ x ex dx cannot be done with simple u-sub (need integration by parts).
- ∫ sin²x dx requires a trig identity, not u-sub.
- ∫ 1/√(a² − x²) dx needs trig substitution or recognizing an arcsin pattern.
These techniques are covered in the upcoming lessons.
Key Takeaways
- u-Substitution reverses the chain rule: look for a composite function and its inner derivative.
- Let u = inner function, compute du, substitute, integrate, back-substitute.
- For definite integrals, change bounds to u-values to avoid back-substitution.
- If du does not appear (even with constant adjustment), try a different technique.
Check Your Understanding
1. Evaluate ∫ cos(3x) dx.
2. Evaluate ∫ x²(x³ + 5)4 dx.
3. Evaluate ∫02 x √(x² + 5) dx.
4. Evaluate ∫ (ln x)² / x dx.