Premed · Premed · Calculus 1
Lecture 21: Substitution Rule
Calculus I — Differential Calculus
Learning Objectives
By the end of this lecture, students will be able to:
- Recognize when u-substitution applies
- Apply the substitution rule for indefinite integrals
- Apply the substitution rule for definite integrals (with limit conversion)
- Choose effective substitutions for various integrand forms
- Use substitution to evaluate integrals involving compositions, products, and chain-rule patterns
Lecture Content
I. Motivation: Reversing the Chain Rule
The chain rule tells us that d/dx [F(g(x))] = F'(g(x)) g'(x) = f(g(x)) g'(x). Running this in reverse gives the integral of f(g(x)) * g'(x) dx = F(g(x)) + C. The substitution rule (u-substitution) provides a systematic way to perform this reversal. The key pattern to recognize is that the integrand contains a function composed with an inner function, and the derivative of the inner function also appears in the integrand (possibly up to a constant factor).
II. The Substitution Rule for Indefinite Integrals
The substitution rule states that if u = g(x) is a differentiable function whose range is an interval I, and f is continuous on I, then integral of f(g(x)) * g'(x) dx = integral of f(u) du.
The procedure has five steps. First, choose u = g(x), typically the "inner function." Second, compute du = g'(x) dx. Third, rewrite the integral entirely in terms of u and du. Fourth, integrate with respect to u. Fifth, substitute back by replacing u with g(x).
III. Examples — Indefinite Integrals
For integral of 2x * cos(x^2) dx, let u = x^2 so that du = 2x dx. The integral becomes integral of cos(u) du = sin(u) + C = sin(x^2) + C.
For integral of x^2 * sqrt(x^3 + 1) dx, let u = x^3 + 1 so that du = 3x^2 dx and x^2 dx = du/3. This gives (1/3) integral of sqrt(u) du = (1/3)(2/3)u^{3/2} + C = (2/9)(x^3 + 1)^{3/2} + C.
For integral of e^{5x} dx, let u = 5x so that du = 5 dx and dx = du/5. The result is (1/5) integral of e^u du = (1/5)e^{5x} + C.
For integral of tan x dx = integral of (sin x / cos x) dx, let u = cos x so that du = -sin x dx. This gives - integral of (1/u) du = -ln|u| + C = -ln|cos x| + C = ln|sec x| + C.
<image>A step-by-step worked example showing u-substitution for integral of 2x*cos(x^2) dx. Step 1: Identify u = x^2 (inner function highlighted in a colored box). Step 2: Compute du = 2x dx (matching the remaining factor in the integrand). Step 3: Rewrite as integral of cos(u) du. Step 4: Integrate to get sin(u) + C. Step 5: Back-substitute to get sin(x^2) + C. Each step has an arrow connecting to the next. Title: "U-Substitution: Step-by-Step."</image>
IV. The Substitution Rule for Definite Integrals
When using substitution with definite integrals, there are two options. Option A is to change the limits of integration to u-values: integral from a to b of f(g(x)) * g'(x) dx = integral from g(a) to g(b) of f(u) du. This allows direct evaluation in u with no need to substitute back. Option B is to compute the indefinite integral, substitute back to get a function of x, and then apply the original limits. Option A is usually cleaner and less error-prone.
V. Examples — Definite Integrals
For integral from 0 to 2 of x * e^{x^2} dx, let u = x^2 so that du = 2x dx and x dx = du/2. When x = 0, u = 0; when x = 2, u = 4. The integral becomes (1/2) integral from 0 to 4 of e^u du = (1/2)[e^u] from 0 to 4 = (1/2)(e^4 - 1).
For integral from 1 to 2 of 1/(3x - 1) dx, let u = 3x - 1 so that du = 3 dx. When x = 1, u = 2; when x = 2, u = 5. The result is (1/3) integral from 2 to 5 of (1/u) du = (1/3)(ln 5 - ln 2) = (1/3) ln(5/2).
For integral from 0 to pi/2 of cos x * sin^3 x dx, let u = sin x so that du = cos x dx. When x = 0, u = 0; when x = pi/2, u = 1. The integral becomes integral from 0 to 1 of u^3 du = [u^4/4] from 0 to 1 = 1/4.
<image>A side-by-side comparison of Option A (change limits) and Option B (back-substitute) for the integral from 0 to 2 of x*e^{x^2} dx. Left panel (Option A): shows the substitution u = x^2, new limits u = 0 to u = 4, and direct evaluation in u. Right panel (Option B): shows finding the indefinite integral (1/2)e^{x^2} + C, then applying the original x-limits 0 to 2. Both arrive at the same answer: (1/2)(e^4 - 1). Title: "Two approaches to definite integrals with substitution."</image>
VI. How to Choose the Right Substitution
The most effective strategy is to look for a composite function and choose the inner function as u. Then check that du (or a constant multiple of it) appears in the integrand. If the extra factor matches g'(x), the substitution will work cleanly.
Several common patterns serve as reliable guides. For integrals of the form f(ax + b) dx, let u = ax + b. For f(x^n) x^{n-1} dx, let u = x^n. For f(sin x) cos x dx, let u = sin x. For f(e^x) e^x dx, let u = e^x. For f(ln x) (1/x) dx, let u = ln x. If the first choice of substitution does not work, try another.
VII. Integrals of Symmetric Functions
Symmetry can dramatically simplify definite integrals over symmetric intervals. If f is an even function (f(-x) = f(x)), then integral from -a to a of f(x) dx = 2 * integral from 0 to a of f(x) dx. If f is an odd function (f(-x) = -f(x)), then integral from -a to a of f(x) dx = 0.
For example, integral from -1 to 1 of x^3 cos x dx = 0, because the integrand x^3 cos x is odd (x^3 is odd and cos x is even, so their product is odd). Recognizing symmetry can save significant computation.
<image>Two panels illustrating symmetry properties. Panel A: An even function f(x) = x^2 on [-2, 2] with the area from -2 to 0 (shaded blue) equal to the area from 0 to 2 (shaded green), so the integral is 2 times the integral from 0 to 2. Panel B: An odd function f(x) = x^3 on [-1, 1] with the negative area below the x-axis on [-1, 0] (shaded red) exactly canceling the positive area above the x-axis on [0, 1] (shaded green), so the integral is 0. Title: "Symmetry shortcuts for definite integrals."</image>
