Difficulty: Intermediate | Prerequisites: Sections 1–4 of Chapter 4, chain rule from Chapter 3
The Substitution Rule is the integration counterpart of the chain rule for derivatives. Many integrands do not match a basic antiderivative formula directly, but after a well-chosen change of variable, they simplify into something you already know how to integrate. This is the single most-used integration technique in Calculus I and the foundation for more advanced methods in Calculus II. You need the chain rule and your basic antiderivative table before starting.
U-substitution rewrites a complicated integral in terms of a new variable u, chosen so the integrand simplifies. For indefinite integrals, you substitute, integrate, and change back. For definite integrals, you can also convert the limits of integration to u-values and never change back. Even and odd function symmetry gives you shortcuts on symmetric intervals.
Substitution rule (u-substitution)
A technique for evaluating integrals by replacing a composite expression with a single new variable u. In simple terms, it is the chain rule run in reverse.
Inner function (u)
The part of the integrand you choose to call u. A good choice is usually the "inside" function of a composition, or the expression whose derivative also appears in the integrand.
Differential (du)
If u = g(x), then du = g'(x) dx. You replace dx in the integral with du/g'(x), or equivalently rewrite g'(x) dx as du.
Change of limits
When applying substitution to a definite integral, you convert the original x-limits to u-limits: if x = a maps to u = g(a) and x = b maps to u = g(b), the new integral runs from g(a) to g(b).
Even function
A function satisfying f(−x) = f(x). Its graph is symmetric about the y-axis. Examples: x², cos x, |x|.
Odd function
A function satisfying f(−x) = −f(x). Its graph has rotational symmetry about the origin. Examples: x³, sin x, tan x.
If u = g(x) is differentiable and f is continuous on the range of g, then:
∫ f(g(x)) · g'(x) dx = ∫ f(u) du
You treat dx and du as if they were differentials: set u = g(x), compute du = g'(x) dx, rewrite the integral entirely in terms of u, integrate, then substitute back.
Look for a composite structure f(g(x)) where g'(x) (or a constant multiple of it) also appears in the integrand. Common patterns:
A function raised to a power with its derivative alongside: ∫ x²√(x³ + 1) dx, let u = x³ + 1.
A function inside a trig function with its derivative present: ∫ sin(2x + 1) dx, let u = 2x + 1.
A denominator raised to a power: ∫ dt/(1 − 3t)⁵, let u = 1 − 3t.
∫ x²√(x³ + 1) dx: Let u = x³ + 1, du = 3x² dx, so x² dx = du/3. The integral becomes (1/3) ∫ √u du = (1/3)(2/3)u^(3/2) + C = (2/9)(x³ + 1)^(3/2) + C.
∫ dt/(1 − 3t)⁵: Let u = 1 − 3t, du = −3 dt, so dt = −du/3. The integral becomes (−1/3) ∫ u⁻⁵ du = (−1/3)(u⁻⁴/(−4)) + C = 1/(12(1 − 3t)⁴) + C.
∫ sin x cos x dx: Let u = sin x, du = cos x dx. The integral becomes ∫ u du = u²/2 + C = sin²(x)/2 + C.
∫ x³√(x² + 1) dx (the harder pattern): Let u = x² + 1, so x² = u − 1 and du = 2x dx. Rewrite x³ dx = x² · x dx = (u − 1)(du/2). The integral becomes (1/2) ∫ (u − 1)√u du = (1/2) ∫ (u^(3/2) − u^(1/2)) du. Expand and integrate term by term, then substitute back.
If g'(x) is continuous on [a, b] and f is continuous on the range of u = g(x), then:
∫ₐᵇ f(g(x)) g'(x) dx = ∫_{g(a)}^{g(b)} f(u) du
You convert the limits to u-values so there is no need to substitute back to x at the end.
∫₀¹ ³√(1 + 7x) dx: Let u = 1 + 7x, du = 7 dx. When x = 0, u = 1; when x = 1, u = 8. The integral becomes (1/7) ∫₁⁸ u^(1/3) du = (1/7)(3/4)[u^(4/3)]₁⁸ = (3/28)(8^(4/3) − 1) = (3/28)(16 − 1) = 45/28.
∫₀^√π x cos(x²) dx: Let u = x², du = 2x dx. When x = 0, u = 0; when x = √π, u = π. The integral becomes (1/2) ∫₀^π cos u du = (1/2)[sin u]₀^π = (1/2)(0 − 0) = 0.
∫₀^(π/3) cos²x · sin x dx: Let u = cos x, du = −sin x dx. When x = 0, u = 1; when x = π/3, u = 1/2. The integral becomes −∫₁^(1/2) u² du = ∫(1/2)¹ u² du = [u³/3](1/2)¹ = 1/3 − 1/24 = 7/24.
These properties give you shortcuts when integrating over a symmetric interval [−a, a].
If f is even (f(−x) = f(x)), then:
∫₋ₐᵃ f(x) dx = 2 ∫₀ᵃ f(x) dx
The area from −a to 0 mirrors the area from 0 to a, so you can double the right half.
If f is odd (f(−x) = −f(x)), then:
∫₋ₐᵃ f(x) dx = 0
The positive area on one side exactly cancels the negative area on the other.
∫₋π/₄^(π/4) (1 + x + x² tan x) dx: split into 1 + x² tan x (which needs checking) and x.
1 is even, x is odd, x² tan x is odd (x² is even, tan x is odd, even × odd = odd).
The odd parts integrate to 0 on [−π/4, π/4].
Only 1 survives: ∫₋π/₄^(π/4) 1 dx = π/2.
\int f(g(x))\,g'(x)\,dx = \int f(u)\,du \quad \text{where } u = g(x)\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du\text{If } f \text{ is even: } \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx\text{If } f \text{ is odd: } \int_{-a}^{a} f(x)\,dx = 0Substitution makes it possible to integrate the kinds of rate functions that appear in physics and engineering. A velocity given by v(t) = t/√(1 + t²) requires substitution to find displacement. Any time a rate is expressed as a composition of functions, u-substitution is the tool that converts it into a basic integral.
Students often choose u but forget to convert every part of the integrand (including dx) into u. If any x remains after substitution, something has gone wrong.
When changing limits for a definite integral, students sometimes convert the limits but then also substitute back to x at the end. Do one or the other, not both.
A common error is forgetting the factor that comes from du. If u = x³ + 1 and du = 3x² dx, the 1/3 factor must appear in front of the integral.
Students sometimes try to substitute when the derivative of u is not present (even as a constant multiple). Substitution only works when g'(x) dx can be expressed as du (up to a constant).
⚠️ U-substitution problems appear on every MTH 132 exam. You should be able to handle them quickly and accurately.
⚠️ WeBWorK problems require you to change the limits of integration for definite integrals. Practise this form even if you prefer substituting back.
⚠️ Even/odd shortcuts save time on exam problems with symmetric intervals. Check for symmetry before integrating.
⚠️ The "sadness maximus" pattern (x³√(x² + 1)) where you must express leftover x terms in terms of u is a common exam challenge. Be ready to write x² = u − 1 (or similar) after setting u.
True or false: If u = 1 − 3t, then du = −3 dt. (True.)
Fill in the blank: To evaluate ∫₀¹ ³√(1 + 7x) dx by substitution with u = 1 + 7x, the new limits are u = ___ to u = ___. (1 to 8.)
True or false: ∫₋₂² x³ dx = 0. (True; x³ is odd.)
Fill in the blank: ∫ sin(2x + 1) dx = ___ + C. (−(1/2) cos(2x + 1).)
Q: Evaluate ∫ x³ / √(4 + 11x⁴) dx.
A: Let u = 4 + 11x⁴, du = 44x³ dx. So x³ dx = du/44. The integral becomes (1/44) ∫ u^(−1/2) du = (1/44)(2)√u + C = √(4 + 11x⁴)/22 + C.
Q: Evaluate ∫ sec²(7x) tan²(7x) dx.
A: Let u = tan(7x), du = 7 sec²(7x) dx. The integral becomes (1/7) ∫ u² du = u³/21 + C = tan³(7x)/21 + C.
Q: Evaluate ∫₀³ 1/√(25 − 3x) dx by changing the bounds.
A: Let u = 25 − 3x, du = −3 dx. When x = 0, u = 25; when x = 3, u = 16. The integral becomes (−1/3) ∫₂₅¹⁶ u^(−1/2) du = (1/3) ∫₁₆²⁵ u^(−1/2) du = (1/3)[2√u]₁₆²⁵ = (2/3)(5 − 4) = 2/3.
Q: If f is continuous and ∫₀⁹ f(x) dx = 4, find ∫₀³ x f(x²) dx.
A: Let u = x², du = 2x dx. When x = 0, u = 0; when x = 3, u = 9. The integral becomes (1/2) ∫₀⁹ f(u) du = (1/2)(4) = 2. Answer: C.
Q: Evaluate ∫ sin³(2x) cos(2x) dx.
A: Let u = sin(2x), du = 2 cos(2x) dx. The integral becomes (1/2) ∫ u³ du = u⁴/8 + C = sin⁴(2x)/8 + C.
U-substitution is the first of several integration techniques. In Calculus II, it becomes one tool among many: integration by parts, trigonometric substitution, and partial fractions all build on the same idea of rewriting an integral into a simpler form. The even/odd properties connect back to symmetry concepts from precalculus and appear again when you study Fourier series.
u-substitution, substitution rule, change of variable, inner function, differential, du, change of limits, even function, odd function, symmetric interval, reverse chain rule, indefinite integral, definite integral, MTH 132, Calculus I, Michigan State University