Difficulty: Intermediate | Prerequisites: Calculus I (derivatives, basic antiderivatives, chain rule)
U-substitution is the integration counterpart of the chain rule from Calc I. If you can spot a composite function and its inner derivative sitting inside an integral, you can simplify the whole thing into a much friendlier form. It is the single most-used technique in Calculus II and a prerequisite for every other integration method you will meet (integration by parts, partial fractions, trig substitution). If the chain rule felt mechanical going forward, u-sub is learning to run it in reverse.
When an integrand contains a function and its derivative, let u equal the inner function, replace everything in terms of u, integrate, then substitute back. For definite integrals, either convert the limits to u-values or back-substitute and use the original limits.
U-substitution (u-sub)
A technique for evaluating integrals by replacing a composite expression with a single variable u, turning a complicated integral into a simpler one. Think of it as "un-doing" the chain rule.
Integrand
The function being integrated, the expression between the integral sign and dx. In simple terms, it is the thing you are trying to find the antiderivative of.
Inner function, g(x)
The function you choose to set equal to u. Picking the right inner function is the core skill of u-substitution.
Differential (du)
The derivative of your chosen u with respect to x, multiplied by dx. You need du (or a constant multiple of it) to appear in the original integrand for the substitution to work.
Indefinite integral
An antiderivative with no specified bounds. The result includes a constant of integration, +C.
Definite integral
An integral evaluated between two specific bounds (limits of integration). The result is a number, not a function.
Constant of integration (+C)
The arbitrary constant added to every indefinite integral, representing the family of all antiderivatives. You drop it when evaluating a definite integral because it cancels.
If u = g(x), then:
\int f(g(x))\, g'(x)\, dx = \int f(u)\, duYou spot g(x) inside a composite function, check that g'(x) (or a constant multiple of it) is also present, set u = g(x), compute du = g'(x) dx, rewrite everything in terms of u, integrate, and substitute back.
Choose u. Pick the inner function, the part whose derivative also appears in the integrand.
Find du. Differentiate u with respect to x and write du = (something) dx.
Rewrite the integral. Replace all x-expressions and dx with u and du. If a constant factor is left over, pull it out front.
Integrate in u. Apply standard antiderivative rules.
Back-substitute. Replace u with the original expression in x. Add +C.
Example 1: Evaluate the integral of 8(x² – 2)⁷(2x) dx.
Let u = x² – 2, so du = 2x dx.
The integral becomes the integral of 8u⁷ du = u⁸ + C.
Back-substitute: (x² – 2)⁸ + C.
Example 2: Evaluate the integral of cos(x³) · 3x² dx.
Let u = x³, so du = 3x² dx.
The integral becomes the integral of cos(u) du = sin(u) + C.
Back-substitute: sin(x³) + C.
Example 3: Evaluate the integral of t² sin(2 – t³) dt.
Let u = 2 – t³, so du = –3t² dt, meaning t² dt = –(1/3) du.
The integral becomes –(1/3) integral of sin(u) du = –(1/3)(–cos u) + C.
Back-substitute: (1/3) cos(2 – t³) + C.
Example 4: Evaluate the integral of 8x(x² + 5)³ dx.
Let u = x² + 5, so du = 2x dx, meaning 8x dx = 4 du.
The integral becomes 4 integral of u³ du = u⁴ + C.
Back-substitute: (x² + 5)⁴ + C.
Example 5: Evaluate the integral of (sin x)⁷ cos x dx.
Let u = sin x, so du = cos x dx.
The integral becomes the integral of u⁷ du = (1/8)u⁸ + C.
Back-substitute: (1/8)(sin x)⁸ + C.
The u-substitution rule (indefinite)
\int f(g(x))\, g'(x)\, dx = \int f(u)\, du \quad \text{where } u = g(x)Key antiderivatives you will need
\int u^n\, du = \frac{u^{n+1}}{n+1} + C \quad (n \neq -1)\int \cos u\, du = \sin u + C\int \sin u\, du = -\cos u + C\int \sec^2 u\, du = \tan u + C\int \frac{1}{u}\, du = \ln|u| + C\int e^u\, du = e^u + CU-substitution is not just an exam technique. Any time a physical quantity is described by a composite function (temperature as a function of position, position as a function of time), finding accumulated totals, areas, or averages requires exactly this kind of integral. Engineers computing work done by a variable force, or physicists finding charge from a current that changes over time, use u-substitution constantly.
Students often forget to account for the constant multiplier when du does not match exactly. If du = 3x² dx but only x² dx appears, you must multiply and divide by 3, not ignore it.
Students sometimes leave the answer in terms of u. Always back-substitute so the final answer is in the original variable.
When dealing with powers, students confuse the power rule with the log rule. The integral of u⁻¹ du is ln|u| + C, not u⁰/0.
Students occasionally pick the wrong piece of the integrand for u. A good heuristic: u is typically the "inside" function whose derivative (or a constant multiple of it) also appears in the integrand.
⚠️ U-substitution appears on virtually every Calc II exam. Expect at least 2-3 problems requiring it.
⚠️ Trig integrals combined with u-sub are a favourite exam combo (e.g. integral of sin⁷(x) cos(x) dx).
⚠️ Definite integral problems will test whether you correctly convert limits. A wrong limit is a wrong answer, full stop.
⚠️ Partial-credit tip: even if you cannot finish the integral, showing the correct u-substitution setup often earns marks.
True or false: If u = x² + 1, then du = x dx. False. du = 2x dx.
Fill in the blank: The integral of cos(5x) dx requires the substitution u = ___. 5x.
True or false: When using u-sub on a definite integral, you must always convert the limits to u-values. False. You can also back-substitute and use the original x-limits (Method 1).
Fill in the blank: The integral of e^u du = ___. e^u + C.
True or false: The integral of 1/u du = u⁰/0 + C. False. It is ln|u| + C.
Q: Evaluate the integral of 3x / (x² + 3) dx.
A: Let u = x² + 3, du = 2x dx. The integral becomes (3/2) integral of (1/u) du = (3/2) ln|x² + 3| + C.
Q: Evaluate the integral of cos(x) · sin(sin(x)) dx.
A: Let u = sin x, du = cos x dx. The integral becomes the integral of sin(u) du = –cos(u) + C = –cos(sin x) + C.
Q: Evaluate the integral of (x³ – 1)¹⁰⁰ · x² dx.
A: Let u = x³ – 1, du = 3x² dx, so x² dx = (1/3) du. The integral becomes (1/3) integral of u¹⁰⁰ du = (1/3)(u¹⁰¹/101) + C = (x³ – 1)¹⁰¹ / 303 + C.
Q: Evaluate the integral of sec²(5x + 1) dx.
A: Let u = 5x + 1, du = 5 dx, so dx = (1/5) du. The integral becomes (1/5) integral of sec²(u) du = (1/5) tan(u) + C = (1/5) tan(5x + 1) + C.
Q: Evaluate the integral of 1/(e^x + 1) dx using u = e^x + 1.
A: du = e^x dx, and e^x = u – 1. The integral becomes the integral of 1/(u(u – 1)) du. Using partial fractions: 1/u – ln u = 1/(e^x + 1) – ln|e^x + 1| + C.
U-substitution is the reverse of the chain rule (Calc I). Once you are comfortable with it, integration by parts (the reverse of the product rule) is the natural next technique. Many trig substitution and partial fractions problems begin with or reduce to a u-sub step, so this skill stays relevant throughout the entire course.
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