Difficulty: Intermediate | Prerequisites: U-Substitution for Indefinite Integrals (Part 1 notes), Fundamental Theorem of Calculus
Once you can perform u-substitution on indefinite integrals, the next step is applying the same technique to definite integrals, where you are computing an actual numerical value between two bounds. The twist is that your limits of integration are in terms of x, but your substitution puts everything in terms of u. You have two clean ways to handle this, and both give the same answer. Knowing both methods is worth it because some problems are faster one way, some the other, and exams will test both.
When using u-sub on a definite integral, either convert the x-limits to u-limits before integrating (Method 2, usually quicker), or find the antiderivative in u, back-substitute to x, and evaluate at the original limits (Method 1). Both methods produce the same number.
Limits of integration (bounds)
The two values (lower and upper) that define the interval over which you are integrating. In a definite integral from a to b, a is the lower limit and b is the upper limit.
Converting limits to u-coordinates
Plugging the original x-limits into the equation u = g(x) to get corresponding u-limits. If u = x² + 1 and x goes from 0 to 1, then u goes from 1 to 2.
Back-substitution (Method 1)
Finding the antiderivative F(u), replacing u with g(x), then evaluating F(g(b)) – F(g(a)) using the original x-limits.
Fundamental Theorem of Calculus (FTC)
The theorem that connects antiderivatives to definite integrals. In simple terms, to evaluate a definite integral, find an antiderivative and subtract its values at the two endpoints.
Write the integral of f(g(x)) g'(x) dx = integral of f(u) du.
Find the antiderivative F(u).
Substitute u = g(x) back in to get F(g(x)).
Evaluate F(g(b)) – F(g(a)) using the original x-limits.
This is the safer method if you are worried about converting limits incorrectly.
Write the integral of f(g(x)) g'(x) dx = integral of f(u) du.
Convert bounds: plug the original x-limits into u = g(x) to get new u-limits.
Evaluate the definite integral entirely in u. No back-substitution needed.
This is typically faster and cleaner.
Example 1 (Method 2): Evaluate the integral from 0 to 1 of x²√(x³ + 1) dx.
Let u = x³ + 1, du = 3x² dx, so x² dx = (1/3) du.
Convert limits: u(0) = 0³ + 1 = 1, u(1) = 1³ + 1 = 2.
The integral becomes (1/3) integral from 1 to 2 of u^(1/2) du.
= (1/3) · (2/3) u^(3/2) evaluated from 1 to 2 = (2/9)(2^(3/2) – 1).
Answer: (2/9)(2^(3/2) – 1).
Example 2 (Method 2): Evaluate the integral from 0 to π/4 of 2 sin(4t) dt.
Let u = 4t, du = 4 dt, so dt = (1/4) du.
Convert limits: u(0) = 0, u(π/4) = π.
The integral becomes (1/2) integral from 0 to π of sin(u) du.
= (1/2)[–cos(u)] from 0 to π = –(1/2)(cos π – cos 0) = –(1/2)(–1 – 1) = 1.
Answer: 1.
Example 3 (Method 2): Evaluate the integral from 0 to 1 of x(x² + 1)^(4/5) dx.
Let u = x² + 1, du = 2x dx, so x dx = (1/2) du.
Convert limits: u(0) = 1, u(1) = 2.
The integral becomes (1/2) integral from 1 to 2 of u^(4/5) du.
= (1/2) · (5/9) u^(9/5) evaluated from 1 to 2 = (5/18)(2^(9/5) – 1).
Answer: (5/18)(2^(9/5) – 1) = (18/5) · 2^(9/5) – (18/5). (Corrected: (5/18) · 2^(9/5) – 5/18.)
Example 4: Evaluate the integral from 0 to 1 of e^x / (e^x + 15)³ dx.
Let u = e^x + 15, du = e^x dx.
Convert limits: u(0) = 16, u(1) = e + 15.
The integral becomes the integral from 16 to e+15 of u⁻³ du.
= [–1/(2u²)] from 16 to e+15 = –1/(2(e + 15)²) + 1/16².
Answer: –1/(2(e + 15)²) + 1/512.
Example 5: Evaluate the integral from 1 to 5 of (x + 3)(x – 1)^(1/2) dx.
Let u = x – 1, du = dx, and x + 3 = u + 4.
Convert limits: u(1) = 0, u(5) = 4.
The integral becomes the integral from 0 to 4 of u^(1/2)(u + 4) du = integral of (u^(3/2) + 4u^(1/2)) du.
= [(2/5)u^(5/2) + (8/3)u^(3/2)] from 0 to 4 = (2/5)(32) + (8/3)(8) = 64/5 + 64/3 = 512/15.
Answer: 512/15.
Example 6: Evaluate the integral from 0 to π/4 of tan(x) dx.
Rewrite tan(x) as sin(x)/cos(x). Let u = cos x, du = –sin x dx.
Convert limits: u(0) = 1, u(π/4) = √2/2.
The integral becomes –integral from 1 to √2/2 of (1/u) du = –ln|u| from 1 to √2/2.
= –ln(√2/2) + ln(1) = –ln(√2/2) = ln(2/√2) = (1/2) ln 2.
Answer: –ln(√2/2), equivalently (1/2) ln 2.
Method 1 (back-substitute)
\int_a^b f(g(x))\, g'(x)\, dx = F(g(x))\Big|_a^b = F(g(b)) - F(g(a))Method 2 (convert limits)
\int_a^b f(g(x))\, g'(x)\, dx = \int_{g(a)}^{g(b)} f(u)\, duStudents sometimes convert the limits to u but then also back-substitute at the end. Pick one method and stick with it. If you convert the limits, you evaluate entirely in u and never go back to x.
When limits flip (the new upper u-limit is smaller than the lower), students panic. This is fine. Either evaluate as-is (you will get a negative sign naturally if appropriate) or swap the limits and negate the integral.
Students forget to convert both limits. Converting only one gives nonsense.
A common error is applying u-sub to the integrand but leaving dx unchanged. Every piece of the original integral, including dx, must be expressed in terms of u and du.
⚠️ Definite integral u-sub problems are nearly guaranteed on a Calc II final. Expect at least one.
⚠️ Showing your limit conversion clearly (writing u(a) = ... and u(b) = ...) is worth partial credit and prevents errors.
⚠️ Problems that combine u-sub with trig functions and definite limits (e.g. integral of sin(4t) from 0 to π/4) are a favourite exam format.
⚠️ The integral of tan(x) dx is a classic exam question. Know it cold: the answer is ln|sec x| + C, derived via u = cos x.
True or false: When converting limits in Method 2, you plug the x-limits into u = g(x). True.
Fill in the blank: If u = x² + 1 and x goes from 0 to 3, then u goes from ___ to ___. 1 to 10.
True or false: In Method 2, you must back-substitute u = g(x) before evaluating. False. You evaluate entirely in u.
True or false: The integral of tan(x) dx = –ln|cos x| + C. True. This is equivalent to ln|sec x| + C.
Fill in the blank: If u = 4t and you are integrating from t = 0 to t = π/4, the u-limits are ___ to ___. 0 to π.
Q: Evaluate the integral of 3x/(x² + 3) dx (indefinite).
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(sin x) + C.
Q: Evaluate the integral of 1/(e^x + 1) dx, letting u = e^x + 1.
A: du = e^x dx, and e^x = u – 1. Rewrite as the integral of 1/(u(u – 1)) du. By partial fractions this becomes (1/(u – 1) – 1/u) du, giving ln|u – 1| – ln|u| + C = x – ln|e^x + 1| + C, since ln|e^x| = x.
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)(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. The integral becomes (1/5) tan(5x + 1) + C.
Q: Evaluate the integral from 0 to π/4 of tan(x) dx using u = cos x.
A: du = –sin x dx, u(0) = 1, u(π/4) = √2/2. The integral becomes –integral from 1 to √2/2 of (1/u) du = –[ln(√2/2) – ln 1] = –ln(√2/2) = (1/2) ln 2.
Definite integral u-sub feeds directly into applications of integration (area between curves, volumes of revolution, arc length), all of which require evaluating definite integrals that rarely come in a tidy form. Trig substitution, covered later in M408D, is an extension of the same idea: a more specialised change of variable for integrands involving square roots of quadratics.
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