Difficulty: Intermediate | Prerequisites: Differentiation (chain rule, power rule, trig derivatives), basic antiderivatives, definite integral notation.
U-substitution is the integration counterpart of the chain rule from differentiation. If you can spot a composite function inside an integral and match its inner derivative to something already in the integrand, you can simplify the whole thing into a basic antiderivative problem. This technique appears everywhere in Integral Calculus, from evaluating definite integrals with trig functions to finding antiderivatives of polynomials raised to powers. You should be comfortable with the chain rule and basic antiderivative formulas before diving in.
U-substitution reverses the chain rule: you pick an inner function u, express du in terms of dx, rewrite the integral in terms of u, integrate, then substitute back. For definite integrals, you can convert the limits of integration to u-values and skip the back-substitution entirely. Master this and you unlock a huge proportion of the integrals you will meet on exams.
U-substitution (integration by substitution)
A method for evaluating integrals by introducing a new variable u = g(x) to simplify the integrand. In simple terms, you are "undoing" the chain rule: if the integrand looks like f(g(x)) · g'(x), setting u = g(x) collapses it into f(u) du.
Integrand
The function being integrated, the expression that sits between the integral sign and dx. Think of it as the thing you are trying to find the antiderivative of.
Differential (du)
The derivative of u with respect to x, multiplied by dx. When you set u = g(x), du = g'(x) dx. You need to express all the remaining x-terms and dx in terms of u and du for the substitution to work.
Definite integral
An integral with specific upper and lower limits of integration (e.g. from 0 to pi/4). It produces a number, not a family of functions. When using substitution on a definite integral, you can convert the x-limits to u-limits and evaluate directly.
Indefinite integral
An integral with no limits, producing a general antiderivative plus an arbitrary constant C. After substituting back from u to x, you must include + C.
Constant of integration (C)
The arbitrary constant added to every indefinite integral. It accounts for the fact that many different functions share the same derivative. When solving initial-value problems (e.g. f(2) = 3), you use the given condition to pin down C.
Look for a composite structure in the integrand: something of the form f(g(x)) · g'(x) dx, or close to it. The inner function g(x) is your candidate for u. Common patterns:
A polynomial raised to a power, with its derivative (or a constant multiple of it) sitting outside. For example, in ∫ 8x(x²/4 - 2)³ dx, the inner function is u = x²/4 - 2 and du = (2x/4) dx = (x/2) dx.
A trig function composed with another function, such as sin(2θ) or cos(x) paired with sin(x) dx.
An expression under a square root, like √(2x² + 1), where the derivative of the expression under the root appears as a factor outside.
Set u = g(x).
Compute du = g'(x) dx.
Rewrite every part of the integrand in terms of u and du. If there is a leftover constant (e.g. you need x dx but du = 2x dx), adjust by multiplying both sides of the du equation.
Integrate with respect to u.
Substitute back to express the result in terms of x (for indefinite integrals). Add + C.
Determine ∫ 8x(x²/4 - 2)³ dx.
Set u = x²/4 - 2. Then du = (2x/4) dx = (x/2) dx, so x dx = 2 du.
The integral becomes ∫ 8 · u³ · 2 du = 16 ∫ u³ du = 16 · u⁴/4 + C = 4u⁴ + C.
Substitute back: I = 4(x²/4 - 2)⁴ + C.
Determine ∫ (x - 4) / (x² - 8x - 3)⁶ dx.
Set u = x² - 8x - 3. Then du = 2(x - 4) dx, so (x - 4) dx = du/2.
The integral becomes ∫ (1/2) u⁻⁶ du = (1/2) · u⁻⁵/(-5) + C = -1/(10u⁵) + C.
Substitute back: I = -1 / [10(x² - 8x - 3)⁵] + C.
When evaluating a definite integral using substitution, you have two options:
Convert the limits from x-values to u-values and evaluate entirely in u. This is usually cleaner.
Integrate in u, substitute back to x, then apply the original x-limits.
The first approach saves a step and is preferred on exams.
When you set u = g(x), the lower limit x = a becomes u = g(a), and the upper limit x = b becomes u = g(b). You then evaluate the antiderivative at these new limits.
Evaluate ∫₀^{π/4} (1/cos²θ - sin 2θ) dθ.
Rewrite 1/cos²θ as sec²θ. The integral becomes ∫₀^{π/4} (sec²θ - sin 2θ) dθ.
The antiderivative of sec²θ is tan θ. The antiderivative of sin 2θ is -cos(2θ)/2.
So I = [tan θ + (1/2)cos 2θ] evaluated from 0 to π/4.
At θ = π/4: tan(π/4) + (1/2)cos(π/2) = 1 + 0 = 1.
At θ = 0: tan(0) + (1/2)cos(0) = 0 + 1/2 = 1/2.
I = 1 - 1/2 = 1/2.
Evaluate ∫₁⁹ 2 / [√x · (√x + 3)²] dx.
Set u² = x, so 2u du = dx. When x = 1, u = 1. When x = 9, u = 3.
√x = u, so the integrand becomes 2 / [u · (u + 3)²] and dx = 2u du.
I = ∫₁³ [2 / (u(u+3)²)] · 2u du = 4 ∫₁³ 1/(u+3)² du.
The antiderivative of 1/(u+3)² is -1/(u+3).
I = 4[-1/(u+3)] from 1 to 3 = 4(-1/6 + 1/4) = 4(1/12) = 1/3.
Evaluate ∫₀^{π/4} (4 tan x - 1) / cos²x dx.
Since 1/cos²x = sec²x and d/dx(tan x) = sec²x, set u = tan x, du = sec²x dx.
When x = 0, u = 0. When x = π/4, u = 1.
I = ∫₀¹ (4u - 1) du = [2u² - u] from 0 to 1 = 2 - 1 = 1.
Given ∫₀⁶ f(x) dx = 15, find ∫₀² 7f(3x) dx.
Set u = 3x, so du = 3 dx and dx = du/3. When x = 0, u = 0. When x = 2, u = 6.
I = ∫₀⁶ 7f(u) · (du/3) = (7/3) ∫₀⁶ f(u) du = (7/3)(15) = 35.
Some problems combine substitution with an initial condition such as f(0) = 3 or f(1) = 5. In these cases you integrate, apply the substitution, find the general antiderivative, then solve for C using the given point.
Given df/dx = 2x√(2x² + 1) and f(2) = 3, find the y-intercept.
Set u = 2x² + 1, du = 4x dx, so x dx = du/4.
f(x) = ∫ 2x√(2x² + 1) dx = (1/2) ∫ u^{1/2} du = (1/3) u^{3/2} + C = (1/3)(2x² + 1)^{3/2} + C.
Apply f(2) = 3: (1/3)(9)^{3/2} + C = 3, so (1/3)(27) + C = 3, giving 9 + C = 3, C = -6.
f(x) = (1/3)(2x² + 1)^{3/2} - 6. At x = 0: f(0) = (1/3)(1) - 6 = 1/3 - 6 = -17/3.
The y-intercept is -17/3.
Given df/dx = -18x/√(4 - 3x²) and f(1) = -2, find the y-intercept.
Set u = 4 - 3x², du = -6x dx.
f(x) = -∫ 18x / √(4 - 3x²) dx = 3 ∫ u^{-1/2} du = 6u^{1/2} + C = 6√(4 - 3x²) + C.
Apply f(1) = -2: 6√1 + C = -2, so 6 + C = -2, C = -8.
f(x) = 6√(4 - 3x²) - 8. At x = 0: f(0) = 6(2) - 8 = 4.
The y-intercept is 4.
Evaluate ∫₀^{π/2} sin x {2f'(cos x) - 3} dx when f(0) = 3 and f(1) = 5.
Set u = cos x, du = -sin x dx. When x = 0, u = 1. When x = π/2, u = 0.
I = -∫₁⁰ {2f'(u) - 3} du = ∫₀¹ {2f'(u) - 3} du = [2f(u) - 3u] from 0 to 1.
I = (2f(1) - 3) - (2f(0) - 0) = (10 - 3) - 6 = 1.
The key insight: the antiderivative of f'(u) is f(u), so you can evaluate these integrals using the given endpoint values of f.
When you see (1/x) · h(ln x), set u = ln x, du = (1/x) dx.
Evaluate ∫₁ᵉ (5/x)(ln x)² dx.
Set u = ln x, du = (1/x) dx. When x = 1, u = 0. When x = e, u = 1.
I = ∫₀¹ 5u² du = [5u³/3] from 0 to 1 = 5/3.
When you see cos x multiplied by a function of sin x, set u = sin x (or u = a + b·sin x if there is a linear expression). Similarly, sin x · f(cos x) suggests u = cos x.
Evaluate ∫₀^{π/6} cos x / (5 + 2 sin x) dx.
Set u = 5 + 2 sin x, du = 2 cos x dx. When x = 0, u = 5. When x = π/6, u = 6.
I = (1/2) ∫₅⁶ (1/u) du = (1/2)[ln u] from 5 to 6 = (1/2) ln(6/5).
When eˣ appears under a composition, set u = 1 + eˣ or u = eˣ depending on the structure.
Evaluate ∫₀^{ln 4} eˣ / √(1 + eˣ) dx.
Set u = 1 + eˣ, du = eˣ dx. When x = 0, u = 2. When x = ln 4, u = 1 + 4 = 5.
I = ∫₂⁵ u^{-1/2} du = [2√u] from 2 to 5 = 2(√5 - √2).
Evaluate ∫₁⁴ (3 - x^{1/2}) / x^{1/2} dx.
Set u = x^{1/2}, du = 1/(2x^{1/2}) dx, so dx/x^{1/2} = 2 du. When x = 1, u = 1. When x = 4, u = 2.
I = 2 ∫₁² (3 - u) du = 2[3u - u²/2] from 1 to 2 = 2[(6 - 2) - (3 - 1/2)] = 2[4 - 5/2] = 2(3/2) = 3.
\int f(g(x)) \cdot g'(x)\, dx = \int f(u)\, du \quad \text{where } u = g(x)\int_a^b f(g(x)) \cdot g'(x)\, dx = \int_{g(a)}^{g(b)} f(u)\, duTrig identities used repeatedly in these problems:
sec²θ = 1/cos²θ
d/dθ (tan θ) = sec²θ
d/dx (sin x) = cos x
d/dx (cos x) = -sin x
d/dx (ln x) = 1/x
d/dx (eˣ) = eˣ
Power rule for integration:
\int u^n\, du = \frac{u^{n+1}}{n+1} + C \quad (n \neq -1)\int \frac{1}{u}\, du = \ln|u| + CU-substitution is how engineers and physicists evaluate integrals that arise from modelling real systems. For instance, computing the work done by a variable force along a curved path, or finding the total charge accumulated in a circuit with a time-varying current, both reduce to integrals that require substitution. Any time a rate of change involves a composition of functions (temperature as a function of position, which is itself a function of time), integration by substitution is the tool that makes the integral tractable.
Students often forget to change the limits of integration when using substitution on a definite integral. If you keep the old x-limits but evaluate in u, you will get the wrong answer.
Students sometimes forget the constant of integration (+C) on indefinite integrals and then cannot solve initial-value problems correctly.
A common error is choosing u but forgetting to account for the entire du. For example, if u = 2x² + 1 then du = 4x dx, not 2x dx. Missing the constant multiplier throws off the whole calculation.
Students occasionally try to substitute only part of the integrand, leaving behind a mix of x and u terms. Every x and every dx must be expressed in terms of u and du for the substitution to be valid.
⚠️ Substitution problems appear on virtually every Integral Calculus exam. Expect at least 2-3 questions requiring u-substitution.
⚠️ Problems that give you f(0) and f(1) then ask you to evaluate an integral involving f'(g(x)) are a favourite exam pattern. The trick is recognising that the antiderivative of f'(u) is f(u).
⚠️ "Determine the integral" (indefinite) requires + C. "Evaluate the integral" (definite) gives a number. Read the question carefully.
⚠️ Rescaling problems (e.g. given ∫₀⁶ f(x) dx = 15, find ∫₀² 7f(3x) dx) test whether you can change variables cleanly. These are quick marks if you practise them.
⚠️ Watch for integrals where the substitution produces a factor that cancels or simplifies the integrand (e.g. the √x in both numerator and denominator). The answer often comes out surprisingly clean.
True or False: When using substitution on a definite integral, you must always substitute back to x before evaluating at the limits.
Answer: False. You can convert the limits to u-values and evaluate directly in u.
Fill in the blank: If u = x² + 5, then du = ____.
Answer: 2x dx.
True or False: ∫ u⁻¹ du = u⁰/0 + C.
Answer: False. ∫ (1/u) du = ln|u| + C. The power rule does not apply when n = -1.
Fill in the blank: If u = cos x, then du = ____.
Answer: -sin x dx.
True or False: Every indefinite integral must include + C.
Answer: True.
Q: Evaluate ∫₀^{π/4} (sec²θ - sin 2θ) dθ.
A: Use the antiderivatives tan θ and -(1/2)cos 2θ. Evaluate at the limits: [tan θ + (1/2)cos 2θ] from 0 to π/4 = (1 + 0) - (0 + 1/2) = 1/2.
Q: If f'(x) = 2x√(2x² + 1) and f(2) = 3, what is f(0)?
A: Integrate using u = 2x² + 1 to get f(x) = (1/3)(2x² + 1)^{3/2} + C. Apply f(2) = 3 to find C = -6. Then f(0) = 1/3 - 6 = -17/3.
Q: Given ∫₀⁶ f(x) dx = 15, find ∫₀² 7f(3x) dx.
A: Substitute u = 3x, du = 3 dx. The integral becomes (7/3)∫₀⁶ f(u) du = (7/3)(15) = 35.
Q: Evaluate ∫₀^{π/6} cos x / (5 + 2 sin x) dx.
A: Set u = 5 + 2 sin x, du = 2 cos x dx. Limits: u = 5 to u = 6. I = (1/2) ln(6/5).
Q: Evaluate ∫₀^{ln 4} eˣ / √(1 + eˣ) dx.
A: Set u = 1 + eˣ, du = eˣ dx. Limits: u = 2 to u = 5. I = 2(√5 - √2).
Q: Evaluate ∫₀¹ x{2f'(x²) - 3} dx when f(0) = 1, f(1) = 4.
A: Set u = x², du = 2x dx. I = (1/2)[2f(u) - 3u] from 0 to 1 = (f(1) - f(0)) - 3/2 = 3 - 3/2 = 3/2.
U-substitution connects directly to the chain rule from differential calculus: every chain-rule derivative you learned to compute gives you a substitution you can run in reverse. This technique is also the foundation for more advanced integration methods such as integration by parts and trigonometric substitution, both of which build on the same idea of transforming the integrand into something more manageable. The area-between-curves problems later in this unit rely on evaluating definite integrals, many of which require substitution to compute.
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