Practice Maths

← Back to Integration Techniques›Specialist Maths

Integration by Substitution

Key Terms

The substitution rule
If u = g(x), then du = g′(x) dx, and ∫f(g(x))g′(x) dx = ∫f(u) du.
Purpose
Substitution simplifies a complex integral by replacing the inner function with a single variable u, reducing it to a standard form.
Identifying the substitution
Look for a composite function where one part is (almost) the derivative of another. The inner function of a composition is typically the best choice for u.
Standard types
   — Power of a function: ∫[g(x)]n g′(x) dx → u = g(x)
   — Exponential: ∫eg(x) g′(x) dx → u = g(x)
   — Trigonometric: ∫f(ax+b) dx → u = ax+b
   — Logarithmic: ∫g′(x)/g(x) dx → u = g(x), giving ln|u|
Definite integrals
When changing the variable, also change the limits: if u = g(x), then when x = a, u = g(a); when x = b, u = g(b). Do not convert back to x — evaluate directly in terms of u.
Key Formula:
  • Substitution: ∫f(g(x))g′(x) dx = ∫f(u) du   where u = g(x), du = g′(x) dx
  • Definite: ∫ab f(g(x))g′(x) dx = ∫g(a)g(b) f(u) du
  • Special case: ∫[g(x)]ng′(x) dx = [g(x)]n+1 / (n+1) + C   (n ≠ −1)
Hot Tip After choosing u, compute du/dx and solve for dx. Then replace every x-expression in the integrand with u-expressions. If you cannot eliminate all x-terms, your choice of u may need to change. For definite integrals, changing the limits avoids the need to substitute back at the end — this saves time and reduces error.

Worked Example 1 — Power of a function

Find ∫2x(x² + 1)³ dx.

Let u = x² + 1. Then du/dx = 2x, so du = 2x dx.

∫2x(x² + 1)³ dx = ∫u³ du = u&sup4;/4 + C = (x² + 1)&sup4; / 4 + C

Worked Example 2 — Trigonometric

Find ∫sin(3x) dx.

Let u = 3x. Then du = 3 dx, so dx = du/3.

∫sin(3x) dx = ∫sin(u) × du/3 = −(1/3)cos(u) + C = −(1/3)cos(3x) + C

Worked Example 3 — Exponential with definite limits

Evaluate ∫01 2xex² dx.

Let u = x². Then du = 2x dx. Change limits: x=0 → u=0; x=1 → u=1.

∫01 2xex² dx = ∫01 eu du = [eu]01 = e1 − e0 = e − 1

Why Substitution Works: The Reverse Chain Rule

Integration by substitution is essentially the chain rule for differentiation run in reverse. Recall that the chain rule gives d/dx[F(g(x))] = F′(g(x)) × g′(x). Integrating both sides: ∫F′(g(x))g′(x) dx = F(g(x)) + C. If we write f = F′ and u = g(x), this becomes ∫f(u) du = F(u) + C. Substitution simply performs this transformation systematically.

The Mechanics: A Step-by-Step Process

Given an integral, the substitution method proceeds as follows:

  1. Choose u: Identify the inner function of a composite expression. Often u is chosen so that du appears (up to a constant multiple) elsewhere in the integrand.
  2. Compute du: Differentiate u with respect to x to find du = (du/dx) dx, then solve for dx.
  3. Substitute: Replace all x-terms with u-terms in the integrand and dx.
  4. Integrate: Evaluate the resulting integral in terms of u.
  5. Back-substitute: Replace u with g(x) to express the answer in terms of x (for indefinite integrals only).

Handling the “Missing” Constant Factor

Often the derivative g′(x) differs from what appears in the integrand by only a constant factor. For example, in ∫(x²+1)³ dx, there is no factor of 2x visible. In this case, substitution does not directly apply — the integral must be expanded or a different technique used. However, in ∫x(x²+1)³ dx, the factor x is present and is proportional to the derivative of x²+1. We compensate for the missing constant: if the derivative is 2x and only x appears, write ∫x(x²+1)³ dx = (1/2)∫2x(x²+1)³ dx = (1/2) × u&sup4;/4 + C.

Changing Limits for Definite Integrals

For a definite integral ∫ab f(g(x))g′(x) dx, substituting u = g(x) transforms the limits: the lower limit a becomes g(a) and the upper limit b becomes g(b). The result is ∫g(a)g(b) f(u) du, which is evaluated directly without converting back to x. This approach is cleaner and less prone to error than converting back after integration.

Common Pitfalls

  • Forgetting to change the limits in definite integrals.
  • Not replacing all x-terms — if any x remains in the integrand after substitution, check your work.
  • Sign errors when du involves a negative: if u = −x, then du = −dx, so dx = −du.
Exam Tip: Always write the substitution explicitly at the start: “Let u = ..., du = ... dx”. Show the limit change for definite integrals. Examiners expect to see working that justifies the substitution, not just the final answer.

Mastery Practice

  1. Find ∫2x(x² + 3)&sup4; dx using the substitution u = x² + 3. Fluency

  2. Find ∫6x²(x³ + 1)² dx. Fluency

  3. Find ∫cos(2x) dx. Fluency

  4. Find ∫(3x + 1)&sup5; dx. Fluency

  5. Evaluate ∫01 2xex² dx, changing the limits of integration. Understanding

  6. Find ∫ x / √(x² + 4) dx. Understanding

  7. Find ∫sin²(x)cos(x) dx using u = sin x. Understanding

  8. Find ∫(ln x) / x dx. Understanding

  9. Evaluate ∫0π/4 tan x dx. Write tan x = sin x / cos x and use an appropriate substitution. Problem Solving

  10. A curve has dy/dx = 3x²ex³. Find y given that y = 2 when x = 0. Problem Solving

See Answers ➔