Integration by substitution changes the variable to make an integral simpler. Let u = g(x), find du/dx and rewrite dx in terms of du, so the whole integral is in u. Integrate, then substitute back. For a definite integral, change the limits to u-values instead, so you never need to substitute back.
The substitution method
- Choose u = g(x) and find du/dx, so du = g′(x) dx.
- Rewrite every part of the integral, including dx, in terms of u.
- Integrate with respect to u.
- Substitute u = g(x) back, or, for a definite integral, use the new u-limits.
If an x is left over, rearrange the substitution to write it in terms of u.
How does integration by substitution work?
Substitution is a formal change of variable. Choose u to be the inner part of the expression, usually a bracket or the inside of a root, an exponential or a power. Differentiate it to link du and dx, then rewrite the integral entirely in terms of u. If the substitution is a good one, the new integral is a standard result.
Exam questions usually tell you which substitution to use. Follow it exactly.
Dealing with a leftover x
Sometimes an x remains after substituting the bracket and dx. Rearrange the substitution to express it in u: if u = x + 1, then x = u − 1. Replace it, expand, and integrate term by term. Forgetting the leftover x is the classic error in this topic.
Definite integrals: change the limits
For a definite integral, convert both limits into u-values using u = g(x). Then evaluate the u-integral between the new limits; there is no need to go back to x. Never mix x-limits with an integral written in u.
Trig and exponential substitutions
Substitution handles trig and exponential integrands too, with choices such as u = sin x, u = cos x or u = eˣ. With u = sin x, du = cos x dx, so a factor of cos x is absorbed into du. Odd powers of sine or cosine can be integrated by peeling off one factor for du and converting the rest with sin²x + cos²x = 1.
When to use substitution rather than the reverse chain rule
The reverse chain rule is quicker when the derivative of the inside is present, apart from a constant. Substitution is needed when it is not quite present, such as x(x + 2)³ or x√(x + 3), where a leftover x must be rewritten in terms of u.
Worked examples
Example 1. Find ∫ 2x(x² + 1)³ dx.
- Let u = x² + 1, so du = 2x dx.
- The 2x dx becomes du: ∫ u³ du.
- Integrate: u⁴/4.
- Substitute back.
Answer: ¼(x² + 1)⁴ + c.
Marks: M1 for u and du, M1 for ∫ u³ du, A1 for the answer.
Example 2. Find ∫ x(x + 2)³ dx.
- Let u = x + 2, so x = u − 2 and dx = du.
- ∫ (u − 2)u³ du = ∫ (u⁴ − 2u³) du.
- Integrate: u⁵/5 − u⁴/2.
- Substitute back.
Answer: (x + 2)⁵/5 − (x + 2)⁴/2 + c.
Marks: M1 for x = u − 2, M1 for expanding and integrating, A1 for the answer.
Example 3. Find ∫ 2x(x² + 1)³ dx from x = 0 to x = 1.
- u = x² + 1, so du = 2x dx.
- Change the limits: x = 0 gives u = 1, and x = 1 gives u = 2.
- ∫ u³ du from 1 to 2 = [u⁴/4] from 1 to 2 = (16 − 1)/4.
Answer: 15/4.
Marks: M1 for the new limits, M1 for the u-integral, A1 for 15/4.
Example 4. Find ∫ sin⁴x cos x dx.
- Let u = sin x, so du = cos x dx.
- The integral becomes ∫ u⁴ du = u⁵/5.
Answer: ⅕ sin⁵x + c.
Marks: M1 for u = sin x, A1 for the answer.
Example 5. Use the substitution u = cos x to find the exact value of ∫ cos³x sin x dx from 0 to π/2.
- u = cos x, so du = −sin x dx, which means sin x dx = −du.
- Limits: x = 0 gives u = 1, and x = π/2 gives u = 0.
- The integral becomes ∫ −u³ du from 1 to 0, which equals ∫ u³ du from 0 to 1.
- [u⁴/4] from 0 to 1 = ¼.
Answer: ¼.
Marks: M1 for the substitution and the minus sign, M1 for the new limits, A1 for ¼.
Common mistakes
- Leaving some x in the integral. Every x, including dx and any leftover x, must be written in terms of u.
- Forgetting to convert dx. Use du = g′(x) dx; dx is not the same as du.
- Using x-limits with a u-integral. Convert the limits when you change variable.
- Losing a minus sign. With u = cos x, du = −sin x dx. Keep the sign or swap the limits.
- Not substituting back for an indefinite integral. The final answer must be in terms of x.
How integration by substitution is examined
Integration by substitution is a regular Pure 4 (WMA14) topic for Edexcel International A-Level students, and part of second-year Pure for 9MA0. Questions normally:
- state the substitution to use
- ask you to show that an integral equals a given expression
- ask for an exact definite integral with changed limits
- use the result to find an area or a volume
Find substitution questions, with mark schemes, among the Pure 4 integration exam questions. The Pure 4 integration worksheet and the integration revision engine give more practice.
Substitution generalises the reverse chain rule. For products that do not suit it, use integration by parts.
Frequently asked questions
When should I use integration by substitution?
When an integral contains a function and something close to its derivative, or when a bracket or root can be simplified by naming it u. In exams, the substitution is usually given.
Do I need to change the limits?
For a definite integral, either change the limits to u-values or substitute back to x before using the original limits. Changing the limits is usually quicker.
What do I do with a leftover x?
Rearrange your substitution to write x in terms of u, for example x = u − 2 when u = x + 2.