Integration by substitution

Integration by substitution means replacing part of an integral with a new variable, u, so the integral becomes one you can do. You choose u, find du in terms of dx, rewrite everything in u, integrate, then substitute back. It is the reverse of the chain rule. Below are the method, six worked examples and a set of integration by substitution questions with answers.

How to integrate by substitution

1. Choose u. Usually the inside of a bracket, root or power, or a function whose derivative also appears. Exam questions often tell you the substitution.
2. Differentiate. Find du/dx, then rearrange to get dx in terms of du.
3. Rewrite the integral in u. Every x must go, including any left over outside the bracket: use the substitution to write x in terms of u.
4. Integrate with respect to u.
5. Substitute back to give the answer in x, and add + C. (For a definite integral, change the limits instead; see Example 3.)

Why it works: if F is an integral of f, the chain rule gives d/dx[F(u)] = f(u) × du/dx. So integrating f(u) du/dx with respect to x gives F(u) + C, and the substitution just makes that pattern visible.

Worked examples

Example 1: the derivative is already there

1. Let u = x³ − 2. Then du/dx = 3x², so dx = du ÷ 3x².
2. Substitute: ∫ 6x² u⁵ × du ÷ 3x² = ∫ 2u⁵ du. The x² cancels.
3. Integrate: 2u⁶ ÷ 6 = u⁶ ÷ 3.
4. Substitute back: (x³ − 2)⁶ ÷ 3 + C.

When the derivative of the inside is already there, as here, you can also spot the answer with the reverse chain rule. Substitution is the method that still works when it is not.

Example 2: an x left over

1. Let u = x + 1, so du/dx = 1 and dx = du.
2. The x outside the root must also be replaced: x = u − 1.
3. The integral becomes ∫ (u − 1)u^1/2 du = ∫ (u^3/2 − u^1/2) du.
4. Integrate: (2/5)u^5/2 − (2/3)u^3/2.
5. Substitute back: (2/5)(x + 1)^5/2 − (2/3)(x + 1)^3/2 + C.

Example 3: a definite integral, changing the limits

1. Let u = x + 1, so dx = du and x = u − 1.
2. Change the limits: when x = 0, u = 1; when x = 3, u = 4.
3. The integral becomes ∫ from 1 to 4 of (u − 1) ÷ u^1/2 du = ∫ from 1 to 4 of (u^1/2 − u^−1/2) du.
4. Integrate: [(2/3)u^3/2 − 2u^1/2] from 1 to 4.
5. At u = 4: (2/3)(8) − 2(2) = 16/3 − 4 = 4/3. At u = 1: 2/3 − 2 = −4/3.
6. Answer: 4/3 − (−4/3) = 8/3.

Because the limits are now values of u, there is no need to substitute back. You can check definite integrals like this one in the Integral Calculator, which also shades the area.

Example 4: an exponential

1. Let u = eˣ + 1, so du/dx = eˣ and dx = du ÷ eˣ.
2. The integral becomes ∫ (eˣ ÷ u) × du ÷ eˣ = ∫ 1/u du = ln|u|.
3. Since eˣ + 1 is always positive: ln(eˣ + 1) + C.

Example 5: a substitution given in the question

Use the substitution x = sin θ to find the exact value of:

1. x = sin θ, so dx/dθ = cos θ and dx = cos θ dθ.
2. Limits: x = 0 gives θ = 0; x = 1/2 gives θ = π/6.
3. √(1 − sin²θ) = √(cos²θ) = cos θ, which is positive for θ between 0 and π/6.
4. The integral becomes ∫ from 0 to π/6 of (cos θ ÷ cos θ) dθ = ∫ from 0 to π/6 of 1 dθ.
5. Answer: π/6.

Example 6: a logarithm inside

1. Let u = ln x, so du/dx = 1/x and dx = x du.
2. Limits: x = e gives u = 1; x = e² gives u = 2.
3. The integral becomes ∫ from 1 to 2 of (1 ÷ (x u)) × x du = ∫ from 1 to 2 of 1/u du.
4. Answer: ln 2 − ln 1 = ln 2.

Practise the basic integration rules these examples finish with in this question from the Calculus course.

How to choose u

Integration by substitution questions

Try these, then check your answers below. Questions 1 to 3 are indefinite, so include + C.

1. Find ∫ 4x(2x² + 1)³ dx.
2. Use u = x + 3 to find ∫ x(x + 3)⁵ dx.
3. Use u = tan x to find ∫ sec²x tan³x dx.
4. Use u = 1 − x to find the exact value of ∫ from 0 to 1 of x(1 − x)⁴ dx.
5. Use u = x − 1 to find the exact value of ∫ from 1 to 5 of x√(x − 1) dx.
6. Use u = x² to find the exact value of ∫ from 0 to 2 of x e^x² dx.

Answers

1. u = 2x² + 1, du = 4x dx: ∫ u³ du = (2x² + 1)⁴ ÷ 4 + C.
2. x = u − 3: ∫ (u − 3)u⁵ du = u⁷/7 − u⁶/2, so (x + 3)⁷/7 − (x + 3)⁶/2 + C.
3. du = sec²x dx: ∫ u³ du = tan⁴x ÷ 4 + C.
4. x = 1 − u, dx = −du, limits 1 to 0: ∫ from 0 to 1 of (1 − u)u⁴ du = 1/5 − 1/6 = 1/30.
5. x = u + 1, limits 0 to 4: ∫ (u + 1)u^1/2 du = [(2/5)u^5/2 + (2/3)u^3/2] from 0 to 4 = 64/5 + 16/3 = 272/15.
6. du = 2x dx, limits 0 to 4: (1/2)∫ eᵘ du = (e⁴ − 1) ÷ 2.

Common mistakes

• Leaving some x in the integral. After substituting, only u (and du) should remain. Use x = u − 1, or similar, for any leftover x.
• Forgetting to replace dx. dx is not du unless du/dx = 1.
• Keeping the x limits. Either change the limits to u values, or substitute back to x before using the original limits.
• Forgetting + C on indefinite integrals.
• Dropping a minus sign when u = 1 − x, so dx = −du. Swapping the limits back cancels it.

Practise with a short timed paper

Integration by substitution is in the second year of AQA A-level Maths (H5), alongside integration by parts. The daily A-level Maths: Calculus paper is a short timed paper of 8 questions, with a new one every day.

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