Integration by Substitution: A Step-by-Step Guide (A-Level & AP)

Quick answer: Integration by substitution reverses the chain rule. Spot a function-inside-a-function whose inner derivative is also sitting in the integrand, let u equal that inner function, replace dx using du = (du/dx) dx so nothing in x is left, then integrate in u. For a definite integral, change the limits to u-values too — forgetting that is the single most common lost mark.

When I first met substitution in A-Level Pure, I could do the algebra fine and still drop marks — because I changed the integrand but left the limits exactly as they were. That one habit costs more marks than the actual calculus does. So this is the guide I wish I'd had: how to see when substitution is the move, how to pick u, the dx step everyone rushes, and the limits trap. It's the same method for Cambridge and Edexcel/AQA/OCR A-Level, and for AP Calculus AB and BC.

When substitution is the right move

The trigger: a composite function — something inside a bracket, root, exponent, or trig function — with the derivative of that inside part also present as a factor, give or take a constant.

Two you'll recognise instantly:

• ∫ x (x^2 + 1)^4 dx — the inside is x^2 + 1, and its derivative 2x is basically the x sitting out front.
• ∫ cos(x) e^(sin x) dx — inside the exponent is sin x, and its derivative cos x is right there next to it.

If the derivative of the inner function is nowhere in the integrand, substitution usually won't clear the x's and you need a different method (parts, a standard result, and so on). That single check — "is the inner derivative here?" — is how I decide in about five seconds.

Step 1: choose u

Let u be the inner function — the thing wrapped inside. In ∫ x (x^2 + 1)^4 dx the bracket is raised to a power, so u = x^2 + 1. Pick the messiest inside part; if choosing u kills the awkward bit, you've chosen well.

Step 2: change dx (the step everyone rushes)

This is where marks leak. Once you have u, differentiate it:

u = x^2 + 1 gives du/dx = 2x, so du = 2x dx.

Now rearrange to swap out dx. You often only have a single x, so isolate exactly what you need: from du = 2x dx you get x dx = du/2. The goal is a new integral with no x anywhere — if an x survives, you haven't finished substituting.

Constants are fine to juggle (that /2), but you can never absorb a variable this way. If the leftover factor is a variable that isn't part of du, substitution as-is won't work.

Worked example 1 — indefinite integral

Find ∫ x (x^2 + 1)^4 dx.

1. Choose u. u = x^2 + 1. (inner function)
2. Differentiate. du/dx = 2x, so x dx = du/2. (change dx)
3. Substitute. ∫ x (x^2 + 1)^4 dx = ∫ (x^2 + 1)^4 · x dx = ∫ u^4 · (du/2) = (1/2) ∫ u^4 du. (all in u now)
4. Integrate. (1/2) · u^5/5 = u^5/10. (one step)
5. Back to x, add C. = (x^2 + 1)^5 / 10 + C.

Check by differentiating: (1/10) · 5(x^2 + 1)^4 · 2x = x (x^2 + 1)^4. Straight back to where we started, so it's right. Differentiating your answer is a free check — I do it every single time.

Worked example 2 — definite integral, and the limits trap

Evaluate ∫ (from 0 to 1) 2x (x^2 + 1)^3 dx.

1. Choose u and change dx. u = x^2 + 1, du = 2x dx — and notice 2x dx is exactly what's already in the integral.
2. Change the limits. This is the step people forget. The 0 and 1 are x-values; once you're in u you need u-values. When x = 0, u = 0^2 + 1 = 1. When x = 1, u = 1^2 + 1 = 2.
3. Rewrite fully in u. ∫ (from u = 1 to u = 2) u^3 du.
4. Integrate and evaluate. [u^4/4] from 1 to 2 = 16/4 - 1/4 = 15/4.

If you had left the limits as 0 and 1 and plugged those into u^4/4, you'd get 1/4 - 0 = 1/4 — completely wrong, and on most A-Level mark schemes that's a lost accuracy mark even though the integration itself was perfect.

There are two safe ways to handle a definite integral, and you must pick one and stay consistent:

• Change the limits to u-values (as above) and evaluate straight away — do not convert back to x.
• Or keep the x-limits, integrate, substitute u back to x first, then put in 0 and 1: (x^2 + 1)^4 / 4 from 0 to 1 = 16/4 - 1/4 = 15/4. Same answer.

Mixing them — a u-limit with an x-expression — is the classic disaster. In the exam I always change the limits and write the new ones right next to the integral sign, so there's no temptation to sub back. If you want to see the limit change laid out cleanly, run one through Math solver and it shows the du line and the new limits explicitly.

What the mark scheme is really rewarding

Across Cambridge and Edexcel/AQA/OCR A-Level, a substitution question usually splits into a mark for a correct du (the dx change), a mark for a fully-in-u integrand, a mark for integrating, and — on a definite integral — a mark for the correct new limits or correct back-substitution. AP Calculus AB and BC score u-substitution the same way on free-response: the antiderivative and the correctly-applied bounds each carry points. Everywhere, examiners pay for the process — so write the du line and the limit change out explicitly, even when you could do them in your head.

Test yourself

Try these, then check your working line by line, not just the final answer:

1. Find ∫ 6x (3x^2 + 5)^3 dx. (indefinite)
2. Evaluate ∫ (from 0 to 2) x (x^2 + 4)^2 dx. (change the limits!)
3. Find ∫ cos(x) (sin x)^4 dx. (spot the inner function)

Quick answers: (1) u = 3x^2 + 5, du = 6x dx, so ∫ u^3 du = u^4/4 = (3x^2 + 5)^4 / 4 + C. (2) u = x^2 + 4, so x dx = du/2; x = 0 gives u = 4 and x = 2 gives u = 8; (1/2) ∫ (from 4 to 8) u^2 du = (1/6)(512 - 64) = 224/3. (3) u = sin x, du = cos x dx, so ∫ u^4 du = (sin x)^5 / 5 + C.

Want each line marked in your board's style? Paste your working into Math solver to check every substitution step, or mark my answer to see exactly where a method mark would drop. To drill the "spot the u" reflex fast, generate a set with the quiz tool. Ten of these and it stops feeling like guesswork.

And if you are aiming at a specific grade, see the GCSE maths pass mark and boundaries.

FAQ

How do I know which part to make u?

Pick the inner function — whatever sits inside a bracket, root, power, exponent, or trig function — provided its derivative also appears in the integrand. If choosing u removes the ugliest part and the leftover matches du up to a constant, you've chosen right.

Do I always have to change the limits?

Only for definite integrals, and only if you evaluate in terms of u. The alternative is to substitute u back to x first and use the original x-limits. Either works; just never mix a u-limit with an x-expression.

What if the derivative of my inner function isn't there?

Then plain substitution won't clear the x's. Check whether you're only missing a constant (fine — juggle it) or a variable factor (not fine). If a variable is genuinely missing, reach for integration by parts or a standard result instead.

Is u-substitution (AP) the same as integration by substitution (A-Level)?

Yes — different name, identical method. AP Calculus AB and BC call it u-substitution; Cambridge and UK A-Level call it integration by substitution. The steps, and the rule about changing the limits, are exactly the same.

In short: substitution is the reverse chain rule — let u be the inner function, use du = (du/dx) dx to replace dx until no x remains, integrate in u, and for a definite integral change the limits to u-values instead of leaving them in x. Nail the dx step and the limits, and these turn into some of the most reliable marks in the whole Pure paper.