IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~12 min read

Integration by Substitution

Substitution is reverse chain rule made formal — when you can’t spot f′(g(x)) · g′(x) by inspection, you set u = g(x), change variables, and let the algebra reveal the pattern. Two pitfalls dominate this topic: forgetting to change the limits when the integral is definite, and forgetting to substitute back to x when it’s indefinite. Get those right and substitution unlocks a huge class of integrals that would otherwise be intractable.

📘 What you need to know

Substitution = reverse chain rule, written out. Set u = g(x), find du/dx, rewrite the entire integral in u, integrate, then either back-substitute (indefinite) or use u-limits (definite).
Choosing u: look for the inner function. For composite f(g(x)), choose u = g(x) (the inside). For quotients, u = denominator often works. For powers and roots, u = expression in brackets / under the root.
Differential relationship: du/dx = g′(x), so du = g′(x) dx. Treat du/dx like a fraction — it’s a substitution shortcut that always gives the right answer.
For DEFINITE integrals: ALWAYS change the limits. If x runs from a to b, then u runs from g(a) to g(b). After changing limits, you never need to substitute back to x.
For INDEFINITE integrals: ALWAYS back-substitute. Your final answer must be in terms of the original variable x, plus c. Leaving the answer in u costs marks.
If a stray x remains after substitution, solve u = g(x) backwards to express x in terms of u. Common with substitutions like u = 2x+1 inside an integrand x·√(2x+1).
“Adjust and compensate”: if du = k·g′(x) dx but the integrand has only g′(x) dx, multiply outside the integral by (1/k) to compensate.
Substitution often equals f′/f spotted formally. If you saw the f′/f pattern from the previous note, substitution gives the same answer — it’s just the long-form derivation when the pattern isn’t obvious.

The change-of-variable formula

Integration by substitution — formula booklet ∫ f(x) dx = ∫ f(x) dxdu du where u = g(x), dudx = g′(x)

In practice you don’t write the formula like this — you just identify u, compute du/dx, and rearrange to swap every x and dx for u and du. The example below shows the full effect: an integral that looks awkward in x becomes a clean evaluation in u.

The integrand y = x/√(x²+1) looks unfriendly in x, but substitution u = x² + 1 reduces it to ½ · 1/√u, which integrates to √u. Limits transform from [0, √3] in x to [1, 4] in u, giving the clean integer 1. The same area, in two different parameterizations — that’s what substitution preserves.

Substitution and f′/f are the same idea in disguise. The integral ∫ x/(x²+1) dx is f′/f with f = x²+1 (almost — numerator is half of f′). The integral ∫ x/√(x²+1) dx is the same setup but the outer function is √u instead of 1/u. Substitution handles BOTH; f′/f is just the shortcut for the specific case where the outer function is 1/u.

Two flavours of substitution

Substitution given in the question

“using u = 1+2x, evaluate …”

Follow the 5-step procedure mechanically. Even if you’d choose a different u, USE the one given — mark schemes mark on the given substitution.

Spotting your own substitution

“find ∫ (ln x)² / x dx“

No hint given. Look for an inner function whose derivative also appears in the integrand. Here ln x has derivative 1/x — both pieces are present, so u = ln x.

🧭 Recipe — integration by substitution (the 5 steps)

Name the integral and identify u: name it I to save rewriting. If the question gives u, use it. Otherwise pick u = inner function, denominator, expression under root, or argument of a composite.
Differentiate to find du/dx, then rearrange to express dx in terms of du (and possibly x if needed). Treat du/dx as a fraction for this manipulation.
For DEFINITE integrals, change the limits. Substitute the original x-limits into u = g(x) to get u-limits. From now on the integral lives entirely in u — no need to back-substitute.
Rewrite the integral entirely in u. Substitute every x-expression. If a stray x remains (common when u = ax+b inside x·f(ax+b)), solve x = (something in u) and substitute.
Integrate, then finish: for indefinite, integrate in u and substitute back to x, adding +c. For definite, integrate in u and apply the u-limits directly — no +c, no back-substitution.

Worked examples

WE 1

Given substitution — polynomial in a power

Use the substitution u = x² + 3 to find ∫ x(x² + 3)⁴ dx.

Step 1 — name and identify I = ∫ x/(x² + 3)⁴ dx; let u = x² + 3 Step 2 — differentiate and rearrange du/dx = 2x ⟹ du = 2x dx ⟹ x dx = du/2 Step 3 — limits? indefinite — skip Step 4 — rewrite entirely in u I = ∫ 1/u⁴ · (du/2) = (1/2) ∫ u⁻⁴ du Step 5 — integrate, back-substitute, add +c = (1/2) · u⁻³/(-3) + c = -1/(6 u³) + c = -1/(6 (x² + 3)³) + c I = − 16(x² + 3)³ + c verify: d/dx[-1/(6(x²+3)³)] = -(1/6)(-3)(x²+3)⁻⁴(2x) = x/(x²+3)⁴. ✓ note how the x in the integrand pairs cleanly with du = 2x dx — that’s why u = x² + 3 was a good choice.

WE 2

Spot your own substitution — log inner function

Find ∫ (ln x)²x dx.

Spotting u — what’s the inner function whose derivative also appears? (ln x)² is composite — inner function is ln x derivative of ln x is 1/x — also in the integrand! ⟹ let u = ln x Step 2 — differential du/dx = 1/x ⟹ du = (1/x) dx Step 4 — rewrite in u I = ∫ (ln x)² · (1/x) dx = ∫ u² du Step 5 — integrate, back-substitute, add +c = u³/3 + c = (ln x)³/3 + c ∫ (ln x)²/x dx = (ln x)³3 + c classic example of “look for the inner function whose derivative is hiding in plain sight”. The (ln x)² and the 1/x in the integrand aren’t accidentally paired — they’re SET UP for substitution. Examiners love this pattern.

WE 3

Trig substitution — sin³ · cos with u = sin x

Find ∫ sin³(x) cos(x) dx.

Step 1 — spot the inner function sin³x is composite — inner is sin x derivative of sin x is cos x — present in the integrand! ⟹ let u = sin x Step 2 — differential du/dx = cos x ⟹ du = cos x dx Step 4 — rewrite in u I = ∫ sin³x · cos x dx = ∫ u³ du Step 5 — integrate, back-substitute = u⁴/4 + c = sin⁴(x)/4 + c ∫ sin³x cos x dx = sin⁴x4 + c general principle for trig substitution: when sin and cos appear together with one as the “outer” factor and one as the “inner”, set u equal to the inner. Same problem with ∫ cos³x sin x dx would give u = cos x and answer -cos⁴x/4 + c (mind the minus from du = -sin x dx).

WE 4

Definite with limit change — sin × e^(cos x)

Evaluate ∫₀^π/2 sin(x) e^{cos x} dx, giving your answer in exact form.

Step 1 — identify u e^(cos x) is composite — inner is cos x derivative of cos x is -sin x — present (up to sign)! ⟹ let u = cos x Step 2 — differential (mind the minus) du/dx = -sin x ⟹ du = -sin x dx ⟹ sin x dx = -du Step 3 — change limits (this is a DEFINITE integral) x = 0: u = cos(0) = 1 x = π/2: u = cos(π/2) = 0 Step 4 — rewrite in u (note new limits and the -du) ∫ from 0 to π/2 of sin x · e^(cos x) dx = ∫ from 1 to 0 of e^u · (-du) = ∫ from 0 to 1 of e^u du (flip limits, absorb the minus) Step 5 — integrate and evaluate = [e^u] from 0 to 1 = e – 1 ∫ from 0 to π/2 of sin x · e^(cos x) dx = e − 1 two cleanup tricks worth knowing: (1) when du has a minus sign, you can either keep the – and integrate, or flip the limits and absorb the minus — both work, flipping is often cleaner; (2) limits going from larger to smaller (here 1 → 0) flip naturally when the minus sign is absorbed.

WE 5

x in terms of u required — stray factor

Find ∫ x √(2x + 1) dx using u = 2x + 1.

Step 1 — set up u u = 2x + 1 Step 2 — differential and dx du/dx = 2 ⟹ dx = du/2 Step 4 — the stray x problem After substituting √(2x+1) → √u and dx → du/2, we still have x left over — must express x in terms of u from u = 2x+1: x = (u-1)/2 ⟹ I = ∫ (u-1)/2 · √u · (du/2) = (1/4) ∫ (u-1)√u du Step 5a — expand and integrate (u-1)√u = u^(3/2) – u^(1/2) I = (1/4) ∫ (u^(3/2) – u^(1/2)) du = (1/4) [(2/5) u^(5/2) – (2/3) u^(3/2)] + c = (1/10) u^(5/2) – (1/6) u^(3/2) + c Step 5b — back-substitute and factor = (1/10)(2x+1)^(5/2) – (1/6)(2x+1)^(3/2) + c factor out (2x+1)^(3/2): common denominator 30 = (2x+1)^(3/2) [(3(2x+1) – 5)/30] + c = (2x+1)^(3/2) (6x – 2)/30 + c = (3x – 1)(2x+1)^(3/2) / 15 + c ∫ x √(2x+1) dx = (3x − 1)(2x + 1)^3/215 + c the unfactored form (1/10)(2x+1)^(5/2) – (1/6)(2x+1)^(3/2) + c is also fully correct — examiners accept either. Factor only if “simplify” is asked. The key technique here is solving u = 2x+1 backwards to get x = (u-1)/2 — needed whenever the integrand has a stray x not consumed by du.

WE 6

Definite — spot u from the structure of the integrand

Evaluate ∫₀^√3 x√(x² + 1) dx.

Spotting u — inner function under the root u = x² + 1 (inside the square root) du/dx = 2x — matches the 2x part of the numerator (numerator is x = (1/2)·2x) Step 2 — differential du = 2x dx ⟹ x dx = du/2 Step 3 — change limits x = 0: u = 0 + 1 = 1 x = √3: u = 3 + 1 = 4 Step 4 — rewrite in u (no stray x — clean) ∫ from 0 to √3 of x/√(x²+1) dx = ∫ from 1 to 4 of (1/√u) · (du/2) = (1/2) ∫ from 1 to 4 of u^(-1/2) du Step 5 — integrate and apply u-limits = (1/2) [2 u^(1/2)] from 1 to 4 = [√u] from 1 to 4 = √4 – √1 = 2 – 1 = 1 ∫ from 0 to √3 of x/√(x²+1) dx = 1 corresponds to the SVG above — the area under x/√(x²+1) from 0 to √3 is exactly 1. The clean integer answer is the examiner’s hint that substitution is the right tool. Limits chosen so √(u) at u=4 gives the nice integer 2.

💡 Top tips

Use the given substitution even if you’d choose differently — mark schemes follow the question’s u. If the question gives one, USE it; if not, pick the most obvious inner function.
For definite integrals, change limits FIRST, then work in u. This avoids the very common mistake of forgetting to back-substitute or applying x-limits to a u-expression.
For indefinite integrals, ALWAYS back-substitute to x. Leaving the answer in u costs marks, and the +c still belongs in the final answer.
The “stray x” trick: if there’s a leftover x after substituting dx, solve u = g(x) for x in terms of u. Don’t try to skip this — guessing leads to wrong answers.
Check by differentiating. The chain rule should reproduce the integrand exactly. If it doesn’t, your substitution or differential has an error.

⚠ Common mistakes

Forgetting to change limits in definite integrals: leaving x-limits on an integral that’s now in u gives the wrong answer. Either change the limits (recommended), or do indefinite first and apply x-limits at the end after back-substituting.
Leaving the final answer in u for an indefinite integral. The original variable is x; the answer must be too.
Mishandling the differential: forgetting the constant from du = k dx. If du = 2x dx and you have x dx in the integrand, you need du/2, not du.
Stray-x blindness: substituting incompletely and leaving an x mixed in with u‘s. The integral must be 100% in u before you integrate — solve x in terms of u if needed.
Reaching for substitution when f′/f would do. ∫ 2x/(x²+5) dx = ln(x²+5) + c in one line; the full substitution procedure gives the same answer but takes four. Pattern-recognition first, substitution only when needed.

Up next: Integration by Parts. This is the reverse-product-rule companion to reverse-chain-rule (substitution). When the integrand is a product of two functions, and neither is the derivative of the other (so substitution doesn’t help), integration by parts is the tool. You’ll learn to pick u and dv/dx (note the new meaning of u!), apply the parts formula ∫u dv = uv − ∫v du, and sometimes repeat the process. Some tricky integrals require trapping the original integral on both sides and solving algebraically.

Need help with Calculus?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →