IB Maths AA HL
Topic 5 — Calculus
Paper 1 & 2
HL
~10 min read
Integration by Substitution
When reverse chain rule is hard to spot or you need to handle a more awkward integrand, substitution gives a step-by-step algorithm. Pick an inner function and call it u; differentiate to get du; replace EVERYTHING (the integrand AND the dx, AND the limits if it’s a definite integral) in terms of u. The result is usually a simpler integral you already know how to do.
📘 What you need to know
- Pick u = inside function — usually the inside of a composite (bracket, denominator, exponent, root, trig argument).
- Differentiate to get du: du/dx = g′(x) → du = g′(x) dx. Rearrange to substitute for dx in the integral.
- Replace EVERYTHING in the integrand by u and du. No x should remain after substitution.
- For definite integrals, change the limits too: x = a → u = g(a); x = b → u = g(b).
- Integrate in u, then either substitute x back (indefinite) or evaluate at u-limits (definite).
- Sometimes you need to rearrange for x: e.g. if u = 2x − 1, then x = (u + 1)/2 — useful when the integrand has an extra x outside the composite.
- Substitution always works where reverse chain rule works — slower, but more systematic.
- GDC can evaluate definite integrals directly, but for “show that” or working-required questions you must use substitution by hand.
The substitution formula
Indefinite — substitution
∫ f(g(x)) · g′(x) dx = ∫ f(u) du, where u = g(x)
Definite — substitution with new limits
∫x=ax=b f(g(x)) · g′(x) dx = ∫u=g(a)u=g(b) f(u) du
Mental shortcut: treat du/dx like a fraction. If du/dx = 6x², then “multiplying by dx” gives du = 6x² dx. Rearrange as needed: x² dx = du/6. (Mathematicians get squeamish about this but it works.)
🧭 Recipe — integration by substitution
- Identify u: pick the inside function (usually inside a bracket, root, exponent, or denominator).
- Differentiate: du/dx = g′(x) → write du = g′(x) dx, then rearrange for dx.
- Substitute everything: replace each x-expression by its u-equivalent. If x remains in the integrand, solve u = g(x) for x and substitute.
- For definite integrals, change the limits: new lower limit = g(a), new upper limit = g(b).
- Integrate in u, then back-substitute x (indefinite) or evaluate at the u-limits (definite).
Worked examples
WE 1Polynomial inside — clean substitution
Find ∫6x²(2x³ + 5)⁴ dx.
Step 1 — pick u = 2x³ + 5 (the inside function)
Step 2 — differentiate
du/dx = 6x² → du = 6x² dx
Step 3 — substitute everything
∫6x²(2x³ + 5)⁴ dx = ∫(2x³ + 5)⁴ · 6x² dx
= ∫u⁴ · du
Step 4 — integrate in u and back-substitute
∫u⁴ du = u⁵/5 + c
= (2x³ + 5)⁵/5 + c
∫6x²(2x³ + 5)⁴ dx = (2x³ + 5)⁵/5 + c
the 6x² dx in the integrand was EXACTLY du — perfect match, no compensation needed
WE 2Trig — substitute the inner trig function
Find ∫sin³(x) cos(x) dx.
Step 1 — pick u = sin x
(because sin³ x is a power of u, and cos x is its derivative)
Step 2 — differentiate
du/dx = cos x → du = cos x dx
Step 3 — substitute
∫sin³ x cos x dx = ∫(sin x)³ · cos x dx
= ∫u³ du
Step 4 — integrate and back-substitute
∫u³ du = u⁴/4 + c = sin⁴(x)/4 + c
∫sin³(x) cos(x) dx = (1/4) sin⁴(x) + c
the cos x dx “becomes” du. If the integrand had been sin³(x) · 2 cos(x), we’d factor out the 2 first
WE 3Solve for x back-substitution — extra x outside the root
Find ∫x √(2x − 1) dx.
Step 1 — pick u = 2x − 1
Step 2 — differentiate and solve for x
du/dx = 2 → du = 2 dx → dx = du/2
also: u = 2x − 1 → x = (u + 1)/2 ← need this because integrand has a loose x
Step 3 — substitute EVERYTHING (the loose x, the √(2x−1), and dx)
∫x √(2x − 1) dx = ∫ (u + 1)/2 · √u · du/2
= (1/4) ∫(u + 1) u^(1/2) du
= (1/4) ∫(u^(3/2) + u^(1/2)) du
Step 4 — integrate term by term
= (1/4) [(2/5) u^(5/2) + (2/3) u^(3/2)] + c
= (1/10) u^(5/2) + (1/6) u^(3/2) + c
Back-substitute u = 2x − 1
= (1/10)(2x − 1)^(5/2) + (1/6)(2x − 1)^(3/2) + c
∫x √(2x − 1) dx = (1/10)(2x − 1)^(5/2) + (1/6)(2x − 1)^(3/2) + c
the extra “x” in the integrand wasn’t part of the chain rule — we had to solve u = 2x − 1 for x to replace it
WE 4Exponential — substitute the exponent
Find ∫x² ex³ − 4 dx.
Step 1 — pick u = x³ − 4 (the exponent)
Step 2 — differentiate
du/dx = 3x² → du = 3x² dx → x² dx = du/3
Step 3 — substitute
∫x² e^(x³ − 4) dx = ∫e^u · (du/3)
= (1/3) ∫e^u du
Step 4 — integrate and back-substitute
(1/3) ∫e^u du = (1/3) e^u + c
= (1/3) e^(x³ − 4) + c
∫x² e^(x³ − 4) dx = (1/3) e^(x³ − 4) + c
the integrand has x² but we need 3x² for the substitution — so x² dx = du/3, and the 1/3 comes outside
WE 5Definite integral with limit change
Evaluate ∫01 3x²(x³ + 2)⁴ dx.
Step 1 — pick u = x³ + 2
Step 2 — differentiate
du/dx = 3x² → du = 3x² dx
Step 3 — change limits
when x = 0: u = 0 + 2 = 2
when x = 1: u = 1 + 2 = 3
Substitute (including the limits)
∫₀¹ 3x²(x³ + 2)⁴ dx = ∫₂³ u⁴ du
Step 4 — integrate and evaluate at the u-limits
∫₂³ u⁴ du = [u⁵/5]₂³
= 3⁵/5 − 2⁵/5
= 243/5 − 32/5
= 211/5
∫₀¹ 3x²(x³ + 2)⁴ dx = 211/5
no need to back-substitute u → x for a definite integral — just use the u-limits directly
WE 6Definite integral with trig substitution — exact value
Evaluate ∫0π/2 cos(x) sin²(x) dx, giving an exact answer.
Step 1 — pick u = sin x
Step 2 — differentiate
du/dx = cos x → du = cos x dx
Step 3 — change limits
when x = 0: u = sin 0 = 0
when x = π/2: u = sin(π/2) = 1
Substitute
∫₀^(π/2) cos(x) sin²(x) dx = ∫₀^(π/2) sin²(x) · cos(x) dx
= ∫₀¹ u² du
Step 4 — integrate and evaluate
∫₀¹ u² du = [u³/3]₀¹ = 1/3 − 0 = 1/3
∫₀^(π/2) cos(x) sin²(x) dx = 1/3
a GDC gives 0.3333… — recognise as 1/3. For “exact” questions, by-hand substitution is essential
💡 Top tips
- Pick u = the inside function — almost always inside brackets, in a denominator, in an exponent, or under a root.
- Treat du/dx like a fraction — multiply both sides by dx to free up du, then rearrange for dx as needed.
- For definite integrals, CHANGE THE LIMITS — saves time on back-substitution and is the cleanest approach.
- If a stray x remains after substituting, solve u = g(x) for x and substitute everywhere.
- GDC check — even when working is required, you can use the GDC to verify the numerical value of your final answer.
⚠ Common mistakes
- Forgetting to change the limits on a definite integral after substituting — if the limits are still x-values, you have to back-substitute to x before evaluating.
- Forgetting dx — every substitution must replace dx in terms of du. Not optional.
- Leaving an x in the u-integral — your integral after substitution must be PURELY in u. If x survives, you haven’t substituted everything (solve u = g(x) for x).
- Sign / coefficient error in du — for u = cos x, du = −sin x dx (with a MINUS sign). Many students lose marks here.
- Wrong choice of u — if your substitution doesn’t simplify the integral, undo it and try a different u (usually a deeper inside function).
Up next: Definite Integrals. We’ll consolidate the F(b) − F(a) workflow, look at the properties of definite integrals (constant factors, term-by-term, splitting intervals, swapping limits, horizontal translations), and use those properties to evaluate integrals without doing the antiderivative at all. Very useful for “given ∫f = 12, find ∫6f(x + 5) =”-type questions.
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