IB Maths AA SLTopic 5 — CalculusPaper 1 & 2~9 min read
Integration by Substitution
When reverse chain rule is too messy to spot, substitution always works. The idea: replace the awkward “inside” function with a single new variable u, integrate in u-world, then translate back. It’s slower than RCR but completely systematic — once you’ve done a few, the steps become automatic.
📘 What you need to know
Use substitution when the integrand is composite but RCR is hard to spot.
Pick u = the “inside” function (the secondary one in the composite).
Differentiate to find du = (du/dx) dx and replace every x term — including dx.
For definite integrals: either change the limits to u-values, or substitute x back in at the end.
On Paper 2, your GDC computes definite integrals directly — but you may still need to show the substitution working.
When (and why) to use substitution
RCR is the quick option when you can spot the pattern. Substitution is the slower, safer method that always works for the same kind of integrals.
Think of it this way: the integrand has an awkward “inside” function (like 3x² + 5x − 1). You give that whole inside a new name — u — and rewrite the entire integral so it only involves u. Suddenly it looks much simpler.
If you can spot RCR, use it — it’s faster. If you can’t, switch to substitution and trust the steps. Both methods give the same answer.
The substitution flow
x-world → u-world → integrate → back
The five-step method
Substitution recipe
Pick u. Let u = inside function — usually the “messy” bracket.
Differentiate u. Find du/dx, then rearrange to du = (…)dx.
Replace everything. Every x, every dx — gone. Only u and du remain. (For definite integrals, also change the limits.)
Integrate in u. Should now be a simple standard integral.
Translate back — sub u back to x (indefinite), or plug in u-limits (definite). Add + c if indefinite.
Five steps, in order. The trickiest is step 3 — every single x must vanish. If you still see x’s after step 3, you’ve either picked the wrong u or missed a substitution.
Definite integrals: change the limits
For definite integrals, you have a choice:
Option A — change the x-limits into u-limits, then evaluate directly in u-world (no need to translate back).
Option B — integrate in u, translate back to x, then plug in the original x-limits.
Option A is usually faster. Just remember to actually change the limits.
Changing limits — example
x-limits
x = 1
x = 2
→
u-limits
u = 3(1)² + 5(1) − 1 = 7
u = 3(2)² + 5(2) − 1 = 21
whatever u equals, plug in each x-limit to get the matching u-limit
📍
Paper 2 GDC shortcut
If it’s a calculator paper and the question doesn’t demand working, your GDC will compute the definite integral directly — no substitution needed. Use this to check your answer even when working is required.
Worked examples
WE 1
A classic — same one we did with RCR
Find ∫ 2x cos(x²) dx using substitution.
step 1 — pick uLet u = x² (the inside function)step 2 — differentiatedu/dx = 2x → du = 2x dxstep 3 — replace∫ cos(x²) · (2x dx) = ∫ cos u dustep 4 — integrate= sin u + cstep 5 — translate back= sin(x²) + csame answer as RCR — substitution is just a different route to the same place!
WE 2
Awkward fraction — substitution wins
Find ∫6x + 5(3x² + 5x − 1)3 dx.
step 1 — pick ucomposite is (3x² + 5x − 1)³let u = 3x² + 5x − 1step 2 — differentiatedu/dx = 6x + 5 → du = (6x + 5) dxstep 3 — replaceI = ∫ du / u³ = ∫ u⁻³ dustep 4 — integrate= u⁻²/(−2) + c = −1/(2u²) + cstep 5 — back to xI = −1 / [2(3x² + 5x − 1)²] + cnotice how (6x + 5) was exactly du — no adjusting needed!
substitution as beforeu = 3x² + 5x − 1, du = (6x + 5) dxchange the limitsx = 1 → u = 3 + 5 − 1 = 7x = 2 → u = 12 + 10 − 1 = 21integral becomesI = ∫(7 to 21) u⁻³ du = [−½u⁻²] from 7 to 21evaluate= −½(1/441) − (−½(1/49))= −1/882 + 9/882 = 8/882I = 4/441limits change with u — never mix x-limits with a u-integrand!
WE 4
When du needs scaling
Find ∫x(x² + 3)5 dx.
step 1 — pick ulet u = x² + 3step 2 — differentiatedu/dx = 2x → du = 2x dxwe have x dx (not 2x dx) → use ½ dux dx = ½ dustep 3 — replaceI = ∫ u⁵ · (½ du) = ½ ∫ u⁵ dustep 4 — integrate= ½ · u⁶/6 = u⁶/12step 5 — back to xI = (x² + 3)⁶ / 12 + cif du doesn’t match exactly, scale it — solve “x dx = ?” in terms of du.
💡 Top tips
u = the messy bracket. Almost always, u is the inside of a composite — the thing raised to a power, inside a sin/cos, or in an exponent.
Treat du/dx like a fraction. Multiply by dx to get du = (…)dx. It’s not technically a fraction, but it works here.
Every x must vanish. If x’s are still there after step 3, your u was wrong.
Definite integrals: change the limits. Saves you from translating back at the end.
Check on GDC — for definite integrals, your GDC gives the numerical answer instantly.
RCR vs substitution: same answer, different route. Use whichever is faster for that question.
⚠ Common mistakes
Forgetting to replace dx. dx must become du (or some multiple of du) — leaving “dx” in a u-integral is wrong.
Mixing x-limits with a u-integrand. If your integrand is in u, the limits must be in u too.
Picking the wrong u. If you can’t make every x disappear, your u was wrong — try the inside of a different bracket.
Bad scaling. If du = 6x dx but you only have x dx, then x dx = (1/6) du — not 6 du.
Forgetting + c on indefinite integrals.
Not translating back. For an indefinite integral, the final answer must be in terms of x, not u.
Next up: Definite Integrals — the Fundamental Theorem of Calculus, the properties (split, swap, scale), and how to use them to save time on Paper 1.
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