Spot the 'inside' and its derivative: The chain rule turns f(inside) into f'(inside) × (inside)'. Substitution runs that backwards.
When an integral has an inner function u sitting next to (a multiple of) its own derivative, let u = that inner function. Then du = u' dx, so u' dx becomes du, and the whole thing collapses to an easy integral in u.
Picture peeling an onion: name the inside u, and the outer layer matches du.
IB-style question — power of an inner function
A curve has gradient given by an expression involving (x² + 1).
Find ∫ 2x(x² + 1)⁴ dx.
Step by step
- The inside is x² + 1, and 2x is exactly its derivative — let u = x² + 1.
- Replace x² + 1 by u and 2x dx by du — the integral becomes a clean power.
- Integrate using the power rule (add 1 to the index, divide by it).
- Swap u back to the original expression.
Final answer
∫ 2x(x² + 1)⁴ dx = (x² + 1)⁵/5 + C.
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Two ways to finish — prefer changing the limits: For a definite integral you can either substitute back to x and use the original x-limits, OR — cleaner — convert the limits to u-values and evaluate entirely in u.
When x = a, work out u; when x = b, work out u. Then you never have to switch back.
IB-style question — trig substitution with new limits
A region is described by the integral below, where the inside is sin x.
Evaluate ∫₀π/2 sin³x cos x dx.
Step by step
- The inside is sin x and cos x is its derivative — let u = sin x.
- Change the limits: x = 0 → u = sin 0 = 0; x = π/2 → u = sin(π/2) = 1.
- Rewrite the whole integral in u (sin³x = u³, cos x dx = du) with the new limits.
- Integrate and evaluate — no need to switch back to x.
Final answer
∫₀π/2 sin³x cos x dx = 1/4.
IB-style question — a linear inner function
Evaluate ∫₀¹ (2x + 1)³ dx.
Step by step
- Let u be the bracket; its derivative is the constant 2.
- Change limits: x = 0 → u = 1; x = 1 → u = 3.
- Pull the 1/2 out and integrate u³.
- Evaluate: u⁴ goes 81 then 1.
Final answer
∫₀¹ (2x + 1)³ dx = 10.