Key Idea: This topic integrates harder functions in two ways: substitution for a function inside another, and integration by parts for products. Both are staples of the long Paper 1 calculus question.
Integration by substitution
- Choose u — u = the inner function.
- Swap dx — du = u′ dx. Balance a missing constant, e.g. x dx = ½ du.
- Integrate — Work only in u.
- Finish — Indefinite: put x back, + C. Definite: change the limits instead.
Integration by parts
- ∫ u dv = uv − ∫ v du.
- Choose u by LIATE: Log, Inverse-trig, Algebra, Trig, Exponential.
- A power xⁿ needs parts n times.
- ∫ ln x dx: u = ln x, dv = dx, giving x ln x − x + C.
If the new integral gets harder, swap your choice of u.
∫ x√(x² + 3) dx? ⅓(x² + 3)³/² + C.
∫₀¹ x eˣ dx? [eˣ(x − 1)] from 0 to 1 = 1.
In x sin x, which part is u? x (Algebra comes before Trig).
How many rounds of parts for x²eˣ? Two.