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Integration by substitution and by parts

The chain rule and the product rule each run backwards into a method. Substitution relabels an integral until it looks like something the shelf recognises, and parts trades one integral for a hopefully easier one, with the logarithm's own integral as its most famous conquest.

Year 12-13EDEXCEL 9MA0 8.5

Builds on Integrating standard functions and The product, quotient and chain rules.

IN THIS TOPIC

  • Spot reverse chain rule patterns, f'/f and f' times a power of f, at sight.
  • Run full substitutions, converting integrand, dx and limits together.
  • Integrate by parts, choosing which factor to differentiate, including ∫ln x dx.

WHAT YOU PROBABLY THINK

∫uv dx = ∫u dx × ∫v dx.

Reverse chain patterns

The chain rule leaves fingerprints. When an integrand contains a function and its derivative standing alongside, the integral undoes a chain in one step, and three patterns cover most exam appearances,

∫ f'(x)f(x) dx = ln|f(x)| + cIN THE FORMULAE BOOKLET

together with f' times a power of f reversing to the next power up, and the famous special case ∫tan x dx = −ln|cos x| + c, which is the first pattern with sin and cos named.

Three reverse chain rule patterns: 2x over x squared plus 1 integrates to the log of the bottom, sine cubed x cos x to a quarter of sine to the fourth, and tan x, being sine over cos, to minus log of cos2x/(x² + 1)ln(x² + 1) + cspot f’/fsin³x cos x¼ sin⁴x + cspot f’ fⁿtan x−ln|cos x| + cf’/f in itthe inside function's derivative, standing alongside, is the tell
FIG. 1The three tells: derivative-over-function makes a log, derivative-times-power reverses the chain, and tan x is the first pattern again.

WORKED EXAMPLE

A full substitution, limits and all

Use the substitution u = 2x + 1 to find ∫x√(2x + 1) dx.

From u = 2x + 1: x = (u − 1)/2 and dx = du/2, so the integral becomes ∫((u − 1)/2)√u (du/2) = ¼∫(u3/2 − u1/2) du.

Integrating: ¼((2/5)u5/2 − (2/3)u3/2) = u5/2/10 − u3/2/6.

Back in x: (2x + 1)5/2/10 − (2x + 1)3/2/6 + c.

Three things convert or nothing does: the integrand, the dx, and, on definite integrals, the limits. Forgetting the dx factor of ½ is this method's signature casualty.

YOUR TURN

Spot, do not substitute

Find ∫sin3 x cos x dx by inspection, before opening the working.

Show the working

cos x is the derivative of sin x, standing right beside its powers: the f' fⁿ pattern.

Reverse the chain: ¼ sin4 x + c, and differentiating back confirms it in one line.

A formal substitution u = sin x reaches the same place; inspection is the same route, travelled lighter.

Integration by parts

The product rule reverses into parts, printed in the booklet as

∫ u dvdx dx = uv − ∫ v dudx dxIN THE FORMULAE BOOKLET

a trade: one factor gets differentiated, the other integrated, and the original integral becomes a new one that had better be easier. Choose to differentiate the factor that simplifies, x becoming 1, ln x becoming 1/x. And the opening lie dies by counter example: integrating x and ex separately gives (x2/2)ex, whose derivative is nowhere near xex.

WORKED EXAMPLE

The standard parts question

Find ∫xex dx.

Differentiate the x (it simplifies to 1) and integrate the ex (it costs nothing): parts gives xex − ∫ex dx.

∫xex dx = (x − 1)ex + c.

The trade turned a product integral into the easiest integral in the course. When the roles are chosen the other way round, the new integral is harder than the old, which is the diagnostic for a wrong choice.

Integration by parts applied to the log: writing ln x as 1 times ln x, differentiating the log and integrating the 1 turns the integral into x ln x minus the integral of 1, which is x ln x minus x plus c∫ ln x dx = ∫ (1 × ln x) dxintegrate the 1 → xdifferentiate ln x → 1/x= x ln x − ∫ x × (1/x) dx = x ln x − ∫ 1 dx= x ln x − x + cthe specification asks for this one by name
FIG. 2The specification's named target: ∫ln x dx by parts, writing ln x as 1 × ln x. Differentiate the log, integrate the 1, and x ln x − x + c falls out.

TRY IT UNSEEN

The logarithm's own integral

Find ∫ln x dx.

Show the working

Write the integrand as 1 × ln x, then differentiate the ln x and integrate the 1.

Parts gives x ln x − ∫x × (1/x) dx = x ln x − ∫1 dx.

∫ln x dx = x ln x − x + c, and differentiating back, x/x + ln x − 1 = ln x, seals it.

The invisible factor of 1 is the whole trick, and the specification names this integral because the trick generalises to arcsin and friends beyond this course.

THE EXAM BIT

  • Scan for a function and its derivative side by side before reaching for substitution; the patterns are faster and carry the same marks.
  • In a substitution convert integrand, dx and limits together, and say so; unconverted limits are the classic definite-integral casualty.
  • In parts differentiate the factor that simplifies; if the new integral is harder, the roles were backwards.
  • ∫tan x dx = −ln|cos x| + c is quotable, and its one-line derivation from f'/f is worth knowing cold.
  • Check every antiderivative by differentiating; substitution and parts both reverse cleanly and instantly.

CHECK YOURSELF

Find the exact value of ∫01 xex dx.

Show a hint

Parts, then substitute both limits into (x − 1)ex.

Show the answer

Parts gives the antiderivative (x − 1)ex.

Evaluating: (1 − 1)e − (0 − 1) × 1 = 0 + 1 = 1.

An area of exactly 1 under xex from 0 to 1, and every step exact; the bracket at the lower limit is where the sign errors hide.

A function beside its own derivative is a reversed chain; relabel or just read it off.

Parts trades one integral for another: differentiate what simplifies, and 1 × ln x is fair game.

CHECK YOUR PROGRESS

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  • Spot reverse chain rule patterns, f'/f and f' times a power of f, at sight.
  • Run full substitutions, converting integrand, dx and limits together.
  • Integrate by parts, choosing which factor to differentiate, including ∫ln x dx.

No animated video for this topic yet; these notes stand alone.