A-Level H2 Mathematics: Integration Techniques — Reading Differentiation in Reverse

Why integration matters for the A-Level exam

If your child has already met differentiation, integration will feel familiar in one important way: integration is differentiation in reverse. Where differentiation asks "what is the rate of change of this function?", integration asks "which function would I have to differentiate to get this?". That single idea — running the rules backwards — sits behind almost every technique in the syllabus.

Integration is unavoidable in H2 Mathematics. It appears as standalone questions, but it also powers areas under curves, volumes of revolution, and large parts of the differential equations and Maclaurin series topics. In Paper 1 and Paper 2 a student who is fluent with integration can pick up easy marks quickly; a student who is shaky tends to lose marks across several questions at once. That is why we treat it as a high-priority skill rather than "just another chapter".

The good news is that the techniques are finite and learnable. Once your child can recognise which of the standard methods a question is asking for, most integrals become routine. The skill being tested is really pattern recognition: spotting the structure of the integrand and matching it to the right tool.

The core techniques

1. Standard integrals and the reverse chain rule

Every integration starts from a small set of standard results. Differentiation rules read backwards give us ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + c (for n ≠ −1), ∫ e^x dx = e^x + c, ∫ (1)/(x) dx = ln|x| + c, ∫ sin x dx = −cos x + c, and ∫ cos x dx = sin x + c.

The MF26 formula list gives the integrals of the less familiar functions — tan x, sec x, cosec x, cot x, 1/(a² + x²), 1/(a² − x²), 1/√(a² − x²) and so on, which produce inverse-trig and logarithm answers. Your child does not need to memorise those; they must, however, be able to read the list quickly and match the form. The basic polynomial, exponential and sine/cosine integrals above are not on MF26 and must be known cold.

The reverse chain rule handles two extremely common patterns. The first is the f'(x)/f(x) form: ∫ (f'(x))/(f(x)) dx = ln|f(x)| + c. So ∫ (2x)/(x² + 1) dx = ln(x² + 1) + c, because the top is exactly the derivative of the bottom. The second is the linear substitution shortcut: when the inside of a function is linear (ax + b), you integrate as normal then divide by a. For example ∫ e^3x+1 dx = (1)/(3) e^3x+1 + c, and ∫ (2x − 5)⁴ dx = ((2x − 5)⁵)/(10) + c. This only works for linear inside functions — a frequent source of errors when students apply it to non-linear cases.

2. Integration by substitution

When the integrand is more tangled, substitution lets you swap the variable for something simpler. The recipe is: let u equal a chosen part of the expression, find du/dx, replace dx in terms of du, and rewrite the whole integral in terms of u. Crucially, every x must disappear before you integrate.

In H2, the substitution to use is usually given in the question, so the skill is executing it cleanly rather than inventing it. For definite integrals, the smart move is to change the limits to u-values as you go, so you never have to substitute back. A typical worked line: for ∫ 2x √(x² + 1) dx with u = x² + 1, we get du = 2x dx, so the integral becomes ∫ √u du = (2)/(3) u^3/2 + c = (2)/(3)(x² + 1)^3/2 + c.

3. Integration by parts

Integration by parts is the reverse of the product rule, and the formula ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx is given on MF26. Use it when you are integrating a product of two different types of function, such as x e^x or x cos x.

The whole game is choosing which factor is u. A reliable guide is LIATE — pick u in this order of preference: Logarithm, Inverse trig, Algebraic (powers of x), Trigonometric, Exponential. Whatever comes first in that list becomes u (the part you differentiate); the rest is dv/dx (the part you integrate). The aim is to make u simpler when differentiated, so the new integral ∫ v (du/dx) dx is easier than the one you started with.

Two special cases are worth knowing. First, to integrate ln x on its own, write it as 1 × ln x and take u = ln x, dv/dx = 1. Second, the "by parts twice" / recurring-integral trick: for integrals like ∫ e^x cos x dx, applying by parts twice brings back a multiple of the original integral. You then treat the original integral as an unknown, call it I, and solve the resulting equation algebraically for I.

4. Integrating using partial fractions

A fraction with a factorisable polynomial denominator is usually impossible to integrate directly, but easy once split. Partial fractions rewrite something like (1)/((x − 1)(x + 2)) as (A)/(x − 1) + (B)/(x + 2). Each simple piece then integrates to a logarithm. The cover-up rule and substituting convenient x-values are the fastest ways to find A and B.

Watch for the standard variations the syllabus expects: repeated factors need a term over both (x − 1) and (x − 1)², and an irreducible quadratic factor like (x² + 1) needs a numerator of the form (Bx + C). If the top degree is greater than or equal to the bottom, do polynomial long division first.

5. Integrating trigonometric expressions

Many trig functions are not directly integrable as written, so the trick is to rewrite them into an integrable form first. The two workhorses are:

Double-angle formulae for even powers. Since cos 2x = 1 − 2sin²x = 2cos²x − 1, we get sin²x = (1 − cos 2x)/(2) and cos²x = (1 + cos 2x)/(2). These turn a squared term you cannot integrate into a linear-angle term you can.
Factor formulae (product-to-sum) for products like sin 3x cos x. These are on MF26 and convert a product into a sum of two simple sines or cosines, each easy to integrate.

For odd powers such as ∫ sin³x dx, split off one factor and use sin²x = 1 − cos²x, then substitute. The double-angle identities themselves are on MF26, but recognising when to deploy them is the examinable skill.

6. Definite integrals and applications

A definite integral evaluates the antiderivative at the upper and lower limits and subtracts: ∫_a^b f(x) dx = F(b) − F(a). No "+ c" is needed here — it cancels. The two headline applications are:

Area under a curve: the area between y = f(x) and the x-axis from x = a to x = b is ∫_a^b y dx. If part of the curve dips below the axis, that part returns a negative value, so split the integral at the x-intercepts and take the modulus of each piece. Area between two curves is ∫ (upper − lower) dx.
Volume of revolution: rotating about the x-axis gives V = π ∫_a^b y² dx; rotating about the y-axis gives V = π ∫_c^d x² dy. These formulae are not on MF26 and must be memorised — including remembering to square the function.

Worked exam-style examples

Example 1 — by parts

Find ∫ x cos 2x dx.

By LIATE the algebraic term wins, so u = x and dv/dx = cos 2x. Then du/dx = 1 and v = (1)/(2) sin 2x. Applying the formula: ∫ x cos 2x dx = x · (1)/(2) sin 2x − ∫ (1)/(2) sin 2x dx = (x)/(2) sin 2x + (1)/(4) cos 2x + c. Notice how choosing u = x made the leftover integral trivial — the test of a good choice.

Example 2 — partial fractions, definite integral

Evaluate ∫₂³ (5)/((x − 1)(x + 4)) dx.

Split: (5)/((x − 1)(x + 4)) = (A)/(x − 1) + (B)/(x + 4). Multiplying out gives 5 = A(x + 4) + B(x − 1). Setting x = 1: 5 = 5A, so A = 1. Setting x = −4: 5 = −5B, so B = −1. The integral becomes ∫₂³ [ (1)/(x − 1) − (1)/(x + 4) ] dx = [ ln|x − 1| − ln|x + 4| ]₂³. Evaluating: (ln 2 − ln 7) − (ln 1 − ln 6) = ln 2 − ln 7 + ln 6 = ln (12)/(7).

Example 3 — trig identity then area

Find the exact area bounded by y = sin²x, the x-axis, and the lines x = 0 and x = π/2.

You cannot integrate sin²x directly, so rewrite it: sin²x = (1 − cos 2x)/(2). The area is ∫₀^π/2 (1 − cos 2x)/(2) dx = (1)/(2) [ x − (1)/(2) sin 2x ]₀^π/2. At x = π/2: (1)/(2)(π/2 − (1)/(2) sin π) = (1)/(2)(π/2 − 0) = π/4. At x = 0 the bracket is 0. So the area is π/4 square units. The lesson: simplify the integrand before you integrate, never fight it as it stands.

Common exam traps and mistakes

Forgetting the + c. Every indefinite integral needs the constant of integration. It is an easy mark to drop and examiners do penalise it.
Choosing u badly in by parts. If your leftover integral looks harder than the original, you almost certainly picked u the wrong way round — go back to LIATE.
Not simplifying before integrating. Trying to integrate sin²x, cos²x or a product of trig functions directly leads nowhere. Apply double-angle or factor formulae first.
Misusing the linear-substitution shortcut. Dividing by the derivative only works when the inside function is linear. ∫ e^x² dx is not (1)/(2x) e^x².
Area sign errors. A curve below the x-axis gives a negative integral. For total area, split at the roots and take the modulus of each section.
Forgetting to square in volume questions, or mixing up dx and dy for the two axes of rotation.
Limits not changed after substitution. If you substitute u for x in a definite integral, either change the limits to u-values or substitute back before applying the original x-limits — never mix the two.
Modulus in logarithms. Indefinite integrals of f'(x)/f(x) give ln|f(x)|; leaving out the modulus loses precision.

How to practise and revise

Integration rewards drilling, but smart drilling beats volume. We suggest this approach:

Build the recognition reflex. Take a mixed set of integrals and, without solving them, label each one with the method it needs. Being able to say "that's by parts, that's partial fractions, that's reverse chain rule" at a glance is half the battle in the exam.
Master MF26. Spend a session simply locating the relevant integrals and formulae on the list so you are fast under pressure. Then make a separate list of what is not on MF26 (basic integrals, the volume formulae, the double-angle relations) and commit those to memory.
Always check by differentiating. Because integration is differentiation in reverse, you can verify any indefinite answer by differentiating it back. This catches sign slips and missing factors instantly.
Work past-year and prelim papers under timed conditions, writing every step. Marks in H2 are awarded for method, so train yourself to show the substitution, the by-parts formula line, and the limit evaluation rather than jumping to the answer.
Keep an error log. Note every careless slip — a dropped + c, a sign, an unchanged limit — and review it before the next practice session so the same mistake does not recur.

With consistent, structured practice, integration becomes one of the most reliable mark-earners in the paper, and at Intuitional we are always glad to help your child build that fluency step by step.