Pure mathematics

Integration: The Three Main Tricks

Differentiation has algorithms that always work. Integration does not. Every differentiation rule has an inverse that, in principle, gives an integration rule — but spotting which rule to apply, and in what order, is a genuine skill that takes practice to develop. The good news is that the vast majority of integrals you will encounter at A-level reduce to one of three core techniques, or a combination of them.

The techniques below are not mutually exclusive, and it is not always obvious from the outset which one to reach for. That is normal. Integration rewards persistence more than inspiration.

The three tricks

Substitution — the reverse of the chain rule. Introduce a new variable to simplify the integrand.
Integration by parts — the reverse of the product rule. Trade one integral for a (simpler) one.
Changing the form — rewrite the integrand into something you already know how to integrate. This is the most varied technique: partial fractions, trigonometric identities, completing the square, and much more.

Trick 1 — Substitution

A-level Maths (compulsory)EdexcelAQAOCROCR MEI

Substitution is the integration counterpart of the chain rule. If you can spot a function and (roughly) its derivative both present in the integrand, introducing a new variable u often collapses the integral into a standard form.

The key step is exchanging dx for du using the relation du = g′(x) dx, which means dx = du / g′(x). When g′(x) is already a factor of the integrand, it cancels cleanly — that is the sign that substitution is the right tool.

Example 1 — a composite power

Notice that the derivative of x² + 1 is 2x, and 2x appears as a factor:

Let u = x² + 1, so du = 2x dx:

Try it yourself

Evaluate ∫4x(x² + 5)² dx.

Show answer

Let u = x² + 5, du = 2x dx, so 4x dx = 2 du:

Example 2 — a composite exponential

The derivative of sin x is cos x, and cos x is present as a factor:

Let u = sin x, so du = cos x dx:

Try it yourself

Evaluate ∫sin x · e^{−cos x} dx.

Show answer

Let u = −cos x, du = sin x dx:

Example 3 — the logarithm trick

When the numerator is (proportional to) the derivative of the denominator, the result is always a logarithm:

Let u = x² + 3, so du = 2x dx:

This pattern — numerator is the derivative of the denominator — is worth recognising on sight. It appears constantly and the answer is always ln|denominator| + C.

Try it yourself

Evaluate ∫3x²/(x³ + 7) dx.

Show answer

The numerator 3x² is the derivative of the denominator x³ + 7:

Definite integrals: change the limits too

With a definite integral, you can either change the limits of integration into u-values (cleaner), or substitute back to x before evaluating. Never evaluate a u-expression at x-limits.

Try it yourself

Evaluate ∫₁² 2x(x² + 3)³ dx. Change the limits as well as the variable.

Show answer

Let u = x² + 3, du = 2x dx. When x = 1, u = 4; when x = 2, u = 7:

Trick 2 — Integration by parts

A-level Maths (compulsory)EdexcelAQAOCROCR MEI

Integration by parts is the inverse of the product rule. It replaces one integral with another, simpler one. The formula is:

The choice of which factor to call u and which to call dv/dx determines whether the resulting integral is simpler or harder. The standard guide — sometimes called LIATE — ranks which functions should preferably be chosen as u:

Priority for u	Type of function	Examples
1st	Logarithms	ln x, log x
2nd	Inverse trig	arctan x, arcsin x
3rd	Algebraic (polynomials)	x, x², x³
4th	Trigonometric	sin x, cos x
5th	Exponential	eˣ, 2ˣ

LIATE is a guide, not a rule. There are cases where reversing it works better, and you will develop a feel for this with practice.

Example 4 — polynomial × exponential

Choose u = x (algebraic, higher priority) and dv/dx = eˣ:

Try it yourself

Evaluate ∫x²eˣ dx. You will need to apply integration by parts twice.

Show answer

First: u = x², dv = eˣ dx → du = 2x, v = eˣ:

Example 5 — logarithm alone (a useful trick)

∫ ln x dx looks like it has only one factor. Write it as 1 · ln x, then choose u = ln x and dv/dx = 1:

The same trick works for arctan x, arcsin x — any function that you cannot integrate directly but can differentiate.

Try it yourself

Evaluate ∫arctan x dx. Write it as 1 · arctan x and use integration by parts.

Show answer

u = arctan x, dv = 1 dx → du = 1/(1+x²), v = x:

Example 6 — applying by parts twice, then recovering the original

Some integrals require by parts twice. Occasionally, the original integral reappears on the right-hand side — a happy accident that lets you solve for it algebraically.

Choose u = sin x, dv/dx = eˣ. First application:

Apply by parts again to the remaining integral (u = cos x, dv/dx = eˣ):

The original integral I appears on the right. Solve for it:

Try it yourself

Evaluate ∫eˣ cos x dx using the same cyclic technique.

Show answer

Let I = ∫eˣ cos x dx. First parts: u = cos x, dv = eˣ dx:

Second parts on the remaining integral (u = sin x, dv = eˣ dx):

Trick 3 — Changing the form

The most varied of the three techniques. The idea is to rewrite the integrand into an equivalent form that you already know how to integrate — without introducing a new variable. How you do that depends entirely on the type of integrand. Several different tools fall under this umbrella.

3a — Partial fractions

A-level Maths (compulsory)EdexcelAQAOCROCR MEI

A rational function — a polynomial over a polynomial — cannot usually be integrated directly, but once split into partial fractions each piece integrates as a logarithm or a standard rational form.

Example 7 — distinct linear factors

Decompose:

Try it yourself

Evaluate ∫1/((x+1)(x−3)) dx.

Show answer

Example 8 — repeated factor

A repeated linear factor (x − a)² requires an extra term with denominator (x − a)²:

The B term integrates to −1/(x − 1) rather than a logarithm — a different standard form, but still elementary.

Try it yourself

Find the constants A, B, C in the partial fractions decomposition of 3/((x + 2)(x − 1)²).

Show answer

x = −2: 3 = 9A → A = 1/3. x = 1: 3 = 3C → C = 1. x = 0: 3/(2·1) = A/2 − B − C → 3/2 = 1/6 − B − 1 → B = −1/3.

3b — Trigonometric identities

A-level Maths (compulsory)EdexcelAQAOCROCR MEI

Powers of trigonometric functions cannot be integrated directly. The double-angle and product-to-sum identities reduce them to standard integrable forms.

Example 9 — sin²x and cos²x

Neither sin²x nor cos²x has a direct antiderivative. Use the double-angle identity:

Try it yourself

Evaluate ∫sin²x dx.

Show answer

Example 10 — an odd power of cosine

For an odd power, peel off one factor of cos x to pair with dx for a substitution, and convert the rest using sin²x + cos²x = 1:

Now let u = sin x, du = cos x dx:

This example uses two tricks in combination: a trig identity (changing the form) and then substitution. Tricks combine freely, and the skill is knowing when to hand off from one to another.

Try it yourself

Evaluate ∫sin³x dx. Peel off one factor of sin x and convert the rest using sin²x + cos²x = 1.

Show answer

Let u = cos x, du = −sin x dx:

3c — Completing the square

A-level Maths (compulsory)EdexcelAQAOCROCR MEI

Integrals involving a quadratic in the denominator often become standard arctan or logarithm forms once the quadratic is rewritten as a perfect square plus a constant.

Example 11

Complete the square: x² + 4x + 5 = (x + 2)² + 1. Then substitute u = x + 2:

Try it yourself

Evaluate ∫1/(x² + 6x + 10) dx.

Show answer

Complete the square: x² + 6x + 10 = (x + 3)² + 1. Then u = x + 3:

3d — Algebraic manipulation

A-level Maths (compulsory)

Sometimes the integrand simply needs rearranging before it is recognisable. Splitting a fraction, multiplying top and bottom by a conjugate, or performing polynomial long division are all fair game.

Example 12 — improper rational function (degree of numerator ≥ denominator)

Partial fractions only apply when the degree of the numerator is strictly less than the degree of the denominator. If not, divide first:

Try it yourself

Evaluate ∫(x² + 5)/(x² + 1) dx.

Show answer

Example 13 — rationalising the denominator

Multiplying numerator and denominator by e^−x puts the integral into the logarithm pattern from Example 3.

Try it yourself

Evaluate ∫eˣ/(1 + eˣ) dx.

Show answer

The numerator eˣ is the derivative of the denominator 1 + eˣ:

3e — Further Maths: identities from De Moivre's theorem

A-level Further Maths — Further Pure (compulsory)Edexcel (Core Pure 1 & 2)AQA (Further Pure 1 & 2)OCR (Pure Core 1 & 2)OCR MEI (Core Pure A & B)

In Further Maths the range of available identities expands considerably. De Moivre's theorem — (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ — can be used to express high powers of sin and cos as combinations of multiple-angle functions, which are directly integrable.

The two key results, derived by writing z = e^iθ:

Example 14 — cos⁴x in terms of multiple angles

Write 2cos θ = z + z⁻¹ and raise to the fourth power:

Group using cos nθ = (zⁿ + z⁻ⁿ)/2:

This is now directly integrable:

Try it yourself

Use De Moivre's theorem to express sin⁴θ in terms of multiple angles, then evaluate ∫₀^π sin⁴x dx.

Show answer

Write 2i sin θ = z − z⁻¹ and raise to the fourth power:

Compare this with the A-level approach for cos³x above: for even powers, the trig identity approach is algebraically awkward, while De Moivre's theorem reduces any power to a direct sum of cosines. For higher powers like cos⁶x or cos⁸x, De Moivre is essentially the only practical route.

Example 15 — sin⁵x

Write 2i sin θ = z − z⁻¹ and raise to the fifth power:

Group using sin nθ = (zⁿ − z⁻ⁿ) / 2i:

Try it yourself

Evaluate ∫cos⁵x dx using the trig identity method (not De Moivre). Peel off one cos x and use sin²x + cos²x = 1.

Show answer

Let u = sin x, du = cos x dx:

When things don't work — and they often won't

Integration is genuinely hard. Even professional mathematicians try wrong approaches before finding one that works. This is not a sign of failure — it is the nature of integration. There is no single strategy that always works, and the absence of one is not a gap in your knowledge.

If your first attempt leads nowhere, a few things are worth trying:

Try a different substitution. If u = f(x) made things worse, try u = g(x). Sometimes the "obvious" substitution is not the productive one.
Switch techniques entirely. If substitution is going nowhere, try changing the form first, and then substituting into the simplified integrand.
See whether your partial attempt can be salvaged. Sometimes an integration by parts leaves a remainder that itself responds to substitution. The techniques are not independent — they combine.
Look at the structure of the answer. If you expect a logarithm and you are getting a polynomial, something has gone wrong earlier than you think. Differentiate your partial answer and compare it to the integrand.
Try a specific numerical value. If you have two candidate antiderivatives and are unsure which is correct, differentiate both and compare to the integrand at x = 1. The one that matches is right.

In an examination, if you are completely stuck on an integration, write down the technique you would try and why — method marks are available even without a correct final answer. And if you do get an answer, differentiate it: there is no easier check in all of mathematics.

Summary — what to reach for when

What you see	Try first	Notes
f(g(x)) · g′(x)	Substitution u = g(x)	Classic chain-rule reversal
Numerator ≈ derivative of denominator	Substitution → ln	Result is always ln\|denominator\|
Product of unlike functions	Integration by parts	Use LIATE to choose u
ln x or arctan x alone	By parts with u = ln x (or arctan)	Write as 1 · ln x first
Rational function (poly/poly)	Partial fractions	Divide first if degree top ≥ degree bottom
sin²x, cos²x, even powers	Double-angle identity	cos⁴x, sin⁴x etc: De Moivre in FM
Odd power of sin or cos	Trig identity + substitution	Peel off one factor to pair with dx
High even power (cos⁶x, sin⁸x …)	De Moivre (Further Maths)	Reduces to sum of multiple-angle terms
Quadratic in denominator	Complete the square → arctan	Or partial fractions if it factorises
eˣ sin x, eˣ cos x	By parts twice, solve for I	The integral reappears and you solve algebraically