Integration

📚The mathematics

The standard integrals

Each is the reverse of the corresponding derivative, with a \( \tfrac{1}{a} \) to undo the chain rule. Two details cost marks routinely: the minus sign on the integral of \( \sin \), and the modulus in \( \ln|ax+b| \), which matters because the logarithm of a negative number does not exist.

P3 The inverse-tangent integral

Paper 3 only, and the reverse of the derivative in P2/3 4. The pattern to recognise is a constant over a sum of squares with no \(x\) in the numerator — that is what separates it from the logarithmic case below, where the numerator is the derivative of the denominator.

The denominator often needs rearranging into the right shape first. For \( \displaystyle\int rac{1}{2 + 3x^{2}}\,\mathrm{d}x \), take the 3 out: \( rac{1}{3}\displaystyle\int rac{1}{rac{2}{3} + x^{2}}\,\mathrm{d}x \), so \( a^{2} = frac{2}{3} \) and \( a = \sqrt{ frac{2}{3}} \).

P3 The logarithmic pattern

Whenever the numerator is the derivative of the denominator — or a constant multiple of it — the integral is a logarithm. So \( \displaystyle\int \frac{2x}{x^{2}+1}\,\mathrm{d}x = \ln\left(x^{2}+1\right) + c \), and \( \displaystyle\int \tan x\,\mathrm{d}x = \int \frac{\sin x}{\cos x}\,\mathrm{d}x = -\ln|\cos x| + c \), the minus arising because the derivative of \( \cos x \) is \( -\sin x \).

Spotting this pattern saves a full substitution, and it is worth actively looking for it before starting any other method.

Integrating \( \sin^{2} \) and \( \cos^{2} \)

These cannot be integrated directly, so use the double angle rearrangements from P2/3 3:

so that \( \displaystyle\int \cos^{2}x\,\mathrm{d}x = \tfrac{1}{2}x + \tfrac{1}{4}\sin 2x + c \). This turns up constantly in volumes of revolution, where squaring a sine or cosine is unavoidable.

P3 Integration by substitution

Substitute \( u = \mathrm{g}(x) \), replace \( \mathrm{d}x \) using \( \dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{g}'(x) \), and integrate in \(u\). Cambridge normally tells you the substitution to use.

The whole integrand must end up in terms of \(u\) — if an \(x\) survives, either the substitution is wrong or you need to express that \(x\) in terms of \(u\) as well.

P3 Integration by parts

This reverses the product rule. Everything depends on choosing \(u\) to be the part that gets simpler when differentiated:

• A polynomial with \( \mathrm{e}^{x} \), \( \sin x \) or \( \cos x \) — let \(u\) be the polynomial, since differentiating reduces its degree.
• Anything with \( \ln x \) — let \( u = \ln x \), because \( \ln x \) is awkward to integrate but differentiates to \( \dfrac1x \).
• \( \displaystyle\int \ln x\,\mathrm{d}x \) — take \( u = \ln x \) and \( \dfrac{\mathrm{d}v}{\mathrm{d}x} = 1 \). The result is \( x\ln x - x + c \), which is worth knowing.

Sometimes parts must be applied twice. With \( \displaystyle\int \mathrm{e}^{x}\sin x\,\mathrm{d}x \) the original integral reappears, and you then solve for it algebraically rather than continuing.

P3 Integration with partial fractions

A rational function that cannot be integrated directly usually can be once split into partial fractions — see P2/3 1. Each simple piece is then a logarithm or a power:

Note the repeated factor gives a power, not a logarithm — only the first power in the denominator produces \( \ln \). Combining logarithms at the end using the laws often produces a much tidier final answer, and questions frequently ask for it in that form.

P2 The trapezium rule

This is the one topic that runs the other way: it is on Paper 2 and not on Paper 3. If you are taking the AS pure route you need it; if you are taking the A Level you do not.

The area is estimated by \(n\) trapezia of equal width. The two end ordinates are counted once and every interior ordinate twice — because each interior one is shared by the trapezium on its left and the one on its right.

Note that \(n\) strips need \(n+1\) ordinates. A question asking for “four intervals” requires five \(y\)-values; asking for “five ordinates” means four strips. Read which of the two is being counted before working out \(h\).

Over- or under-estimate? Sketch the curve over the interval and compare it with the straight chords:

• Convex (curving upwards, holding water) — each chord lies above the curve, so the rule over-estimates.
• Concave (curving downwards) — the chords lie below the curve, so it under-estimates.

For example, four strips on \( \displaystyle\int_0^1 \sqrt{1+x^{3}}\,\mathrm{d}x \) give \( h = 0.25 \) and ordinates \( 1,\ 1.00778,\ 1.06066,\ 1.19242,\ 1.41421 \), so

The true value is 1.111, so this is an over-estimate — as it must be, since the curve is convex on \( [0, 1] \). Answering the over/under part from a sketch, rather than from the number, is what the mark is for.