OxfordAQA International AS and A-level Mathematics: Integration by inspection and substitution (9660)

This guide covers the first part of section P2.7 (Integration) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2: the standard integrals of eˣ, 1/x, sin x and cos x, the reverse chain rule by inspection and by substitution, and integrals of the form f′(x)/f(x). The specification places this content in International A-level (A2) Unit P2 (Pure Maths); it is not examined at International AS. Integration by parts, partial fractions and volumes of revolution are the other parts of P2.7 and are not covered here. Pair this guide with the revision notes and the practice questions. The basics of integrating powers are in the P1.4 integration guide. The course home is the OxfordAQA A-level Mathematics hub, with a printable checklist. If you are unsure where you stand, take a free diagnostic first.

What this unit asks you to do

Specification content	What you must be able to do
Integration of eˣ, 1/x, sin x, cos x	Integrate these, and the versions with a linear argument such as e^(kx), sin kx, cos kx
Reverse chain rule by inspection	Spot that the integrand looks like a derivative from the chain rule, write down the answer and adjust the constant factor
Integration by substitution	Choose u, rewrite the whole integral in u, integrate, and return to x (or change the limits)
f′(x)/f(x)	Recognise it, and write the integral as ln|f(x)| + c

The specification treats inspection and substitution as the reverse process of the chain rule. If you are rusty on the chain rule, revise the differentiation guide first. Every result below can be checked by differentiating your answer, and you should get into the habit of doing so.

Standard integrals, including a linear argument

Angles are in radians throughout. In each row, k and a are non-zero constants.

Integrand	Integral
eˣ	eˣ + c
e^(kx)	e^(kx)/k + c
1/x	ln|x| + c
1/(ax + b)	(1/a) ln|ax + b| + c
sin kx	−cos(kx)/k + c
cos kx	sin(kx)/k + c

The factor 1/k appears because differentiating e^(kx) brings down a factor k. To undo that, divide by k. The modulus signs in ln|x| are needed because 1/x exists for negative x, where ln x does not.

Example 1: a mixed integral

Find ∫ (10e^(−x/2) − 7 sin 2x + 4/x) dx, for x > 0.

∫ 10e^(−x/2) dx = 10 × e^(−x/2)/(−½) = −20e^(−x/2)
∫ 7 sin 2x dx   = 7 × (−cos(2x)/2) = −(7/2) cos 2x
∫ 4/x dx        = 4 ln x

The middle term has a minus sign in front of it, so it contributes +(7/2) cos 2x. The integral is −20e^(−x/2) + (7/2) cos 2x + 4 ln x + c.

Check: differentiating −20e^(−x/2) gives 10e^(−x/2), and differentiating (7/2) cos 2x gives −7 sin 2x. Both match.

Example 2: definite integrals

Evaluate (a) ∫ from 0 to π/6 of cos 3x dx and (b) ∫ from 1 to 4 of 1/(2x − 1) dx.

(a) The antiderivative is sin(3x)/3.

[sin(3x)/3] from 0 to π/6 = sin(π/2)/3 − sin 0/3 = 1/3

(b) The antiderivative is ½ ln|2x − 1|. On the interval from 1 to 4, 2x − 1 runs from 1 to 7 and stays positive.

[½ ln(2x − 1)] from 1 to 4 = ½ ln 7 − ½ ln 1 = ½ ln 7

The exact answer is ½ ln 7 (about 0.973). Keep answers exact when the question says “exact” or “in terms of ln”.

The reverse chain rule by inspection

The chain rule says that d/dx [F(g(x))] = F′(g(x)) × g′(x). Reading it backwards:

∫ F′(g(x)) × g′(x) dx = F(g(x)) + c

You are looking for an integrand made of an “inside” function g(x) and a copy of its derivative g′(x) (possibly missing a constant factor). The method has four steps.

1. Spot the inside function g(x) and check that its derivative appears, up to a constant.
2. Write down the answer as if the derivative were present.
3. Differentiate your guess and compare it with the integrand.
4. Divide or multiply by the constant needed to match.

Example 3: power of a bracket

Find ∫ x(4x² − 5)³ dx.

The inside function is g = 4x² − 5, with g′ = 8x. The integrand has x, which is g′/8.

Guess: (4x² − 5)⁴
Differentiate: 4(4x² − 5)³ × 8x = 32x(4x² − 5)³

That is 32 times too large, so divide by 32.

∫ x(4x² − 5)³ dx = (4x² − 5)⁴/32 + c

Example 4: exponential and trigonometric inside functions

(a) Evaluate ∫ from 0 to 1 of x e^(5 − 2x²) dx.

The inside function is 5 − 2x², with derivative −4x. Differentiating e^(5 − 2x²) gives −4x e^(5 − 2x²), so the antiderivative of x e^(5 − 2x²) is −¼ e^(5 − 2x²).

[−¼ e^(5 − 2x²)] from 0 to 1 = −¼ e³ + ¼ e⁵ = (e⁵ − e³)/4

(b) Find ∫ sin x cos³x dx.

The inside function is cos x, with derivative −sin x. Differentiating cos⁴x gives −4cos³x sin x, so

∫ sin x cos³x dx = −cos⁴x/4 + c

Integrals of the form f′(x)/f(x)

Differentiating ln|f(x)| gives f′(x)/f(x). So

∫ f′(x)/f(x) dx = ln|f(x)| + c

The numerator must be the derivative of the denominator, up to a constant factor. Take the constant outside the integral and apply the result.

Example 5: three recognitions

(a) ∫ (6x − 4)/(3x² − 4x + 7) dx. The derivative of the denominator is 6x − 4, which is exactly the numerator. The answer is ln(3x² − 4x + 7) + c. The quadratic has a negative discriminant, so it is always positive and the modulus signs can be dropped.

(b) ∫ x²/(x³ + 5) dx. The derivative of x³ + 5 is 3x², so the numerator is one third of it. The answer is ⅓ ln|x³ + 5| + c.

(c) ∫ from π/6 to π/3 of cot x dx, where cot x = cos x/sin x. The derivative of sin x is cos x, so the numerator is exactly the derivative of the denominator.

∫ cot x dx = ln|sin x| + c
[ln(sin x)] from π/6 to π/3 = ln(√3/2) − ln(1/2) = ln √3 = ½ ln 3

Integration by substitution

When the match is not obvious, substitute. The method is the same each time.

1. Choose u, usually the inside function or the awkward bracket.
2. Differentiate: du/dx = …, then write dx = du/(du/dx) or du = (…) dx.
3. Replace every x in the integrand, including dx. If an x is left over, rearrange u = … to write x in terms of u.
4. For a definite integral, change the limits using u = … at each x-limit.
5. Integrate with respect to u. For an indefinite integral, put x back at the end.

If you change the limits, you do not go back to x. If you keep the original limits, you must write the answer in x before using them. Mixing the two is a common way to lose marks.

Example 6: a leftover x

Find ∫ x(3x − 2)⁴ dx using u = 3x − 2.

u = 3x − 2  →  du = 3 dx  →  dx = du/3,   x = (u + 2)/3
∫ x(3x − 2)⁴ dx = ∫ ((u + 2)/3) × u⁴ × (du/3)
                = (1/9) ∫ (u⁵ + 2u⁴) du
                = (1/9)(u⁶/6 + 2u⁵/5) + c
                = u⁶/54 + 2u⁵/45 + c

Replace u by 3x − 2 to finish: (3x − 2)⁶/54 + 2(3x − 2)⁵/45 + c.

Example 7: a definite integral with changed limits

Evaluate ∫ from 1 to 6 of x/√(x + 3) dx using u = x + 3.

u = x + 3  →  du = dx,   x = u − 3
Limits: x = 1 → u = 4;   x = 6 → u = 9
∫ x/√(x + 3) dx = ∫ from 4 to 9 of (u − 3) u^(−1/2) du
                = ∫ (u^(1/2) − 3u^(−1/2)) du
                = [ (2/3)u^(3/2) − 6u^(1/2) ] from 4 to 9
At u = 9: (2/3)(27) − 6(3) = 18 − 18 = 0
At u = 4: (2/3)(8) − 6(2) = 16/3 − 12 = −20/3
0 − (−20/3) = 20/3

The answer is 20/3.

Example 8: a trigonometric substitution

Evaluate ∫ from 0 to π/2 of cos x √(1 + 3 sin x) dx using u = 1 + 3 sin x.

du = 3 cos x dx  →  cos x dx = du/3
Limits: x = 0 → u = 1;   x = π/2 → u = 4
∫ from 1 to 4 of √u × (du/3) = ⅓ [ (2/3)u^(3/2) ] from 1 to 4
                             = (2/9)(8 − 1) = 14/9

This one could also be done by inspection. Both methods must agree, which makes a useful check.

Choosing between the methods

Try inspection first when the integrand is one function of g(x) multiplied by g′(x), or divided by g(x). Use substitution when a factor of x is left over that is not the derivative of the inside function, as in Examples 6 and 7. When a question says “use the substitution u = …”, follow it exactly, because the working is usually what earns the credit. A “show that” question needs every step of the substitution written out.

Common errors

• Writing k e^(kx) as the integral of e^(kx), when the factor has to be 1/k.
• Writing ∫ 1/(3x + 2) dx = ln|3x + 2| and leaving out the factor ⅓.
• Applying the f′/f rule when the numerator is not a multiple of the derivative. For example, ∫ 1/(x² + 1) dx is not ln(x² + 1), and ∫ x/(x + 1) dx is not a ln integral either.
• Giving cos(kx)/k for ∫ sin kx dx. The correct result is −cos(kx)/k.
• Leaving x in the integrand after substituting, or forgetting to replace dx.
• Changing the limits to u-values, then substituting x-values into the answer (or the reverse).
• Using degrees. These integrals assume radians.
• Missing + c on an indefinite integral.

Next steps

Try the practice questions, then use the revision notes to recall the results quickly. Tick off the topic on the checklist.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA. This guide follows the content and additional information listed under P2.7.