Integration by substitution - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 17 exam-style questions on Integration by substitution for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

17 questions
2–11 marks each
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17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

Find
\[\int \sec^2 2x\tan^3 2x\,dx\]
[You may use \(\dfrac{d}{dx}(\tan kx) = k\sec^2 kx\)]

Model answer

Let \(u = \tan 2x\), so \(\dfrac{du}{dx} = 2\sec^2 2x\), i.e. \(\sec^2 2x\,dx = \dfrac{1}{2}\,du\).
\(\displaystyle\int \sec^2 2x\tan^3 2x\,dx = \int \dfrac{1}{2}u^3\,du = \dfrac{1}{8}u^4 + c\)
\(= \dfrac{1}{8}\tan^4 2x + c\)

Explanation

\(\sec^2 2x\) is half the derivative of \(\tan 2x\), so the integral is a reverse chain rule: with \(u = \tan 2x\), \(du = 2\sec^2 2x\,dx\) and the integral becomes \(\dfrac{1}{2}\displaystyle\int u^3\,du = \dfrac{1}{8}u^4\). Check by differentiating: \(\dfrac{d}{dx}\left(\dfrac{1}{8}\tan^4 2x\right) = \dfrac{1}{8} \times 4\tan^3 2x \times 2\sec^2 2x = \sec^2 2x\tan^3 2x\). Common errors are missing the factor \(\dfrac{1}{2}\), integrating to a logarithm and leaving the answer in terms of \(u\).

Mark scheme

1 mark
M1 Integrates to \(k\tan^4 2x\) (reverse chain rule), or uses \(u = \tan 2x\), \(du = 2\sec^2 2x\,dx\), to reach \(k\displaystyle\int u^3\,du\) and integrates to \(ku^4\); or a correct final answer of \(\dfrac{1}{8}\tan^4 2x + c\) with no working
An answer of \(\dfrac{1}{8}\tan^4 2x + c\) with no working scores 3 marks. Alternative: \(u = \sec 2x\), with \(\tan^2 2x = \sec^2 2x - 1\), gives \(\dfrac{1}{8}\sec^4 2x - \dfrac{1}{4}\sec^2 2x + c\), which differs from \(\dfrac{1}{8}\tan^4 2x\) by a constant and is marked in the same way. Integrating to a logarithm, or integrating \(\sec^2 2x\) and \(\tan^3 2x\) separately and multiplying the results, is M0.
1 mark
A1 \(\dfrac{1}{8}\tan^4 2x\) oe (or \(\dfrac{1}{8}u^4\) with \(u = \tan 2x\))
A common error is \(\dfrac{1}{4}\tan^4 2x\), from missing the factor \(\dfrac{1}{2}\) that comes from \(du = 2\sec^2 2x\,dx\): A0.
1 mark
A1 \(\dfrac{1}{8}\tan^4 2x + c\), in terms of \(x\) and including the constant
Depends on the M mark. Accept \(\dfrac{\tan^4 2x}{8} + c\) or \(\dfrac{1}{8}\sec^4 2x - \dfrac{1}{4}\sec^2 2x + c\). An answer left in terms of \(u\) scores A0 for this mark.

Question 2

4 marks

Amir is asked to use the substitution \(u = x^2 + 16\) to find the exact value of
\[\int_{0}^{3} 2x\sqrt{x^2 + 16}\,dx\]
His working is shown below.
Line 1: \(u = x^2 + 16\), \(du = 2x\,dx\)
Line 2: \(\displaystyle\int_0^3 2x\sqrt{x^2 + 16}\,dx = \int_0^3 u^{\frac{1}{2}}\,du\)
Line 3: \(= \left[\dfrac{3}{2}u^{\frac{3}{2}}\right]_0^3\)
Line 4: \(= \dfrac{3}{2} \times 3\sqrt{3} - 0 = \dfrac{9\sqrt{3}}{2}\)

Identify the two errors in Amir's working.

Find the correct exact value of the integral.

Model answer

Line 2 is wrong because the limits are still the \(x\)-limits: when \(x = 0\), \(u = 16\), and when \(x = 3\), \(u = 25\).
Line 3 is wrong because \(\displaystyle\int u^{\frac{1}{2}}\,du = \dfrac{2}{3}u^{\frac{3}{2}}\), not \(\dfrac{3}{2}u^{\frac{3}{2}}\).
Correct value: \(\displaystyle\int_{16}^{25} u^{\frac{1}{2}}\,du = \left[\dfrac{2}{3}u^{\frac{3}{2}}\right]_{16}^{25} = \dfrac{2}{3}(125 - 64) = \dfrac{122}{3}\)

Explanation

When a definite integral is rewritten in terms of \(u\), the limits must be changed to \(u\)-values (or the answer must be put back in terms of \(x\) before the original limits are used). Raising the power by 1 and dividing by the new power gives \(\dfrac{2}{3}u^{\frac{3}{2}}\). These are both errors the examiners' reports describe as common: limits left unchanged and slips in integrating fractional powers.

Mark scheme

1 mark
B1 Error in Line 2: the limits have not been changed; in terms of \(u\) they should be \(u = 16\) and \(u = 25\). A correct value of \(\dfrac{122}{3}\) with no working earns only the M1A1 below.
An answer of \(\dfrac{122}{3}\) with no working scores 2 marks (M1A1); the two B marks need the errors to be identified. Accept "the limits are still values of \(x\)" with the correct limits stated.
1 mark
B1 Error in Line 3: \(u^{\frac{1}{2}}\) integrates to \(\dfrac{2}{3}u^{\frac{3}{2}}\), not \(\dfrac{3}{2}u^{\frac{3}{2}}\)
Accept "he multiplied by \(\dfrac{3}{2}\) instead of dividing by \(\dfrac{3}{2}\)". Do not accept vague answers such as "the integration is wrong" without saying what it should be.
1 mark
M1 Correct method: \(\dfrac{2}{3}\left(25^{\frac{3}{2}} - 16^{\frac{3}{2}}\right)\), or \(\left[\dfrac{2}{3}(x^2 + 16)^{\frac{3}{2}}\right]_0^3\)
1 mark
A1 \(\dfrac{122}{3}\)
From \(\dfrac{2}{3}(125 - 64)\). Accept \(40\tfrac{2}{3}\); a decimal (40.7) on its own scores A0.

Question 3

6 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

The constant \(k\) is positive. Use the substitution \(u = \sin kx\) to show that
\[\int \cot kx\,dx = \dfrac{1}{k}\ln\lvert\sin kx\rvert + c\]
Hence find the exact value of
\[\int_{\frac{\pi}{9}}^{\frac{\pi}{6}} \cot 3x\,dx\]
giving your answer in the form \(a\ln b\), where \(a\) and \(b\) are rational numbers.

Model answer

\(\cot kx = \dfrac{\cos kx}{\sin kx}\). Let \(u = \sin kx\), so \(du = k\cos kx\,dx\).
\(\displaystyle\int \cot kx\,dx = \int \dfrac{1}{u} \times \dfrac{1}{k}\,du = \dfrac{1}{k}\ln\lvert u\rvert + c = \dfrac{1}{k}\ln\lvert\sin kx\rvert + c\)
With \(k = 3\): \(\displaystyle\int_{\frac{\pi}{9}}^{\frac{\pi}{6}} \cot 3x\,dx = \left[\dfrac{1}{3}\ln\lvert\sin 3x\rvert\right]_{\frac{\pi}{9}}^{\frac{\pi}{6}} = \dfrac{1}{3}\ln 1 - \dfrac{1}{3}\ln\dfrac{\sqrt{3}}{2}\)
\(= -\dfrac{1}{3} \times \dfrac{1}{2}\ln\dfrac{3}{4} = \dfrac{1}{6}\ln\dfrac{4}{3}\)

Explanation

Writing \(\cot kx\) as \(\dfrac{\cos kx}{\sin kx}\) shows a function over \(\dfrac{1}{k}\) times its derivative, so \(u = \sin kx\) gives \(\dfrac{1}{k}\ln\lvert\sin kx\rvert\), the result printed in the formula booklet. With \(k = 3\) the limits give \(\sin\dfrac{\pi}{2} = 1\) and \(\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}\), and \(\ln\dfrac{\sqrt{3}}{2} = \dfrac{1}{2}\ln\dfrac{3}{4}\) removes the surd. Common errors are a factor of \(k\) instead of \(\dfrac{1}{k}\) and leaving \(\sqrt{3}\) inside the logarithm.

Mark scheme

1 mark
M1 Writes \(\cot kx = \dfrac{\cos kx}{\sin kx}\) and uses \(du = k\cos kx\,dx\) to reach \(\dfrac{1}{k}\displaystyle\int \dfrac{1}{u}\,du\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the first part is a show-that and the question demands working.
1 mark
A1* \(\dfrac{1}{k}\ln\lvert u\rvert + c = \dfrac{1}{k}\ln\lvert\sin kx\rvert + c\) cso
The final line must be in terms of \(x\), with the constant.
1 mark
M1 Uses the result with \(k = 3\) and substitutes the limits: \(\dfrac{1}{3}\left(\ln\sin\dfrac{\pi}{2} - \ln\sin\dfrac{\pi}{3}\right)\)
The given result may be used even if it was not shown.
1 mark
A1 \(\dfrac{1}{3}\left(\ln 1 - \ln\dfrac{\sqrt{3}}{2}\right) = -\dfrac{1}{3}\ln\dfrac{\sqrt{3}}{2}\) oe
1 mark
M1 (dM1) Uses laws of logarithms to remove the surd, e.g. \(-\dfrac{1}{3}\ln\dfrac{\sqrt{3}}{2} = -\dfrac{1}{6}\ln\dfrac{3}{4}\)
Depends on the second M mark.
1 mark
A1 \(\dfrac{1}{6}\ln\dfrac{4}{3}\) or \(-\dfrac{1}{6}\ln\dfrac{3}{4}\)
Either form is acceptable (\(a = \dfrac{1}{6}\), \(b = \dfrac{4}{3}\), or \(a = -\dfrac{1}{6}\), \(b = \dfrac{3}{4}\)). \(\dfrac{1}{3}\ln\dfrac{2}{\sqrt{3}}\) is not in the required form because \(b\) is not rational. A decimal (0.0479) on its own scores A0.

Question 4

9 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Use the substitution \(x = 2\sec\theta\) to show that
\[\int_{2}^{4} \dfrac{\sqrt{x^2 - 4}}{x^2}\,dx = \int_{0}^{\frac{\pi}{3}} \tan\theta\sin\theta\,d\theta\]
Hence find the exact value of \(\displaystyle\int_{2}^{4} \dfrac{\sqrt{x^2 - 4}}{x^2}\,dx\).

[You may use \(\dfrac{d}{d\theta}(\sec\theta) = \sec\theta\tan\theta\) and \(\displaystyle\int \sec\theta\,d\theta = \ln\lvert\sec\theta + \tan\theta\rvert + c\)]

Model answer

\(x = 2\sec\theta \Rightarrow dx = 2\sec\theta\tan\theta\,d\theta\); \(\sqrt{x^2 - 4} = \sqrt{4\sec^2\theta - 4} = 2\sqrt{\sec^2\theta - 1} = 2\tan\theta\) (as \(\tan\theta \geqslant 0\) here)
When \(x = 2\), \(\sec\theta = 1\), so \(\theta = 0\); when \(x = 4\), \(\sec\theta = 2\), so \(\theta = \dfrac{\pi}{3}\).
\(\displaystyle\int_{2}^{4} \dfrac{\sqrt{x^2 - 4}}{x^2}\,dx = \int_{0}^{\frac{\pi}{3}} \dfrac{2\tan\theta}{4\sec^2\theta} \times 2\sec\theta\tan\theta\,d\theta = \int_{0}^{\frac{\pi}{3}} \dfrac{\tan^2\theta}{\sec\theta}\,d\theta = \int_{0}^{\frac{\pi}{3}} \tan\theta \times \dfrac{\sin\theta}{\cos\theta} \times \cos\theta\,d\theta = \int_{0}^{\frac{\pi}{3}} \tan\theta\sin\theta\,d\theta\)
\(\tan\theta\sin\theta = \dfrac{\sin^2\theta}{\cos\theta} = \dfrac{1 - \cos^2\theta}{\cos\theta} = \sec\theta - \cos\theta\)
\(\displaystyle\int_{0}^{\frac{\pi}{3}} \left(\sec\theta - \cos\theta\right)d\theta = \left[\ln\lvert\sec\theta + \tan\theta\rvert - \sin\theta\right]_{0}^{\frac{\pi}{3}}\)
\(= \left(\ln\left(2 + \sqrt{3}\right) - \dfrac{\sqrt{3}}{2}\right) - (\ln 1 - 0) = \ln\left(2 + \sqrt{3}\right) - \dfrac{\sqrt{3}}{2}\)

Explanation

With \(x = 2\sec\theta\), the identity \(\sec^2\theta - 1 = \tan^2\theta\) removes the square root, and \(dx = 2\sec\theta\tan\theta\,d\theta\) combines with it to give \(\tan^2\theta\cos\theta = \tan\theta\sin\theta\); the limits come from \(\sec\theta = 1\) and \(\sec\theta = 2\). Writing \(\sin^2\theta\) as \(1 - \cos^2\theta\) splits the integrand into \(\sec\theta - \cos\theta\), and the integral of \(\sec\theta\) is in the formula booklet. Common errors are not converting \(dx\), leaving the limits as 2 and 4, and trying to integrate \(\tan\theta\sin\theta\) as a product of separate integrals.

Mark scheme

1 mark
B1 \(\dfrac{dx}{d\theta} = 2\sec\theta\tan\theta\). An answer with no working scores 0 marks.
An answer of \(\ln\left(2 + \sqrt{3}\right) - \dfrac{\sqrt{3}}{2}\) with no working scores 0 marks: the question demands working.
1 mark
M1 Substitutes fully: \(\displaystyle\int \dfrac{\sqrt{4\sec^2\theta - 4}}{4\sec^2\theta} \times 2\sec\theta\tan\theta\,d\theta\)
1 mark
M1 (dM1) Uses \(\sec^2\theta - 1 = \tan^2\theta\), so that \(\sqrt{4\sec^2\theta - 4} = 2\tan\theta\), and simplifies to \(\dfrac{k\tan^2\theta}{\sec\theta}\) or \(k\tan\theta\sin\theta\)
Depends on the previous M mark.
1 mark
B1 The limits \(\theta = 0\) and \(\theta = \dfrac{\pi}{3}\), from \(\sec\theta = 1\) and \(\sec\theta = 2\)
Limits in degrees score B0.
1 mark
A1* \(\displaystyle\int_{0}^{\frac{\pi}{3}} \tan\theta\sin\theta\,d\theta\), with the integral sign, the limits and the \(d\theta\), and no errors
Depends on both M marks. A step such as \(\dfrac{\tan^2\theta}{\sec\theta} = \tan\theta \times \dfrac{\sin\theta}{\cos\theta} \times \cos\theta\) must be seen.
1 mark
M1 Writes \(\tan\theta\sin\theta = \dfrac{\sin^2\theta}{\cos\theta} = \dfrac{1 - \cos^2\theta}{\cos\theta} = \sec\theta - \cos\theta\) and integrates to the form \(\alpha\ln\lvert\sec\theta + \tan\theta\rvert + \beta\sin\theta\)
The given integral may be used even if it was not shown. Integrating \(\tan\theta\) and \(\sin\theta\) separately and multiplying the results is M0.
1 mark
A1 \(\ln\lvert\sec\theta + \tan\theta\rvert - \sin\theta\)
1 mark
M1 (dM1) Substitutes the limits \(\dfrac{\pi}{3}\) and 0 and subtracts, using \(\sec\dfrac{\pi}{3} = 2\), \(\tan\dfrac{\pi}{3} = \sqrt{3}\), \(\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}\) and \(\ln 1 = 0\)
Depends on the third M mark.
1 mark
A1 \(\ln\left(2 + \sqrt{3}\right) - \dfrac{\sqrt{3}}{2}\) cso
Accept any exact equivalent, e.g. \(-\ln\left(2 - \sqrt{3}\right) - \dfrac{1}{2}\sqrt{3}\). A decimal (0.451) on its own scores A0.

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