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

Practise 16 exam-style questions on Integration by parts 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.

16 questions
3–13 marks each
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16 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

5 marks

Find
\[\int \left(x^2 + 2x\right)\cos x\,dx\]

[Integration by parts: \(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\)]

Model answer

\(u = x^2 + 2x\), \(\dfrac{dv}{dx} = \cos x\), \(\dfrac{du}{dx} = 2x + 2\), \(v = \sin x\)
\(\displaystyle\int \left(x^2 + 2x\right)\cos x\,dx = \left(x^2 + 2x\right)\sin x - \int (2x + 2)\sin x\,dx\)
\(u = 2x + 2\), \(\dfrac{dv}{dx} = \sin x\), \(v = -\cos x\): \(\displaystyle\int (2x + 2)\sin x\,dx = -(2x + 2)\cos x + \int 2\cos x\,dx = -(2x + 2)\cos x + 2\sin x\)
So \(\displaystyle\int \left(x^2 + 2x\right)\cos x\,dx = \left(x^2 + 2x\right)\sin x + (2x + 2)\cos x - 2\sin x + c\)
\(= \left(x^2 + 2x - 2\right)\sin x + 2(x + 1)\cos x + c\)

Explanation

Take \(u = x^2 + 2x\) so that each application of parts lowers the power of \(x\); two applications leave only \(\displaystyle\int 2\cos x\,dx\). The second integral is subtracted and \(v = -\cos x\) in the second application, so the signs need care. Common errors are choosing \(u = \cos x\), integrating \(\cos x\) as \(-\sin x\) and losing a sign in the second application.

Mark scheme

1 mark
M1 Integrates by parts with \(u = x^2 + 2x\) and \(\dfrac{dv}{dx} = \cos x\), reaching \(\left(x^2 + 2x\right)\sin x - \displaystyle\int (\alpha x + \beta)\sin x\,dx\); a correct final answer of \(\left(x^2 + 2x - 2\right)\sin x + 2(x + 1)\cos x + c\) with no working scores 2 marks (special case)
An answer of \(\left(x^2 + 2x - 2\right)\sin x + 2(x + 1)\cos x + c\) with no working scores 2 marks: a special case, as answers without working may not gain full credit when an integration method is needed. The tabular (DI) method is acceptable if its rows are shown. Choosing \(u = \cos x\) is M0.
1 mark
A1 \(\left(x^2 + 2x\right)\sin x - \displaystyle\int (2x + 2)\sin x\,dx\)
1 mark
M1 (dM1) Integrates \((2x + 2)\sin x\) by parts again, reaching \(\pm(2x + 2)\cos x \pm \displaystyle\int \gamma\cos x\,dx\)
Depends on the first M mark.
1 mark
A1 \(\displaystyle\int (2x + 2)\sin x\,dx = -(2x + 2)\cos x + 2\sin x\)
1 mark
A1 \(\left(x^2 + 2x - 2\right)\sin x + 2(x + 1)\cos x + c\) oe
Depends on both M marks. Accept any correct unsimplified form, e.g. \(\left(x^2 + 2x\right)\sin x + (2x + 2)\cos x - 2\sin x + c\), isw. The constant is required.

Question 2

6 marks

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

Find the exact value of
\[\int_{0}^{\frac{\pi}{6}} x\cos 3x\,dx\]

[Integration by parts: \(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\)]

Model answer

\(u = x\), \(\dfrac{dv}{dx} = \cos 3x\), \(\dfrac{du}{dx} = 1\), \(v = \dfrac{1}{3}\sin 3x\)
\(\displaystyle\int x\cos 3x\,dx = \dfrac{1}{3}x\sin 3x - \int \dfrac{1}{3}\sin 3x\,dx = \dfrac{1}{3}x\sin 3x + \dfrac{1}{9}\cos 3x\)
\(\displaystyle\int_0^{\frac{\pi}{6}} x\cos 3x\,dx = \left(\dfrac{\pi}{18} \times 1 + \dfrac{1}{9} \times 0\right) - \left(0 + \dfrac{1}{9}\right) = \dfrac{\pi}{18} - \dfrac{1}{9}\)

Explanation

Take \(u = x\) so that its derivative is 1, leaving \(\dfrac{1}{3}\displaystyle\int \sin 3x\,dx\) to integrate. At the upper limit \(3x = \dfrac{\pi}{2}\), so the cosine term is 0 and the sine term is 1; at the lower limit only the cosine term survives. Common errors are a sign slip when integrating \(\sin 3x\), forgetting \(\cos 0 = 1\), and working in degrees.

Mark scheme

1 mark
M1 Integrates by parts the right way round (\(u = x\)), reaching \(\alpha x\sin 3x - \beta\displaystyle\int \sin 3x\,dx\). An answer with no working scores 0 marks.
An answer of \(\dfrac{\pi}{18} - \dfrac{1}{9}\) with no working scores 0 marks: the question demands working.
1 mark
A1 \(\dfrac{1}{3}x\sin 3x - \dfrac{1}{3}\displaystyle\int \sin 3x\,dx\)
1 mark
M1 (dM1) Integrates \(\sin 3x\) to \(k\cos 3x\)
Depends on the first M mark.
1 mark
A1 \(\dfrac{1}{3}x\sin 3x + \dfrac{1}{9}\cos 3x\)
1 mark
M1 (ddM1) Substitutes the limits \(\dfrac{\pi}{6}\) and 0 and subtracts, using \(\sin\dfrac{\pi}{2} = 1\), \(\cos\dfrac{\pi}{2} = 0\) and \(\cos 0 = 1\)
Depends on both previous M marks. Both limits and the subtraction must be seen.
1 mark
A1 \(\dfrac{\pi}{18} - \dfrac{1}{9}\) cso
Accept \(\dfrac{\pi - 2}{18}\). A decimal (0.0634) 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.

Figure 1 shows part of the curve \(C\) with equation
\[y = (x - 1)\sin x, \quad 0 \leqslant x \leqslant \pi\]
where \(x\) is in radians. The curve meets the \(x\)-axis at the point \(A\) and at the point \(B(\pi, 0)\), and lies above the \(x\)-axis between \(A\) and \(B\).
The finite region \(R\), shaded in Figure 1, is bounded by \(C\) and the \(x\)-axis between \(A\) and \(B\).

Find the exact area of \(R\), giving your answer in the form \(p\pi + q + r\sin 1\), where \(p\), \(q\) and \(r\) are integers.

[Integration by parts: \(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\)]

Figure 1: the curve C starts at the origin, dips a little below the x-axis, crosses the x-axis at A, rises to a maximum and comes back down to meet the x-axis at B, where π is marked. The region R between C and the x-axis from A to B, above the axis, is shaded.

Model answer

For \(0 < x < \pi\), \(\sin x ≠ 0\), so \((x - 1)\sin x = 0\) only when \(x = 1\): \(A\) is \((1, 0)\).
\(u = x - 1\), \(\dfrac{dv}{dx} = \sin x\), \(\dfrac{du}{dx} = 1\), \(v = -\cos x\)
\(\displaystyle\int (x - 1)\sin x\,dx = -(x - 1)\cos x + \int \cos x\,dx = -(x - 1)\cos x + \sin x\)
Area of \(R = \displaystyle\int_1^{\pi} (x - 1)\sin x\,dx = \left[-(x - 1)\cos x + \sin x\right]_1^{\pi}\)
\(= \left(-(\pi - 1)(-1) + 0\right) - (0 + \sin 1)\)
\(= \pi - 1 - \sin 1\)

Explanation

The curve meets the \(x\)-axis where \(x - 1 = 0\), so \(R\) runs from \(x = 1\) to \(x = \pi\). Integrate by parts with \(u = x - 1\) and \(\dfrac{dv}{dx} = \sin x\), which gives \(-(x - 1)\cos x + \sin x\); then \(\cos \pi = -1\) gives the exact area \(\pi - 1 - \sin 1\). Common errors are choosing \(u = \sin x\), sign slips with \(-\cos x\), and using the limits 0 and \(\pi\), which include a region below the axis.

Mark scheme

1 mark
M1 Integrates \((x - 1)\sin x\) by parts with \(u = x - 1\), reaching the form \(\pm(x - 1)\cos x \pm \displaystyle\int \cos x\,dx\). An answer with no working scores 0 marks.
An answer of \(\pi - 1 - \sin 1\) with no working scores 0 marks: the question demands working. Choosing \(u = \sin x\) is M0. Expanding to \(x\sin x - \sin x\) and integrating \(x\sin x\) by parts is equally acceptable; mark it in the same way.
1 mark
A1 \(-(x - 1)\cos x + \displaystyle\int \cos x\,dx\)
1 mark
M1 (dM1) Integrates \(\cos x\) to \(\sin x\) to complete the integral
Depends on the first M mark.
1 mark
A1 \(-(x - 1)\cos x + \sin x\) oe
e.g. \(-x\cos x + \cos x + \sin x\).
1 mark
M1 (ddM1) Finds \(x = 1\) at \(A\) (from \(x - 1 = 0\), as \(\sin x ≠ 0\) for \(0 < x < \pi\)) and substitutes the limits \(\pi\) and 1, subtracting: \(\left((\pi - 1) + 0\right) - (0 + \sin 1)\)
Depends on both previous M marks. Using the limits 0 and \(\pi\) (the whole interval) is M0: the curve is below the \(x\)-axis for \(0 < x < 1\).
1 mark
A1 \(\pi - 1 - \sin 1\) cso
Accept \(p = 1\), \(q = -1\), \(r = -1\). A decimal (1.30) on its own scores A0.

Question 4

10 marks

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

Figure 1 shows the curve \(C\) with parametric equations
\[x = 4\sin\left(\dfrac{t}{2}\right), \quad y = t^2, \quad 0 \leqslant t \leqslant \pi\]
The curve starts at the origin \(O\) and ends at the point \(D(4, \pi^2)\). The region \(R\), shaded in Figure 1, is bounded by \(C\), the \(x\)-axis and the line with equation \(x = 4\).

Show that the area of \(R\) is given by
\[\int_{0}^{\pi} 2t^2\cos\left(\dfrac{t}{2}\right)\,dt\]
Hence, using integration by parts, find the exact area of \(R\).

[Integration by parts: \(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\)]

Figure 1: the curve C starts at the origin O and rises, slowly at first and then very steeply, to the point D at the top of a vertical line x = 4. The region R under the curve, above the x-axis and to the left of the line x = 4, is shaded.

Model answer

\(\dfrac{dx}{dt} = 4 \times \dfrac{1}{2}\cos\left(\dfrac{t}{2}\right) = 2\cos\left(\dfrac{t}{2}\right)\). At \(O\), \(t = 0\); \(x = 4\) when \(\sin\left(\dfrac{t}{2}\right) = 1\), so \(t = \pi\).
Area \(= \displaystyle\int_0^{\pi} y\,\dfrac{dx}{dt}\,dt = \int_0^{\pi} t^2 \times 2\cos\left(\dfrac{t}{2}\right)\,dt = \int_0^{\pi} 2t^2\cos\left(\dfrac{t}{2}\right)\,dt\)
\(u = 2t^2\), \(\dfrac{dv}{dt} = \cos\left(\dfrac{t}{2}\right)\), \(v = 2\sin\left(\dfrac{t}{2}\right)\): \(\displaystyle\int 2t^2\cos\left(\dfrac{t}{2}\right)\,dt = 4t^2\sin\left(\dfrac{t}{2}\right) - \int 8t\sin\left(\dfrac{t}{2}\right)\,dt\)
\(u = 8t\), \(\dfrac{dv}{dt} = \sin\left(\dfrac{t}{2}\right)\), \(v = -2\cos\left(\dfrac{t}{2}\right)\): \(\displaystyle\int 8t\sin\left(\dfrac{t}{2}\right)\,dt = -16t\cos\left(\dfrac{t}{2}\right) + \int 16\cos\left(\dfrac{t}{2}\right)\,dt = -16t\cos\left(\dfrac{t}{2}\right) + 32\sin\left(\dfrac{t}{2}\right)\)
Area \(= \left[4t^2\sin\left(\dfrac{t}{2}\right) + 16t\cos\left(\dfrac{t}{2}\right) - 32\sin\left(\dfrac{t}{2}\right)\right]_0^{\pi} = \left(4\pi^2 + 0 - 32\right) - 0 = 4\pi^2 - 32\)

Explanation

The area is \(\displaystyle\int y\,\dfrac{dx}{dt}\,dt\) with limits in \(t\): \(t = 0\) at \(O\) and \(t = \pi\) where \(x = 4\). Parts is needed twice, each time differentiating the power of \(t\), and each integration of \(\cos\left(\dfrac{t}{2}\right)\) or \(\sin\left(\dfrac{t}{2}\right)\) brings in a factor of 2. The examiners' reports single out sign errors in repeated parts and limits not shown being substituted; dividing by 2 instead of multiplying when integrating \(\cos\left(\dfrac{t}{2}\right)\) is a common slip here.

Mark scheme

1 mark
B1 \(\dfrac{dx}{dt} = 2\cos\left(\dfrac{t}{2}\right)\). An answer with no working scores 0 marks.
An answer of \(4\pi^2 - 32\) with no working scores 0 marks: the question demands working.
1 mark
M1 Uses area \(= \displaystyle\int y\,\dfrac{dx}{dt}\,dt = \int t^2 \times 2\cos\left(\dfrac{t}{2}\right)\,dt\)
1 mark
B1 The limits \(t = 0\) (at \(O\)) and \(t = \pi\) (at \(x = 4\), where \(\sin\left(\dfrac{t}{2}\right) = 1\))
Using the \(x\)-limits 0 and 4 is B0.
1 mark
A1* \(\displaystyle\int_0^{\pi} 2t^2\cos\left(\dfrac{t}{2}\right)\,dt\), with the integral sign, the limits and the \(dt\), and no errors
Depends on the M mark.
1 mark
M1 First application of parts with \(u = 2t^2\): \(4t^2\sin\left(\dfrac{t}{2}\right) - \displaystyle\int 8t\sin\left(\dfrac{t}{2}\right)\,dt\) (signs and coefficients may be slightly wrong)
The given integral may be used even if it was not shown. Choosing \(u = \cos\left(\dfrac{t}{2}\right)\) is M0.
1 mark
A1 \(4t^2\sin\left(\dfrac{t}{2}\right) - \displaystyle\int 8t\sin\left(\dfrac{t}{2}\right)\,dt\)
1 mark
M1 (dM1) Second application of parts: \(\displaystyle\int 8t\sin\left(\dfrac{t}{2}\right)\,dt = -16t\cos\left(\dfrac{t}{2}\right) + \int 16\cos\left(\dfrac{t}{2}\right)\,dt\)
Depends on the previous M mark (the first application of parts).
1 mark
A1 \(4t^2\sin\left(\dfrac{t}{2}\right) + 16t\cos\left(\dfrac{t}{2}\right) - 32\sin\left(\dfrac{t}{2}\right)\)
1 mark
M1 (ddM1) Substitutes the limits \(\pi\) and 0, using \(\sin\dfrac{\pi}{2} = 1\), \(\cos\dfrac{\pi}{2} = 0\) and \(\sin 0 = 0\), and subtracts
Depends on the M marks for both applications of parts. Both limits and the subtraction must be seen.
1 mark
A1 \(4\pi^2 - 32\) cso
From \(\left(4\pi^2 + 0 - 32\right) - (0 + 0 - 0)\). A decimal (7.48) on its own scores A0.

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