The shape of the lesson
By the end of the lesson, learners can
- The formula and choosing u: Integrate the product rule \(\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\) and rearrange to get the formula.
- Integrating ln x: the hidden 1: You cannot integrate \(\ln x\) directly, but you can differentiate it to \(\dfrac{1}{x}\).
- Parts twice, and inverse tan: With \(x^2\) as \(u\), one application leaves \(x\) times something, so apply parts again.
- Formula: \(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\).
- u is \(\ln x\) or \(\tan^{-1}x\) if present, otherwise the power of \(x\).
- Lone ln x: take \(\dfrac{dv}{dx} = 1\).
- Twice for \(x^2\) times an exponential or trig function; bracket the second result.
- Check by differentiating your answer.
One picture every idea returns to
\(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\). Choose \(u\) to be the factor that gets simpler when differentiated (a power of \(x\), or \(\ln x\) and \(\tan^{-1}x\), which you cannot integrate directly), and \(\dfrac{dv}{dx}\) to be the factor you can integrate.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, The product rule, run backwards.. Paper 3 only. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The formula and choosing u | Integrate the product rule \(\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\) and rearrange to get the formula. Model the worked example Power times exponential on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Integrating ln x: the hidden 1 | You cannot integrate \(\ln x\) directly, but you can differentiate it to \(\dfrac{1}{x}\). Model the worked example ln x on its own on the board, one line at a time. Your turn: Find \(\displaystyle\int x\ln x\,dx\). | Pairs |
| 21 to 30 min | Teach and check: Parts twice, and inverse tan | With \(x^2\) as \(u\), one application leaves \(x\) times something, so apply parts again. Model the worked example Parts twice on the board, one line at a time. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 30 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Parts twice, and inverse tan, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Choosing \(u = e^{2x}\) and \(\dfrac{dv}{dx} = x\) makes things worse: the new integral contains \(\dfrac{x^2}{2}e^{2x}\). | If the integral you produce is harder than the original, swap \(u\) and \(\dfrac{dv}{dx}\). |
| With \(\ln x\) never choose \(\dfrac{dv}{dx} = \ln x\): you would need to integrate \(\ln x\) to find \(v\), which is the very thing you cannot yet do. | Return to Integrating ln x: the hidden 1: ask a pair to state the correct version and show it on the lab before moving on. |
| The minus sign in front of the second integral applies to EVERYTHING it produces. | When parts is used twice, bracket the second result: \(-x^2e^{-x} + \left(-2xe^{-x} - 2e^{-x}\right)\). Sign slips here cost the accuracy marks. |
Integration by parts: worksheet
6 questions, 30 marks. Written in the style of the papers; no past paper question is reproduced.
Find \(\displaystyle\int xe^{3x}\,dx\).
Find the exact value of \(\displaystyle\int_0^{\pi} x\sin x\,dx\).
Show that \(\displaystyle\int_1^{2} x^2\ln x\,dx = \dfrac{8}{3}\ln 2 - \dfrac{7}{9}\).
Show that \(\displaystyle\int_0^1 x^2e^{x}\,dx = e - 2\).
Find the exact value of \(\displaystyle\int_0^1 \tan^{-1}x\,dx\).
The curve \(y = xe^{-x/2}\) has a maximum point \(M\).
(a) Find the exact coordinates of \(M\).
(b) Find the exact area of the region bounded by the curve, the \(x\)-axis and the line through \(M\) parallel to the \(y\)-axis, and give it to 3 significant figures.
Answer key and marking notes
- \(u = x\), \(v = \dfrac{1}{3}e^{3x}\).
- \(\dfrac{x}{3}e^{3x} - \displaystyle\int \dfrac{1}{3}e^{3x}\,dx = \dfrac{x}{3}e^{3x} - \dfrac{1}{9}e^{3x} + c\).
- \(u = x\), \(v = -\cos x\): \(\displaystyle\int x\sin x\,dx = -x\cos x + \int \cos x\,dx = -x\cos x + \sin x\).
- \(\left[-x\cos x + \sin x\right]_0^{\pi} = (-\pi\cos\pi + 0) - 0 = \pi\).
- \(u = \ln x\), \(\dfrac{du}{dx} = \dfrac{1}{x}\); \(\dfrac{dv}{dx} = x^2\), \(v = \dfrac{x^3}{3}\).
- \(\dfrac{x^3}{3}\ln x - \displaystyle\int \dfrac{x^3}{3}\cdot\dfrac{1}{x}\,dx = \dfrac{x^3}{3}\ln x - \dfrac{x^3}{9}\).
- \(\left(\dfrac{8}{3}\ln 2 - \dfrac{8}{9}\right) - \left(0 - \dfrac{1}{9}\right) = \dfrac{8}{3}\ln 2 - \dfrac{7}{9}\).
- \(u = x^2\), \(v = e^{x}\): \(x^2e^{x} - \displaystyle\int 2xe^{x}\,dx\).
- \(\displaystyle\int 2xe^{x}\,dx = 2xe^{x} - 2e^{x}\).
- So \(\displaystyle\int x^2e^{x}\,dx = e^{x}(x^2 - 2x + 2)\).
- \(\left[e^{x}(x^2 - 2x + 2)\right]_0^1 = e(1) - 1(2) = e - 2\).
- \(u = \tan^{-1}x\), \(\dfrac{dv}{dx} = 1\), so \(v = x\).
- \(x\tan^{-1}x - \displaystyle\int \dfrac{x}{1+x^2}\,dx = x\tan^{-1}x - \dfrac{1}{2}\ln(1+x^2)\).
- \(\left[x\tan^{-1}x - \dfrac{1}{2}\ln(1+x^2)\right]_0^1 = \dfrac{\pi}{4} - \dfrac{1}{2}\ln 2\).
- (a) \(\dfrac{dy}{dx} = e^{-x/2} - \dfrac{x}{2}e^{-x/2} = e^{-x/2}\left(1 - \dfrac{x}{2}\right)\).
- \(\dfrac{dy}{dx} = 0 \Rightarrow x = 2\), \(y = 2e^{-1}\). So \(M = \left(2, \dfrac{2}{e}\right)\).
- (b) Area \(= \displaystyle\int_0^2 xe^{-x/2}\,dx\). Parts: \(u = x\), \(v = -2e^{-x/2}\).
- \(\displaystyle\int xe^{-x/2}\,dx = -2xe^{-x/2} + \int 2e^{-x/2}\,dx = -2xe^{-x/2} - 4e^{-x/2}\).
- \(\left[-2xe^{-x/2} - 4e^{-x/2}\right]_0^2 = \left(-\dfrac{4}{e} - \dfrac{4}{e}\right) - (-4) = 4 - \dfrac{8}{e}\).
- \(= 1.06\) square units (3 s.f.).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
(Paper 3 only) Find \(\displaystyle\int xe^{3x}\,dx\).
- A\(\dfrac{x}{3}e^{3x} - \dfrac{1}{9}e^{3x} + c\)
- B\(\dfrac{x}{3}e^{3x} + \dfrac{1}{9}e^{3x} + c\)
- C\(\dfrac{x}{3}e^{3x} - \dfrac{1}{3}e^{3x} + c\)
- D\(\dfrac{x^2}{6}e^{3x} + c\)
(Paper 3 only) Find the exact value of \(\displaystyle\int_0^{\frac{\pi}{2}} x\cos x\,dx\).
- A\(\dfrac{\pi}{2} - 1\)
- B\(\dfrac{\pi}{2} + 1\)
- C\(\dfrac{\pi}{2}\)
- D\(\dfrac{\pi^2}{8}\)
(Paper 3 only) Find \(\displaystyle\int x\ln x\,dx\).
- A\(\dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + c\)
- B\(\dfrac{x^2}{2}\ln x - \dfrac{x^2}{2} + c\)
- C\(\dfrac{x^2}{2}\ln x + \dfrac{x^2}{4} + c\)
- D\(x\ln x - x + c\)
(Paper 3 only) To find \(\displaystyle\int x^2\ln x\,dx\) by parts, which choice should you make?
- A\(u = \ln x\), \(\dfrac{dv}{dx} = x^2\)
- B\(u = x^2\), \(\dfrac{dv}{dx} = \ln x\)
- C\(u = x^2\ln x\), \(\dfrac{dv}{dx} = 1\)
- D\(u = x\), \(\dfrac{dv}{dx} = x\ln x\)
(Paper 3 only) Find the exact value of \(\displaystyle\int_1^e x^2\ln x\,dx\).
- A\(\dfrac{2e^3 + 1}{9}\)
- B\(\dfrac{2e^3 - 1}{9}\)
- C\(\dfrac{2e^3}{9}\)
- D\(\dfrac{1}{3}\)
(Paper 3 only) Find the exact value of \(\displaystyle\int_0^1 x^2e^{x}\,dx\).
- A\(e - 2\)
- B\(5e - 2\)
- C\(e\)
- D\(2 - 3e\)
(Paper 3 only) Find the exact value of \(\displaystyle\int_0^1 \tan^{-1}x\,dx\).
- A\(\dfrac{\pi}{4} - \dfrac{1}{2}\ln 2\)
- B\(\dfrac{\pi}{4} + \dfrac{1}{2}\ln 2\)
- C\(\dfrac{\pi}{4} - \ln 2\)
- D\(\dfrac{\pi}{4}\)
(Paper 3 only) Find the exact value of \(\displaystyle\int_1^4 \dfrac{\ln x}{\sqrt{x}}\,dx\).
- A\(8\ln 2 - 4\)
- B\(8\ln 2 - 8\)
- C\(8\ln 2 + 4\)
- D\(2\ln 2 - 1\)
Quiz key
\(\dfrac{x}{3}e^{3x} - \displaystyle\int \dfrac{1}{3}e^{3x}\,dx = \dfrac{x}{3}e^{3x} - \dfrac{1}{9}e^{3x} + c\). Check: differentiating gives \(\dfrac{1}{3}e^{3x} + xe^{3x} - \dfrac{1}{3}e^{3x} = xe^{3x}\). The tempting \(+\dfrac{1}{9}\) ignores the minus sign in the formula. \(-\dfrac{1}{3}e^{3x}\) forgets to divide by 3 the second time, and \(\dfrac{x^2}{6}e^{3x}\) integrates each factor separately and multiplies, which is not a rule.
\(\left[x\sin x\right]_0^{\pi/2} - \displaystyle\int_0^{\pi/2} \sin x\,dx = \dfrac{\pi}{2} - \left[-\cos x\right]_0^{\pi/2} = \dfrac{\pi}{2} - (0 + 1) = \dfrac{\pi}{2} - 1\). The tempting \(\dfrac{\pi}{2} + 1\) integrates \(\sin x\) to \(+\cos x\). \(\dfrac{\pi}{2}\) forgets that \(\cos 0 = 1\) at the lower limit, and \(\dfrac{\pi^2}{8}\) uses \(\dfrac{x^2}{2}\sin x\), integrating both factors.
\(v = \dfrac{x^2}{2}\), \(\dfrac{du}{dx} = \dfrac{1}{x}\). So \(\dfrac{x^2}{2}\ln x - \displaystyle\int \dfrac{x^2}{2} \cdot \dfrac{1}{x}\,dx = \dfrac{x^2}{2}\ln x - \int \dfrac{x}{2}\,dx = \dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + c\). The tempting \(-\dfrac{x^2}{2}\) forgets to integrate \(\dfrac{x}{2}\) (it just multiplies back). \(+\dfrac{x^2}{4}\) is a sign slip, and \(x\ln x - x\) is \(\displaystyle\int \ln x\,dx\), not \(\int x\ln x\,dx\).
\(\ln x\) has no simple integral to use as \(v\), but its derivative \(\dfrac{1}{x}\) cancels with a power of \(x\). So \(u = \ln x\), \(v = \dfrac{x^3}{3}\), and the new integral is \(\displaystyle\int \dfrac{x^2}{3}\,dx\). The tempting \(u = x^2\) needs \(v = \displaystyle\int \ln x\,dx\), making things harder. Taking \(\dfrac{dv}{dx} = 1\) leaves you differentiating the whole product, and the last choice still needs \(\displaystyle\int x\ln x\,dx\) for \(v\).
At \(x = e\): \(\dfrac{e^3}{3} - \dfrac{e^3}{9} = \dfrac{2e^3}{9}\). At \(x = 1\): \(0 - \dfrac{1}{9}\). Subtracting: \(\dfrac{2e^3}{9} + \dfrac{1}{9} = \dfrac{2e^3 + 1}{9}\). The tempting \(\dfrac{2e^3 - 1}{9}\) subtracts \(\dfrac{1}{9}\) instead of \(-\dfrac{1}{9}\). \(\dfrac{2e^3}{9}\) assumes the lower limit gives 0, and \(\dfrac{1}{3}\) integrates \(\dfrac{x^2}{3}\) as \(\dfrac{x^3}{3}\), forgetting to divide by 3.
\(\displaystyle\int x^2e^x\,dx = x^2e^x - \int 2xe^x\,dx = x^2e^x - 2(xe^x - e^x) = e^x(x^2 - 2x + 2)\). So \(\left[e^x(x^2 - 2x + 2)\right]_0^1 = e(1) - 1(2) = e - 2\). The tempting \(5e - 2\) loses the minus sign of the second application, giving \(x^2 + 2x + 2\). \(e\) forgets the lower limit (it gives 2, not 0), and \(2 - 3e\) has a sign slip in the last term, \(x^2 - 2x - 2\).
\(\left[x\tan^{-1}x\right]_0^1 - \displaystyle\int_0^1 \dfrac{x}{1 + x^2}\,dx = \dfrac{\pi}{4} - \left[\dfrac{1}{2}\ln(1 + x^2)\right]_0^1 = \dfrac{\pi}{4} - \dfrac{1}{2}\ln 2\). The tempting \(+\dfrac{1}{2}\ln 2\) ignores the minus sign in the parts formula. \(-\ln 2\) forgets the \(\dfrac{1}{2}\) because the top is \(x\), not \(2x\), and \(\dfrac{\pi}{4}\) leaves out the second integral.
\(\left[2\sqrt{x}\ln x\right]_1^4 - \displaystyle\int_1^4 2x^{-\frac{1}{2}}\,dx = 4\ln 4 - \left[4\sqrt{x}\right]_1^4 = 4\ln 4 - (8 - 4) = 8\ln 2 - 4\) (about 1.55). The tempting \(8\ln 2 - 8\) forgets the lower limit of \(\left[4\sqrt{x}\right]\). \(8\ln 2 + 4\) is a sign slip in the formula, and \(2\ln 2 - 1\) integrates \(x^{-\frac{1}{2}}\) to \(\dfrac{1}{2}x^{\frac{1}{2}}\), multiplying by the power instead of dividing.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.