The shape of the lesson
By the end of the lesson, learners can
- Exponentials and logarithms: divide by a: Each result below is a derivative read backwards.
- Sine, cosine and secant squared: watch the signs: The trigonometric derivatives \(\dfrac{d}{dx}\sin x = \cos x\), \(\dfrac{d}{dx}\cos x = -\sin x\) and \(\dfrac{d}{dx}\tan x = \sec^2 x\) only hold in radians, so these integrals do too.
- Paper 3 only: the inverse tan integral: In Pure 3 you meet \(\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}\).
- Recognise the standard form: \(e^{ax+b}\), \(\dfrac{1}{ax+b}\), \(\sin\), \(\cos\), \(\sec^2\) or \(\dfrac{1}{x^2+a^2}\).
- Divide by the coefficient of \(x\) inside the bracket.
- Signs: \(\int \sin = -\cos\), \(\int \cos = +\sin\).
- Radians for every trigonometric limit.
- Exact answers: tidy logs with the log laws and keep \(\pi\), \(e\) and surds.
One picture every idea returns to
If \(\dfrac{d}{dx}F(x) = f(x)\), then \(\displaystyle\int f(ax+b)\,dx = \dfrac{1}{a}F(ax+b) + c\). Integrate as if the bracket were just \(x\), then divide by \(a\), the coefficient of \(x\). Angles are always in radians.
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, Reverse the chain, divide by a.. Papers 2 and 3. 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: Exponentials and logarithms: divide by a | Each result below is a derivative read backwards. Model the worked example Indefinite integral on the board, one line at a time. Your turn: Find \(\displaystyle\int \left(e^{x} + 1\right)^{2} dx\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: Sine, cosine and secant squared: watch the signs | The trigonometric derivatives \(\dfrac{d}{dx}\sin x = \cos x\), \(\dfrac{d}{dx}\cos x = -\sin x\) and \(\dfrac{d}{dx}\tan x = \sec^2 x\) only hold in radians, so these integrals do too. Model the worked example Sine with a definite integral on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Paper 3 only: the inverse tan integral | In Pure 3 you meet \(\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}\). Model the worked example Straight from the formula on the board, one line at a time. | Cooperative Concept Mapping |
| 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, 22 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 Paper 3 only: the inverse tan integral, 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.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
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 |
|---|---|
| \(\displaystyle\int \dfrac{1}{2x+1}\,dx\) is NOT \(\ln|2x+1|\). | Differentiate \(\ln|2x+1|\) and you get \(\dfrac{2}{2x+1}\), twice too big, so the answer needs \(\dfrac{1}{2}\) in front. Equally, never multiply by \(a\): that is differentiating, not integrating. |
| \(\displaystyle\int \sin x\,dx = -\cos x\), with a MINUS. | The most common lost mark is writing \(+\cos x\). A quick check: differentiate your answer and see whether you get back exactly what you started with. |
| Do not forget the \(\dfrac{1}{a}\) in front of \(\tan^{-1}\), and do not leave \(\tan^{-1}1\) in degrees: \(\tan^{-1}1 = \dfrac{\pi}{4}\), never 45, inside a calculus answer. | Return to Paper 3 only: the inverse tan integral: ask a pair to state the correct version and show it on the lab before moving on. |
Integrating exponentials, 1/(ax + b) and trigonometric functions: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Find \(\displaystyle\int \left(e^{-2x} + \dfrac{5}{2x-1}\right) dx\).
Find \(\displaystyle\int \left(4\cos\dfrac{x}{2} + \sec^2 4x\right) dx\).
Show that \(\displaystyle\int_0^{\ln 3} \left(e^{x} + e^{-x}\right)^{2} dx = \dfrac{40}{9} + 2\ln 3\).
Given that \(k > 1\) and \(\displaystyle\int_1^{k} \dfrac{4}{2x-1}\,dx = \ln 25\), find the value of \(k\).
(Paper 3) Find the exact value of \(\displaystyle\int_0^{2/3} \dfrac{1}{4+9x^2}\,dx\).
The curve \(y = e^{2x} - \dfrac{6}{2x+1}\) lies above the \(x\)-axis for \(1 \le x \le 2\).
(a) Verify that \(y > 0\) when \(x = 1\).
(b) Find the exact area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 2\), and give its value to 3 significant figures.
Answer key and marking notes
- \(\displaystyle\int e^{-2x}\,dx = -\dfrac{1}{2}e^{-2x}\).
- \(\displaystyle\int \dfrac{5}{2x-1}\,dx = \dfrac{5}{2}\ln|2x-1|\).
- Answer: \(-\dfrac{1}{2}e^{-2x} + \dfrac{5}{2}\ln|2x-1| + c\).
- \(\displaystyle\int 4\cos\dfrac{x}{2}\,dx = 4 \times 2\sin\dfrac{x}{2} = 8\sin\dfrac{x}{2}\).
- \(\displaystyle\int \sec^2 4x\,dx = \dfrac{1}{4}\tan 4x\).
- Answer: \(8\sin\dfrac{x}{2} + \dfrac{1}{4}\tan 4x + c\).
- Expand: \((e^{x} + e^{-x})^2 = e^{2x} + 2 + e^{-2x}\).
- Integrate: \(\left[\dfrac{1}{2}e^{2x} + 2x - \dfrac{1}{2}e^{-2x}\right]_0^{\ln 3}\).
- Upper: \(e^{2\ln 3} = 9\) and \(e^{-2\ln 3} = \dfrac{1}{9}\), so \(\dfrac{9}{2} + 2\ln 3 - \dfrac{1}{18}\).
- Lower: \(\dfrac{1}{2} + 0 - \dfrac{1}{2} = 0\).
- \(\dfrac{9}{2} - \dfrac{1}{18} = \dfrac{81 - 1}{18} = \dfrac{40}{9}\), so the value is \(\dfrac{40}{9} + 2\ln 3\).
- \(\displaystyle\int \dfrac{4}{2x-1}\,dx = 2\ln(2x-1)\).
- \(\left[2\ln(2x-1)\right]_1^{k} = 2\ln(2k-1) - 2\ln 1 = 2\ln(2k-1)\).
- \(2\ln(2k-1) = \ln 25 \Rightarrow \ln(2k-1)^2 = \ln 25 \Rightarrow (2k-1)^2 = 25\).
- \(2k - 1 > 0\), so \(2k - 1 = 5\) and \(k = 3\).
- \(\dfrac{1}{4+9x^2} = \dfrac{1}{9}\times\dfrac{1}{x^2+\frac{4}{9}}\), so \(a = \dfrac{2}{3}\).
- \(\displaystyle\int \dfrac{1}{4+9x^2}\,dx = \dfrac{1}{9}\times\dfrac{3}{2}\tan^{-1}\dfrac{3x}{2} = \dfrac{1}{6}\tan^{-1}\dfrac{3x}{2}\).
- At \(x = \dfrac{2}{3}\): \(\dfrac{1}{6}\tan^{-1}1 = \dfrac{1}{6}\times\dfrac{\pi}{4}\). At 0 the value is 0.
- Answer: \(\dfrac{\pi}{24}\).
- (a) At \(x = 1\): \(y = e^{2} - \dfrac{6}{3} = e^2 - 2 = 5.39 > 0\).
- (b) \(\displaystyle\int \left(e^{2x} - \dfrac{6}{2x+1}\right) dx = \dfrac{1}{2}e^{2x} - 3\ln(2x+1)\).
- Upper: \(\dfrac{1}{2}e^{4} - 3\ln 5\). Lower: \(\dfrac{1}{2}e^{2} - 3\ln 3\).
- Area \(= \dfrac{1}{2}(e^{4} - e^{2}) - 3\ln 5 + 3\ln 3 = \dfrac{1}{2}(e^{4} - e^{2}) - 3\ln\dfrac{5}{3}\).
- \(= 23.6045 - 1.5325 = 22.1\) 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.
Find \(\displaystyle\int e^{4x - 1}\,dx\).
- A\(\dfrac{1}{4}e^{4x - 1} + c\)
- B\(4e^{4x - 1} + c\)
- C\(\dfrac{e^{4x - 1}}{4x - 1} + c\)
- D\(e^{4x - 1} + c\)
Find the exact value of \(\displaystyle\int_0^1 \dfrac{6}{3x + 1}\,dx\).
- A\(2\ln 4\)
- B\(6\ln 4\)
- C\(18\ln 4\)
- D\(\ln 4\)
Find the exact value of \(\displaystyle\int_0^{\frac{\pi}{6}} \cos 3x\,dx\).
- A\(\dfrac{1}{3}\)
- B\(3\)
- C\(-\dfrac{1}{3}\)
- D\(\dfrac{1}{6}\)
Find the exact value of \(\displaystyle\int_0^{\frac{\pi}{8}} \sec^2 2x\,dx\).
- A\(\dfrac{1}{2}\)
- B\(2\)
- C\(1\)
- D\(\dfrac{\sqrt{2} - 1}{2}\)
Find the exact value of \(\displaystyle\int_0^{\frac{\pi}{6}} (\cos 2x + \sin 3x)\,dx\).
- A\(\dfrac{\sqrt{3}}{4} + \dfrac{1}{3}\)
- B\(\dfrac{\sqrt{3}}{4}\)
- C\(\dfrac{\sqrt{3}}{4} - \dfrac{1}{3}\)
- D\(\sqrt{3} + 3\)
(Paper 3 only) Find the exact value of \(\displaystyle\int_0^2 \dfrac{1}{x^2 + 4}\,dx\).
- A\(\dfrac{\pi}{8}\)
- B\(\dfrac{\pi}{4}\)
- C\(\dfrac{\pi}{2}\)
- D\(\ln 2\)
(Paper 3 only) Find \(\displaystyle\int \dfrac{1}{9 + 4x^2}\,dx\).
- A\(\dfrac{1}{6}\tan^{-1}\left(\dfrac{2x}{3}\right) + c\)
- B\(\dfrac{2}{3}\tan^{-1}\left(\dfrac{2x}{3}\right) + c\)
- C\(\dfrac{1}{12}\tan^{-1}\left(\dfrac{x}{3}\right) + c\)
- D\(\dfrac{1}{3}\tan^{-1}\left(\dfrac{2x}{3}\right) + c\)
Given that \(\displaystyle\int_0^k e^{2x}\,dx = 4\), find the exact value of \(k\).
- A\(k = \ln 3\)
- B\(k = \dfrac{1}{2}\ln 5\)
- C\(k = \dfrac{1}{2}\ln 8\)
- D\(k = \ln 9\)
Quiz key
\(\displaystyle\int e^{ax + b}\,dx = \dfrac{1}{a}e^{ax + b} + c\), so the answer is \(\dfrac{1}{4}e^{4x - 1} + c\). Check: differentiating gives \(\dfrac{1}{4} \times 4e^{4x - 1}\). The tempting \(4e^{4x - 1}\) multiplies by 4, which is what DIFFERENTIATING does. Dividing by \(4x - 1\) treats the exponential like a power, and \(e^{4x - 1}\) forgets the chain-rule factor altogether.
\(\left[\dfrac{6}{3}\ln(3x + 1)\right]_0^1 = 2\ln 4 - 2\ln 1 = 2\ln 4\) (also \(= \ln 16 = 4\ln 2\)). The tempting \(6\ln 4\) forgets to divide by the coefficient 3. \(18\ln 4\) multiplies by 3 instead, and \(\ln 4\) loses the factor 6 in the numerator.
\(\left[\dfrac{1}{3}\sin 3x\right]_0^{\pi/6} = \dfrac{1}{3}\sin\dfrac{\pi}{2} - 0 = \dfrac{1}{3}\). The tempting 3 multiplies by 3 instead of dividing. \(-\dfrac{1}{3}\) integrates \(\cos\) to \(-\sin\), a sign slip, and \(\dfrac{1}{6}\) evaluates \(\sin\dfrac{\pi}{6}\) instead of \(\sin\left(3 \times \dfrac{\pi}{6}\right)\).
\(\left[\dfrac{1}{2}\tan 2x\right]_0^{\pi/8} = \dfrac{1}{2}\tan\dfrac{\pi}{4} - 0 = \dfrac{1}{2}\). The tempting 2 multiplies by the coefficient instead of dividing. 1 forgets the \(\dfrac{1}{2}\), and \(\dfrac{\sqrt{2} - 1}{2} = \dfrac{1}{2}\tan\dfrac{\pi}{8}\) substitutes \(x\) instead of \(2x\).
\(\left[\dfrac{1}{2}\sin 2x - \dfrac{1}{3}\cos 3x\right]_0^{\pi/6} = \left(\dfrac{1}{2} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{1}{3}\cos\dfrac{\pi}{2}\right) - \left(0 - \dfrac{1}{3}\right) = \dfrac{\sqrt{3}}{4} + \dfrac{1}{3}\). The tempting \(\dfrac{\sqrt{3}}{4}\) assumes the lower limit gives 0, but \(-\dfrac{1}{3}\cos 0 = -\dfrac{1}{3}\). \(\dfrac{\sqrt{3}}{4} - \dfrac{1}{3}\) integrates \(\sin 3x\) to \(+\dfrac{1}{3}\cos 3x\), and \(\sqrt{3} + 3\) multiplies by the coefficients instead of dividing.
\(\left[\dfrac{1}{2}\tan^{-1}\dfrac{x}{2}\right]_0^2 = \dfrac{1}{2}\tan^{-1}1 - 0 = \dfrac{1}{2} \times \dfrac{\pi}{4} = \dfrac{\pi}{8}\). The tempting \(\dfrac{\pi}{4}\) forgets the factor \(\dfrac{1}{a}\). \(\dfrac{\pi}{2}\) multiplies by \(a\) instead, and \(\ln 2 = \left[\ln(x^2 + 4)\right]_0^2\) treats the integrand as \(\dfrac{f'(x)}{f(x)}\), which it is not (there is no \(2x\) on top).
\(\dfrac{1}{4}\displaystyle\int \dfrac{1}{x^2 + \left(\frac{3}{2}\right)^2}\,dx = \dfrac{1}{4} \cdot \dfrac{2}{3}\tan^{-1}\left(\dfrac{2x}{3}\right) + c = \dfrac{1}{6}\tan^{-1}\left(\dfrac{2x}{3}\right) + c\). Check by differentiating: \(\dfrac{1}{6} \cdot \dfrac{2/3}{1 + 4x^2/9} = \dfrac{1}{9 + 4x^2}\). The tempting \(\dfrac{2}{3}\tan^{-1}\) forgets the \(\dfrac{1}{4}\) taken out. \(\dfrac{1}{12}\tan^{-1}\dfrac{x}{3}\) uses \(a = 3\) without dividing \(9\) by 4, and \(\dfrac{1}{3}\) halves instead of quartering.
\(\left[\dfrac{1}{2}e^{2x}\right]_0^k = \dfrac{1}{2}e^{2k} - \dfrac{1}{2} = 4\), so \(e^{2k} = 9\), \(2k = \ln 9\) and \(k = \dfrac{1}{2}\ln 9 = \ln 3\). The tempting \(\dfrac{1}{2}\ln 5\) forgets the \(\dfrac{1}{2}\) from integrating (\(e^{2k} - 1 = 4\)). \(\dfrac{1}{2}\ln 8\) forgets the lower limit (\(e^0 = 1\), not 0), and \(\ln 9\) forgets to halve \(2k = \ln 9\).
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.