The shape of the lesson
By the end of the lesson, learners can
- The method: convert everything: Treat \(\dfrac{du}{dx}\) like a fraction to swap \(dx\) for \(du\).
- Trigonometric substitutions: spot the derivative: With \(u = \sin x\) the group \(\cos x\,dx\) becomes \(du\); with \(u = \cos x\) the group \(\sin x\,dx\) becomes \(-du\).
- Exponential substitutions and back-substituting: For \(u = e^{x} + 1\), \(du = e^{x}\,dx\).
- Differentiate the substitution and express \(dx\) (or a group like \(\cos x\,dx\)) in \(du\).
- Rewrite any leftover \(x\) in terms of \(u\).
- Change the limits for a definite integral.
- Integrate in \(u\); simplify powers first.
- Back-substitute for an indefinite integral.
One picture every idea returns to
With a substitution \(u = g(x)\), replace three things: the integrand (in terms of \(u\)), \(dx\) (using \(\dfrac{du}{dx}\)), and, for a definite integral, the limits (\(x\)-values become \(u\)-values). Then the whole integral is in \(u\) and no \(x\) may remain.
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, Change the variable.. 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 method: convert everything | Treat \(\dfrac{du}{dx}\) like a fraction to swap \(dx\) for \(du\). Model the worked example A linear substitution on the board, one line at a time. Your turn: Use the substitution \(u = x + 1\) to find the exact value of \(\displaystyle\int_0^3 \dfrac{x}{\sqrt{x+1}}\,dx\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: Trigonometric substitutions: spot the derivative | With \(u = \sin x\) the group \(\cos x\,dx\) becomes \(du\); with \(u = \cos x\) the group \(\sin x\,dx\) becomes \(-du\). Model the worked example The syllabus example on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Exponential substitutions and back-substituting | For \(u = e^{x} + 1\), \(du = e^{x}\,dx\). Model the worked example Splitting the exponential on the board, one line at a time. | Rally Coach |
| 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, 31 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 Exponential substitutions and back-substituting, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| Forgetting to replace \(dx\) is the most expensive slip: with \(u = 2x + 1\), \(dx = \dfrac{1}{2}du\), not \(du\). | The second is mixing old and new limits: once the integral is in \(u\), the limits must be \(u\)-values. |
| With \(u = \cos x\), \(du = -\sin x\,dx\). | The minus sign is easy to drop; it is usually absorbed by swapping the limits, because \(x = 0\) gives \(u = 1\), the larger value. |
| An indefinite answer left in \(u\) loses the final mark. | A definite answer evaluated with the old \(x\)-limits in a \(u\)-integral is simply wrong. |
Integration by substitution: worksheet
6 questions, 31 marks. Written in the style of the papers; no past paper question is reproduced.
Use the substitution \(u = x + 2\) to find \(\displaystyle\int \dfrac{x}{(x+2)^3}\,dx\), giving your answer in terms of \(x\).
Use the substitution \(u = 1 + \sin x\) to find the exact value of \(\displaystyle\int_0^{\pi/2} \cos x\sqrt{1+\sin x}\,dx\).
Use the substitution \(u = \sqrt{x}\) to show that \(\displaystyle\int_1^4 \dfrac{1}{\sqrt{x}\left(1+\sqrt{x}\right)}\,dx = 2\ln\dfrac{3}{2}\).
Use the substitution \(u = \cos x\) to show that \(\displaystyle\int_0^{\pi/3} \sin^3 x\,dx = \dfrac{5}{24}\).
Use the substitution \(u = 1 + x^2\) to find the exact value of \(\displaystyle\int_0^{\sqrt{3}} \dfrac{x^3}{\sqrt{1+x^2}}\,dx\).
The curve \(y = x\sqrt{1-x}\), for \(0 \le x \le 1\), has a maximum point \(M\).
(a) Find the exact coordinates of \(M\).
(b) Use the substitution \(u = 1 - x\) to find the exact area of the region between the curve and the \(x\)-axis.
Answer key and marking notes
- \(dx = du\) and \(x = u - 2\).
- \(\displaystyle\int \dfrac{u-2}{u^3}\,du = \int \left(u^{-2} - 2u^{-3}\right) du = -u^{-1} + u^{-2}\).
- \(-\dfrac{1}{x+2} + \dfrac{1}{(x+2)^2} + c\).
- \(du = \cos x\,dx\). Limits: \(x = 0 \Rightarrow u = 1\); \(x = \dfrac{\pi}{2} \Rightarrow u = 2\).
- \(\displaystyle\int_1^2 u^{1/2}\,du = \left[\dfrac{2}{3}u^{3/2}\right]_1^2\).
- \(= \dfrac{2}{3}\left(2\sqrt{2} - 1\right) = \dfrac{4\sqrt{2} - 2}{3}\).
- \(x = u^2\), so \(dx = 2u\,du\). Limits: \(x = 1 \Rightarrow u = 1\); \(x = 4 \Rightarrow u = 2\).
- Integrand: \(\dfrac{1}{u(1+u)}\times 2u\,du = \dfrac{2}{1+u}\,du\).
- \(\left[2\ln(1+u)\right]_1^2 = 2\ln 3 - 2\ln 2 = 2\ln\dfrac{3}{2}\).
- \(\sin^3 x = \sin x(1 - \cos^2 x)\), and \(du = -\sin x\,dx\).
- Limits: \(x = 0 \Rightarrow u = 1\); \(x = \dfrac{\pi}{3} \Rightarrow u = \dfrac{1}{2}\).
- \(\displaystyle -\int_1^{1/2} (1 - u^2)\,du = \int_{1/2}^1 (1 - u^2)\,du = \left[u - \dfrac{u^3}{3}\right]_{1/2}^1\).
- \(= \dfrac{2}{3} - \left(\dfrac{1}{2} - \dfrac{1}{24}\right) = \dfrac{16}{24} - \dfrac{11}{24} = \dfrac{5}{24}\).
- \(du = 2x\,dx\), and \(x^3\,dx = x^2\cdot x\,dx = (u - 1)\cdot\dfrac{1}{2}du\). Limits: 1 to 4.
- \(\dfrac{1}{2}\displaystyle\int_1^4 \left(u^{1/2} - u^{-1/2}\right) du = \dfrac{1}{2}\left[\dfrac{2}{3}u^{3/2} - 2u^{1/2}\right]_1^4\).
- \(= \dfrac{1}{2}\left[\left(\dfrac{16}{3} - 4\right) - \left(\dfrac{2}{3} - 2\right)\right] = \dfrac{1}{2}\times\dfrac{8}{3} = \dfrac{4}{3}\).
- (a) Product rule: \(\dfrac{dy}{dx} = \sqrt{1-x} - \dfrac{x}{2\sqrt{1-x}} = \dfrac{2(1-x) - x}{2\sqrt{1-x}} = \dfrac{2 - 3x}{2\sqrt{1-x}}\).
- \(\dfrac{dy}{dx} = 0 \Rightarrow x = \dfrac{2}{3}\); then \(y = \dfrac{2}{3}\sqrt{\dfrac{1}{3}} = \dfrac{2\sqrt{3}}{9}\). So \(M = \left(\dfrac{2}{3}, \dfrac{2\sqrt{3}}{9}\right)\).
- (b) \(dx = -du\), \(x = 1 - u\), limits \(x = 0 \to u = 1\), \(x = 1 \to u = 0\).
- Area \(= \displaystyle\int_0^1 (1-u)u^{1/2}\,du = \left[\dfrac{2}{3}u^{3/2} - \dfrac{2}{5}u^{5/2}\right]_0^1\).
- \(= \dfrac{2}{3} - \dfrac{2}{5} = \dfrac{4}{15}\) square units.
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) Using the substitution \(u = \sin x\), the integral \(\displaystyle\int_0^{\frac{\pi}{2}} \sin^3 x\cos x\,dx\) becomes which integral?
- A\(\displaystyle\int_0^1 u^3\,du\)
- B\(\displaystyle\int_0^{\frac{\pi}{2}} u^3\,du\)
- C\(\displaystyle\int_0^1 u^3\cos x\,du\)
- D\(-\displaystyle\int_0^1 u^3\,du\)
(Paper 3 only) Use the substitution \(u = x^2 + 1\) to find the exact value of \(\displaystyle\int_0^1 x(x^2 + 1)^3\,dx\).
- A\(\dfrac{15}{8}\)
- B\(\dfrac{15}{4}\)
- C\(\dfrac{15}{2}\)
- D\(\dfrac{1}{8}\)
(Paper 3 only) Use the substitution \(u = \cos x\) to find the exact value of \(\displaystyle\int_0^{\frac{\pi}{3}} \sin x\cos^2 x\,dx\).
- A\(\dfrac{7}{24}\)
- B\(-\dfrac{7}{24}\)
- C\(\dfrac{7}{8}\)
- D\(\dfrac{1}{24}\)
(Paper 3 only) Use the substitution \(u = 1 + e^{x}\) to find \(\displaystyle\int \dfrac{e^{x}}{(1 + e^{x})^2}\,dx\).
- A\(-\dfrac{1}{1 + e^{x}} + c\)
- B\(\dfrac{1}{1 + e^{x}} + c\)
- C\(\ln(1 + e^{x}) + c\)
- D\(-\dfrac{1}{3(1 + e^{x})^3} + c\)
(Paper 3 only) Use the substitution \(u = x - 1\) to find the exact value of \(\displaystyle\int_1^2 x\sqrt{x - 1}\,dx\).
- A\(\dfrac{16}{15}\)
- B\(\dfrac{2}{5}\)
- C\(-\dfrac{4}{15}\)
- D\(\dfrac{2}{3}\)
(Paper 3 only) Use the substitution \(x = \tan\theta\) to find the exact value of \(\displaystyle\int_0^{\sqrt{3}} \dfrac{1}{(1 + x^2)^{\frac{3}{2}}}\,dx\).
- A\(\dfrac{\sqrt{3}}{2}\)
- B\(\dfrac{1}{2}\)
- C\(\dfrac{\pi}{3}\)
- D\(-\dfrac{\sqrt{3}}{2}\)
(Paper 3 only) Use the substitution \(u = \sqrt{x}\) to find the exact value of \(\displaystyle\int_1^4 \dfrac{1}{x + \sqrt{x}}\,dx\).
- A\(2\ln\dfrac{3}{2}\)
- B\(\ln\dfrac{3}{2}\)
- C\(2\ln\dfrac{5}{2}\)
- D\(\ln\dfrac{4}{3}\)
(Paper 3 only) Use the substitution \(u = \sin x\) to find the exact value of \(\displaystyle\int_0^{\frac{\pi}{2}} \cos^3 x\,dx\).
- A\(\dfrac{2}{3}\)
- B\(\dfrac{4}{3}\)
- C\(\dfrac{1}{3}\)
- D\(\dfrac{1}{4}\)
Quiz key
\(du = \cos x\,dx\), so \(\cos x\,dx\) becomes \(du\) and \(\sin^3 x\) becomes \(u^3\). Limits: \(x = 0 \to u = 0\), \(x = \dfrac{\pi}{2} \to u = 1\). Result: \(\displaystyle\int_0^1 u^3\,du = \dfrac{1}{4}\). The tempting option with limits \(0\) to \(\dfrac{\pi}{2}\) forgets to change the limits. Leaving \(\cos x\) in the integrand mixes \(x\) and \(u\), and the minus sign belongs to \(u = \cos x\), not \(u = \sin x\).
\(\displaystyle\int_1^2 \dfrac{1}{2}u^3\,du = \dfrac{1}{2}\left[\dfrac{u^4}{4}\right]_1^2 = \dfrac{1}{8}(16 - 1) = \dfrac{15}{8}\). The tempting \(\dfrac{15}{4}\) forgets the \(\dfrac{1}{2}\) from \(x\,dx = \dfrac{1}{2}du\). \(\dfrac{15}{2}\) multiplies by 2 instead, and \(\dfrac{1}{8}\) keeps the old limits 0 and 1 in the \(u\) integral.
\(\displaystyle\int_1^{1/2} -u^2\,du = \int_{1/2}^1 u^2\,du = \left[\dfrac{u^3}{3}\right]_{1/2}^1 = \dfrac{1}{3}\left(1 - \dfrac{1}{8}\right) = \dfrac{7}{24}\). The answer must be positive, since the integrand is positive on the interval. The tempting \(-\dfrac{7}{24}\) keeps the minus sign AND swaps the limits, using the same minus sign twice. \(\dfrac{7}{8}\) forgets to divide by 3, and \(\dfrac{1}{24}\) takes \(\cos 0\) to be 0.
\(\displaystyle\int u^{-2}\,du = -u^{-1} + c = -\dfrac{1}{1 + e^x} + c\). Check: differentiating gives \(\dfrac{e^x}{(1 + e^x)^2}\). The tempting \(+\dfrac{1}{1 + e^x}\) loses the minus sign from \(\dfrac{u^{-1}}{-1}\). \(\ln(1 + e^x)\) treats the integrand as \(\dfrac{f'(x)}{f(x)}\), but the denominator is squared, and the last option lowers the power instead of raising it.
\(\displaystyle\int_0^1 (u + 1)u^{\frac{1}{2}}\,du = \int_0^1 \left(u^{\frac{3}{2}} + u^{\frac{1}{2}}\right)du = \dfrac{2}{5} + \dfrac{2}{3} = \dfrac{16}{15}\). The tempting \(\dfrac{2}{5}\) replaces the \(x\) outside the root with \(u\) instead of \(u + 1\). \(-\dfrac{4}{15}\) uses \(x = u - 1\), and \(\dfrac{2}{3}\) drops the factor \(x\) altogether.
\((1 + x^2)^{\frac{3}{2}} = \sec^3\theta\), so the integrand becomes \(\dfrac{\sec^2\theta}{\sec^3\theta} = \cos\theta\). Limits: \(\theta = 0\) to \(\dfrac{\pi}{3}\). \(\left[\sin\theta\right]_0^{\pi/3} = \dfrac{\sqrt{3}}{2}\). The tempting \(\dfrac{1}{2}\) uses the upper limit \(\dfrac{\pi}{6}\) (\(\tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}\), not \(\sqrt{3}\)). \(\dfrac{\pi}{3}\) forgets the integrand and just finds the length of the \(\theta\) interval, and \(-\dfrac{\sqrt{3}}{2}\) integrates \(\cos\theta\) to \(-\sin\theta\).
\(\displaystyle\int_1^2 \dfrac{2u}{u(u + 1)}\,du = \int_1^2 \dfrac{2}{u + 1}\,du = \left[2\ln(u + 1)\right]_1^2 = 2\ln 3 - 2\ln 2 = 2\ln\dfrac{3}{2}\). The tempting \(\ln\dfrac{3}{2}\) forgets the 2 in \(dx = 2u\,du\). \(2\ln\dfrac{5}{2}\) keeps the old limits 1 and 4, and \(\ln\dfrac{4}{3}\) replaces \(dx\) with \(du\), integrating \(\dfrac{1}{u(u + 1)}\).
\(\displaystyle\int_0^1 (1 - u^2)\,du = \left[u - \dfrac{u^3}{3}\right]_0^1 = \dfrac{2}{3}\). The tempting \(\dfrac{4}{3}\) uses \(\cos^2 x = 1 + \sin^2 x\). \(\dfrac{1}{3}\) replaces \(\cos^2 x\) with \(u^2\), confusing sine and cosine, and \(\dfrac{1}{4}\) uses the power rule on \(\cos^3 x\) as if it were \(x^3\), giving \(\left[-\dfrac{1}{4}\cos^4 x\right]\).
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.