The shape of the lesson
By the end of the lesson, learners can
- Squares of sine and cosine: lower the power: Rearrange \(\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1\) to express each square in terms of \(\cos 2x\).
- Tan squared and products: find the hidden identity: Other integrands become standard once you spot the right identity.
- Expand first, then use the identities: A bracket squared often contains \(\sin^2 x + \cos^2 x\), which is 1, and a cross term \(2\sin x\cos x\), which is \(\sin 2x\).
- Squares of sin or cos: use \(\cos 2\theta\) to lower the power; the angle doubles.
- \(\tan^2\): replace by \(\sec^2 - 1\).
- Products: \(\sin x\cos x = \dfrac{1}{2}\sin 2x\).
- Brackets: expand, then use \(\sin^2 + \cos^2 = 1\).
- Evaluate in radians and keep \(\pi\) and surds exact.
One picture every idea returns to
Squares of sine and cosine are integrated by lowering the power with the double-angle formula: \(\sin^2 x = \dfrac{1}{2}(1 - \cos 2x)\) and \(\cos^2 x = \dfrac{1}{2}(1 + \cos 2x)\). For \(\tan^2 x\) use \(\tan^2 x = \sec^2 x - 1\).
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, Rewrite it, then integrate.. 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: Squares of sine and cosine: lower the power | Rearrange \(\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1\) to express each square in terms of \(\cos 2x\). Model the worked example Integrate sine squared on the board, one line at a time. Your turn: Find the exact value of \(\displaystyle\int_0^{\pi/3} 4\sin^2\dfrac{x}{2}\,dx\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: Tan squared and products: find the hidden identity | Other integrands become standard once you spot the right identity. Model the worked example Tan squared on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Expand first, then use the identities | A bracket squared often contains \(\sin^2 x + \cos^2 x\), which is 1, and a cross term \(2\sin x\cos x\), which is \(\sin 2x\). Model the worked example A squared bracket 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, 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 Expand first, then use the identities, 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 |
|---|---|
| \(\displaystyle\int \sin^2 x\,dx\) is NOT \(\dfrac{\sin^3 x}{3}\). | The power rule only works on \(x^{n}\); differentiating \(\dfrac{\sin^3 x}{3}\) gives \(\sin^2 x\cos x\), not \(\sin^2 x\). And when the angle is \(2x\), the identity brings in \(\cos 4x\), not \(\cos 2x\). |
| \(\tan^2 x = \sec^2 x - 1\), not \(\sec^2 x + 1\). | Check with \(x = 0\): \(\tan^2 0 = 0\) and \(\sec^2 0 - 1 = 0\). The wrong sign turns \(1 - \dfrac{\pi}{4}\) into \(1 + \dfrac{\pi}{4}\). |
| When you square \(\dfrac{1 + \cos 2x}{2}\), square the 2 in the denominator as well: the factor in front is \(\dfrac{1}{4}\), not \(\dfrac{1}{2}\). | Return to Expand first, then use the identities: ask a pair to state the correct version and show it on the lab before moving on. |
Integration using trigonometric identities: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Find \(\displaystyle\int \cos^2 x\,dx\).
Find the exact value of \(\displaystyle\int_0^{\pi/4} \sin^2 x\,dx\).
Find the exact value of \(\displaystyle\int_0^{\pi/4} \left(2 + \tan^2 x\right) dx\).
(a) Show that \(\cos^4 x - \sin^4 x = \cos 2x\).
(b) Hence find the exact value of \(\displaystyle\int_0^{\pi/8} \left(\cos^4 x - \sin^4 x\right) dx\).
Find the exact value of \(\displaystyle\int_0^{\pi/2} \sin^2 x\cos^2 x\,dx\).
The region under the curve \(y = \sin x\) between \(x = 0\) and \(x = \pi\) is rotated through \(360^\circ\) about the \(x\)-axis to form a solid shaped like a rugby ball.
Find the exact volume of the solid, and give it to 3 significant figures.
Answer key and marking notes
- \(\cos^2 x = \dfrac{1}{2}(1 + \cos 2x)\).
- \(\displaystyle\int \cos^2 x\,dx = \dfrac{x}{2} + \dfrac{1}{4}\sin 2x + c\).
- \(\displaystyle\int \sin^2 x\,dx = \dfrac{x}{2} - \dfrac{1}{4}\sin 2x\).
- Upper: \(\dfrac{\pi}{8} - \dfrac{1}{4}\sin\dfrac{\pi}{2} = \dfrac{\pi}{8} - \dfrac{1}{4}\). Lower: 0.
- Answer: \(\dfrac{\pi}{8} - \dfrac{1}{4}\).
- \(2 + \tan^2 x = 2 + \sec^2 x - 1 = 1 + \sec^2 x\).
- \(\left[x + \tan x\right]_0^{\pi/4} = \dfrac{\pi}{4} + 1\).
- (a) Difference of two squares: \(\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)\).
- \(\cos^2 x + \sin^2 x = 1\) and \(\cos^2 x - \sin^2 x = \cos 2x\), so the expression equals \(\cos 2x\).
- (b) \(\displaystyle\int_0^{\pi/8} \cos 2x\,dx = \left[\dfrac{1}{2}\sin 2x\right]_0^{\pi/8} = \dfrac{1}{2}\sin\dfrac{\pi}{4}\).
- \(= \dfrac{1}{2}\times\dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{2}}{4}\).
- \(\sin x\cos x = \dfrac{1}{2}\sin 2x\), so \(\sin^2 x\cos^2 x = \dfrac{1}{4}\sin^2 2x\).
- \(\sin^2 2x = \dfrac{1}{2}(1 - \cos 4x)\), so the integrand is \(\dfrac{1}{8}(1 - \cos 4x)\).
- \(\left[\dfrac{x}{8} - \dfrac{1}{32}\sin 4x\right]_0^{\pi/2} = \dfrac{\pi}{16} - \dfrac{1}{32}\sin 2\pi = \dfrac{\pi}{16}\).
- Volume \(= \pi\displaystyle\int_0^{\pi} y^2\,dx = \pi\displaystyle\int_0^{\pi} \sin^2 x\,dx\).
- \(\sin^2 x = \dfrac{1}{2}(1 - \cos 2x)\), so \(\displaystyle\int_0^{\pi} \sin^2 x\,dx = \left[\dfrac{x}{2} - \dfrac{1}{4}\sin 2x\right]_0^{\pi} = \dfrac{\pi}{2}\).
- Volume \(= \pi\times\dfrac{\pi}{2} = \dfrac{\pi^2}{2}\).
- \(\dfrac{\pi^2}{2} = 4.93\) cubic 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 \(\displaystyle\int \cos^2 x\,dx\).
- A\(\dfrac{x}{2} + \dfrac{1}{4}\sin 2x + c\)
- B\(\dfrac{x}{2} - \dfrac{1}{4}\sin 2x + c\)
- C\(\dfrac{1}{3}\cos^3 x + c\)
- D\(\dfrac{x}{2} + \dfrac{1}{2}\sin 2x + c\)
Find the exact value of \(\displaystyle\int_0^{\frac{\pi}{4}} \sin^2 x\,dx\).
- A\(\dfrac{\pi}{8} - \dfrac{1}{4}\)
- B\(\dfrac{\pi}{8} + \dfrac{1}{4}\)
- C\(\dfrac{\pi}{4} - \dfrac{1}{2}\)
- D\(\dfrac{\sqrt{2}}{12}\)
Find \(\displaystyle\int \tan^2 x\,dx\).
- A\(\tan x - x + c\)
- B\(\tan x + x + c\)
- C\(\dfrac{1}{3}\tan^3 x + c\)
- D\(2\tan x\sec^2 x + c\)
Find the exact value of \(\displaystyle\int_0^{\frac{\pi}{6}} \sin x\cos x\,dx\).
- A\(\dfrac{1}{8}\)
- B\(\dfrac{1}{4}\)
- C\(-\dfrac{1}{8}\)
- D\(\dfrac{1}{2}\)
Find the exact value of \(\displaystyle\int_0^{\frac{\pi}{12}} \cos^2 3x\,dx\).
- A\(\dfrac{\pi}{24} + \dfrac{1}{12}\)
- B\(\dfrac{\pi}{24} + \dfrac{1}{4}\)
- C\(\dfrac{\pi}{24} - \dfrac{1}{12}\)
- D\(\dfrac{\pi}{12} + \dfrac{1}{6}\)
Find the exact value of \(\displaystyle\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^2\,dx\).
- A\(\dfrac{\pi}{4} - \dfrac{1}{2}\)
- B\(\dfrac{\pi}{4} + \dfrac{1}{2}\)
- C\(\dfrac{\pi}{4}\)
- D\(\dfrac{\pi}{4} - 1\)
Find the exact value of \(\displaystyle\int_0^{\frac{\pi}{3}} \dfrac{3}{1 + \cos 2x}\,dx\).
- A\(\dfrac{3\sqrt{3}}{2}\)
- B\(3\sqrt{3}\)
- C\(\dfrac{\sqrt{3}}{2}\)
- D\(-\dfrac{3\sqrt{3}}{2}\)
Find the exact value of \(\displaystyle\int_0^{\frac{\pi}{2}} \sin^2 x\cos^2 x\,dx\).
- A\(\dfrac{\pi}{16}\)
- B\(\dfrac{\pi}{8}\)
- C\(\dfrac{\pi}{4}\)
- D\(\dfrac{\pi^2}{16}\)
Quiz key
\(\displaystyle\int \dfrac{1}{2}(1 + \cos 2x)\,dx = \dfrac{x}{2} + \dfrac{1}{2} \cdot \dfrac{1}{2}\sin 2x + c = \dfrac{x}{2} + \dfrac{1}{4}\sin 2x + c\). The tempting minus sign uses the \(\sin^2 x\) identity. \(\dfrac{1}{3}\cos^3 x\) applies the power rule to a function of \(x\), which does not work, and \(\dfrac{1}{2}\sin 2x\) forgets to divide by 2 when integrating \(\cos 2x\).
\(\left[\dfrac{x}{2} - \dfrac{1}{4}\sin 2x\right]_0^{\pi/4} = \dfrac{\pi}{8} - \dfrac{1}{4}\sin\dfrac{\pi}{2} = \dfrac{\pi}{8} - \dfrac{1}{4}\). The tempting \(\dfrac{\pi}{8} + \dfrac{1}{4}\) uses the cosine identity with a plus sign. \(\dfrac{\pi}{4} - \dfrac{1}{2}\) forgets the \(\dfrac{1}{2}\) in the identity, and \(\dfrac{\sqrt{2}}{12}\) is \(\left[\dfrac{1}{3}\sin^3 x\right]\), the power rule wrongly applied.
\(\displaystyle\int (\sec^2 x - 1)\,dx = \tan x - x + c\). The tempting \(\tan x + x\) uses \(\tan^2 x = \sec^2 x + 1\), the identity the wrong way round. \(\dfrac{1}{3}\tan^3 x\) uses the power rule on a function, and \(2\tan x\sec^2 x\) is the DERIVATIVE of \(\tan^2 x\).
\(\dfrac{1}{2}\displaystyle\int_0^{\pi/6} \sin 2x\,dx = \dfrac{1}{2}\left[-\dfrac{1}{2}\cos 2x\right]_0^{\pi/6} = \dfrac{1}{2}\left(-\dfrac{1}{4} + \dfrac{1}{2}\right) = \dfrac{1}{8}\). The tempting \(\dfrac{1}{4}\) writes \(\sin x\cos x\) as \(\sin 2x\), forgetting the \(\dfrac{1}{2}\). \(-\dfrac{1}{8}\) integrates \(\sin 2x\) to \(+\dfrac{1}{2}\cos 2x\), and \(\dfrac{1}{2}\) forgets both halves.
\(\displaystyle\int \dfrac{1}{2}(1 + \cos 6x)\,dx = \dfrac{x}{2} + \dfrac{1}{12}\sin 6x\). At \(\dfrac{\pi}{12}\): \(\dfrac{\pi}{24} + \dfrac{1}{12}\sin\dfrac{\pi}{2} = \dfrac{\pi}{24} + \dfrac{1}{12}\); at 0 it is 0. The tempting \(\dfrac{1}{4}\) divides by 2 instead of 6 when integrating \(\cos 6x\) (as if it were \(\cos 2x\)). The minus sign uses the \(\sin^2\) identity, and \(\dfrac{\pi}{12} + \dfrac{1}{6}\) forgets the \(\dfrac{1}{2}\) in the identity.
\((\cos x - \sin x)^2 = 1 - \sin 2x\). \(\left[x + \dfrac{1}{2}\cos 2x\right]_0^{\pi/4} = \left(\dfrac{\pi}{4} + 0\right) - \left(0 + \dfrac{1}{2}\right) = \dfrac{\pi}{4} - \dfrac{1}{2}\). The tempting \(\dfrac{\pi}{4} + \dfrac{1}{2}\) expands with \(+2\sin x\cos x\). \(\dfrac{\pi}{4}\) forgets the cross term entirely (\((a - b)^2 \ne a^2 + b^2\)), and \(\dfrac{\pi}{4} - 1\) integrates \(\sin 2x\) to \(-\cos 2x\), forgetting to divide by 2.
\(1 + \cos 2x = 2\cos^2 x\), so the integrand is \(\dfrac{3}{2}\sec^2 x\). \(\left[\dfrac{3}{2}\tan x\right]_0^{\pi/3} = \dfrac{3}{2}\sqrt{3}\). The tempting \(3\sqrt{3}\) uses \(1 + \cos 2x = \cos^2 x\), losing the 2. \(\dfrac{\sqrt{3}}{2}\) evaluates \(\tan\dfrac{\pi}{6}\) instead of \(\tan\dfrac{\pi}{3}\), and the negative answer integrates \(\sec^2 x\) to \(-\tan x\).
\(\sin^2 x\cos^2 x = \dfrac{1}{4}\sin^2 2x = \dfrac{1}{4} \cdot \dfrac{1}{2}(1 - \cos 4x) = \dfrac{1}{8}(1 - \cos 4x)\). \(\left[\dfrac{x}{8} - \dfrac{1}{32}\sin 4x\right]_0^{\pi/2} = \dfrac{\pi}{16}\). The tempting \(\dfrac{\pi}{8}\) forgets to square the \(\dfrac{1}{2}\) (using \(\dfrac{1}{2}\sin^2 2x\)). \(\dfrac{\pi}{4}\) drops the \(\dfrac{1}{2}\) altogether, and \(\dfrac{\pi^2}{16}\) integrates \(\sin^2 x\) and \(\cos^2 x\) separately and multiplies, which is not allowed.
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.