A2 June 2024 Q6
6.
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
\[I_n = \int \frac{\cos(nx)}{\sin x}\,\mathrm{d}x \qquad n \geqslant 1\]| Scheme | Marks | AO |
|---|---|---|
| \(I_{n+2} = \int \dfrac{\cos(n+2)x}{\sin x}\,\mathrm{d}x = \int \dfrac{\cos(nx)\cos 2x - \sin(nx)\sin 2x}{\sin x}\,\mathrm{d}x\) | M1 | 3.1a |
| \(= \int \dfrac{\cos(nx)(1 - 2\sin^2 x) - 2\sin x\cos x\sin(nx)}{\sin x}\,\mathrm{d}x\) | M1 A1 | 1.1b 2.1 |
| \(= \int \tfrac{\cos(nx)}{\sin x}\,\mathrm{d}x - \int \tfrac{\cos(nx)(2\sin^2 x) + 2\sin x\cos x\sin(nx)}{\sin x}\,\mathrm{d}x\) \(= I_n - 2\int \cos(nx)\sin x + \cos x\sin(nx)\,\mathrm{d}x\) | dM1 | 1.1b |
| \(= I_n - 2\int \sin(n+1)x\,\mathrm{d}x\) | ddM1 | 2.2a |
| \(I_{n+2} = 2\dfrac{\cos(n+1)x}{n+1} + I_n\ *\) | A1* | 2.1 |
| (6) |
Notes
M1: Applies the compound angle formula as shown.
M1: Applies both double angle formulae \(\cos 2x = 1 - 2\sin^2 x\) and \(\sin 2x = 2\sin x\cos x\) to the expression.
A1: Correct intermediate step reached, need not be simplified.
dM1: Dependent on previous method mark. Splits the integrand to identify \(I_n\) and cancels the \(\sin x\) term in the remaining integral.
ddM1: Dependent on both previous method marks. Applies the compound angle formula to combine terms.
A1*: Correct completion to the given result. No errors seen.
Alternative 1
| Scheme | Marks | AO |
|---|---|---|
| \(I_n = \int \dfrac{\cos(n+1-1)x}{\sin x}\,\mathrm{d}x = \int \dfrac{\cos(n+1)x\cos x + \sin(n+1)\sin x}{\sin x}\,\mathrm{d}x\) | M1 | 3.1a |
| \(= \int \dfrac{\cos(n+2)x + \cos nx}{2\sin x} + \sin(n+1)x\,\mathrm{d}x\) | M1 A1 | 1.1b 2.1 |
| \(I_n = \dfrac{1}{2}I_{n+2} + \dfrac{1}{2}I_n + \int \sin(n+1)x\,\mathrm{d}x\) | dM1 | 1.1b |
| \(I_n = \dfrac{1}{2}I_{n+2} + \dfrac{1}{2}I_n - \dfrac{1}{n+1}\cos(n+1)x\,\mathrm{d}x \Rightarrow I_{n+2} = \ldots\) | ddM1 | 2.2a |
| \(I_{n+2} = 2\dfrac{\cos(n+1)x}{n+1} + I_n\ *\) | A1* | 2.1 |
| (6) |
M1: Applies the compound angle formula as shown.
M1: Applies the sum product formula \(\cos P\cos Q = \dfrac{1}{2}\left[\cos(P + Q) + \cos(P - Q)\right]\)
A1: Correct intermediate step reached, need not be simplified.
dM1: Dependent on previous method mark. Splits the integrand to identify \(I_n\) and. \(I_{n+2}\)
ddM1: Dependent on both previous method marks. Integrates \(\int \sin(n+1)x\) and rearranges to make \(I_{n+2}\)
A1*: Correct completion to the given result. No errors seen.
Alternative 2
| Scheme | Marks | AO |
|---|---|---|
| \(I_{n+2} = \int \dfrac{\cos(n+2)x}{\sin x}\,\mathrm{d}x = \int \dfrac{\cos(n+1)x\cos x - \sin(n+1)x\sin x}{\sin x}\,\mathrm{d}x\) | M1 | 3.1a |
| \(= \int \tfrac{\cos(n+2)x + \cos nx}{2\sin x} - \sin(n+1)x\,\mathrm{d}x\) Or \(= \int \tfrac{(\cos nx\cos x - \sin nx\sin x)\cos x}{\sin x} - \sin(n+1)x\,\mathrm{d}x\) \(= \int \tfrac{\cos nx(1 - \sin^2 x) - \sin nx\sin x\cos x}{\sin x} - \sin(n+1)x\,\mathrm{d}x\) \(= \int \tfrac{\cos nx}{\sin x} - \cos nx\sin x - \sin nx\cos x - \sin(n+1)x\,\mathrm{d}x\) \(= \int \tfrac{\cos nx}{\sin x} - 2\sin(n+1)x\,\mathrm{d}x\) | M1 A1 | 1.1b 2.1 |
| \(I_{n+2} = \tfrac{1}{2}I_{n+2} + \tfrac{1}{2}I_n - \int \sin(n+1)x\,\mathrm{d}x\) Or \(I_{n+2} = I_n - 2\int \sin(n+1)x\,\mathrm{d}x\) | dM1 | 1.1b |
| \(I_{n+2} = \tfrac{1}{2}I_{n+2} + \tfrac{1}{2}I_n + \tfrac{1}{n+1}\cos(n+1)x\,\mathrm{d}x \Rightarrow I_{n+2} = \ldots\) Or \(I_{n+2} = I_n + \tfrac{2}{n+1}\cos(n+1)x\,\mathrm{d}x \Rightarrow I_{n+2} = \ldots\) | ddM1 | 2.2a |
| \(I_{n+2} = 2\dfrac{\cos(n+1)x}{n+1} + I_n\) | A1* | 2.1 |
| (6) |
M1: Applies the compound angle formula as shown.
M1: Applies the sum product formula \(\cos P\cos Q = \dfrac{1}{2}\left[\cos(P + Q) + \cos(P - Q)\right]\)
Alternatively uses \(\cos(A + B) = \cos A\cos B - \sin A\sin B\), \(\cos^2 x = 1 - \sin^2 x\) and \(\sin A\cos B + \cos A\sin B = \sin(A + B)\) in an attempt to simplify.
A1: Correct intermediate step reached, need not be simplified.
dM1: Dependent on previous method mark. Splits the integrand to identify \(I_n\) and. \(I_{n+2}\)
ddM1: Dependent on both previous method marks. Integrates \(\int \sin(n+1)x\) and rearranges to make \(I_{n+2}\)
A1*: Correct completion to the given result. No errors seen.
6(a) Alt 3
| Scheme | Marks | AO |
|---|---|---|
| \(I_{n+2} - I_n = \int \dfrac{\cos(n+2)x - \cos nx}{\sin x}\,\mathrm{d}x\) | M1 | 3.1a |
| \(= -\int \dfrac{2\sin\dfrac{1}{2}\left((n+2)x + nx\right)\sin\dfrac{1}{2}\left((n+2)x - nx\right)}{\sin x}\,\mathrm{d}x\) | M1 | 1.1b |
| \(= -\int \dfrac{2\sin(n+1)x\sin x}{\sin x}\,\mathrm{d}x\) | A1 | 2.1 |
| \(= -2\int \sin(n+1)x\,\mathrm{d}x\) | dM1 | 1.1b |
| \(= 2\dfrac{\cos(n+1)x}{n+1} \Rightarrow I_{n+2} = \ldots\) | ddM1 | 2.2a |
| \(\Rightarrow I_{n+2} = 2\dfrac{\cos(n+1)x}{n+1} + I_n\ *\) | A1* | 2.1 |
| (6) |
M1: Attempts the difference \(I_{n+2} - I_n = \ldots\) and combines to a single fraction
M1: Applies the difference of two cosines terms formula.
A1: Correct expression.
dM1: Dependent on previous method mark. Cancels the \(\sin x\) term in the integral.
ddM1: Dependent on both previous method marks. Integrates.
A1*: Correct completion to the given result. No errors seen.
| Scheme | Marks | AO |
|---|---|---|
| \([I_1] = \int \cot x\,\mathrm{d}x = [\ln\sin x]\) may be seen in an expression for \(I_5\) | M1 | 1.1b |
| \(= \ln\dfrac{\sqrt{3}}{2} - \ln\dfrac{\sqrt{2}}{2}\) may be seen in an expression for \(I_5\) | A1 | 2.2a |
| \(I_5 = 2\dfrac{\cos 4x}{4} + I_3\) or \(I_3 = 2\dfrac{\cos 2x}{2} + I_1\) | M1 | 1.1b |
| \(\left[I_5\right]_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}} = \left[\tfrac{\cos 4x}{2} + \cos 2x + I_1\right]_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}}\) or \(\left[\tfrac{\cos 4x}{2} + \cos 2x + \ln\sin x\right]_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}}\) \(= \tfrac{1}{2}\cos\tfrac{4\pi}{3} + \cos\tfrac{2\pi}{3} - \tfrac{1}{2}\cos\pi - \cos\tfrac{\pi}{2} + \text{“}\ln\tfrac{\sqrt{3}}{2} - \ln\tfrac{\sqrt{2}}{2}\text{”}\) | M1 | 1.1b |
| \(\left(= -\dfrac{1}{4} - \dfrac{1}{2} + \dfrac{1}{2} - 0 + \ln\dfrac{\sqrt{3}}{2} - \ln\dfrac{\sqrt{2}}{2}\right) = -\dfrac{1}{4} + \dfrac{1}{2}\ln\dfrac{3}{2}\) (oe) | A1 | 2.1 |
| (5) | ||
| (11 marks) |
Notes
M1: Attempts \(I_1\), look for \(K\ln\sin x\) anywhere in their solution
A1: A correct expression, not necessarily simplified, for \(I_1\). May be seen as part of the final answer.
M1: Any one correct application of the reduction formula applied to the question (either working down or working up).
M1: Applies the reduction formula twice and substitutes the limits.
A1: Correct answer in the form required but accept equivalents in this form. E.g. accept \(\dfrac{\ln\dfrac{9}{4} - 1}{4}\) Logs must have been combined.