A2 June 2022 Q9
9.
\[I_n = \int_0^{\frac{\pi}{2}} \sin^n 2x \,\mathrm{d}x\]| Scheme | Marks | AO |
|---|---|---|
| \(I_n = \int \sin^n 2x\,\mathrm{d}x = \int \sin^{n-1} 2x \sin 2x\,\mathrm{d}x\) Leading to \(I_n = \left[\lambda \sin^{n-1} 2x \cos 2x\right] - \mu\int \sin^{n-2} 2x \cos^2 2x\,\mathrm{d}x\) | M1 | 2.1 |
| \(I_n = \left[-\dfrac{1}{2}\sin^{n-1} 2x \cos 2x\right] + \int (n-1)\sin^{n-2} 2x \cos^2 2x\,\mathrm{d}x\) | A1 | 1.1b |
| \(I_n = 0 + (n-1)\int \sin^{n-2} 2x\left(1 - \sin^2 2x\right)\,\mathrm{d}x\) \(= (n-1)\int \sin^{n-2} 2x\,\mathrm{d}x - (n-1)\int \sin^n 2x\,\mathrm{d}x\) | dM1 | 1.1b |
| \(nI_n = (n-1)I_{n-2} \Rightarrow I_n = \dfrac{n-1}{n}I_{n-2}\) * | A1* | 2.1 |
| (4) |
Notes
M1: Writes \(\sin^n 2x\) as \(\sin^{n-1} 2x \sin 2x\) and integrates using by parts to the form \(I_n = \left[\lambda \sin^{n-1} 2x \cos 2x\right] - \mu\int \sin^{n-2} 2x \cos^2 2x\,\mathrm{d}x\)
A1: Correct integration, may be unsimplified
dM1: Substitutes the limits of 0 and \(\frac{\pi}{2}\) into ‘\(uv\)’, this may be implied by 0. Replaces \(\cos^2 2x = 1 - \sin^2 2x\) and multiplies out into separate integrals.
A1*: Achieves the printed answer following a correct intermediate line and no errors. Cso
Alternative
M1: Writes \(\sin^n 2x\) as \(\sin^{n-2} 2x \sin^2 2x = \sin^{n-2} 2x\,(1 - \cos^2 2x)\), writes as \(= \int \sin^{n-2} 2x\,\mathrm{d}x - \int \sin^{n-2} 2x \cos^2 2x\,\mathrm{d}x\) and attempts to integrate
dM1: Integrates using by parts to the form \(I_n = I_{n-2} - \left[\lambda \sin^{n-1} 2x \cos 2x\right] + \mu\int \sin 2x \sin^{n-1} 2x\,\mathrm{d}x\)
A1: Correct integration, may be unsimplified
A1*: Achieves the printed answer following a correct intermediate line and no errors. cso
(corrected from the printed mark scheme: some lines are printed in a garbled symbol font, e.g. ò for the integral sign; they are typed here as intended)
Alternative: On epen MAMA make sure marks are recorded in the correct place
| Scheme | Marks | AO |
|---|---|---|
| \(I_n = \int \sin^n 2x\,\mathrm{d}x = \int \sin^{n-2} 2x \sin^2 2x\,\mathrm{d}x\) \(= \int \sin^{n-2} 2x\,(1 - \cos^2 2x)\,\mathrm{d}x\) \(= \int \sin^{n-2} 2x\,\mathrm{d}x - \int \sin^{n-2} 2x \cos^2 2x\,\mathrm{d}x\) Leading to an attempt at integration | M1 | 1.1b |
| \(I_n = I_{n-2} - \int (\sin^{n-2} 2x \cos 2x)(\cos 2x)\,\mathrm{d}x\) \(I_n = I_{n-2} - \left[\lambda \sin^{n-1} 2x \cos 2x\right] + \mu\int \sin 2x \sin^{n-1} 2x\,\mathrm{d}x\) | dM1 | 2.1 |
| \(I_n = I_{n-2} - \left[\dfrac{1}{2(n-1)}\sin^{n-1} 2x \cos 2x\right] - \dfrac{1}{n-1}\int \sin^n 2x\,\mathrm{d}x\) | A1 | 1.1b |
| \(I_n = I_{n-2} - \dfrac{1}{n-1}I_n \Rightarrow (n-1)I_n = (n-1)I_{n-2} - I_n \Rightarrow I_n = \dfrac{n-1}{n}I_{n-2}\)* | A1* | 2.1 |
| (4) |
| Scheme | Marks | AO |
|---|---|---|
| \(\int 64\sin^5 x \cos^5 x\,\mathrm{d}x = \int A\sin^5 2x\,\mathrm{d}x\) Note \(A = 2\) | M1 | 2.1 |
| \(I_5 = \frac{4}{5}I_3,\ I_3 = \frac{2}{3}I_1\) and \(I_1 = \int_0^{\frac{\pi}{2}} \sin 2x\,\mathrm{d}x = [\alpha\cos 2x]_0^{\frac{\pi}{2}} = \ldots\) | M1 | 1.1b |
| \(= 2 \times \left(\dfrac{4}{5}\right) \times \left(\dfrac{2}{3}\right) \times 1 = \dfrac{16}{15}\) | A1 | 1.1b |
| (3) | ||
| (7 marks) |
Notes
M1: Uses the identity \(\sin 2x = 2\sin x\cos x\) in an attempt to write the integral as \(\int A\sin^5 2x\,\mathrm{d}x\)
M1: Uses the answer to part (a) to find a value for \(I_5\) and \(I_3\) and finds \(I_1 = \int_0^{\frac{\pi}{2}} \sin 2x\,\mathrm{d}x = \ldots\)
A1: \(\frac{16}{15}\) cso