FP3 June 2016 Q7
7. Given that \[I_n = \int\frac{\sin nx}{\sin x}\,\mathrm{d}x, \quad n \geqslant 1\]
| Scheme | Marks |
|---|---|
| \(\dfrac{\sin nx}{\sin x} - \dfrac{\sin(n - 2)x}{\sin x} = \dfrac{\sin nx - \sin nx\cos 2x + \cos nx\sin 2x}{\sin x}\) Expands \(\sin(n - 2)x\) correctly | M1 |
| \(= \dfrac{\sin nx - \sin nx\left(1 - 2\sin^2 x\right) + 2\sin x\cos x\cos nx}{\sin x}\) Replaces \(\cos 2x\) and \(\sin 2x\) by the correct trigonometric identities | M1 |
| \(= 2\sin nx\sin x + 2\cos nx\cos x\) | |
| \(= 2\cos(n - 1)x\) | |
| \(\left(\therefore I_n - I_{n-2}\right) = \displaystyle\int 2\cos(n - 1)x\,\mathrm{d}x\) * | A1* |
| (3) |
Notes
A1*: Correct completion with no errors. The \(I_n - I_{n-2}\) does not need to be seen explicitly but \(\int 2\cos(n - 1)x\,\mathrm{d}x\) must seen, including the integral sign.
(a) Way 2 (factor formula)
| Scheme | Marks |
|---|---|
| \(\dfrac{\sin nx}{\sin x} - \dfrac{\sin(n - 2)x}{\sin x} = \dfrac{2\cos\left(\frac{nx + nx - 2x}{2}\right)\sin\left(\frac{nx - nx + 2x}{2}\right)}{\sin x}\) Use of the correct factor formula | M1 |
| \(= \dfrac{2\cos(nx - x)\sin x}{\sin x}\) | M1 |
| \(= 2\cos(n - 1)x\) | |
| \(\left(I_n - I_{n-2}\right) = \displaystyle\int 2\cos(n - 1)x\,\mathrm{d}x\) * | A1* |
M1: Attempts to replaces \(nx + nx - 2x\) with \(2nx - 2x\) and attempts to replace \(nx - nx + 2x\) with \(2x\)
A1*: Correct completion with no errors. The \(I_n - I_{n-2}\) does not need to be seen explicitly but \(\int 2\cos(n - 1)x\,\mathrm{d}x\) must seen, including the integral sign.
(a) Way 3
| Scheme | Marks |
|---|---|
| \(I_n = \displaystyle\int\frac{\sin((n - 1)x + x)}{\sin x}\,\mathrm{d}x\) | M1 |
| \(= \displaystyle\int\frac{\sin(n - 1)x\cos x + \sin x\cos(n - 1)x}{\sin x}\,\mathrm{d}x\) | M1 |
| \(= \dfrac{1}{2}\displaystyle\int\frac{\sin nx + \sin(n - 2)x}{\sin x}\,\mathrm{d}x + \int\cos(n - 1)x\,\mathrm{d}x\) | |
| \(= \dfrac{1}{2}I_n + \dfrac{1}{2}I_{n-2} + \displaystyle\int\cos(n - 1)x\,\mathrm{d}x\) | |
| \(\therefore I_n - I_{n-2} = \displaystyle\int 2\cos(n - 1)x\,\mathrm{d}x\) * | A1* |
M1: Uses \(\sin nx = \sin((n - 1)x + x)\)
M1: Expands \(\sin((n - 1)x + x)\) correctly
A1*: Correct completion with no errors.
(a) Way 4
| Scheme | Marks |
|---|---|
| \(\dfrac{\sin nx}{\sin x} = \dfrac{\sin((n - 2)x + 2x)}{\sin x}\) | M1 |
| \(= \dfrac{\sin(n - 2)x\left(1 - 2\sin^2 x\right) + 2\sin x\cos x\cos(n - 2)x}{\sin x}\) Replaces \(\cos 2x\) and \(\sin 2x\) by the correct trigonometric identities | M1 |
| \(= \dfrac{\sin(n - 2)x}{\sin x} - 2\sin x\sin(n - 2)x + 2\cos x\cos(n - 2)x\) | |
| \(= \dfrac{\sin(n - 2)x}{\sin x} + 2\cos((n - 2)x + x)\) | |
| \(I_n = I_{n-2} + 2\displaystyle\int\cos(n - 1)x\,\mathrm{d}x\) | |
| \(\therefore I_n - I_{n-2} = \displaystyle\int 2\cos(n - 1)x\,\mathrm{d}x\) * | A1* |
M1: Uses \(\sin nx = \sin((n - 2)x + 2x)\)
A1*: Correct completion with no errors.
(a) Way 5
| Scheme | Marks |
|---|---|
| \(\sin nx = \sin((n - 1)x + x)\) and \(\sin(n - 2)x = \sin((n - 1)x - x)\) | M1 |
| \(\dfrac{\sin nx}{\sin x} - \dfrac{\sin(n - 2)x}{\sin x} = \dfrac{\sin(n - 1)x\cos x + \cos(n - 1)x\sin x - \left(\sin(n - 1)x\cos x - \sin x\cos(n - 1)x\right)}{\sin x}\) Replaces \(\sin((n - 1)x + x)\) with \(\sin(n - 1)x\cos x + \cos(n - 1)x\sin x\) and Replaces \(\sin((n - 1)x - x)\) with \(\sin(n - 1)x\cos x - \cos(n - 1)x\sin x\) | M1 |
| \(\dfrac{\sin nx}{\sin x} - \dfrac{\sin(n - 2)x}{\sin x} = \dfrac{2\sin x\cos(n - 1)x}{\sin x}\) | |
| \(\left(\therefore I_n - I_{n-2}\right) = \displaystyle\int 2\cos(n - 1)x\,\mathrm{d}x\) * | A1 |
A1: Correct completion with no errors. The \(I_n - I_{n-2}\) does not need to be seen explicitly but \(\int 2\cos(n - 1)x\,\mathrm{d}x\) must seen, including the integral sign.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int\cos 4x\,\mathrm{d}x = k\sin 4x\) or \(\displaystyle\int\cos 2x\,\mathrm{d}x = k\sin 2x\) | M1 |
| \(2\displaystyle\int\cos 4x\,\mathrm{d}x = \frac{1}{2}\sin 4x\) and \(2\displaystyle\int\cos 2x\,\mathrm{d}x = \sin 2x\) | A1 |
| \(\displaystyle\int\frac{\sin 5x}{\sin x}\,\mathrm{d}x = \frac{2\sin(4x)}{4} + I_3\) or \(\displaystyle\int\frac{\sin 3x}{\sin x}\,\mathrm{d}x = \frac{2\sin(2x)}{2} + I_1\) | M1 |
| \(\displaystyle\int\frac{\sin 5x}{\sin x}\,\mathrm{d}x = \frac{2\sin(4x)}{4} + I_3\) and \(\displaystyle\int\frac{\sin 3x}{\sin x}\,\mathrm{d}x = \frac{2\sin(2x)}{2} + I_1\) | M1 |
| \(I_1 = \dfrac{\pi}{12}\) | B1 |
| \(\left[\dfrac{2\sin(4x)}{4} + \dfrac{2\sin(2x)}{2}\right]_{\frac{\pi}{12}}^{\frac{\pi}{6}} = \dfrac{\sqrt{3}}{4} + \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{4} - \dfrac{1}{2}\) | M1 |
| \(\displaystyle\int_{\frac{\pi}{12}}^{\frac{\pi}{6}}\frac{\sin 5x}{\sin x}\,\mathrm{d}x = \frac{1}{12}\left(\pi + 6\sqrt{3} - 6\right)\) | A1 |
| (7) | |
| (10 marks) |
Notes
M1: \(\cos 4x\) integrated to \(\pm k\sin 4x\) or \(\cos 2x\) integrated to \(\pm k\sin 2x\)
A1: Both \(2\cos 4x\) and \(2\cos 2x\) integrated correctly with the correct (possibly un-simplified) coefficients
M1: One application of reduction formula. This may appear in any form and there does not need to be any integration e.g. \(I_5 = \int 2\cos 4x\,\mathrm{d}x + I_3\) or e.g. \(I_3 = \int 2\cos 2x\,\mathrm{d}x + I_1\)
M1: Two applications of reduction formula. This may appear in any form and there does not need to be any integration e.g. \(I_5 = \int 2\cos 4x\,\mathrm{d}x + I_3\) and e.g. \(I_3 = \int 2\cos 2x\,\mathrm{d}x + I_1\)
Note that \(\displaystyle\int\frac{\sin 3x}{\sin x}\,\mathrm{d}x\) may be attempted using trig. Identities and can score full marks as long as use of the reduction formula is seen at least once.
B1: (Could be implied by their final answer)
M1: Correct use of the given limits at least once on an expression of the form \(\pm k\sin 4x\) or \(\pm k\sin 2x\)
A1: cao
Note that correct work leading to \(\left[\dfrac{2\sin(4x)}{4} + \dfrac{2\sin(2x)}{2} + x\right]\) or equivalent could score the first 4 marks