FP3 June 2015 Q7
7. \[I_n = \int\sin^n x\,\mathrm{d}x, \quad n \geqslant 0\]
Given that \(n\) is an odd number, \(n \geqslant 3\)
| Scheme | Marks |
|---|---|
| \(\displaystyle I_n = \int\sin^{n-1}x\sin x\,\mathrm{d}x\) | M1 |
| \(\displaystyle I_n = \sin^{n-1}x(-\cos x) + \int(n - 1)\sin^{n-2}x\cos^2 x\,\mathrm{d}x\) | dM1 |
| \(I_n = -\sin^{n-1}x\cos x + (n - 1)(I_{n-2} - I_n)\) | A1 |
| \(I_n = -\sin^{n-1}x\cos x + (n - 1)I_{n-2} - nI_n + I_n\) | |
| \(I_n = \dfrac{1}{n}\left(-\sin^{n-1}x\cos x + (n - 1)I_{n-2}\right)\) * | A1* |
| (4) |
Notes
M1: Split into \(\sin^{n-1}x\) and \(\sin x\)
dM1: Integration by parts in the right direction (if the method is unclear or formula not quoted only allow sign errors) Dependent on the first method mark.
A1: Obtains \(I_n\) correctly in terms of \(I_{n-2}\) and \(I_n\)
A1*: Printed answer obtained with at least one intermediate step and no errors seen (condone the occasional \(x\) lost along the way but the final answer must be exactly as printed)
Condone omission of “\(\mathrm{d}x\)” throughout in both methods
Alternative
| Scheme | Marks |
|---|---|
| \(\displaystyle = \int\sin^{n-2}x\left(1 - \cos^2 x\right)\mathrm{d}x\) | M1 |
| \(\displaystyle = I_{n-2} - \left\{\frac{\sin^{n-1}x\cos x}{n - 1} + \int\frac{\sin^n x}{n - 1}\,\mathrm{d}x\right\}\) | dM1 |
| \(= I_{n-2} - \dfrac{\sin^{n-1}x\cos x}{n - 1} - \dfrac{1}{n - 1}I_n\) | A1 |
| \((n - 1)I_n = (n - 1)I_{n-2} - \sin^{n-1}x\cos x - I_n\) | |
| \(I_n = \dfrac{1}{n}\left(-\sin^{n-1}x\cos x + (n - 1)I_{n-2}\right)\) * | A1* |
M1: Splits into \(\sin^{n-2}x\) and \(\sin^2 x\) and uses \(\sin^2 x = 1 - \cos^2 x\)
dM1: Integration by parts in the right direction (if the method is unclear or formula not quoted only allow sign errors). Dependent on the first method mark.
A1: Obtains \(I_n\) correctly in terms of \(I_{n-2}\) and \(I_n\)
A1*: Printed answer obtained with at least one intermediate step and no errors seen (condone the occasional \(x\) lost along the way but the final answer must be exactly as printed)
| Scheme | Marks |
|---|---|
| \(I_n = \dfrac{1}{n}\left(\left[-\sin^{n-1}x\cos x\right]_0^{\frac{\pi}{2}} + (n - 1)I_{n-2}\right)\) | M1 |
| \(I_n = \frac{n-1}{n}I_{n-2}\) | A1 |
| \(n\) odd, \(\displaystyle I_1 = \int_0^{\frac{\pi}{2}}\sin x\,\mathrm{d}x = \left[-\cos x\right]_0^{\frac{\pi}{2}} = 1\) | |
| \(I_n = \dfrac{(n - 1)}{n}I_{n-2} = \dfrac{(n - 1)}{n}\dfrac{(n - 3)}{n - 2}I_{n-4} = \ldots\) | M1 |
| \(I_n = \dfrac{(n - 1)(n - 3)\ldots 6.4.2}{n(n - 2)(n - 4)\ldots 7.5.3}\) ** | A1** |
| (4) |
Notes
M1: Use part (a) with limits
A1: Sight of the expression could score M1A1
An attempt at \(I_1\) must be seen before any more marks are awarded
M1: Attempts \(I_1\) and at least 2 fractions in terms of \(n\)
A1**: Cso. Note this may be awarded for ‘extra’ brackets top and bottom provided all previous marks are scored.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^{\frac{\pi}{2}}\sin^5 x\cos^2 x\,\mathrm{d}x = \int_0^{\frac{\pi}{2}}\sin^5 x(1 - \sin^2 x)\ \mathrm{d}x\) | M1 |
| \(= I_5 - I_7 = \dfrac{4 \times 2}{5 \times 3} - \dfrac{6 \times 4 \times 2}{7 \times 5 \times 3}\) | A1 |
| \(= \dfrac{8}{105}\) | A1 |
| (3) | |
| (11 marks) |
Notes
M1: Uses \(\cos^2 x = 1 - \sin^2 x\)
A1: Correct numerical expression
A1: Cao (accept awrt 0.0761)
Correct answer only with no working would generally score no marks