FP3 June 2017 Q7
7. \[I_n = \int_0^{\ln 2}\cosh^n x\,\mathrm{d}x, \quad n \geqslant 0\]
| Scheme | Marks |
|---|---|
| \(\displaystyle I_n = \int_0^{\ln 2}\cosh^n x\,\mathrm{d}x\) | |
| \(\displaystyle I_n = \int\cosh^{n-1}x\cosh x\,\mathrm{d}x\) | |
| \(\displaystyle I_n = \int\cosh^{n-1}x\cosh x\,\mathrm{d}x = \sinh x\cosh^{n-1}x - \int(n - 1)\cosh^{n-2}x\sinh^2 x\,\mathrm{d}x\) | M1A1 |
| \(\displaystyle = \sinh x\cosh^{n-1}x - \int(n - 1)\cosh^{n-2}x\left(\cosh^2 x - 1\right)\mathrm{d}x\) | dM1 |
| \(\displaystyle = \sinh x\cosh^{n-1}x - (n - 1)\int\cosh^n x\,\mathrm{d}x + (n - 1)\int\cosh^{n-2}x\,\mathrm{d}x\) | |
| \(= \sinh x\cosh^{n-1}x - (n - 1)I_n + (n - 1)I_{n-2}\) | ddM1 |
| \(\left[\sinh x\cosh^{n-1}x\right]_0^{\ln 2} = \sinh(\ln 2)\cosh^{n-1}(\ln 2)(-0)\) \(\left(= \left(\dfrac{3}{4}\right)\left(\dfrac{5}{4}\right)^{n-1}\right)\) | M1 |
| \(I_n = \dfrac{3 \times 5^{n-1}}{n \times 4^n} + \dfrac{(n - 1)}{n}I_{n-2}\) * | A1* |
| (6) |
Notes
M1: Integration by parts in the correct direction. If the formula is quoted it must be correct otherwise look for an expression of the form \(\pm\sinh x\cosh^{n-1}x \pm k\int\cosh^{n-2}x\sinh^2 x\,\mathrm{d}x\)
A1: Correct expression
dM1: Replaces \(\sinh^2 x\) with \(\pm\cosh^2 x \pm 1\) on the “integration part” to obtain an expression in \(\cosh x\) only. Dependent on the first method mark.
ddM1: Introduces \(I_n\) and \(I_{n-2}\). Dependent on both previous method marks.
M1: Use of given limits on their \(\sinh x\cosh^{n-1}x\). Does not need to be evaluated but note that \(\cosh(\ln 2) = \dfrac{5}{4},\ \sinh(\ln 2) = \dfrac{3}{4}\)
A1*: cao
(a) Way 2
| Scheme | Marks |
|---|---|
| \(\displaystyle I_n = \int\cosh^{n-2}x\cosh^2 x\,\mathrm{d}x = \int\cosh^{n-2}x\,\mathrm{d}x + \int\cosh^{n-2}x\sinh^2 x\,\mathrm{d}x\) Writes \(\cosh^n x\) as \(\cosh^{n-2}x\cosh^2 x\) and uses \(\sinh^2 x = \pm\cosh^2 x \pm 1\) | M1 |
| \(\displaystyle\int\cosh^{n-2}x\sinh^2 x\,\mathrm{d}x = \left[\frac{\sinh x\cosh^{n-1}x}{n - 1}\right] - \frac{1}{n - 1}\int\cosh^n x\,\mathrm{d}x\) M1: Integration by parts in the correct direction. If the formula is quoted it must be correct otherwise look for an expression of the form \(\displaystyle p\sinh x\cosh^{n-1}x \pm q\int\cosh^n x\,\mathrm{d}x\) A1: Correct expression | dM1A1 |
| \((n - 1)I_n = (n - 1)I_{n-2} + \left[\sinh x\cosh^{n-1}x\right] - I_n\) | ddM1 |
| \(\left[\sinh x\cosh^{n-1}x\right]_0^{\ln 2} = \sinh(\ln 2)\cosh^{n-1}(\ln 2)(-0)\) \(\left(= \left(\dfrac{3}{4}\right)\left(\dfrac{5}{4}\right)^{n-1}\right)\) | M1 |
| \(I_n = \dfrac{3 \times 5^{n-1}}{n \times 4^n} + \dfrac{(n - 1)}{n}I_{n-2}\) * | A1* |
ddM1: Introduces \(I_n\) and \(I_{n-2}\). Dependent on both previous method marks.
M1: Use of given limits on their \(\sinh x\cosh^{n-1}x\). Does not need to be evaluated but note that \(\cosh(\ln 2) = \dfrac{5}{4},\ \sinh(\ln 2) = \dfrac{3}{4}\)
A1*: cao
You can condone the occasional missing \(x\), \(\mathrm{d}x\) and limits along the way and “invisible” brackets may be recovered.
Do not allow e.g. an obvious sign error that gets “corrected” later – withhold the final A1 in such cases.
| Scheme | Marks |
|---|---|
| \(I_4 = \dfrac{3 \times 5^3}{4 \times 4^4} + \dfrac{3}{4}I_2\) or \(\dfrac{3 \times a^3}{4 \times b^4} + \dfrac{3}{4}I_2\) | M1 |
| \(= \dfrac{3 \times 5^3}{4 \times 4^4} + \dfrac{3}{4}\left(\dfrac{3 \times 5}{2 \times 4^2} + \dfrac{1}{2}I_0\right)\) or \(\dfrac{3 \times a^3}{4 \times b^4} + \dfrac{3}{4}\left(\dfrac{3 \times a}{2 \times b^2} + \dfrac{1}{2}I_0\right)\) | M1 |
| \(I_0 = \ln 2\) | B1 |
| \(I_4 = \dfrac{735}{1024} + \dfrac{3}{8}\ln 2\) | A1 |
| (4) | |
| (10 marks) |
Notes
M1: Correct first application of their or the given reduction formula
M1: Correct second application of their or the given reduction formula that is consistent with the formula used in the first application to obtain \(I_4\) in terms of \(I_0\)
A1: Cao (Allow equivalent exact forms e.g. may be factorised but fractions must be collected)
Note that candidates may work from the “other end” e.g.
\(I_0 = \ln 2\) B1
\(I_2 = \dfrac{3 \times 5}{2 \times 4^2} + \dfrac{1}{2}I_0\) M1 \(I_2\) in terms of \(I_0\)
\(I_4 = \dfrac{3 \times 5^3}{4 \times 4^4} + \dfrac{3}{4}\left(\dfrac{3 \times 5}{2 \times 4^2} + \dfrac{1}{2}I_0\right)\) M1 \(I_4\) in terms of \(I_0\)
\(I_4 = \dfrac{735}{1024} + \dfrac{3}{8}\ln 2\) A1
Cao (Allow equivalent exact forms e.g. may be factorised but fractions must be collected)
(b) Way 2
| Scheme | Marks |
|---|---|
| \(I_4 = \dfrac{3 \times 5^3}{4 \times 4^4} + \dfrac{3}{4}I_2\) | M1 |
| \(\displaystyle I_2 = \int_0^{\ln 2}\cosh^2 x\,\mathrm{d}x = \int_0^{\ln 2}\left(\frac{1}{2} + \frac{1}{2}\cosh 2x\right)\mathrm{d}x\) | |
| \(\displaystyle\int\left(\frac{1}{2} + \frac{1}{2}\cosh 2x\right)\mathrm{d}x = \frac{x}{2} + \frac{1}{4}\sinh 2x\) | B1 |
| \(I_2 = \left[\dfrac{x}{2} + \dfrac{1}{4}\sinh 2x\right]_0^{\ln 2} = \dfrac{1}{2}\ln 2 + \dfrac{15}{32}\) | M1 |
| \(I_4 = \dfrac{3 \times 5^3}{4 \times 4^4} + \dfrac{3}{4}\left(\dfrac{1}{2}\ln 2 + \dfrac{15}{32}\right)\) | |
| \(I_4 = \dfrac{735}{1024} + \dfrac{3}{8}\ln 2\) | A1 |
M1: Correct application of their reduction formula
B1: Correct integration
M1: Correct use of limits on an expression of the form \(\alpha x + \beta\sinh 2x\)
A1: Cao (Allow equivalent exact forms e.g. may be factorised but fractions must be collected)
(b) Way 3
| Scheme | Marks |
|---|---|
| \(\displaystyle I_4 = \int_0^{\ln 2}\cosh^4 x\,\mathrm{d}x = \int_0^{\ln 2}\left(\frac{1}{2} + \frac{1}{2}\cosh 2x\right)^2\mathrm{d}x\) | |
| \(\displaystyle\int_0^{\ln 2}\left(\frac{1}{4} + \frac{1}{2}\cosh 2x + \frac{1}{4}\cosh^2 2x\right)\mathrm{d}x\) | B1 |
| \(\displaystyle\frac{1}{4}\int_0^{\ln 2}\left(1 + 2\cosh 2x + \frac{1}{2}(1 + \cosh 4x)\right)\mathrm{d}x\) | M1 |
| \(\dfrac{1}{4}\left[\dfrac{3x}{2} + \sinh 2x + \dfrac{1}{8}\sinh 4x\right]_0^{\ln 2}\) | M1 |
| \(I_4 = \dfrac{735}{1024} + \dfrac{3}{8}\ln 2\) | A1 |
B1: \(\cosh^4 x = \dfrac{1}{4} + \dfrac{1}{2}\cosh 2x + \dfrac{1}{4}\cosh^2 2x\)
M1: \(\cosh^2 2x = \pm\dfrac{1}{2} \pm \dfrac{1}{2}\cosh 4x\) and attempt to integrate
M1: Correct use of correct limits
A1: Cao (Allow equivalent exact forms e.g. may be factorised but fractions must be collected)
(b) Way 4
| Scheme | Marks |
|---|---|
| \(\displaystyle I_4 = \int_0^{\ln 2}\cosh^4 x\,\mathrm{d}x = \int_0^{\ln 2}\left(\frac{e^x + e^{-x}}{2}\right)^4\mathrm{d}x\) | |
| \(\displaystyle = \int_0^{\ln 2}\left(\frac{e^x + e^{-x}}{2}\right)^4\mathrm{d}x = \left(\frac{1}{16}\right)\int_0^{\ln 2}\left(e^{4x} + 4e^{2x} + 6 + 4e^{-2x} + e^{-4x}\right)\mathrm{d}x\) Correct expansion | B1 |
| \(= \left(\dfrac{1}{16}\right)\left[\dfrac{e^{4x}}{4} + 2e^{2x} + 6x - 2e^{-2x} - \dfrac{e^{-4x}}{4}\right]_0^{\ln 2}\) | M1 |
| \(\left(\dfrac{1}{16}\right)\left[\left(4 + 8 + 6\ln 2 - \dfrac{1}{2} - \dfrac{1}{64}\right) - (0)\right]_0^{\ln 2}\) | M1 |
| \(I_4 = \dfrac{735}{1024} + \dfrac{3}{8}\ln 2\) | A1 |
M1: Attempts to integrate their expansion
M1: Correct use of correct limits
A1: Cao (Allow equivalent exact forms e.g. may be factorised but fractions must be collected)