A2 June 2021 Paper 2 Q12
12 The integral \(S_n\) is defined by
\[S_n = \int_0^a x^n\sinh x\,\mathrm{d}x \qquad (n \geqslant 0)\](a) Show that for \(n \geqslant 2\)\[S_n = n(n - 1)S_{n-2} + a^n\cosh a - na^{n-1}\sinh a\] [7 marks]
(b) Hence show that\[\int_0^1 x^4\sinh x\,\mathrm{d}x = \frac{9}{2}\mathrm{e} + \frac{65}{2}\mathrm{e}^{-1} - 24\] [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Selects a method to find the required result by integrating by parts | M1 | 1.1a |
| Obtains the correct expressions for \(u^{\prime}\) and \(v\) when integrating the first time | A1 | 1.1b |
| Correctly applies integration by parts formula the first time | A1 | 1.1b |
| Correctly applies integration by parts formula to an integral of the form \(\int_0^a x^r\cosh x\,\mathrm{d}x\) | M1 | 3.1a |
| Obtains correct result from 2nd integration by parts and substitutes limits correctly in all terms | A1 | 1.1b |
| Obtains an expression for \(S_n\) in terms of \(S_{n-2}\) | M1 | 1.1a |
| Completes a rigorous argument to show the required result, including correct use of limits throughout | R1 | 2.1 |
| (7) |
Typical solution
\[S_n = \int_0^a x^n\sinh x\,\mathrm{d}x\]\[u = x^n \qquad v^{\prime} = \sinh x\]\[u^{\prime} = nx^{n-1} \qquad v = \cosh x\]\[\begin{aligned} S_n &= \left[x^n\cosh x\right]_0^a - \int_0^a nx^{n-1}\cosh x\,\mathrm{d}x \\ &= a^n\cosh a - n\int_0^a x^{n-1}\cosh x\,\mathrm{d}x \end{aligned}\]\[u = x^{n-1} \qquad v^{\prime} = \cosh x\]\[u^{\prime} = (n - 1)x^{n-2} \qquad v = \sinh x\]\[S_n = a^n\cosh a - n\left(\left[x^{n-1}\sinh x\right]_0^a - \int_0^a (n - 1)x^{n-2}\sinh x\,\mathrm{d}x\right)\]\[S_n = a^n\cosh a - n\left(a^{n-1}\sinh a - (n - 1)S_{n-2}\right)\]\[S_n = a^n\cosh a - na^{n-1}\sinh a + n(n - 1)S_{n-2}\]\[S_n = n(n - 1)S_{n-2} + a^n\cosh a - na^{n-1}\sinh a\]| Scheme | Marks | AO |
|---|---|---|
| Deduces that the integral is \(S_4\) with \(a = 1\) and requires \(S_2\) and \(S_0\) | M1 | 2.2a |
| Uses the reduction formula once | M1 | 3.1a |
| Uses the reduction formula a second time and finds \(S_0\) | M1 | 3.1a |
| Converts both hyperbolic expressions to exponentials | M1 | 1.1a |
| Completes a rigorous argument to show the required result | R1 | 2.1 |
| (5) | ||
| (12 marks) |
Typical solution
The required integral is \(S_4\) with \(a = 1\)
\[\begin{aligned} S_0 &= \int_0^1 \sinh x\,\mathrm{d}x = \left[\cosh x\right]_0^1 \\ &= \cosh 1 - 1 \end{aligned}\]\[\begin{aligned} S_2 &= (2)(1)S_0 + \cosh 1 - 2\sinh 1 \\ &= 2(\cosh 1 - 1) + \cosh 1 - 2\sinh 1 \\ &= 3\cosh 1 - 2\sinh 1 - 2 \end{aligned}\]\[\begin{aligned} S_4 &= (4)(3)S_2 + \cosh 1 - 4\sinh 1 \\ &= 12(3\cosh 1 - 2\sinh 1 - 2) + \cosh 1 - 4\sinh 1 \\ &= 37\cosh 1 - 28\sinh 1 - 24 \\ &= \frac{37}{2}(\mathrm{e} + \mathrm{e}^{-1}) - 14(\mathrm{e} - \mathrm{e}^{-1}) - 24 \\ &= \frac{9}{2}\mathrm{e} + \frac{65}{2}\mathrm{e}^{-1} - 24 \end{aligned}\]