A2 June 2024 Paper 1 Q17
17 By making a suitable substitution, show that
\[\int_{-2}^{1} \sqrt{x^2 + 6x + 8}\,\mathrm{d}x = 2\sqrt{15} - \frac{1}{2}\cosh^{-1}(4)\][7 marks]
| Scheme | Marks | AO |
|---|---|---|
| Completes the square to obtain \((x + a)^2 + b\) | M1 | 3.1a |
| Writes \(x + a = k\cosh\theta\) or \(x + a = k\sinh\theta\) | M1 | 2.2a |
| Obtains \(\displaystyle\int \sinh^2\theta\,\mathrm{d}\theta\) | A1 | 1.1b |
| Uses \(\sinh^2\theta = \dfrac{1}{2}\cosh 2\theta - \dfrac{1}{2}\) Condone \(\sinh^2\theta = \pm\dfrac{1}{2}\cosh 2\theta \pm \dfrac{1}{2}\) | M1 | 3.1a |
| Obtains \(\dfrac{1}{2}\left[\dfrac{\sinh 2\theta}{2} - \theta\right]\) OE | A1 | 1.1b |
| Uses \(\cosh^2\theta - \sinh^2\theta = 1\) when \(x = 1\) to obtain \(\sinh\theta = \sqrt{15}\) | B1 | 2.2a |
| Completes a reasoned argument to obtain \(2\sqrt{15} - \dfrac{1}{2}\cosh^{-1}(4)\) AG | R1 | 2.1 |
| (7 marks) |
Typical solution
\[x^2 + 6x + 8 = (x + 3)^2 - 1\]Let \(x + 3 = \cosh\theta\)
Then
\[\frac{\mathrm{d}x}{\mathrm{d}\theta} = \sinh\theta\]Let \(I = \displaystyle\int_{-2}^{1} \sqrt{x^2 + 6x + 8}\,\mathrm{d}x\)
When \(x = -2\), \(\cosh\theta = 1\) and \(\theta = 0\)
When \(x = 1\), \(\cosh\theta = 4\) and \(\theta = \cosh^{-1}(4)\)
\[\begin{aligned}I &= \int_0^{\cosh^{-1}(4)} \sqrt{\sinh^2\theta}\,\sinh\theta\,\mathrm{d}\theta \\ &= \int_0^{\cosh^{-1}(4)} \sinh^2\theta\,\mathrm{d}\theta\end{aligned}\]\[\begin{aligned}I &= \frac{1}{2}\int_0^{\cosh^{-1}(4)} (\cosh 2\theta - 1)\,\mathrm{d}\theta \\ &= \frac{1}{2}\left[\frac{\sinh 2\theta}{2} - \theta\right]_0^{\cosh^{-1}(4)} \\ &= \frac{1}{2}\Big[\sinh\theta\cosh\theta - \theta\Big]_0^{\cosh^{-1}(4)}\end{aligned}\]When \(\cosh\theta = 4\), \(\sinh\theta = \sqrt{4^2 - 1} = \sqrt{15}\)
\[\begin{aligned}I &= \frac{1}{2}\left(4\sqrt{15} - \cosh^{-1}(4)\right) - 0 \\ &= 2\sqrt{15} - \frac{1}{2}\cosh^{-1}(4)\end{aligned}\]