C4 June 2010 Q2
2. Using the substitution \(u = \cos x + 1\), or otherwise, show that \[\int_0^{\frac{\pi}{2}} \mathrm{e}^{\cos x + 1}\sin x\,\mathrm{d}x = \mathrm{e}(\mathrm{e} - 1)\] (6)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = -\sin x\) | B1 |
| \(\displaystyle\int \sin x\,\mathrm{e}^{\cos x + 1}\,\mathrm{d}x = -\int \mathrm{e}^u\,\mathrm{d}u\) | M1 A1 |
| \(= -\mathrm{e}^u\) ft sign error \(= -\mathrm{e}^{\cos x + 1}\) | A1ft |
| \(\Big[-\mathrm{e}^{\cos x + 1}\Big]_0^{\frac{\pi}{2}} = -\mathrm{e}^1 - \left(-\mathrm{e}^2\right)\) or equivalent with \(u\) | M1 |
| \(= \mathrm{e}(\mathrm{e} - 1)\ \ \ast\) cso | A1 |
| (6) | |
| (6 marks) |