FP3 June 2009 Q4
4. Given that \(\quad y = \mathrm{arsinh}\,(\sqrt{x}), \quad x > 0,\)
(a) find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), giving your answer as a simplified fraction. (3)
(b) Hence, or otherwise, find \[\int_{\frac{1}{4}}^{4} \frac{1}{\sqrt{[x(x+1)]}}\,\mathrm{d}x,\] giving your answer in the form \(\ln\left(\dfrac{a + b\sqrt{5}}{2}\right)\), where \(a\) and \(b\) are integers. (6)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \tfrac{1}{2}x^{-\frac{1}{2}} \times \dfrac{1}{\sqrt{1 + (\sqrt{x})^2}}\) | B1, M1 |
| \(\therefore \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{1 + x}} \quad \left(= \dfrac{1}{2\sqrt{x(1 + x)}}\right)\) | A1 |
| (3) |
Notes
Alternative (i) for part (a)
| Scheme | Marks |
|---|---|
| Use \(\sinh y = \sqrt{x}\) and state \(\cosh y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \tfrac{1}{2}x^{-\frac{1}{2}}\) | B1 |
| \(\therefore \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{1 + \sinh^2 y}} = \dfrac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{\left(1 + \left(\sqrt{x}\right)^2\right)}}\) | M1 |
| \(\therefore \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\frac{1}{2}x^{-\frac{1}{2}}}{\sqrt{1 + x}} \quad \left(= \dfrac{1}{2\sqrt{x(1 + x)}}\right)\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\therefore \displaystyle\int_{\frac{1}{4}}^{4} \dfrac{1}{\sqrt{x(x+1)}}\,\mathrm{d}x = \left[2\,\mathrm{arsinh}\sqrt{x}\right]_{\frac{1}{4}}^{4}\) | M1 |
| \(= \left[2\,\mathrm{arsinh}\,2 - 2\,\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right]\) | M1 |
| \(= \left[2\ln(2 + \sqrt{5})\right] - \left[2\ln\left(\tfrac{1}{2} + \sqrt{\tfrac{5}{4}}\right)\right]\) | M1 |
| \(2\ln\dfrac{(2 + \sqrt{5})}{\left(\frac{1}{2} + \sqrt{\left(\frac{5}{4}\right)}\right)} = 2\ln\dfrac{2(2 + \sqrt{5})}{(1 + \sqrt{5})} = 2\ln\dfrac{2(\sqrt{5} + 2)(\sqrt{5} - 1)}{(\sqrt{5} + 1)(\sqrt{5} - 1)} = 2\ln\dfrac{(3 + \sqrt{5})}{2}\) | M1 |
| \(= \ln\dfrac{(3 + \sqrt{5})(3 + \sqrt{5})}{4} = \ln\dfrac{14 + 6\sqrt{5}}{4} = \ln\left(\dfrac{7}{2} + \dfrac{3\sqrt{5}}{2}\right)\) | A1 A1 |
| (6) | |
| (9 marks) |
Notes
Alternative (i) for part (b)
| Scheme | Marks |
|---|---|
| Use \(x = \tan^2\theta\), \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 2\tan\theta\sec^2\theta\) to give \(2\displaystyle\int \sec\theta\,\mathrm{d}\theta = \left[2\ln(\sec\theta + \tan\theta)\right]\) | M1 |
| \(= \left[2\ln(\sec\theta + \tan\theta)\right]_{\tan\theta = \frac{1}{2}}^{\tan\theta = 2}\) i.e. use of limits | M1 |
| then proceed as before from line 3 of scheme |
Alternative (ii) for part (b)
| Scheme | Marks |
|---|---|
| Use \(\displaystyle\int \dfrac{1}{\sqrt{\left[(x + \frac{1}{2})^2 - \frac{1}{4}\right]}}\,\mathrm{d}x = \mathrm{arcosh}\,\dfrac{x + \frac{1}{2}}{\frac{1}{2}}\) | M1 |
| \(= \left[\mathrm{arcosh}\,9 - \mathrm{arcosh}\left(\tfrac{3}{2}\right)\right]\) | M1 |
| \(= \left[\ln(9 + \sqrt{80})\right] - \left[\ln\left(\tfrac{3}{2} + \tfrac{1}{2}\sqrt{5}\right)\right]\) | M1 |
| \(= \ln\dfrac{(9 + \sqrt{80})}{\left(\frac{3}{2} + \frac{1}{2}\sqrt{5}\right)} = \ln\dfrac{2(9 + \sqrt{80})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}\), | M1 |
| \(= \ln\dfrac{2(9 + 4\sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} = \ln\left(\dfrac{7}{2} + \dfrac{3\sqrt{5}}{2}\right)\) | A1 A1 |
| (6) |