FP3 June 2013 Q2
2.
(a) Find \[\int \frac{1}{\sqrt{(4x^2 + 9)}}\,\mathrm{d}x\] (2)
(b) Use your answer to part (a) to find the exact value of \[\int_{-3}^{3} \frac{1}{\sqrt{(4x^2 + 9)}}\,\mathrm{d}x\] giving your answer in the form \(k\ln(a + b\sqrt{5})\), where \(a\) and \(b\) are integers and \(k\) is a constant. (3)
| Scheme | Marks |
|---|---|
| \(k\,\mathrm{arsinh}\left(\dfrac{2x}{3}\right)\ (+c)\) or \(k\ln\left[px + \sqrt{\left(p^2x^2 + \dfrac{9}{4}p^2\right)}\right]\ (+c)\) | M1 |
| \(\dfrac{1}{2}\mathrm{arsinh}\left(\dfrac{2x}{3}\right)\ (+c)\) or \(\dfrac{1}{2}\ln\left[px + \sqrt{\left(p^2x^2 + \dfrac{9}{4}p^2\right)}\right]\ (+c)\) | A1 |
| (2) |
Notes
Alternative for (a)
| Scheme | Marks |
|---|---|
| \(x = \dfrac{3}{2}\sinh u \Rightarrow \displaystyle\int \frac{1}{\sqrt{9\sinh^2 u + 9}} \cdot \frac{3}{2}\cosh u\,\mathrm{d}u = k\,\mathrm{arsinh}\left(\frac{2x}{3}\right)\ (+c)\) | M1 |
| \(\dfrac{1}{2}\mathrm{arsinh}\left(\dfrac{2x}{3}\right)\ (+c)\) | A1 |
| Scheme | Marks |
|---|---|
| So: \(\dfrac{1}{2}\ln\left[6 + \sqrt{45}\right] - \dfrac{1}{2}\ln\left[-6 + \sqrt{45}\right] = \dfrac{1}{2}\ln\left[\dfrac{6 + \sqrt{45}}{-6 + \sqrt{45}}\right]\) Uses correct limits and combines logs | M1 |
| \(= \dfrac{1}{2}\ln\left[\dfrac{6 + \sqrt{45}}{-6 + \sqrt{45}}\right]\left[\dfrac{6 + \sqrt{45}}{6 + \sqrt{45}}\right] = \dfrac{1}{2}\ln\left[\dfrac{(6 + \sqrt{45})^2}{9}\right]\) Correct method to rationalise denominator (may be implied) Method must be clear if answer does not follow their fraction | M1 |
| \(= \ln[2 + \sqrt{5}]\) (or \(\dfrac{1}{2}\ln[9 + 4\sqrt{5}]\)) | A1cso |
| (3) | |
| (5 marks) |
Notes
Note that the last 3 marks can be scored without the need to rationalise e.g.
\(2 \times \dfrac{1}{2}\left[\ln[2x + \sqrt{(4x^2 + 9)}]\right]_0^3 = \ln(6 + \sqrt{45}) - \ln 3 = \ln\left(\dfrac{6 + \sqrt{45}}{3}\right)\)
M1: Uses the limits 0 and 3 and doubles
M1: Combines Logs
A1: \(\ln[2 + \sqrt{5}]\) oe
Alternative for (b)
| Scheme | Marks |
|---|---|
| \(\left[\dfrac{1}{2}\mathrm{arsinh}\left(\dfrac{2x}{3}\right)\right]_{-3}^{3} = \dfrac{1}{2}\mathrm{arsinh}\ 2 - \dfrac{1}{2}\mathrm{arsinh}\ {-2}\) | |
| \(\dfrac{1}{2}\ln(2 + \sqrt{5}) - \dfrac{1}{2}\ln(\sqrt{5} - 2) = \dfrac{1}{2}\ln\left(\dfrac{2 + \sqrt{5}}{\sqrt{5} - 2}\right)\) Uses correct limits and combines logs | M1 |
| \(= \dfrac{1}{2}\ln\left(\dfrac{2 + \sqrt{5}}{\sqrt{5} - 2} \cdot \dfrac{\sqrt{5} + 2}{\sqrt{5} + 2}\right) = \dfrac{1}{2}\ln\left(\dfrac{2\sqrt{5} + 4 + 5 + 2\sqrt{5}}{5 - 4}\right)\) Correct method to rationalise denominator (may be implied) Method must be clear if answer does not follow their fraction | M1 |
| \(= \dfrac{1}{2}\ln[9 + 4\sqrt{5}]\) | A1cso |