A2 October 2020 Paper 1 Q3
3 In this question you must show detailed reasoning.
Find \(\displaystyle\int_0^{\frac{1}{3}} \frac{1}{\sqrt{4 - 9x^2}}\,\mathrm{d}x\), expressing your answer in terms of \(\pi\). [4]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\displaystyle\int_0^{\frac{1}{3}} \frac{\mathrm{d}x}{\sqrt{4 - 9x^2}} = \frac{1}{3}\int_0^{\frac{1}{3}} \frac{\mathrm{d}x}{\sqrt{\frac{4}{9} - x^2}}\) | M1 | 3.1a |
| \(= \left[\dfrac{1}{3}\arcsin\dfrac{3x}{2}\right]_0^{\frac{1}{3}}\) | A1 | 1.1 |
| \(= \tfrac{1}{3}\left(\arcsin\tfrac{1}{2}\,[-\arcsin 0]\right)\) | M1 | 1.1 |
| \(= \dfrac{\pi}{18}\) | A1 | 1.1 |
| [4] |
Notes
M1: must be in the form \(\dfrac{k}{\sqrt{\frac{4}{9} - x^2}}\), \(k \neq 1\)
A1: \(k\arcsin(3x/2)\)
can award M1A1 if integral is fully correct before limits substituted
Alternative method
| Scheme | Marks |
|---|---|
| let \(x = \dfrac{2}{3}\sin\theta,\ \dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \dfrac{2}{3}\cos\theta\) | M1 |
| \(\displaystyle\int_0^{\frac{1}{3}} \frac{\mathrm{d}x}{\sqrt{4 - 9x^2}} = \int_0^{\frac{\pi}{6}} \frac{\frac{2}{3}\cos\theta\,\mathrm{d}\theta}{2\cos\theta}\) | A1 |
| \(= \dfrac{1}{3}\left[\theta\right]_0^{\frac{\pi}{6}} = \dfrac{\pi}{18}\) | M1 A1 |
| [4] |
M1: for suitable substitution
A1: an equivalent expression that can be integrated
M1: substitution of correct limits of their variable