A2 October 2020 Paper 1 Q12
12 Show that \(\displaystyle\int_0^{\frac{1}{\sqrt{3}}} \frac{4}{1 - x^4}\,\mathrm{d}x = \ln\left(a + \sqrt{b}\right) + \frac{\pi}{c}\) where \(a\), \(b\) and \(c\) are integers to be determined. [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{4}{1 - x^4} = \dfrac{A}{1 - x} + \dfrac{B}{1 + x} + \dfrac{Cx + D}{1 + x^2}\) \(\Rightarrow A(1 + x)(1 + x^2) + B(1 - x)(1 + x^2) + (Cx + D)(1 - x^2) = 4\) | M1 | 3.1a |
| \(x = 1 : 4A = 4 \Rightarrow A = 1\) \(x = -1 : 4B = 4 \Rightarrow B = 1\) | A1 | 1.1 |
| \(x = 0 : A + B + D = 4 \Rightarrow D = 2\) \(x^3 : A - B - C = 0 \Rightarrow C = 0\) \(I = \displaystyle\int_0^{\frac{1}{\sqrt{3}}} \left(\frac{1}{1 - x} + \frac{1}{1 + x} + \frac{2}{1 + x^2}\right)\mathrm{d}x\) | A1 | 1.1 |
| \(= \left[\ln\left(\dfrac{1 + x}{1 - x}\right) + 2\tan^{-1}x\right]_0^{\frac{1}{\sqrt{3}}}\) | M1 | 1.1 |
| \(= \ln\left(\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}\right) + 2\tan^{-1}\dfrac{1}{\sqrt{3}}\ (-0) = \ln\left(\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}\right) + \dfrac{\pi}{3}\) | M1 | 1.1 |
| \(= \ln\left(2 + \sqrt{3}\right) + \dfrac{\pi}{3}\) i.e. \(a = 2,\ b = 3,\ c = 3\) | A1 | 1.1 |
| [6] |
Notes
M1: Proper split to produce integrable integrand
A1: For one of the terms /constants
A1: For all terms/constants
Or \(\dfrac{1}{4}\) of these
See below for other possibilities
M1: Correctly integration of their integrand without simplification – ignore limits
M1: Substitution – ignore – 0
A1: Values must be stated
Alternatives
\(\dfrac{D}{1 + x^2}\) is M0 A0 A0 but consider integration for 3 marks.
\(\dfrac{1}{1 - x^4} = \dfrac{1}{2}\left(\dfrac{1}{1 - x^2} + \dfrac{1}{1 + x^2}\right) = \dfrac{1}{2}\left(\dfrac{1}{2}\left(\dfrac{1}{1 - x} + \dfrac{1}{1 + x}\right) + \dfrac{1}{1 + x^2}\right)\) is M1 A1 A1
| Scheme | Marks |
|---|---|
| \(\dfrac{4}{1 - x^4} = \dfrac{A}{1 - x} + \dfrac{B}{1 + x} + \dfrac{D}{1 + x^2}\) | M0 |
| \(\dfrac{4}{1 - x^4} = \dfrac{A}{1 - x^2} + \dfrac{D}{1 + x^2}\) with \(A = D = 2\) by inspection Followed by \(\dfrac{2}{1 + x^2} + \dfrac{1}{1 - x} + \dfrac{1}{1 + x}\) in integration section | M1 A1 A1 |
| \(\dfrac{4}{1 - x^4} = \dfrac{A}{1 - x^2} + \dfrac{D}{1 + x^2}\) \(\Rightarrow \displaystyle\int_0^{\frac{1}{\sqrt{3}}} \frac{4}{1 - x^4}\,\mathrm{d}x = \int_0^{\frac{1}{\sqrt{3}}} \left(\frac{2}{1 - x^2} + \frac{2}{1 + x^2}\right)\mathrm{d}x\) \(= \left[2\tanh^{-1}x + 2\tan^{-1}x\right]_0^{\frac{1}{\sqrt{3}}} = \left(2\tanh^{-1}\dfrac{1}{\sqrt{3}} + 2\tan^{-1}\dfrac{1}{\sqrt{3}}\right)\) \(= \ln\left(\dfrac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}\right) + 2\dfrac{\pi}{6} - 0\) \(= \ln\left(\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}\right) + \dfrac{\pi}{3} = \ln\left(\dfrac{1 + 2\sqrt{3} + 3}{2}\right) + \dfrac{\pi}{3} = \ln\left(\dfrac{4 + 2\sqrt{3}}{2}\right) + \dfrac{\pi}{3}\) \(= \ln\left(2 + \sqrt{3}\right) + \dfrac{\pi}{3}\) | M1 A1 A1 |
M0: But consider last 3 marks for correct integration
M1 A1 A1 (second line): Look to see the second split further on in question
M1 A1 A1 (third line): Look for the integration. If \(\tanh^{-1}\) is used then give full marks here.