A2 June 2023 Q5
5.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{1 + t^2}{2}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{2}{1 + t^2}\) or \(\mathrm{d}x = \dfrac{2\,\mathrm{d}t}{1 + t^2}\) or \(\mathrm{d}t = \dfrac{1 + t^2}{2}\mathrm{d}x\) oe | B1 | 1.1b |
| \(2\sin x - \cos x + 5 = 2\left(\dfrac{2t}{1 + t^2}\right) - \left(\dfrac{1 - t^2}{1 + t^2}\right) + 5\) | M1 | 1.1a |
| \(\displaystyle\int \frac{1}{2\sin x - \cos x + 5}\,\mathrm{d}x = \int \frac{1}{2\left(\dfrac{2t}{1 + t^2}\right) - \left(\dfrac{1 - t^2}{1 + t^2}\right) + 5} \times \frac{2\,\mathrm{d}t}{1 + t^2}\) | M1 | 2.1 |
| \(\displaystyle = \int \frac{2}{4t - 1 + t^2 + 5 + 5t^2} \times \mathrm{d}t = \int \frac{1}{3t^2 + 2t + 2}\,\mathrm{d}t\ *\) | A1* | 2.1 |
| (4) |
Notes
B1: Correct equation in terms of \(\mathrm{d}x\), \(\mathrm{d}t\) and \(t\) – can be implied if seen as part of their substitution.
M1: Express \(2\sin x - \cos x + 5\) in terms of \(t\) using half angle formulae with at least one correct. Allow if e.g. there are slips in sign when substituting or the \(+5\) is missed.
M1: Makes a complete substitution to obtain an integral in terms of \(t\) only. Allow slips with the substitution of “\(\mathrm{d}x\)” but must be \(\mathrm{d}x = \mathrm{f}(t)\,\mathrm{d}t\) where \(\mathrm{f}(t) \ne 1\). This mark is also available if the candidate makes errors when attempting to simplify \(2\left(\dfrac{2t}{1 + t^2}\right) - \left(\dfrac{1 - t^2}{1 + t^2}\right)\) before attempting the substitution. Condone omission of the integral and \(\mathrm{d}t\) if the intent is clear.
A1*: Obtains the printed answer (including \(\mathrm{d}t\)) with at least one intermediate line or aside working showing elimination of the fractions and no errors seen.
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \frac{1}{3t^2 + 2t + 2}\,\mathrm{d}t = \frac{1}{3}\int \frac{1}{t^2 + \frac{2}{3}t + \frac{2}{3}}\,\mathrm{d}t = \frac{1}{3}\int \frac{1}{\left(t + \frac{1}{3}\right)^2 + \ldots}\,\mathrm{d}t\) or \(\displaystyle\int \frac{1}{\left(\sqrt{3}t + \frac{1}{\sqrt{3}}\right)^2 + \ldots}\,\mathrm{d}t\) | M1 | 3.1a |
| \(\displaystyle\frac{1}{3}\int \frac{1}{\left(t + \frac{1}{3}\right)^2 + \frac{5}{9}}\,\mathrm{d}t\) or \(\displaystyle\int \frac{1}{\left(\sqrt{3}t + \frac{\sqrt{3}}{3}\right)^2 + \frac{5}{3}}\,\mathrm{d}t\) (oe) | A1 | 1.1b |
| \(= \dfrac{1}{3} \times \dfrac{1}{\frac{\sqrt{5}}{3}}\tan^{-1}\left(\dfrac{t + \frac{1}{3}}{\frac{\sqrt{5}}{3}}\right)\ (+c) = \dfrac{1}{\sqrt{5}}\tan^{-1}(\mathrm{f}(x))\ (+c)\) (oe) | M1 | 3.1a |
| \(= \dfrac{1}{\sqrt{5}}\tan^{-1}\left(\dfrac{3\tan\left(\frac{x}{2}\right) + 1}{\sqrt{5}}\right)\ (+c)\) (oe) | A1 | 1.1b |
| (4) | ||
| (8 marks) |
Notes
M1: Adopts the correct strategy of completing the square in order to attempt the integration. Must achieve one of the forms shown in scheme. May be seen separately but must be applied to the integral, not just as an attempt to solve the quadratic.
A1: Correct integral with completed square integrand.
M1: For recognising the arctan form for the integration (look for e.g \(k\arctan\left(\dfrac{\text{“}t + \frac{1}{3}\text{”}}{b}\right)\)) and makes further progress by undoing the substitution to obtain an answer in terms of \(x\)
A1: Correct answer. Accept equivalents with simplified (and no nested) fractions. No need for \(c\).