A2 June 2019 Q5
5.
\[I = \int\frac{1}{4\cos x - 3\sin x}\,\mathrm{d}x \qquad 0 \lt x \lt \frac{\pi}{4}\]Use the substitution \(t = \tan\left(\dfrac{x}{2}\right)\) to show that
\[I = \frac{1}{5}\ln\left(\frac{2 + \tan\left(\frac{x}{2}\right)}{1 - 2\tan\left(\frac{x}{2}\right)}\right) + k\]where \(k\) is an arbitrary constant.
(8)
| Scheme | Marks | AO |
|---|---|---|
| \(4\cos x - 3\sin x = 4\left(\dfrac{1 - t^2}{1 + t^2}\right) - 3\left(\dfrac{2t}{1 + t^2}\right)\) | B1 | 1.1a |
| \(\dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{1 + t^2}{2}\) 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 M1 on ePEN | 2.1 |
| \(\displaystyle\int\frac{1}{4\cos x - 3\sin x}\,\mathrm{d}x = \int\frac{1}{4\left(\dfrac{1 - t^2}{1 + t^2}\right) - 3\left(\dfrac{2t}{1 + t^2}\right)} \times \frac{2\,\mathrm{d}t}{1 + t^2}\) | M1 | 2.1 |
| \(\displaystyle = \int\frac{2}{4 - 4t^2 - 6t}(\mathrm{d}t)\) or \(\displaystyle\int\frac{1}{2 - 2t^2 - 3t}(\mathrm{d}t)\) or \(\displaystyle\int\frac{-1}{2t^2 + 3t - 2}(\mathrm{d}t)\) etc. | A1 | 1.1b |
| \(\begin{aligned}&\dfrac{-2}{4t^2 + 6t - 4} = \dfrac{-1}{(t + 2)(2t - 1)} = \dfrac{A}{(t + 2)} + \dfrac{B}{(2t - 1)}\\[6pt] &\dfrac{-1}{(t + 2)(2t - 1)} = \dfrac{1}{5(t + 2)} + \dfrac{2}{5(1 - 2t)}\end{aligned}\) | M1 | 3.1a |
| \(\displaystyle\Rightarrow I = \frac{1}{5}\int\frac{1}{(t + 2)} - \frac{2}{(2t - 1)}(\mathrm{d}t)\) or equivalent | A1 | 1.1b |
| \(\displaystyle = \frac{1}{5}\int\frac{1}{(t + 2)} - \frac{2}{(2t - 1)}\,\mathrm{d}t = \frac{1}{5}\ln(t + 2) - \frac{1}{5}\ln(1 - 2t)\,(+k)\) | A1 | 1.1b |
| \(= \dfrac{1}{5}\ln\left(\dfrac{2 + t}{1 - 2t}\right)(+k) = \dfrac{1}{5}\ln\left(\dfrac{2 + \tan\left(\tfrac{x}{2}\right)}{1 - 2\tan\left(\tfrac{x}{2}\right)}\right) + k\ *\) | A1* | 2.1 |
| (8) | ||
| (8 marks) |
Notes
B1: Uses the correct formulae to express \(4\cos x - 3\sin x\) in terms of \(t\)
B1(M1 on ePEN): Correct equation in terms of \(\mathrm{d}x\), \(\mathrm{d}t\) and \(t\) – can be implied if seen as part of their substitution.
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 \(\mathrm{f}(t) \neq 1\). This mark is also available if the candidate makes errors when attempting to simplify \(4\left(\dfrac{1 - t^2}{1 + t^2}\right) - 3\left(\dfrac{2t}{1 + t^2}\right)\) before attempting the substitution.
A1: For obtaining a fully correct simplified integral with a constant in the numerator and a 3 term quadratic expression in the denominator. (“\(\mathrm{d}t\)” not required)
M1: Realises the need to express the integrand in terms of partial fractions in order to attempt the integration. Must have a 3 term quadratic expression in the denominator and a constant in the numerator.
A1: Correct integral in terms of partial fractions – allow any equivalent correct integral. (“\(\mathrm{d}t\)” not required)
A1: Fully correct integration in terms of \(t\)
A1*: Correct solution with no errors including “\(+\,k\)” (allow “\(+\,c\)”) and with the constant dealt with correctly if necessary. The denominator must also be dealt with correctly. E.g. if it appears as \(2t - 1\) initially and becomes \(1 - 2t\) without justification, this final mark should be withheld.
Alternative for final 4 marks
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle = \int\frac{2}{4 - 4t^2 - 6t}(\mathrm{d}t) = -\frac{1}{2}\int\frac{1}{t^2 + \frac{3}{2}t - 1}(\mathrm{d}t) = -\frac{1}{2}\int\frac{1}{\left(t + \frac{3}{4}\right)^2 - \frac{25}{16}}(\mathrm{d}t)\) or e.g. \(\displaystyle\int\frac{1}{\frac{25}{8} - 2\left(t + \frac{3}{4}\right)^2}(\mathrm{d}t)\) | M1 A1 | 3.1a 1.1b |
| \(-\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{4}{5}\ln\left(\dfrac{t + \frac{3}{4} - \frac{5}{4}}{t + \frac{3}{4} + \frac{5}{4}}\right)(+c)\) | A1 | 1.1b |
| \(\begin{aligned}&-\dfrac{1}{5}\ln\left|\dfrac{\tan\left(\tfrac{x}{2}\right) - \frac{1}{2}}{\tan\left(\tfrac{x}{2}\right) + 2}\right| + c = \dfrac{1}{5}\ln\left|\dfrac{\tan\left(\tfrac{x}{2}\right) + 2}{\tan\left(\tfrac{x}{2}\right) - \frac{1}{2}}\right| + c = \dfrac{1}{5}\ln\left(\dfrac{\tan\left(\tfrac{x}{2}\right) + 2}{\frac{1}{2} - \tan\left(\tfrac{x}{2}\right)}\right) + c\\[6pt] &= \dfrac{1}{5}\ln\left(\dfrac{2\left(\tan\left(\tfrac{x}{2}\right) + 2\right)}{1 - 2\tan\left(\tfrac{x}{2}\right)}\right) + c = \dfrac{1}{5}\ln\left(\dfrac{\left(\tan\left(\tfrac{x}{2}\right) + 2\right)}{1 - 2\tan\left(\tfrac{x}{2}\right)}\right) + \dfrac{1}{5}\ln 2 + c\\[6pt] &= \dfrac{1}{5}\ln\left(\dfrac{\left(\tan\left(\tfrac{x}{2}\right) + 2\right)}{1 - 2\tan\left(\tfrac{x}{2}\right)}\right) + k\end{aligned}\) | A1* | 2.1 |
M1: Realises the need to express the integrand in completed square form in order to attempt the integration. Must have a 3 term quadratic expression in the denominator and a constant in the numerator.
A1: Correct integral with the square completed – allow any equivalent correct integral (“\(\mathrm{d}t\)” not required)
A1: Fully correct integration in terms of \(t\)
A1*: Correct solution with no errors including “\(+\,k\)” (allow “\(+\,c\)”) and with the constant dealt with correctly if necessary as shown in the scheme and with the denominator dealt with correctly if necessary.
Note that it is acceptable for the “\(\mathrm{d}t\)” to appear and disappear throughout the proof as long as the intention is clear.