A2 June 2024 Q7
7.
In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \frac{1}{2\sin\theta + \cos\theta + 2}\,\mathrm{d}\theta = \int \frac{1}{2\left(\frac{2t}{1 + t^2}\right) + \left(\frac{1 - t^2}{1 + t^2}\right) + 2} \times \frac{2}{1 + t^2}\,\mathrm{d}t\) | M1 | 2.1 |
| \(\displaystyle\int \frac{2}{4t + \left(1 - t^2\right) + 2\left(1 + t^2\right)}\,\mathrm{d}t = \int \frac{2}{t^2 + 4t + 3}\,\mathrm{d}t\) | M1 | 1.1b |
| \(\displaystyle\int \frac{2}{(t + 2)^2 - 1}\,\mathrm{d}t\) | A1 | 2.2a |
| (3) |
Notes
M1: Applies the correct substitutions \(\sin\theta = \dfrac{2t}{1 + t^2}\), \(\cos\theta = \dfrac{1 - t^2}{1 + t^2}\) and \(\mathrm{d}\theta = \dfrac{2}{1 + t^2}\,\mathrm{d}t\) The \(\mathrm{d}t\) may be missing for this and the next M.
M1: Simplifies the integrand to the form \(\dfrac{a}{pt^2 + qt + r}\) allowing slips in coefficients but the “\(1 + t^2\)” must be correctly dealt with.
A1: Correct answer in the form required, including the \(\displaystyle\int \ldots\,\mathrm{d}t\)
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \frac{a}{(t + b)^2 - c}\,\mathrm{d}t = \frac{a}{2\sqrt{c}}\ln\left|\frac{(t + b) - \sqrt{c}}{(t + b) + \sqrt{c}}\right|, c \gt 0\) or \(\displaystyle\int \frac{a}{(t + b)^2 + c}\,\mathrm{d}t = \frac{a}{\sqrt{c}}\arctan\left(\frac{t + b}{\sqrt{c}}\right), c \gt 0\) Note correct answer is \(\displaystyle\int \frac{2}{(t + 2)^2 - 1}\,\mathrm{d}t = \frac{2}{2(1)}\ln\left|\frac{(t + 2) - 1}{(t + 2) + 1}\right|\) | M1 | 3.1a |
| Correctly uses the limits \(t = 1\) and \(t = \sqrt{3} \to \ln\left|\dfrac{\sqrt{3} + 1}{\sqrt{3} + 3}\right| - \ln\left|\dfrac{2}{4}\right|\) if correct. | M1 | 1.1b |
| \(= \ln\left(\dfrac{\sqrt{3} + 1}{\sqrt{3} + 3} \div \dfrac{1}{2}\right)\) | M1 | 1.1b |
| \(= \ln\left(2 \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 3} \times \dfrac{\sqrt{3} - 3}{\sqrt{3} - 3}\right) = \ln\left(\dfrac{2\sqrt{3}}{3}\right)\ *\) | A1* | 2.1 |
| (4) | ||
| (7 marks) |
Notes
M1: A correct method for the integration, by use of standard formula. Modulus signs are not necessary. Note that \(-\dfrac{a}{\sqrt{c}}\operatorname{artanh}\left(\dfrac{t + b}{\sqrt{c}}\right) = -\dfrac{a}{2\sqrt{c}}\ln\left|\dfrac{t + b + \sqrt{c}}{t + b - \sqrt{c}}\right|\) is also a correct version. If by error they end up with a positive \(c\) then accept for the arctan integral shown in scheme.
M1: Correctly uses the limits \(t = 1\) and \(t = \sqrt{3}\) in their integral to produce an exact expression. Use of the original limits is M0, must have changed to \(t\) (unless they reverse the substitution).
M1: Correctly combines the ln terms with at least one step showing the result of substituting limits before the final answer, and their final answer must follow this step if no further working is shown. May be seen before or after the rationalisation. (M0 if no ln terms.)
A1*cso: Shows the method to rationalise the denominator and correctly completes to the given answer. Some suitable working must be seen. Note that \(\sqrt{3} + 3 = \sqrt{3}\left(1 + \sqrt{3}\right)\) may be used in the denominator, followed by cancelling.
(b) Alt
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \frac{2}{(t + 2)^2 - 1}\,\mathrm{d}t = \int \frac{2}{(t + 1)(t + 3)}\,\mathrm{d}t\) \(\displaystyle = \int \frac{1}{t + 1} - \frac{1}{t + 3}\,\mathrm{d}t = \ln|t + 1| - \ln|t + 3|\) | M1 | 3.1a |
| Correctly uses the limits \(t = 1\) and \(t = \sqrt{3}\) \(\ln\left(\sqrt{3} + 1\right) - \ln\left(\sqrt{3} + 3\right) - \ln 2 + \ln 4\) | M1 | 1.1b |
| \(= \ln\left(2 \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 3}\right)\) | M1 | 2.1 |
| \(= \ln\left(2 \times \dfrac{\sqrt{3} + 1}{\sqrt{3}\left(1 + \sqrt{3}\right)}\right) = \ln\left(\dfrac{2\sqrt{3}}{3}\right)\ *\) | A1* | 1.1b |
| (4) |
M1: A correct method for the integration, applying partial fractions and integrating to ln terms.
M1: Correctly uses the limits \(t = 1\) and \(t = \sqrt{3}\) (or correct \(\theta\) limits if substitution is reversed) to produce an exact expression.
M1: Correctly combines the ln terms with at least one step showing the result of substituting limits before the final answer, and their final answer must follow this step if no further working is shown. May be seen before or after the rationalisation. (M0 if no ln terms.)
A1*cso: Shows the method to rationalise the denominator and correctly completes to the given answer. Some suitable working must be seen. Note that \(\sqrt{3} + 3 = \sqrt{3}\left(1 + \sqrt{3}\right)\) may be used in the denominator, followed by cancelling.