June 2022 Paper 1 Q15
15
(a) Given that\[y = \operatorname{cosec}\theta\]
(i) Express \(y\) in terms of \(\sin\theta\). [1 mark]
(ii) Hence, prove that\[\frac{\mathrm{d}y}{\mathrm{d}\theta} = -\operatorname{cosec}\theta\cot\theta\] [3 marks]
(iii) Show that\[\frac{\sqrt{y^2 - 1}}{y} = \cos\theta \qquad \text{for } 0 \lt \theta \lt \frac{\pi}{2}\] [3 marks]
(b)
(i) Use the substitution\[x = 2\operatorname{cosec} u\]to show that\[\int \frac{1}{x^2\sqrt{x^2 - 4}}\,\mathrm{d}x \qquad \text{for } x \gt 2\]can be written as\[k\int \sin u\,\mathrm{d}u\]where \(k\) is a constant to be found. [6 marks]
(ii) Hence, show\[\int \frac{1}{x^2\sqrt{x^2 - 4}}\,\mathrm{d}x = \frac{\sqrt{x^2 - 4}}{4x} + c \qquad \text{for } x \gt 2\]where \(c\) is a constant. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| (i) States \((\sin\theta)^{-1}\) or \(\dfrac{1}{\sin\theta}\) \(\sin^{-1}\theta\) scores B0 Ignore \(\sin^{-1}\theta\) if a correct expression has already been written | B1 | 1.2 |
| (1) | ||
| (ii) Uses chain rule or quotient rule to obtain \(\pm k(\sin\theta)^{-2}\cos\theta\) OE or Multiplies and uses product rule and implicit differentiation to obtain \(\pm k(\sin\theta)^{-2}\cos\theta\) This mark can be awarded for using \(\dfrac{1}{\cos\theta}\) and differentiating to obtain \(\pm k(\cos\theta)^{-2}\sin\theta\) OE Ignore wrong or missing angles | M1 | 3.1a |
| Obtains \(-(\sin\theta)^{-2}\cos\theta\) OE | A1 | 1.1b |
| Completes rigorous argument to show the given result. Must either see separated fractions before final line or sight of \(-\dfrac{\cot\theta}{\sin\theta}\) or \(-\dfrac{1}{\tan\theta\sin\theta}\) or \(-\dfrac{\cos\theta}{\sin\theta} \times \operatorname{cosec}\theta\) or Makes clear use of stated identities as part of the solution At some point the solution must have included \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta} =\) AG Condone change of order of functions at the end | R1 | 2.1 |
| (3) | ||
| (iii) Substitutes \(y = \operatorname{cosec}\theta\) OE or Draws a right angled triangle labelling hypotenuse as \(y\) and opposite as 1 PI by obtaining \(y^2\) or \(\dfrac{1}{y^2}\) in terms of \(\cos\theta\) | B1 | 1.1b |
| Uses \(\operatorname{cosec}^2\theta - 1 = \cot^2\theta\) OE or Uses Pythagoras theorem to find missing adjacent side in the right angled triangle or Obtains \(\cos^2\theta\) in terms of \(y\) | M1 | 1.1a |
| Completes rigorous argument to show the given result This must include clear replacement of \(\operatorname{cosec}\theta\) and \(\cot\theta\) within the solution using only sine and cosine functions to complete the argument prior to obtaining the answer given AG | R1 | 2.1 |
| (3) |
Typical solution
(i)
\[\frac{1}{\sin\theta}\](ii)
\[\frac{\mathrm{d}y}{\mathrm{d}\theta} = -(\sin\theta)^{-2}\cos\theta\]\[= -\frac{\cos\theta}{\sin\theta} \times \frac{1}{\sin\theta}\]\[= -\operatorname{cosec}\theta\cot\theta\](iii)
\[\frac{\sqrt{\operatorname{cosec}^2\theta - 1}}{\operatorname{cosec}\theta}\]\[= \frac{\sqrt{\cot^2\theta}}{\operatorname{cosec}\theta}\]\[= \frac{\cot\theta}{\operatorname{cosec}\theta}\]\[= \frac{\cos\theta}{\sin\theta} \times \sin\theta\]\[= \cos\theta\]| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = -2\operatorname{cosec} u\cot u\) OE | B1 | 1.1b |
| Makes complete substitution to obtain integrand of the form \(\dfrac{P\operatorname{cosec} u\cot u}{Q\operatorname{cosec}^2 u\sqrt{R\operatorname{cosec}^2 u - 4}}\) OE or \(\dfrac{P\operatorname{cosec} u\cot u}{Q\operatorname{cosec}^3 u\sqrt{1 - R\sin^2 u}}\) OE Ignore wrong or missing angles | M1 | 1.1a |
| Obtains correct integrand \(\dfrac{-2\operatorname{cosec} u\cot u}{4\operatorname{cosec}^2 u\sqrt{4\operatorname{cosec}^2 u - 4}}\) or \(\dfrac{-2\operatorname{cosec} u\cot u}{8\operatorname{cosec}^3 u\sqrt{1 - \sin^2 u}}\) OE | A1 | 1.1b |
| Uses appropriate Pythagorean-trig identity under the square root. Either \(1 + \cot^2 u = \operatorname{cosec}^2 u\) or \(1 - \sin^2 u = \cos^2 u\) Ignore wrong or missing angles | M1 | 3.1a |
| Obtains \(k\displaystyle\int \sin u\,\mathrm{d}u\) with no errors seen in any trig identities Must have \(u\) and \(\mathrm{d}u\) | A1F | 1.1b |
| Obtains \(k = -\dfrac{1}{4}\) OE CSO | R1 | 2.1 |
| (6) | ||
| (ii) Integrates \(\displaystyle\int \sin u\,\mathrm{d}u\) to obtain \(-\cos u\) | B1 | 1.1b |
| Deduces \(\cos u = \dfrac{\sqrt{\left(\frac{x}{2}\right)^2 - 1}}{\left(\frac{x}{2}\right)}\) OE | M1 | 2.2a |
| Completes reasoned argument to show given result Must have \(+c\) throughout Validation by starting with \(\dfrac{\sqrt{x^2 - 4}}{4x}\) and replacing \(x\) with \(2\operatorname{cosec} u\) to achieve \(\dfrac{1}{4}\cos u\) scores a maximum of B1M1R0 | R1 | 2.1 |
| (3) | ||
| (16 marks) |
Typical solution
(i)
\[\frac{\mathrm{d}x}{\mathrm{d}u} = -2\operatorname{cosec} u\cot u\]\[\mathrm{d}x = -2\operatorname{cosec} u\cot u\,\mathrm{d}u\]\[= \int \frac{-2\operatorname{cosec} u\cot u}{4\operatorname{cosec}^2 u\sqrt{4\operatorname{cosec}^2 u - 4}}\,\mathrm{d}u\]\[= \int \frac{-2\operatorname{cosec} u\cot u}{4\operatorname{cosec}^2 u\sqrt{4\cot^2 u}}\,\mathrm{d}u\]\[= \int \frac{-2\operatorname{cosec} u\cot u}{4\operatorname{cosec}^2 u \times 2\cot u}\,\mathrm{d}u\]\[= \int -\frac{1}{4\operatorname{cosec} u}\,\mathrm{d}u\]\[= -\frac{1}{4}\int \sin u\,\mathrm{d}u\](ii)
\[-\frac{1}{4}\int \sin u\,\mathrm{d}u = \frac{1}{4}\cos u + c\]\[= \frac{1}{4}\,\frac{\sqrt{\left(\frac{x}{2}\right)^2 - 1}}{\left(\frac{x}{2}\right)} + c\]\[= \frac{\sqrt{\frac{x^2 - 4}{4}}}{2x} + c\]\[= \frac{\sqrt{x^2 - 4}}{4x} + c\]