## Question 15(a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\sin\theta)^{-1}$ or $\frac{1}{\sin\theta}$ | B1 | $\sin^{-1}\theta$ scores B0; ignore $\sin^{-1}\theta$ if a correct expression already written |

## Question 15(a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses chain rule or quotient rule to obtain $\pm k(\sin\theta)^{-2}\cos\theta$ | M1 | Also valid via product rule and implicit differentiation; or using $\frac{1}{\cos\theta}$ to get $\pm k(\cos\theta)^{-2}\sin\theta$; ignore wrong/missing angles |
| $\frac{dy}{d\theta} = -(\sin\theta)^{-2}\cos\theta$ | A1 | |
| $= -\frac{\cos\theta}{\sin\theta} \times \frac{1}{\sin\theta} = -\cosec\theta\cot\theta$ | R1 | Rigorous argument required; must see separated fractions before final line, or sight of $-\frac{\cot\theta}{\sin\theta}$ or $-\frac{1}{\tan\theta\sin\theta}$ or $-\frac{\cos\theta}{\sin\theta}\times\cosec\theta$; AG |

## Question 15(a)(iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $y = \cosec\theta$ | B1 | OE; or draws right-angled triangle labelling hypotenuse as $y$ and opposite as 1; PI by obtaining $y^2$ or $\frac{1}{y^2}$ in terms of $\cos\theta$ |
| Uses $\cosec^2\theta - 1 = \cot^2\theta$, giving $\frac{\sqrt{\cot^2\theta}}{\cosec\theta} = \frac{\cot\theta}{\cosec\theta}$ | M1 | OE; or uses Pythagoras theorem to find missing adjacent side in right-angled triangle; or obtains $\cos^2\theta$ in terms of $y$. Ignore wrong/missing angles |
| Completes rigorous argument: $= \frac{\cos\theta}{\sin\theta} \times \sin\theta = \cos\theta$ | R1 | Must include clear replacement of $\cosec\theta$ and $\cot\theta$ within the solution using only sine and cosine functions prior to obtaining the answer. AG |

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## Question 15(b)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{du} = -2\cosec u \cot u$ | B1 | OE |
| Makes complete substitution to obtain integrand of the form $\frac{P\cosec u \cot u}{Q\cosec^2 u\sqrt{R\cosec^2 u - 4}}$ | M1 | OE forms accepted; ignore wrong or missing angles |
| Obtains correct integrand $\frac{-2\cosec u \cot u}{4\cosec^2 u\sqrt{4\cosec^2 u - 4}}$ giving $= \int -\frac{1}{4\cosec u}\,du = -\frac{1}{4}\int \sin u\,du$ | A1 | OE e.g. $\frac{-2\cosec u\cot u}{8\cosec^3 u\sqrt{1-\sin^2 u}}$ |
| Uses appropriate Pythagorean trig identity under the square root: $1 + \cot^2 u = \cosec^2 u$ or $1 - \sin^2 u = \cos^2 u$ | M1 | Ignore wrong or missing angles |
| Obtains $k\int \sin u\,du$ with no errors seen in any trig identities | A1F | Must have $u$ and $du$ |
| Obtains $k = -\frac{1}{4}$ | R1 | OE; CSO |

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## Question 15(b)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrates $\int \sin u\,du$ to obtain $-\cos u$ | B1 | |
| Deduces $\cos u = \dfrac{\sqrt{\left(\frac{x}{2}\right)^2 - 1}}{\left(\frac{x}{2}\right)}$, giving $-\frac{1}{4}\int \sin u\,du = \frac{1}{4}\cos u + c$ | M1 | OE |
| Completes reasoned argument to show $= \dfrac{\sqrt{x^2-4}}{4x} + c$ | R1 | Must have $+c$ throughout. Note: validation by starting with $\frac{\sqrt{x^2-4}}{4x}$ and replacing $x$ with $2\cosec u$ to achieve $\frac{1}{4}\cos u$ scores maximum B1M1R0 |