## Question 4(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cosh u = \frac{e^u + e^{-u}}{2} \Rightarrow \cosh^2 u = \frac{e^{2u} + 2 + e^{-2u}}{4}$ | | |
| $\sinh u = \frac{e^u - e^{-u}}{2} \Rightarrow \sinh^2 u = \frac{e^{2u} - 2 + e^{-2u}}{4}$ | B1 | Numerators of both expressions; Accept other variables |
| $\Rightarrow \cosh^2 u - \sinh^2 u = 1$ | B1(ag) | Completion www |
| **OR** $\cosh u + \sinh u = e^u$; $\cosh u - \sinh u = e^{-u}$ | B1 | Both expressions s.o.i. and multiplication |
| $\Rightarrow \cosh^2 u - \sinh^2 u = e^u \times e^{-u}$ | |  |
| $\Rightarrow \cosh^2 u - \sinh^2 u = 1$ | B1(ag) | Completion www |

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## Question 4(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \text{arsinh}\, x \Rightarrow x = \sinh y$ | | |
| $\Rightarrow \frac{dx}{dy} = \cosh y$ | M1 | $\sinh y = \ldots$ and differentiating w.r.t. $y$ or $x$; Or $\cosh y \frac{dy}{dx} = 1$ or differentiating (*) |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{\cosh y}$ | A1 | o.e. |
| $\Rightarrow \frac{dy}{dx} = (\pm)\frac{1}{\sqrt{1 + \sinh^2 y}} = (\pm)\frac{1}{\sqrt{1+x^2}}$ | A1(ag) | Completion www with valid intermediate step; $\frac{dy}{dx} = \pm\frac{1}{\sqrt{1+x^2}}$ as final answer or $\pm$ not considered scores max 3/4; Or $\cosh y \geq 1$, or $\cosh y > 0$ |
| $y$ is an increasing function so take $+$ sign | B1 | Validly rejecting negative value |
| $x = \sinh y \Rightarrow x = \frac{e^y - e^{-y}}{2}$ | B1 | $x$ in exponential form |
| $\Rightarrow e^y - e^{-y} = 2x$ | | |
| $\Rightarrow e^{2y} - 2xe^y - 1 = 0$ | M1 | Obtaining quadratic in $e^y$ |
| $\Rightarrow (e^y - x)^2 = 1 + x^2$ | | |
| $\Rightarrow e^y = x \pm \sqrt{1+x^2}$ | M1 | Solving to reach $e^y$; Dep. on M1 above; Allow one slip |
| $\Rightarrow y = \ln\!\left(x(\pm)\sqrt{1+x^2}\right)$ (*) | A1(ag) | Completion www |
| $x - \sqrt{1+x^2} < 0$ so take $+$ sign | B1 | Validly rejecting negative root; e.g. $e^y > 0$ |

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## Question 4(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^2 \frac{1}{\sqrt{4+9x^2}}\,dx = \frac{1}{3}\int_0^2 \frac{1}{\sqrt{\frac{4}{9}+x^2}}\,dx$ | M1 | Integral involving arsinh |
| $= \frac{1}{3}\!\left[\text{arsinh}\,\frac{3x}{2}\right]_0^2$ | A1A1 | $\frac{1}{3}$, $\frac{3x}{2}$ o.e. |
| $= \frac{1}{3}\,\text{arsinh}\,3$ | | |
| **OR** $= \frac{1}{3}\!\left[\ln\!\left(x + \sqrt{x^2 + \frac{4}{9}}\right)\right]_0^2$ | M1, A1A1 | Integral in form $\ln\!\left(kx + \sqrt{k^2x^2+\ldots}\right)$; $\frac{1}{3}$, $x + \sqrt{x^2 + \frac{4}{9}}$ or $3x + \sqrt{9x^2+4}$; Or $\frac{3x}{2} + \sqrt{\frac{9x^2}{4}+1}$ |
| **OR** $x = \frac{2}{3}\sinh u \Rightarrow \frac{dx}{du} = \frac{2}{3}\cosh u$ | M1, A1 | Using a sinh substitution; Correct substitution |
| $\int_0^2 \frac{1}{\sqrt{4+9x^2}}\,dx = \int_0^{\ln(3+\sqrt{10})} \frac{1}{3}\,du$ | A1 | $\int \frac{1}{3}\,du$ |
| $= \frac{1}{3}\ln\!\left(3+\sqrt{10}\right)$ | A1(ag) | Completion with valid intermediate step(s); Condone omitted brackets |

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## Question 4(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \frac{1}{\sqrt{1+x^2}}\,\text{arsinh}\,x\,dx$ | | |
| $= \left[(\text{arsinh}\,x)^2\right]_0^1 - \int_0^1 \frac{1}{\sqrt{1+x^2}}\,\text{arsinh}\,x\,dx$ | M1 | Parts with $u = \text{arsinh}\,x$, $v' = \frac{1}{\sqrt{1+x^2}}$; Allow one error; Allow equivalent form |
| **OR** $\int \frac{1}{\sqrt{1+x^2}}\,\text{arsinh}\,x\,dx = \int u\,du$ | M1 | Substitution with $u = \text{arsinh}\,x$ or $x = \sinh u$; Must reach $\int u\,du$ |
| **OR** inspection | M1 | Recognising integrand as $k(\text{arsinh}\,x)^2$, $k \neq 0$ |
| $\Rightarrow \frac{1}{2}(\text{arsinh}\,x)^2$ | A1 | A correct indefinite integral |
| $\Rightarrow I = \frac{1}{2}\!\left(\ln\!\left(1+\sqrt{2}\right)\right)^2$ | A1 | This answer only; Mark final answer |