## Question 8(i):

**Method 1**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sinh(\cosh^{-1}2)=\sinh\beta=\sqrt{\cosh^2\beta-1}=\sqrt{2^2-1}=\sqrt{3}$ | M1 | For appropriate use of $\sinh^2\theta=\cosh^2\theta-1$ |
| | A1 **2** | For correct verification to **AG** |

**Method 2**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sinh^{-1}\sqrt{3}=\ln(\sqrt{3}+2),\;\cosh^{-1}2=\ln(2+\sqrt{3})$ | M1 | For attempted use of ln forms of $\sinh^{-1}x$ and $\cosh^{-1}x$ |
| $\Rightarrow\sinh(\cosh^{-1}2)=\sqrt{3}$ | A1 | For both ln expressions seen |

**Method 3**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cosh^{-1}2=\ln(2+\sqrt{3})$ | M1 | For use of ln form of $\cosh^{-1}x$ and definition of $\sinh x$ |
| $\sinh(\cosh^{-1}2)=\frac{1}{2}\left(e^{\ln(2+\sqrt{3})}-e^{-\ln(2+\sqrt{3})}\right)=\frac{1}{2}(2+\sqrt{3}-(2-\sqrt{3}))=\sqrt{3}$ | A1 | For correct verification to **AG**; Note $\ln(2+\sqrt{3})=-\ln(2-\sqrt{3})$ |

## Question 8(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_n=\int_0^\beta\cosh^n x\,dx$ | M1* | For attempt to integrate $\cosh x\cdot\cosh^{n-1}x$ by parts |
| $=\left[\sinh x\cdot\cosh^{n-1}x\right]_0^\beta-\int_0^\beta\sinh^2 x\cdot(n-1)\cosh^{n-2}x\,dx$ | A1 | For correct first stage of integration (ignore limits) |
| $=\sinh\beta\cdot\cosh^{n-1}\beta-(n-1)\int_0^\beta(\cosh^2 x-1)\cosh^{n-2}x\,dx$ | M1(*dep) | For using $\sinh^2 x=\cosh^2 x-1$ |
| $=2^{n-1}\sqrt{3}-(n-1)(I_n-I_{n-2})$ | A1 | For $(n-1)(I_n-I_{n-2})$ correct |
| | B1 | For $2^{n-1}\sqrt{3}$ correct at any stage |
| $\Rightarrow nI_n=2^{n-1}\sqrt{3}+(n-1)I_{n-2}$ | A1 **6** | For correct result **AG** |

## Question 8(iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_1=\int_0^\beta\cosh x\,dx=\sinh\beta=\sqrt{3}$ | B1 | For correct value |
| $I_3=\frac{1}{3}(2^2\sqrt{3}+2\sqrt{3})=2\sqrt{3}$ | M1 | For using **(ii)** with $n=3$ OR $n=5$ |
| | A1 | For $I_3=\frac{1}{3}(2^2\sqrt{3}+2I_1)$ OR $I_5=\frac{1}{5}(2^4\sqrt{3}+4I_3)$ |
| $I_5=\frac{1}{5}(2^4\sqrt{3}+8\sqrt{3})=\frac{24}{5}\sqrt{3}$ | A1 **4** | For correct value |

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