# Question 4:

## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Divide through by $\cosh^2 t$ to obtain $1 - \tanh^2 t = \text{sech}^2 t$ | B1 | No working required |
| | **[1]** | |

## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tanh y = x$ | B1 | OR M1 for $\frac{d}{dx}\left(\frac{1}{2}(\ln(1+x)-\ln(1-x))\right)$ |
| $\text{sech}^2 y \cdot y' = 1$ oe | M1A1 | M1 for $(\pm)\text{sech}^2 y$; A1A1 for $\frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right)$ |
| $y' = 1/\text{sech}^2 y = 1/(1-\tanh^2 y) = 1/(1-x^2)$ | E1 | Answer given; E1 for $1/(1-x^2)$ |
| | **[4]** | |

## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{1}{1-x^2} = \dfrac{1}{2}\left(\dfrac{1}{1-x}+\dfrac{1}{1+x}\right)$ | M1A1 | M0 if standard formula for integrating $1/(a^2-x^2)$ is used; but B1B1 can still be given |
| Integrate to $\dfrac{1}{2}(\ln(1+x)-\ln(1-x))+c$ | B1 | B0 if no '$+c$' |
| Hence $\dfrac{1}{2}\ln\left(\dfrac{1+x}{1-x}\right)+c$ | | Answer given (without $c$) |
| Use $x=0,\ y=0$ to show $c=0$ | B1 | |
| | **[4]** | |

## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $u=\sqrt{3}x$ to obtain $\dfrac{1}{\sqrt{3}}\displaystyle\int_0^{1/2}\dfrac{1}{1-u^2}\,du$ | M1 | or $\ldots\text{artanh}(x\sqrt{3})$ |
| | A1 | or $(1/\sqrt{3})\text{artanh}(x\sqrt{3})$ |
| Hence $\dfrac{1}{\sqrt{3}}\text{artanh}\dfrac{1}{2}$ | E1 | Answer given |
| Which is $\dfrac{1}{2\sqrt{3}}\ln 3$ | B1 | oe, e.g. $(\sqrt{3}/6)\ln 3$, $(1/\sqrt{3})\ln(\sqrt{3})$ |
| | **[4]** | |

## Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\displaystyle\int 1\times\text{artanh}\,x\,dx = x\,\text{artanh}\,x - \int\dfrac{x}{1-x^2}\,dx$ | M1A1 | |
| Hence $x\,\text{artanh}\,x + \dfrac{1}{2}\ln(1-x^2)+c$ | B1 | |
| Equals $\dfrac{1}{2}x\ln\left(\dfrac{1+x}{1-x}\right)+\dfrac{1}{2}\ln(1-x^2)+c$ | B1B1B1 | Third mark is for $c$ |
| | **[5]** | |

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