Challenging +1.2 This is a structured Further Maths question with clear guidance at each step. Part (i) is a standard derivation of arsinh using the definition of sinh and completing the square—routine for FP2 students. Part (ii) applies the result directly with a simple substitution. Part (iii) uses a given substitution and requires manipulating hyperbolic identities, but the path is prescribed. While it requires competence with hyperbolic functions and careful algebra, there's no novel insight needed, making it moderately above average difficulty.
Given that \(\sinh y = x\), show that
$$y = \ln \left( x + \sqrt { 1 + x ^ { 2 } } \right)$$
Differentiate (*) to show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 1 + x ^ { 2 } } }$$
Find \(\int \frac { 1 } { \sqrt { 25 + 4 x ^ { 2 } } } \mathrm {~d} x\), expressing your answer in logarithmic form.
Use integration by substitution with \(2 x = 5 \sinh u\) to show that
$$\int \sqrt { 25 + 4 x ^ { 2 } } \mathrm {~d} x = \frac { 25 } { 4 } \left( \ln \left( \frac { 2 x } { 5 } + \sqrt { 1 + \frac { 4 x ^ { 2 } } { 25 } } \right) + \frac { 2 x } { 5 } \sqrt { 1 + \frac { 4 x ^ { 2 } } { 25 } } \right) + c$$
where \(c\) is an arbitrary constant.
\section*{OCR}
Finding \(\frac{dx}{du}\) and complete substitution
### 4(i)
$x = \sinh y \Rightarrow x = \frac{e^y - e^{-y}}{2}$
$\Rightarrow e^y - e^{-y} = 2x$
$\Rightarrow e^{2y} - 2xe^y - 1 = 0$
$\Rightarrow (e^y - x)^2 = 1 + x^2$
$\Rightarrow e^y = x \pm \sqrt{1+x^2}$
$\Rightarrow y = \ln\left(x(\pm)\sqrt{1+x^2}\right)$
$x - \sqrt{1+x^2} < 0$ so take + sign
OR $\ln\left(x + \sqrt{1+x^2}\right) = \ln\left(\sinh y + \sqrt{1+\sinh^2 y}\right) = \ln(\sinh y + \cosh y) = \ln(e^y) = y$
$\Rightarrow \frac{dy}{dx} = \frac{1}{x+\sqrt{1+x^2}} \times \frac{d}{dx}\left(x+\sqrt{1+x^2}\right) = \frac{1}{x+\sqrt{1+x^2}} \times \left(1 + \frac{x}{\sqrt{1+x^2}}\right)$
$= \frac{1}{x+\sqrt{1+x^2}} \times \left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right) = \frac{1}{\sqrt{1+x^2}}$ | B1, M1, A1(ag), B1, M1, B1, A1, A1(ag) | x in exponential form | Solving to reach $e^y$ | Completion www | Validly rejecting negative root. Dependent on A1 above | Attempting $\frac{1}{x} \cdot \frac{du}{dx}$ | $\frac{d}{dx}(x+\sqrt{1+x^2}) = 1 + \frac{x}{\sqrt{1+x^2}}$ o.e. | Any correct form of $\frac{dy}{dx}$ in terms of x | Obtained www with valid intermediate step, e.g. (*) | Allow one slip. Ignore variables. Allow unsimplified | $y = \ln|x \pm \sqrt{x^2+1}|$ A0; e.g. $e^y > 0$; $e^y \ge 0$ B0 | Or implicit differentiation of $e^y = x + \sqrt{1+x^2}$ as far as $\frac{dy}{dx} =$
### 4(ii)
$\int\frac{1}{\sqrt{25+4x^2}}dx = \frac{1}{2}\int\frac{1}{\sqrt{\frac{25}{4}+x^2}}dx = \frac{1}{2}\ln\left(x+\sqrt{x^2+\frac{25}{4}}\right)+c$ | M1, A1, A1 | arsinh $kx$ or $\ln\left(kx+\sqrt{k^2x^2+...}\right)$ | $\frac{1}{2}\text{arsinh}\frac{2x}{5}$ or $\ln\left(\frac{2x}{5}+\sqrt{1+\frac{4x^2}{25}}\right)$ o.e. | Fully correct in logarithmic form | Condone omitted c | $\ln\left(2x+\sqrt{4x^2+25}\right)$, $\ln\left(x+\sqrt{x^2+\frac{25}{4}}\right)$
### 4(iii)
$2x = 5 \sinh u \Rightarrow \frac{dx}{du} = \frac{5}{2}\cosh u$
$\int\sqrt{25+4x^2}dx = \int\sqrt{25+25\sinh^2 u} \times \frac{5}{2}\cosh u du$
$= \int\frac{25}{2}\cosh^2 u du = \int\left(\frac{25}{4}\cosh 2u + \frac{25}{4}\right)du$
$= \frac{25}{8}\sinh 2u + \frac{25}{4}u + c$
$= \frac{25}{4}\sinh u\cosh u + \frac{25}{4}u + c$
$= \frac{25}{4} \times \frac{5}{2}x \times \sqrt{1+\frac{4x^2}{25}} + \frac{25}{4} \times \frac{2x}{5} + c$
$= \frac{25}{4}\left(\left(\frac{2x}{5}+\sqrt{1+\frac{4x^2}{25}}+\frac{2x}{5}\sqrt{1+\frac{4x^2}{25}}\right)+c$ | M1, A1, M1*, A2, M1, A1(ag) | Finding $\frac{dx}{du}$ and complete substitution | Substituting for all elements correctly | Simplifying an expression of the form $k\cosh^2u$ to an integrable form | Give A2ñ for $\frac{25}{4}\sinh 2u + \frac{25}{4}u$. Give A1ñ A0 for one error | Using double "angle" formula. Dependent on M1* | Completion www with convincing intermediate step, e.g. reversing terms | Condone "upside-down" substitution for dx | e.g. $\frac{25}{8}e^{2u} + \frac{25}{4} - \frac{25}{8}e^{-2u} + c$; Using exponential definition of sinh 2u and substituting for sinh 2u and substituting for sinh 2u scores this M1 if an expression with a constant denominator is found validly