| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Integrate using hyperbolic substitution |
| Difficulty | Challenging +1.3 This is a structured Further Maths question on hyperbolic substitution with clear guidance. Part (a) involves routine manipulation of hyperbolic identities (cosh²u - sinh²u = 1) and differentiation, while part (b) requires integrating cosh 2u, back-substituting using inverse hyperbolic functions, and evaluating limits. The scaffolding significantly reduces difficulty compared to an unguided integration problem, though it still requires facility with hyperbolic functions beyond standard A-level. |
| Spec | 1.08h Integration by substitution4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dx}{du} = \frac{3}{4}\cosh u\) | B1 | Correct expression |
| \(\int\frac{x^2}{\sqrt{16x^2+9}}\,dx = \int\frac{\frac{9}{16}\sinh^2 u}{\sqrt{16\cdot\frac{9}{16}\sinh^2 u+9}}\cdot\frac{3}{4}\cosh u\, du\) | M1A1 | M1: Complete substitution attempt; A1: Correct expression |
| \(= k\int\sinh^2 u\, du\) | A1 | |
| \(\sinh^2 u = \frac{\cosh 2u - 1}{2}\) | M1 | Use of \(\sinh^2 u = \pm\frac{1}{2}\cosh 2u \pm\frac{1}{2}\); may be implied by integration |
| \(= \frac{9}{128}\int(\cosh 2u - 1)\,du\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x=0 \Rightarrow u=0\), \(x=1 \Rightarrow u = \text{arsinh}\!\left(\frac{4}{3}\right)\) | B1 | Correct limits |
| \(\int(\cosh 2u-1)\,du = \left[\frac{1}{2}\sinh 2u - u\right]\) | M1 | Attempt integration of form \(\alpha\sinh 2u + \beta u\) |
| \(\int_0^1\frac{64x^2}{\sqrt{16x^2+9}}\,dx = \frac{9}{2}\left[\frac{1}{2}\sinh 2u - u\right]_0^{\text{arsinh}(\frac{4}{3})}\) | ||
| \(= \frac{9}{2}\left\{\left(\frac{4}{3}\sqrt{1+\frac{16}{9}}\right) - \ln\!\left(\frac{4}{3}+\sqrt{1+\frac{16}{9}}\right)\right\} - (-0)\) | dM1 | Substitute \(u\) limits and subtract; dependent on previous M mark |
| \(= 10, -\frac{9}{2}\ln 3\) | A1, A1 | cao |
## Question 8(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dx}{du} = \frac{3}{4}\cosh u$ | B1 | Correct expression |
| $\int\frac{x^2}{\sqrt{16x^2+9}}\,dx = \int\frac{\frac{9}{16}\sinh^2 u}{\sqrt{16\cdot\frac{9}{16}\sinh^2 u+9}}\cdot\frac{3}{4}\cosh u\, du$ | M1A1 | M1: Complete substitution attempt; A1: Correct expression |
| $= k\int\sinh^2 u\, du$ | A1 | |
| $\sinh^2 u = \frac{\cosh 2u - 1}{2}$ | M1 | Use of $\sinh^2 u = \pm\frac{1}{2}\cosh 2u \pm\frac{1}{2}$; may be implied by integration |
| $= \frac{9}{128}\int(\cosh 2u - 1)\,du$ | A1 | |
---
## Question 8(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x=0 \Rightarrow u=0$, $x=1 \Rightarrow u = \text{arsinh}\!\left(\frac{4}{3}\right)$ | B1 | Correct limits |
| $\int(\cosh 2u-1)\,du = \left[\frac{1}{2}\sinh 2u - u\right]$ | M1 | Attempt integration of form $\alpha\sinh 2u + \beta u$ |
| $\int_0^1\frac{64x^2}{\sqrt{16x^2+9}}\,dx = \frac{9}{2}\left[\frac{1}{2}\sinh 2u - u\right]_0^{\text{arsinh}(\frac{4}{3})}$ | | |
| $= \frac{9}{2}\left\{\left(\frac{4}{3}\sqrt{1+\frac{16}{9}}\right) - \ln\!\left(\frac{4}{3}+\sqrt{1+\frac{16}{9}}\right)\right\} - (-0)$ | dM1 | Substitute $u$ limits and subtract; dependent on previous M mark |
| $= 10, -\frac{9}{2}\ln 3$ | A1, A1 | cao |
---\begin{enumerate}
\item (a) Show that, under the substitution $x = \frac { 3 } { 4 } \sinh u$,
\end{enumerate}
$$\int \frac { x ^ { 2 } } { \sqrt { 16 x ^ { 2 } + 9 } } \mathrm {~d} x = k \int ( \cosh 2 u - 1 ) \mathrm { d } u$$
where $k$ is a constant to be determined.\\
(b) Hence show that
$$\int _ { 0 } ^ { 1 } \frac { 64 x ^ { 2 } } { \sqrt { 16 x ^ { 2 } + 9 } } \mathrm {~d} x = p + q \ln 3$$
where $p$ and $q$ are rational numbers to be found.\\
\hfill \mbox{\textit{Edexcel F3 2015 Q8 [11]}}