| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Integrate using hyperbolic substitution |
| Difficulty | Standard +0.8 This is a Further Maths question requiring hyperbolic function differentiation and a substitution integral with inverse trigonometric result. Part (a) is routine (using sinh²x + cosh²x = 1 and chain rule), but part (b) requires recognizing the substitution u = cosh²x transforms the integral into arctan form, then careful evaluation of limits. The multi-step nature and need to connect hyperbolic and inverse trig functions makes this moderately challenging but still a standard FP2 technique question. |
| Spec | 1.08h Integration by substitution4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| - \(\frac{du}{dx} = 2\cosh x \sinh x\) | M1 | Any correct method |
| - \(= \sinh 2x\) | A1 | 2 marks; AG |
| Answer | Marks | Guidance |
|---|---|---|
| - \(I = \int_{x=0}^{x=1} \frac{du}{1+u^2}\) | M1A1 | Ignore limits here |
| - \(= \left[\tan^{-1}u\right]_{x=0}\) | A1 | |
| - \(= \left[\tan^{-1}(\cosh^2x)\right]_0\) | A1 | Or A1 for change of limits |
| - \(= \tan^{-1}(\cosh^2 1) - \tan^{-1}(\cosh^2 0)\) | ||
| - \(= \tan^{-1}(\cosh^2 1) - \frac{\pi}{4}\) | A1 | 5 marks; AG |
**(a)**
- $\frac{du}{dx} = 2\cosh x \sinh x$ | M1 | Any correct method
- $= \sinh 2x$ | A1 | 2 marks; AG
- **Total: 2 marks**
**(b)**
- $I = \int_{x=0}^{x=1} \frac{du}{1+u^2}$ | M1A1 | Ignore limits here
- $= \left[\tan^{-1}u\right]_{x=0}$ | A1 |
- $= \left[\tan^{-1}(\cosh^2x)\right]_0$ | A1 | Or A1 for change of limits
- $= \tan^{-1}(\cosh^2 1) - \tan^{-1}(\cosh^2 0)$ | |
- $= \tan^{-1}(\cosh^2 1) - \frac{\pi}{4}$ | A1 | 5 marks; AG
- **Total: 5 marks**
**Question 5 Total: 7 marks**
---5
\begin{enumerate}[label=(\alph*)]
\item Given that $u = \cosh ^ { 2 } x$, show that $\frac { \mathrm { d } u } { \mathrm {~d} x } = \sinh 2 x$.
\item Hence show that
$$\int _ { 0 } ^ { 1 } \frac { \sinh 2 x } { 1 + \cosh ^ { 4 } x } \mathrm {~d} x = \tan ^ { - 1 } \left( \cosh ^ { 2 } 1 \right) - \frac { \pi } { 4 }$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2009 Q5 [7]}}