| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Integrate using hyperbolic substitution |
| Difficulty | Challenging +1.2 This is a structured Further Maths question requiring hyperbolic function manipulation and substitution integration. Part (a) is routine algebraic manipulation using standard definitions. Part (b) follows a guided substitution leading to a standard inverse tan integral. While requiring multiple techniques, the question provides clear scaffolding and uses well-practiced methods for FP2 students, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.08h Integration by substitution4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
6
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { 1 } { 5 \cosh x - 3 \sinh x } = \frac { \mathrm { e } ^ { x } } { m + \mathrm { e } ^ { 2 x } }$, where $m$ is an integer.
\item Use the substitution $u = \mathrm { e } ^ { x }$ to show that
$$\int _ { 0 } ^ { \ln 2 } \frac { 1 } { 5 \cosh x - 3 \sinh x } \mathrm {~d} x = \frac { \pi } { 8 } - \frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2013 Q6 [8]}}