| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Derivative of inverse hyperbolic function |
| Difficulty | Standard +0.8 This is a Further Maths question requiring proof of an inverse hyperbolic derivative, application to integration, and a non-trivial substitution for part (iii). While parts (i) and (ii) are standard FP2 material, part (iii) requires insight to choose an appropriate substitution (likely x = ½cosh u) and careful algebraic manipulation. The multi-step nature and the need for problem-solving beyond routine application places this moderately above average difficulty. |
| Spec | 1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Get \(\sinh y \frac{dy}{dx} = 1\) | M1 | Or equivalent; allow \(\pm\). Allow use of ln equivalent with Chain Rule |
| Replace \(\sinh y = \sqrt{(\cosh^2 y) - 1}\). Justify positive grad. to A.G. | A1, B1 | e.g. sketch |
| (ii) Get \(k \cosh^{-1}2x\). Get \(k=\frac{1}{2}\) | M1, A1 | No need for \(c\) |
| (iii) Sub. \(x = k \cosh u\). Replace all \(x\) to \(\int k_1 \sinh^2 u \, du\). Replace as \(\int k_2(\cosh 2u - 1) \, du\). Integrate correctly. Attempt to replace \(u\) with \(x\) equivalent. Tidy to reasonable form | M1, A1, M1, A1, M1, A1 | Or exponential equivalent. No need for \(c\). In their answer cao (\(\frac{1}{2}x\sqrt{(4x^2-1)} - \frac{1}{4} \cosh^{-1}2x (+c)\)) |
**(i)** Get $\sinh y \frac{dy}{dx} = 1$ | M1 | Or equivalent; allow $\pm$. Allow use of ln equivalent with Chain Rule
Replace $\sinh y = \sqrt{(\cosh^2 y) - 1}$. Justify positive grad. to A.G. | A1, B1 | e.g. sketch
**(ii)** Get $k \cosh^{-1}2x$. Get $k=\frac{1}{2}$ | M1, A1 | No need for $c$
**(iii)** Sub. $x = k \cosh u$. Replace all $x$ to $\int k_1 \sinh^2 u \, du$. Replace as $\int k_2(\cosh 2u - 1) \, du$. Integrate correctly. Attempt to replace $u$ with $x$ equivalent. Tidy to reasonable form | M1, A1, M1, A1, M1, A1 | Or exponential equivalent. No need for $c$. In their answer cao ($\frac{1}{2}x\sqrt{(4x^2-1)} - \frac{1}{4} \cosh^{-1}2x (+c)$)9 (i) Prove that $\frac { \mathrm { d } } { \mathrm { d } x } \left( \cosh ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 2 } - 1 } }$.\\
(ii) Hence, or otherwise, find $\int \frac { 1 } { \sqrt { 4 x ^ { 2 } - 1 } } \mathrm {~d} x$.\\
(iii) By means of a suitable substitution, find $\int \sqrt { 4 x ^ { 2 } - 1 } \mathrm {~d} x$.
\hfill \mbox{\textit{OCR FP2 2008 Q9 [11]}}