| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Integration by parts with inverse trig |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with three guided parts. Part (i) is a standard hyperbolic substitution with clear guidance. Part (ii) is routine implicit differentiation. Part (iii) requires integration by parts combining results from (i) and (ii), but the setup is heavily scaffolded. While it involves advanced topics (inverse trig, hyperbolic functions), the question provides substantial guidance throughout, making it moderately above average difficulty but not requiring significant independent insight. |
| Spec | 1.08i Integration by parts4.08h Integration: inverse trig/hyperbolic substitutions |
**Question 9(i)**
Full elimination of $x$: $I = \displaystyle\int\dfrac{1}{\cosh^2\theta\cdot\sinh\theta}\cdot\sinh\theta\,\text{d}\theta$ **M1**
$\Rightarrow I = \displaystyle\int\text{sech}^2\theta\,\text{d}\theta$ **A1**
$= \tanh\theta\ (+C)$ **A1**
$= \dfrac{\sqrt{x^2-1}}{x}\ (+C)$ from $\dfrac{\sinh\theta}{\cosh\theta}$ **A1** (AG)
**Question 9(ii)**
$\sec y = x \Rightarrow \sec y\tan y\,\dfrac{\text{d}y}{\text{d}x} = 1$ **M1A1**
Use of $\tan y = \sqrt{\sec^2 y - 1}$ **M1**
to get $\dfrac{\text{d}y}{\text{d}x} = \dfrac{1}{x\sqrt{x^2-1}}$ **A1** AG. Ignore lack of reason for taking the $+$ve sq.rt.
**Question 9(iii)**
$\displaystyle\int\sec^{-1}x\cdot\dfrac{1}{x^2}\,\text{d}x$ **M1A2** By parts
$= \sec^{-1}x\cdot\dfrac{-1}{x} - \displaystyle\int\dfrac{-1}{x}\cdot\dfrac{1}{x\sqrt{x^2-1}}\,\text{d}x$
$= \dfrac{-\sec^{-1}x}{x} + \displaystyle\int\dfrac{1}{x^2\sqrt{x^2-1}}\,\text{d}x$
$= \dfrac{-\sec^{-1}x}{x} + \dfrac{\sqrt{x^2-1}}{x}\ (+C)$ **A1** using **(i)**
**Alternative:**
Use $u = \sec^{-1}x \Rightarrow \dfrac{\text{d}u}{\text{d}x} = \dfrac{1}{x\sqrt{x^2-1}}$ **M1**
$\Rightarrow \sec u\tan u\,\text{d}u = \text{d}x$
$\Rightarrow \displaystyle\int\sec^{-1}x\cdot\dfrac{1}{x^2}\,\text{d}x = \displaystyle\int u\sin u\,\text{d}u$ **A1**
2-stage integration by parts: $\displaystyle\int u\sin u\,\text{d}u = -u\cos u + \displaystyle\int\cos u\,\text{d}u = -u\cos u + \sin u\ (+C)$ **M1**
Correctly turning back: $= \dfrac{-\sec^{-1}x}{x} + \dfrac{\sqrt{x^2-1}}{x}\ (+C)$ **A1**
**Total: 11 marks**9 (i) Given that $x \geqslant 1$, use the substitution $x = \cosh \theta$ to show that
$$\int \frac { 1 } { x ^ { 2 } \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x = \frac { \sqrt { x ^ { 2 } - 1 } } { x } + C$$
where $C$ is an arbitrary constant.\\
(ii) By differentiating sec $y = x$ implicitly, show that $\frac { \mathrm { d } } { \mathrm { d } x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } }$ for $x \geqslant 1$.\\
(iii) Use integration by parts to determine $\int \frac { \sec ^ { - 1 } x } { x ^ { 2 } } \mathrm {~d} x$ for $x \geqslant 1$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2017 Q9 [11]}}