| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2023 |
| Session | November |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Hyperbolic substitution to evaluate integral |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring hyperbolic substitution (x = cosh u), integration by parts for the arcosh term, and careful algebraic manipulation. Part (a) is a guided proof requiring multiple steps but follows standard techniques. Part (b) requires integration by parts on x·arcosh(x) combined with the given result, demanding sustained accuracy across 11 marks total. The techniques are all syllabus-standard for FM, but the length and need for algebraic precision place it well above average difficulty. |
| Spec | 1.08i Integration by parts4.07d Differentiate/integrate: hyperbolic functions4.07f Inverse hyperbolic: logarithmic forms |
(a) Use a hyperbolic substitution and calculus to show that
$$\int \frac{x^2}{\sqrt{x^2 - 1}} dx = \frac{1}{2}\left[x\sqrt{x^2 - 1} + \arcosh x\right] + k$$
where $k$ is an arbitrary constant. (6)
\includegraphics{figure_8}
Figure 1 shows a sketch of part of the curve $C$ with equation
$$y = \frac{4}{15}x \arcosh x \quad x \geqslant 1$$
The finite region $R$, shown shaded in Figure 1, is bounded by the curve $C$, the $x$-axis and the line with equation $x = 3$
(b) Using algebraic integration and the result from part (a), show that the area of $R$ is given by
$$\frac{1}{15}\left[17\ln\left(3 + 2\sqrt{2}\right) - 6\sqrt{2}\right]$$ (5)
This is the last question on the paper.
\hfill \mbox{\textit{SPS SPS FM Pure 2023 Q8}}