| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2013 |
| Session | June |
| Marks | 4 |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Completing square then standard inverse trig |
| Difficulty | Standard +0.3 This is a straightforward application of completing the square followed by a standard inverse tan integral. While it requires knowledge of inverse trig integration (a Further Maths topic), the method is direct and well-practiced, making it slightly easier than average overall but standard for Further Maths students. |
| Spec | 4.08h Integration: inverse trig/hyperbolic substitutions |
$x^2 - 6x + 12 \equiv (x-3)^2 + 3$ **B1**
$$\int_2^6 \frac{1}{(\sqrt{3})^2 + (x-3)^2} \,\mathrm{d}x = \left[\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x-3}{\sqrt{3}}\right)\right]_2^6$$
**M1** for $\tan^{-1}$, 1st **A1** for $\left(\frac{x-3}{\sqrt{3}}\right)$
$$= \frac{1}{\sqrt{3}}\left(\frac{\pi}{3} - \left(-\frac{\pi}{6}\right)\right) = \frac{\pi}{2\sqrt{3}}$$
**A1**
**[4]**1 By completing the square, or otherwise, find the exact value of $\int _ { 2 } ^ { 6 } \frac { 1 } { x ^ { 2 } - 6 x + 12 } \mathrm {~d} x$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2013 Q1 [4]}}