| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Trigonometric substitution to simplify integral |
| Difficulty | Standard +0.8 This is a Further Maths question requiring trigonometric substitution with careful manipulation of differentials and trigonometric identities, followed by integration to an inverse tan form and evaluation of definite integral limits. While the steps are systematic once the approach is clear, it requires multiple techniques (substitution mechanics, trig identities, inverse trig integration, limit transformation) making it moderately challenging but still within standard FP2 territory. |
| Spec | 4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = \tan\theta\) | B1 | OE |
| \(dt = \sec^2\theta \, d\theta\) | B1 | OE |
| \(I = \int\frac{dt}{(9\cos^2\theta + \sin^2\theta)\sec^2\theta}\) | M1 | OE |
| \(= \int\frac{dt}{t^2 + 9}\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = \left[\frac{1}{3}\tan^{-1}\frac{t}{3}\right]_0^{\sqrt{3}}\) | M1 | M1 for \(\tan^{-1}\) |
| \(\frac{1}{3}\tan^{-1}\sqrt{3}\) or \(\frac{1}{3}\tan^{-1}\frac{1}{\sqrt{3}}\) | A1 | |
| \(= \frac{\pi}{18}\) | A1 | 3 marks |
### Part (a)
$t = \tan\theta$ | B1 | OE
$dt = \sec^2\theta \, d\theta$ | B1 | OE
$I = \int\frac{dt}{(9\cos^2\theta + \sin^2\theta)\sec^2\theta}$ | M1 | OE
$= \int\frac{dt}{t^2 + 9}$ | A1 | 3 marks | AG
### Part (b)
$I = \left[\frac{1}{3}\tan^{-1}\frac{t}{3}\right]_0^{\sqrt{3}}$ | M1 | M1 for $\tan^{-1}$
$\frac{1}{3}\tan^{-1}\sqrt{3}$ or $\frac{1}{3}\tan^{-1}\frac{1}{\sqrt{3}}$ | A1 |
$= \frac{\pi}{18}$ | A1 | 3 marks | AG6 (a) Show that the substitution $t = \tan \theta$ transforms the integral
$$\int \frac { \mathrm { d } \theta } { 9 \cos ^ { 2 } \theta + \sin ^ { 2 } \theta }$$
into
$$\int \frac { \mathrm { d } t } { 9 + t ^ { 2 } }$$
(b) Hence show that
$$\int _ { 0 } ^ { \frac { \pi } { 3 } } \frac { d \theta } { 9 \cos ^ { 2 } \theta + \sin ^ { 2 } \theta } = \frac { \pi } { 18 }$$
\hfill \mbox{\textit{AQA FP2 2010 Q6 [6]}}