| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Trigonometric power reduction |
| Difficulty | Challenging +1.8 This is a substantial Further Maths reduction formula question requiring multiple techniques: substitution integration, integration by parts to derive the reduction formula, and recursive application. While the steps are guided, it demands careful algebraic manipulation, understanding of the derivative of sec x, and working with exact values. The multi-part structure and need to connect all parts elevates it above standard questions, but it's a classic Further Maths exercise without requiring exceptional insight. |
| Spec | 1.08h Integration by substitution4.08e Mean value of function: using integral8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{du}{dx}=\sec^2 x\) so integral becomes \(\int u^2\, du\) | M1 | Uses substitution correctly |
| \(=\frac{1}{3}\tan^3 x+C\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_n=\int_0^{\frac{\pi}{4}}\sec^{n-2}x\sec^2 x\tan^2 x\) | B1 | Separates into correct structure |
| \(=\frac{1}{3}\left[\sec^{n-2}x\tan^3 x\right]_0^{\frac{\pi}{4}}-\frac{n-2}{3}\int_0^{\frac{\pi}{4}}\sec^{n-2}x\tan^4 x\,dx\) | M1 | Uses integration by parts correctly |
| \(=\frac{2^{\frac{n-2}{2}}}{3}-\frac{n-2}{3}\int_0^{\frac{\pi}{4}}\sec^{n-2}x\tan^2 x(\sec^2 x-1)\,dx\) | M1 | Uses \(\tan^2 x=\sec^2 x-1\) |
| \(=\frac{2^{\frac{n-2}{2}}}{3}-\frac{n-2}{3}I_n+\frac{n-2}{3}I_{n-2}\) | A1 | |
| Thus \((3+n-2)I_n=(\sqrt{2})^{n-2}+(n-2)I_{n-2}\) | A1 | AG |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_2=\left[\frac{\tan^3 x}{3}\right]_0^{\frac{\pi}{4}}=\frac{1}{3}\) | B1 FT | Attempts to find \(I_2\) |
| \(5I_4=2+2I_2\) | M1 | Uses reduction formula |
| Mean value \(=\frac{4}{\pi}I_4=\frac{32}{15\pi}\) | A1 | |
| Total: 3 |
## Question 9(i):
| $\frac{du}{dx}=\sec^2 x$ so integral becomes $\int u^2\, du$ | M1 | Uses substitution correctly |
|---|---|---|
| $=\frac{1}{3}\tan^3 x+C$ | A1 | |
| **Total: 2** | | |
---
## Question 9(ii):
| $I_n=\int_0^{\frac{\pi}{4}}\sec^{n-2}x\sec^2 x\tan^2 x$ | B1 | Separates into correct structure |
|---|---|---|
| $=\frac{1}{3}\left[\sec^{n-2}x\tan^3 x\right]_0^{\frac{\pi}{4}}-\frac{n-2}{3}\int_0^{\frac{\pi}{4}}\sec^{n-2}x\tan^4 x\,dx$ | M1 | Uses integration by parts correctly |
| $=\frac{2^{\frac{n-2}{2}}}{3}-\frac{n-2}{3}\int_0^{\frac{\pi}{4}}\sec^{n-2}x\tan^2 x(\sec^2 x-1)\,dx$ | M1 | Uses $\tan^2 x=\sec^2 x-1$ |
| $=\frac{2^{\frac{n-2}{2}}}{3}-\frac{n-2}{3}I_n+\frac{n-2}{3}I_{n-2}$ | A1 | |
| Thus $(3+n-2)I_n=(\sqrt{2})^{n-2}+(n-2)I_{n-2}$ | A1 | AG |
| **Total: 5** | | |
---
## Question 9(iii):
| $I_2=\left[\frac{\tan^3 x}{3}\right]_0^{\frac{\pi}{4}}=\frac{1}{3}$ | B1 FT | Attempts to find $I_2$ |
|---|---|---|
| $5I_4=2+2I_2$ | M1 | Uses reduction formula |
| Mean value $=\frac{4}{\pi}I_4=\frac{32}{15\pi}$ | A1 | |
| **Total: 3** | | |
---9 (i) Using the substitution $u = \tan x$, or otherwise, find $\int \sec ^ { 2 } x \tan ^ { 2 } x \mathrm {~d} x$.\\
It is given that, for $n \geqslant 0$,
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sec ^ { n } x \tan ^ { 2 } x \mathrm {~d} x$$
(ii) Using the result that $\frac { \mathrm { d } } { \mathrm { d } x } ( \sec x ) = \tan x \sec x$, show that, for $n \geqslant 2$,
$$( n + 1 ) I _ { n } = ( \sqrt { } 2 ) ^ { n - 2 } + ( n - 2 ) I _ { n - 2 }$$
(iii) Hence find the mean value of $\sec ^ { 4 } x \tan ^ { 2 } x$ with respect to $x$ over the interval $0 \leqslant x \leqslant \frac { 1 } { 4 } \pi$, giving your answer in exact form.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q9 [10]}}