| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Trigonometric power reduction |
| Difficulty | Challenging +1.8 This is a standard reduction formula derivation requiring integration by parts with careful manipulation of trigonometric identities, followed by routine application. While it demands technical proficiency and multiple steps, the approach (IBP with cosec^{n-2} × cosec^2) is a well-established technique taught explicitly in Further Pure 3. Part (b) is straightforward application once the formula is established. The question is harder than typical A-level due to the Further Maths content and algebraic manipulation required, but represents a standard exam question within this topic rather than requiring novel insight. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \int \text{cosec}^n x\,dx = \int \text{cosec}^{n-2}x\,\text{cosec}^2x\,dx\) | ||
| \(= -\cot x\,\text{cosec}^{n-2}x - \int(n-2)\text{cosec}^{n-2}x\cot^2x\,dx\) | M1A1 | M1: Parts in correct direction. A1: Correct unsimplified expression |
| Uses \(\cot^2x = \text{cosec}^2x - 1\) (incorrect signs allowed) | dM1 | |
| \(= -\cot x\,\text{cosec}^{n-2}x - (n-2)I_n + (n-2)I_{n-2}\) | ddM1 | Introduces \(I_n\) and \(I_{n-2}\) |
| \(I_n = \frac{n-2}{n-1}I_{n-2} - \frac{1}{n-1}\cot x\,\text{cosec}^{n-2}x\) * | A1* cso | Correct completion with no errors (apart from possible omission of \(dx\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \int \text{cosec}^{n-2}x(1+\cot^2x)\,dx\) | M1 | Uses \(\cot^2x=\text{cosec}^2x-1\) (incorrect signs allowed) |
| \(\int \cot^2x\,\text{cosec}^{n-2}x\,dx = -\frac{1}{(n-2)}\cot x\,\text{cosec}^{n-2}x - \frac{1}{(n-2)}\int\text{cosec}^n x\,dx\) | dM1, A1 | M1: Parts in correct direction. A1: Correct expression (1st A mark on e-PEN) |
| Introduces \(I_n\) and \(I_{n-2}\) | ddM1 | |
| \(I_n = \frac{n-2}{n-1}I_{n-2} - \frac{1}{n-1}\cot x\,\text{cosec}^{n-2}x\) * | A1* cso | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_4 = \frac{2}{3}I_2 - \frac{1}{3}\cot x\,\text{cosec}^2x\) | M1 | Correct application of the given reduction formula |
| \(I_2 = \int \text{cosec}^2x\,dx = -\cot x\) (or \(= -\cot x\,\text{cosec}^0x\)) | B1 | By integration or use of reduction formula |
| \(I_4 = \frac{2}{3}(-\cot x) - \frac{1}{3}\cot x(1+\cot^2x)\) | M1 | Uses their \(I_2\) and \(\text{cosec}^2x = 1+\cot^2x\) (incorrect signs allowed) |
| \(I_4 = -\cot x - \frac{1}{3}\cot^3x\ (+c)\) | A1 cso | \(+c\) not required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\text{cosec}^4x\,dx = \int(1+\cot^2x)\text{cosec}^2x\,dx\) | M1 | \(\text{cosec}^4x=(1+\cot^2x)\text{cosec}^2x\) |
| \(I_4 = -\cot x - \frac{1}{3}\cot^3x\ (+c)\) | B1M1A1 cso | B1: \(\int\text{cosec}^2x\,dx=-\cot x\); M1: \(\int\text{cosec}^2x\cot^2x\,dx=k\cot^3x\); A1: \(\int\text{cosec}^2x\cot^2x\,dx=-\frac{1}{3}\cot^3x\) |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int \text{cosec}^n x\,dx = \int \text{cosec}^{n-2}x\,\text{cosec}^2x\,dx$ | | |
| $= -\cot x\,\text{cosec}^{n-2}x - \int(n-2)\text{cosec}^{n-2}x\cot^2x\,dx$ | M1A1 | M1: Parts in correct direction. A1: Correct unsimplified expression |
| Uses $\cot^2x = \text{cosec}^2x - 1$ (incorrect signs allowed) | dM1 | |
| $= -\cot x\,\text{cosec}^{n-2}x - (n-2)I_n + (n-2)I_{n-2}$ | ddM1 | Introduces $I_n$ and $I_{n-2}$ |
| $I_n = \frac{n-2}{n-1}I_{n-2} - \frac{1}{n-1}\cot x\,\text{cosec}^{n-2}x$ * | A1* cso | Correct completion with no errors (apart from possible omission of $dx$) |
**Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int \text{cosec}^{n-2}x(1+\cot^2x)\,dx$ | M1 | Uses $\cot^2x=\text{cosec}^2x-1$ (incorrect signs allowed) |
| $\int \cot^2x\,\text{cosec}^{n-2}x\,dx = -\frac{1}{(n-2)}\cot x\,\text{cosec}^{n-2}x - \frac{1}{(n-2)}\int\text{cosec}^n x\,dx$ | dM1, A1 | M1: Parts in correct direction. A1: Correct expression (1st A mark on e-PEN) |
| Introduces $I_n$ and $I_{n-2}$ | ddM1 | |
| $I_n = \frac{n-2}{n-1}I_{n-2} - \frac{1}{n-1}\cot x\,\text{cosec}^{n-2}x$ * | A1* cso | Correct completion with no errors |
---
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_4 = \frac{2}{3}I_2 - \frac{1}{3}\cot x\,\text{cosec}^2x$ | M1 | Correct application of the given reduction formula |
| $I_2 = \int \text{cosec}^2x\,dx = -\cot x$ (or $= -\cot x\,\text{cosec}^0x$) | B1 | By integration or use of reduction formula |
| $I_4 = \frac{2}{3}(-\cot x) - \frac{1}{3}\cot x(1+\cot^2x)$ | M1 | Uses their $I_2$ and $\text{cosec}^2x = 1+\cot^2x$ (incorrect signs allowed) |
| $I_4 = -\cot x - \frac{1}{3}\cot^3x\ (+c)$ | A1 cso | $+c$ not required |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\text{cosec}^4x\,dx = \int(1+\cot^2x)\text{cosec}^2x\,dx$ | M1 | $\text{cosec}^4x=(1+\cot^2x)\text{cosec}^2x$ |
| $I_4 = -\cot x - \frac{1}{3}\cot^3x\ (+c)$ | B1M1A1 cso | B1: $\int\text{cosec}^2x\,dx=-\cot x$; M1: $\int\text{cosec}^2x\cot^2x\,dx=k\cot^3x$; A1: $\int\text{cosec}^2x\cot^2x\,dx=-\frac{1}{3}\cot^3x$ |
---5.
$$I _ { n } = \int \operatorname { cosec } ^ { n } x \mathrm {~d} x , \quad 0 < x < \frac { \pi } { 2 } , \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geqslant 2$
$$I _ { n } = \frac { n - 2 } { n - 1 } I _ { n - 2 } - \frac { 1 } { n - 1 } \cot x \operatorname { cosec } ^ { n - 2 } x$$
\item Hence, or otherwise, find
$$\int \operatorname { cosec } ^ { 4 } x \mathrm {~d} x$$
giving your answer in terms of $\cot x$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2017 Q5 [9]}}