| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Method of differences |
| Difficulty | Challenging +1.8 This is a Further Maths FP2 reduction formula question requiring multiple sophisticated techniques: proving a recurrence relation using trigonometric identities and integration by parts, then applying it twice with telescoping to evaluate a definite integral. The proof in part (a) requires recognizing how to manipulate cos((n+2)x) and integrate strategically, while part (b) demands careful application of the reduction formula and exact evaluation at non-standard limits. This is substantially harder than typical A-level questions but follows a standard FP2 pattern, placing it well above average difficulty. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08h Integration by substitution8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(I_{n+2}=\int\frac{\cos(n+2)x}{\sin x}\,dx=\int\frac{\cos(nx)\cos 2x-\sin(nx)\sin 2x}{\sin x}\,dx\) | M1 | 3.1a |
| \(=\int\frac{\cos(nx)(1-2\sin^2x)-2\sin x\cos x\sin(nx)}{\sin x}\,dx\) | M1, A1 | 1.1b, 2.1 |
| \(=\int\frac{\cos(nx)}{\sin x}\,dx-\int\frac{\cos(nx)(2\sin^2x)+2\sin x\cos x\sin(nx)}{\sin x}\,dx\) | dM1 | 1.1b |
| \(=I_n-2\int\cos(nx)\sin x+\cos x\sin(nx)\,dx\) | ||
| \(=I_n-2\int\sin(n+1)x\,dx\) | ddM1 | 2.2a |
| \(I_{n+2}=2\frac{\cos(n+1)x}{n+1}+I_n\) * | A1\* | 2.1 |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([I_1]=\int\cot x\,dx=[\ln\sin x]\) may be seen in an expression for \(I_5\) | M1 | 1.1b |
| \(=\ln\frac{\sqrt{3}}{2}-\ln\frac{\sqrt{2}}{2}\) may be seen in an expression for \(I_5\) | A1 | 2.2a |
| \(I_5=2\frac{\cos 4x}{4}+I_3\) or \(I_3=2\frac{\cos 2x}{2}+I_1\) | M1 | 1.1b |
| \(\left[I_5\right]_{\pi/4}^{\pi/3}=\left[\frac{\cos 4x}{2}+\cos 2x+I_1\right]_{\pi/4}^{\pi/3}\) or \(\left[\frac{\cos 4x}{2}+\cos 2x+\ln\sin x\right]_{\pi/4}^{\pi/3}\) | M1 | 1.1b |
| \(=\frac{1}{2}\cos\frac{4\pi}{3}+\cos\frac{2\pi}{3}-\frac{1}{2}\cos\pi-\cos\frac{\pi}{2}+\text{"}\ln\frac{\sqrt{3}}{2}-\ln\frac{\sqrt{2}}{2}\text{"}\) | ||
| \(\left(=-\frac{1}{4}-\frac{1}{2}+\frac{1}{2}-0+\ln\frac{\sqrt{3}}{2}-\ln\frac{\sqrt{2}}{2}\right)=-\frac{1}{4}+\frac{1}{2}\ln\frac{3}{2}\) (oe) | A1 | 2.1 |
| (5) | ||
| (11 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_{n+2} - I_n = \int \frac{\cos(n+2)x - \cos nx}{\sin x}\, dx\) | M1 | 3.1a |
| \(= -\int \frac{2\sin\frac{1}{2}((n+2)x+nx)\sin\frac{1}{2}((n+2)x-nx)}{\sin x}\, dx\) | M1 | 1.1b |
| \(= -\int \frac{2\sin(n+1)x\sin x}{\sin x}\, dx\) | A1 | 2.1 |
| \(= -2\int \sin(n+1)x\, dx\) | dM1 | 1.1b |
| \(= 2\frac{\cos(n+1)x}{n+1} \Rightarrow I_{n+2} = \ldots\) | ddM1 | 2.2a |
| \(\Rightarrow I_{n+2} = 2\frac{\cos(n+1)x}{n+1} + I_n\) | A1\* | 2.1 |
# Question 6:
## Part (a) - Reduction Formula:
| Working | Mark | Guidance |
|---------|------|----------|
| $I_{n+2}=\int\frac{\cos(n+2)x}{\sin x}\,dx=\int\frac{\cos(nx)\cos 2x-\sin(nx)\sin 2x}{\sin x}\,dx$ | **M1** | 3.1a |
| $=\int\frac{\cos(nx)(1-2\sin^2x)-2\sin x\cos x\sin(nx)}{\sin x}\,dx$ | **M1, A1** | 1.1b, 2.1 |
| $=\int\frac{\cos(nx)}{\sin x}\,dx-\int\frac{\cos(nx)(2\sin^2x)+2\sin x\cos x\sin(nx)}{\sin x}\,dx$ | **dM1** | 1.1b |
| $=I_n-2\int\cos(nx)\sin x+\cos x\sin(nx)\,dx$ | | |
| $=I_n-2\int\sin(n+1)x\,dx$ | **ddM1** | 2.2a |
| $I_{n+2}=2\frac{\cos(n+1)x}{n+1}+I_n$ * | **A1\*** | 2.1 |
| | **(6)** | |
**Notes:**
- **M1:** Applies the compound angle formula as shown
- **M1:** Applies both double angle formulae $\cos 2x=1-2\sin^2x$ and $\sin 2x=2\sin x\cos x$ to the expression
- **A1:** Correct intermediate step reached, need not be simplified
- **dM1:** Dependent on previous method mark. Splits the integrand to identify $I_n$ and cancels the $\sin x$ term
- **ddM1:** Dependent on both previous method marks. Applies the compound angle formula to combine terms
- **A1\*:** Correct completion to the given result. No errors seen.
## Part (b) - Evaluation of $I_5$:
| Working | Mark | Guidance |
|---------|------|----------|
| $[I_1]=\int\cot x\,dx=[\ln\sin x]$ may be seen in an expression for $I_5$ | **M1** | 1.1b |
| $=\ln\frac{\sqrt{3}}{2}-\ln\frac{\sqrt{2}}{2}$ may be seen in an expression for $I_5$ | **A1** | 2.2a |
| $I_5=2\frac{\cos 4x}{4}+I_3$ or $I_3=2\frac{\cos 2x}{2}+I_1$ | **M1** | 1.1b |
| $\left[I_5\right]_{\pi/4}^{\pi/3}=\left[\frac{\cos 4x}{2}+\cos 2x+I_1\right]_{\pi/4}^{\pi/3}$ or $\left[\frac{\cos 4x}{2}+\cos 2x+\ln\sin x\right]_{\pi/4}^{\pi/3}$ | **M1** | 1.1b |
| $=\frac{1}{2}\cos\frac{4\pi}{3}+\cos\frac{2\pi}{3}-\frac{1}{2}\cos\pi-\cos\frac{\pi}{2}+\text{"}\ln\frac{\sqrt{3}}{2}-\ln\frac{\sqrt{2}}{2}\text{"}$ | | |
| $\left(=-\frac{1}{4}-\frac{1}{2}+\frac{1}{2}-0+\ln\frac{\sqrt{3}}{2}-\ln\frac{\sqrt{2}}{2}\right)=-\frac{1}{4}+\frac{1}{2}\ln\frac{3}{2}$ (oe) | **A1** | 2.1 |
| | **(5)** | |
| | **(11 marks)** | |
# Question 6(a) Alt 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_{n+2} - I_n = \int \frac{\cos(n+2)x - \cos nx}{\sin x}\, dx$ | **M1** | 3.1a |
| $= -\int \frac{2\sin\frac{1}{2}((n+2)x+nx)\sin\frac{1}{2}((n+2)x-nx)}{\sin x}\, dx$ | **M1** | 1.1b |
| $= -\int \frac{2\sin(n+1)x\sin x}{\sin x}\, dx$ | **A1** | 2.1 |
| $= -2\int \sin(n+1)x\, dx$ | **dM1** | 1.1b |
| $= 2\frac{\cos(n+1)x}{n+1} \Rightarrow I_{n+2} = \ldots$ | **ddM1** | 2.2a |
| $\Rightarrow I_{n+2} = 2\frac{\cos(n+1)x}{n+1} + I_n$ | **A1\*** | 2.1 |
**Notes (a) Alt 3:**
- **M1:** Attempts the difference $I_{n+2} - I_n = \ldots$ and combines to a single fraction
- **M1:** Applies the difference of two cosines formula
- **A1:** Correct expression
- **dM1:** Dependent on previous method mark. Cancels the $\sin x$ term in the integral
- **ddM1:** Dependent on both previous method marks. Integrates
- **A1\*:** Correct completion to the given result. No errors seen
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}
$$I _ { n } = \int \frac { \cos ( n x ) } { \sin x } \mathrm {~d} x \quad n \geqslant 1$$
(a) Show that, for $n \geqslant 1$
$$I _ { n + 2 } = \frac { 2 \cos ( n + 1 ) x } { n + 1 } + I _ { n }$$
(b) Hence determine the exact value of
$$\int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 3 } } \frac { \cos ( 5 x ) } { \sin x } d x$$
giving the answer in the form $a + b \ln c$ where $a , b$ and $c$ are rational numbers to be found.
\hfill \mbox{\textit{Edexcel FP2 2024 Q6 [11]}}