| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Polynomial times trigonometric |
| Difficulty | Challenging +1.2 This is a standard Further Maths reduction formula question requiring integration by parts twice to establish the recurrence relation, then iterative application to find I₅. While it involves multiple steps and careful algebraic manipulation, the technique is well-practiced in FP2 and follows a predictable pattern without requiring novel insight. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_n = \left[-x^n \cos x\right]_0^\pi + n\int_0^\pi x^{n-1} \cos x \, dx\) | M1 | For attempt to integrate by parts |
| (correct result before limits) | A1 | For correct result before limits |
| \(= \pi^n + n\left\{\left[x^{n-1} \sin x\right]_0^\pi - (n-1)\int_0^\pi x^{n-2} \sin x \, dx\right\}\) | M1 | For attempt at second integration by parts |
| (correct result before limits) | A1 | For correct result before limits |
| \(\Rightarrow I_n = \pi^n - n(n-1)I_{n-2}\) | A1 | For correct result www AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_1 = \left[-x\cos x\right]_0^\pi + \int_0^\pi \cos x \, dx\) | M1 | For integrating by parts for \(I_1\) |
| \(\Rightarrow I_1 = \pi + \left[\sin x\right]_0^\pi = \pi\) | A1 | For correct \(I_1\); SC B1 \(I_1 = \pi\) with no working |
| \(I_3 = \pi^3 - 6I_1\), \(I_5 = \pi^5 - 20I_3\) | M1 | For substituting \(n=3\) or \(5\) in reduction formula |
| \(\Rightarrow I_5 = \pi^5 - 20\pi^3 + 120\pi\) | A1 | For correct result |
## Question 6:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_n = \left[-x^n \cos x\right]_0^\pi + n\int_0^\pi x^{n-1} \cos x \, dx$ | M1 | For attempt to integrate by parts |
| (correct result before limits) | A1 | For correct result before limits |
| $= \pi^n + n\left\{\left[x^{n-1} \sin x\right]_0^\pi - (n-1)\int_0^\pi x^{n-2} \sin x \, dx\right\}$ | M1 | For attempt at second integration by parts |
| (correct result before limits) | A1 | For correct result before limits |
| $\Rightarrow I_n = \pi^n - n(n-1)I_{n-2}$ | A1 | For correct result **www AG** |
**[5]**
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_1 = \left[-x\cos x\right]_0^\pi + \int_0^\pi \cos x \, dx$ | M1 | For integrating by parts for $I_1$ |
| $\Rightarrow I_1 = \pi + \left[\sin x\right]_0^\pi = \pi$ | A1 | For correct $I_1$; SC B1 $I_1 = \pi$ with no working |
| $I_3 = \pi^3 - 6I_1$, $I_5 = \pi^5 - 20I_3$ | M1 | For substituting $n=3$ or $5$ in reduction formula |
| $\Rightarrow I_5 = \pi^5 - 20\pi^3 + 120\pi$ | A1 | For correct result |
**[4]**
---6 It is given that, for non-negative integers $n$,
$$I _ { n } = \int _ { 0 } ^ { \pi } x ^ { n } \sin x \mathrm {~d} x$$
(i) Prove that, for $n \geqslant 2 , I _ { n } = \pi ^ { n } - n ( n - 1 ) I _ { n - 2 }$.\\
(ii) Find $I _ { 5 }$ in terms of $\pi$.
\hfill \mbox{\textit{OCR FP2 2012 Q6 [9]}}