| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Polynomial times trigonometric |
| Difficulty | Challenging +1.2 This is a standard reduction formula question from FP3 requiring integration by parts twice to establish the recurrence relation, then straightforward substitution to find specific values. While it involves multiple steps and careful algebraic manipulation, it follows a well-practiced template that FP3 students drill extensively. The techniques are routine for this level, though the bookwork and arithmetic require care. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_n = \left[x^n(-\frac{1}{2}\cos 2x)\right]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} -\frac{1}{2}nx^{n-1}\cos 2x\, dx\) | M1 A1 | Use of integration by parts, integrating \(\sin 2x\), differentiating \(x^n\); cao |
| \(I_n = \left\langle\left[x^n(-\frac{1}{2}\cos 2x)\right]_0^{\frac{\pi}{4}}\right\rangle + \left[\frac{1}{4}nx^{n-1}\sin 2x\right]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}}\frac{1}{4}n(n-1)x^{n-2}\sin 2x\, dx\) | M1 A1 | Second application of integration by parts, integrating \(\cos 2x\), differentiating \(x^{n-1}\); cao |
| \(I_n = \frac{1}{4}n\left(\frac{\pi}{4}\right)^{n-1} - \frac{1}{4}n(n-1)I_{n-2}\) | A1cso | cso including correct use of \(\frac{\pi}{4}\) and \(0\) as limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_0 = \int_0^{\frac{\pi}{4}}\sin 2x\, dx = \left[-\frac{1}{2}\cos 2x\right]_0^{\frac{\pi}{4}} = \frac{1}{2}\) | M1 A1 | Integrating to find \(I_0\); cao (accept \(I_0 = \frac{1}{2}\)) |
| \(I_2 = \frac{1}{4}\times 2\times\left(\frac{\pi}{4}\right) - \frac{1}{4}\times 2\times I_0\), so \(I_2 = \frac{\pi}{8} - \frac{1}{4}\) | M1 A1 | Finding \(I_2\) in terms of \(\pi\); cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_4 = \left(\frac{\pi}{4}\right)^3 - \frac{1}{4}\times 4\times 3 I_2 = \frac{\pi^3}{64} - 3\left(\frac{\pi}{8} - \frac{1}{4}\right) = \frac{1}{64}(\pi^3 - 24\pi + 48)\) | M1 A1cso | Finding \(I_4\) in terms of \(I_2\) then in terms of \(\pi\); cso |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_n = \left[x^n(-\frac{1}{2}\cos 2x)\right]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} -\frac{1}{2}nx^{n-1}\cos 2x\, dx$ | M1 A1 | Use of integration by parts, integrating $\sin 2x$, differentiating $x^n$; cao |
| $I_n = \left\langle\left[x^n(-\frac{1}{2}\cos 2x)\right]_0^{\frac{\pi}{4}}\right\rangle + \left[\frac{1}{4}nx^{n-1}\sin 2x\right]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}}\frac{1}{4}n(n-1)x^{n-2}\sin 2x\, dx$ | M1 A1 | Second application of integration by parts, integrating $\cos 2x$, differentiating $x^{n-1}$; cao |
| $I_n = \frac{1}{4}n\left(\frac{\pi}{4}\right)^{n-1} - \frac{1}{4}n(n-1)I_{n-2}$ | A1cso | cso including correct use of $\frac{\pi}{4}$ and $0$ as limits |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_0 = \int_0^{\frac{\pi}{4}}\sin 2x\, dx = \left[-\frac{1}{2}\cos 2x\right]_0^{\frac{\pi}{4}} = \frac{1}{2}$ | M1 A1 | Integrating to find $I_0$; cao (accept $I_0 = \frac{1}{2}$) |
| $I_2 = \frac{1}{4}\times 2\times\left(\frac{\pi}{4}\right) - \frac{1}{4}\times 2\times I_0$, so $I_2 = \frac{\pi}{8} - \frac{1}{4}$ | M1 A1 | Finding $I_2$ in terms of $\pi$; cao |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_4 = \left(\frac{\pi}{4}\right)^3 - \frac{1}{4}\times 4\times 3 I_2 = \frac{\pi^3}{64} - 3\left(\frac{\pi}{8} - \frac{1}{4}\right) = \frac{1}{64}(\pi^3 - 24\pi + 48)$ | M1 A1cso | Finding $I_4$ in terms of $I_2$ then in terms of $\pi$; cso |
---4.
$$I _ { n } = \int _ { 0 } ^ { \frac { \pi } { 4 } } x ^ { n } \sin 2 x \mathrm {~d} x , \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Prove that, for $n \geqslant 2$,
$$I _ { n } = \frac { 1 } { 4 } n \left( \frac { \pi } { 4 } \right) ^ { n - 1 } - \frac { 1 } { 4 } n ( n - 1 ) I _ { n - 2 }$$
\item Find the exact value of $I _ { 2 }$
\item Show that $I _ { 4 } = \frac { 1 } { 64 } \left( \pi ^ { 3 } - 24 \pi + 48 \right)$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2012 Q4 [11]}}