| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Trigonometric power reduction |
| Difficulty | Challenging +1.2 This is a standard FP3 reduction formula question with typical structure: prove the formula using integration by parts, apply it to definite integrals to derive a product formula, then use it for a related integral. While it requires multiple techniques (integration by parts, recursive application, pattern recognition), these are well-practiced procedures in Further Maths. Part (c) requires the insight to rewrite cos²x = 1-sin²x, but this is a common trick. The question is harder than average A-level due to being Further Maths content, but it's a textbook example of its type rather than requiring novel problem-solving. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^28.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \int \sin^{n-1}x \sin x\, dx\) | M1 | Split into \(\sin^{n-1}x\) and \(\sin x\) |
| \(I_n = \sin^{n-1}x(-\cos x) + \int(n-1)\sin^{n-2}x\cos^2 x\, dx\) | dM1 | Integration by parts in right direction. Dependent on first M. |
| \(I_n = -\sin^{n-1}x\cos x + (n-1)(I_{n-2} - I_n)\) | A1 | Obtains \(I_n\) correctly in terms of \(I_{n-2}\) and \(I_n\) |
| \(I_n = \frac{1}{n}\left(-\sin^{n-1}x\cos x + (n-1)I_{n-2}\right)\)* | A1* | Printed answer obtained with at least one intermediate step and no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \frac{1}{n}\left[-\sin^{n-1}x\cos x\right]_0^{\pi/2} + (n-1)I_{n-2}\) | M1 | Use part (a) with limits |
| \(I_n = \frac{n-1}{n}I_{n-2}\) | A1 | Sight of expression could score M1A1 |
| \(n\) odd, \(I_1 = \int_0^{\pi/2}\sin x\, dx = [-\cos x]_0^{\pi/2} = 1\) | An attempt at \(I_1\) must be seen before any more marks awarded | |
| \(I_n = \frac{(n-1)}{n}I_{n-2} = \frac{(n-1)(n-3)}{n(n-2)}I_{n-4} = \ldots\) | M1 | Attempts \(I_1\) and at least 2 fractions in terms of \(n\) |
| \(I_n = \frac{(n-1)(n-3)\ldots 6\cdot4\cdot2}{n(n-2)(n-4)\ldots7\cdot5\cdot3}\) | A1 | cso. May be awarded for 'extra' brackets top and bottom provided all previous marks are scored |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_0^{\pi/2}\sin^5 x\cos^2 x\, dx = \int_0^{\pi/2}\sin^5 x(1-\sin^2 x)\, dx\) | M1 | Uses \(\cos^2 x = 1 - \sin^2 x\) |
| \(= I_5 - I_7 = \frac{4\times2}{5\times3} - \frac{6\times4\times2}{7\times5\times3}\) | A1 | Correct numerical expression |
| \(= \frac{8}{105}\) | A1 | cao (accept awrt 0.0761) |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int \sin^{n-1}x \sin x\, dx$ | M1 | Split into $\sin^{n-1}x$ and $\sin x$ |
| $I_n = \sin^{n-1}x(-\cos x) + \int(n-1)\sin^{n-2}x\cos^2 x\, dx$ | dM1 | Integration by parts in right direction. Dependent on first M. |
| $I_n = -\sin^{n-1}x\cos x + (n-1)(I_{n-2} - I_n)$ | A1 | Obtains $I_n$ correctly in terms of $I_{n-2}$ and $I_n$ |
| $I_n = \frac{1}{n}\left(-\sin^{n-1}x\cos x + (n-1)I_{n-2}\right)$* | A1* | Printed answer obtained with at least one intermediate step and no errors seen |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \frac{1}{n}\left[-\sin^{n-1}x\cos x\right]_0^{\pi/2} + (n-1)I_{n-2}$ | M1 | Use part (a) with limits |
| $I_n = \frac{n-1}{n}I_{n-2}$ | A1 | Sight of expression could score M1A1 |
| $n$ odd, $I_1 = \int_0^{\pi/2}\sin x\, dx = [-\cos x]_0^{\pi/2} = 1$ | | An attempt at $I_1$ must be seen before any more marks awarded |
| $I_n = \frac{(n-1)}{n}I_{n-2} = \frac{(n-1)(n-3)}{n(n-2)}I_{n-4} = \ldots$ | M1 | Attempts $I_1$ and at least 2 fractions in terms of $n$ |
| $I_n = \frac{(n-1)(n-3)\ldots 6\cdot4\cdot2}{n(n-2)(n-4)\ldots7\cdot5\cdot3}$** | A1** | cso. May be awarded for 'extra' brackets top and bottom provided all previous marks are scored |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^{\pi/2}\sin^5 x\cos^2 x\, dx = \int_0^{\pi/2}\sin^5 x(1-\sin^2 x)\, dx$ | M1 | Uses $\cos^2 x = 1 - \sin^2 x$ |
| $= I_5 - I_7 = \frac{4\times2}{5\times3} - \frac{6\times4\times2}{7\times5\times3}$ | A1 | Correct numerical expression |
| $= \frac{8}{105}$ | A1 | cao (accept awrt 0.0761) |
---7.
$$I _ { n } = \int \sin ^ { n } x \mathrm {~d} x , \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Prove that for $n \geqslant 2$
$$I _ { n } = \frac { 1 } { n } \left( - \sin ^ { n - 1 } x \cos x + ( n - 1 ) I _ { n - 2 } \right)$$
Given that $n$ is an odd number, $n \geqslant 3$
\item show that
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { n } x \mathrm {~d} x = \frac { ( n - 1 ) ( n - 3 ) \ldots 6.4 .2 } { n ( n - 2 ) ( n - 4 ) \ldots 7.5 .3 }$$
\item Hence find $\int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { 5 } x \cos ^ { 2 } x d x$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2015 Q7 [11]}}