| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Hyperbolic function reduction |
| Difficulty | Challenging +1.8 This is a Further Maths reduction formula question requiring integration by parts with hyperbolic functions, evaluation at specific limits involving ln 2, and recursive application. While the technique is standard for FP3, it demands careful algebraic manipulation and working with exponential forms of cosh, placing it well above average difficulty but within expected Further Maths scope. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.08c Improper integrals: infinite limits or discontinuous integrands8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \int \cosh^{n-1}x \cosh x \, dx = \sinh x \cosh^{n-1}x - \int (n-1)\cosh^{n-2}x \sinh^2 x \, dx\) | M1A1 | M1: Integration by parts in correct direction. Must be of form \(\pm\sinh x\cosh^{n-1}x \pm k\int \cosh^{n-2}x\sinh^2 x\,dx\). A1: Correct expression |
| \(= \sinh x\cosh^{n-1}x - \int(n-1)\cosh^{n-2}x(\cosh^2 x - 1)\,dx\) | dM1 | Replaces \(\sinh^2 x\) with \(\pm\cosh^2 x \pm 1\) on the integration part. Dependent on first method mark |
| \(= \sinh x\cosh^{n-1}x - (n-1)\int\cosh^n x\,dx + (n-1)\int\cosh^{n-2}x\,dx\) | ddM1 | Introduces \(I_n\) and \(I_{n-2}\). Dependent on both previous method marks |
| \(\left[\sinh x\cosh^{n-1}x\right]_0^{\ln 2} = \sinh(\ln 2)\cosh^{n-1}(\ln 2)(−0) = \left(\frac{3}{4}\right)\left(\frac{5}{4}\right)^{n-1}\) | M1 | Use of given limits on \(\sinh x\cosh^{n-1}x\). Note \(\cosh(\ln 2)=\frac{5}{4}\), \(\sinh(\ln 2)=\frac{3}{4}\) |
| \(I_n = \dfrac{3\times 5^{n-1}}{n\times 4^n} + \dfrac{(n-1)}{n}I_{n-2}\) * | A1* | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \int\cosh^{n-2}x\cosh^2 x\,dx = \int\cosh^{n-2}x\,dx + \int\cosh^{n-2}x\sinh^2 x\,dx\) | M1 | Writes \(\cosh^n x\) as \(\cosh^{n-2}x\cosh^2 x\) and uses \(\sinh^2 x = \pm\cosh^2 x \pm 1\) |
| \(\int\cosh^{n-2}x\sinh^2 x\,dx = \left[\dfrac{\sinh x\cosh^{n-1}x}{n-1}\right] - \dfrac{1}{n-1}\int\cosh^n x\,dx\) | dM1A1 | M1: Integration by parts in correct direction, of form \(p\sinh x\cosh^{n-1}x \pm q\int\cosh^n x\,dx\). A1: Correct expression |
| \((n-1)I_n = (n-1)I_{n-2} + \left[\sinh x\cosh^{n-1}x\right] - I_n\) | ddM1 | Introduces \(I_n\) and \(I_{n-2}\). Dependent on both previous method marks |
| \(\left[\sinh x\cosh^{n-1}x\right]_0^{\ln 2} = \left(\frac{3}{4}\right)\left(\frac{5}{4}\right)^{n-1}\) | M1 | Use of given limits. Note \(\cosh(\ln 2)=\frac{5}{4}\), \(\sinh(\ln 2)=\frac{3}{4}\) |
| \(I_n = \dfrac{3\times 5^{n-1}}{n\times 4^n} + \dfrac{(n-1)}{n}I_{n-2}\) * | A1* | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_4 = \dfrac{3\times 5^3}{4\times 4^4} + \dfrac{3}{4}I_2\) | M1 | Correct first application of their or the given reduction formula |
| \(= \dfrac{3\times 5^3}{4\times 4^4} + \dfrac{3}{4}\left(\dfrac{3\times 5}{2\times 4^2} + \dfrac{1}{2}I_0\right)\) | M1 | Correct second application consistent with first application, to obtain \(I_4\) in terms of \(I_0\) |
| \(I_0 = \ln 2\) | B1 | |
| \(I_4 = \dfrac{735}{1024} + \dfrac{3}{8}\ln 2\) | A1 | cao. Allow equivalent exact forms, fractions must be collected |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_4 = \dfrac{3\times 5^3}{4\times 4^4} + \dfrac{3}{4}I_2\) | M1 | Correct application of their reduction formula |
| \(I_2 = \int_0^{\ln 2}\cosh^2 x\,dx = \int_0^{\ln 2}\left(\dfrac{1}{2}+\dfrac{1}{2}\cosh 2x\right)dx\) | ||
| \(\int\left(\dfrac{1}{2}+\dfrac{1}{2}\cosh 2x\right)dx = \dfrac{x}{2}+\dfrac{1}{4}\sinh 2x\) | B1 | Correct integration |
| \(I_2 = \left[\dfrac{x}{2}+\dfrac{1}{4}\sinh 2x\right]_0^{\ln 2} = \dfrac{1}{2}\ln 2 + \dfrac{15}{32}\) | M1 | Correct use of limits on expression of form \(\alpha x + \beta\sinh 2x\) |
| \(I_4 = \dfrac{735}{1024} + \dfrac{3}{8}\ln 2\) | A1 | cao. Allow equivalent exact forms, fractions must be collected |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_0^{\ln 2}\left(\dfrac{1}{4}+\dfrac{1}{2}\cosh 2x+\dfrac{1}{4}\cosh^2 2x\right)dx\) | B1 | \(\cosh^4 x = \dfrac{1}{4}+\dfrac{1}{2}\cosh 2x+\dfrac{1}{4}\cosh^2 2x\) |
| \(\dfrac{1}{4}\int_0^{\ln 2}\left(1+2\cosh 2x+\dfrac{1}{2}(1+\cosh 4x)\right)dx\) | M1 | \(\cosh^2 2x = \pm\dfrac{1}{2}\pm\dfrac{1}{2}\cosh 4x\) and attempt to integrate |
| \(\dfrac{1}{4}\left[\dfrac{3x}{2}+\sinh 2x+\dfrac{1}{8}\sinh 4x\right]_0^{\ln 2}\) | M1 | Correct use of correct limits |
| \(I_4 = \dfrac{735}{1024}+\dfrac{3}{8}\ln 2\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_4 = \int_0^{\ln 2}\left(\dfrac{e^x+e^{-x}}{2}\right)^4 dx = \dfrac{1}{16}\int_0^{\ln 2}(e^{4x}+4e^{2x}+6+4e^{-2x}+e^{-4x})\,dx\) | B1 | Correct expansion |
| \(= \dfrac{1}{16}\left[\dfrac{e^{4x}}{4}+2e^{2x}+6x-2e^{-2x}-\dfrac{e^{-4x}}{4}\right]_0^{\ln 2}\) | M1 | Attempts to integrate their expansion |
| \(= \dfrac{1}{16}\left[\left(4+8+6\ln 2-\dfrac{1}{2}-\dfrac{1}{64}\right)-(0)\right]\) | M1 | Correct use of correct limits |
| \(I_4 = \dfrac{735}{1024}+\dfrac{3}{8}\ln 2\) | A1 | cao |
# Question 7:
## Part (a) – Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int \cosh^{n-1}x \cosh x \, dx = \sinh x \cosh^{n-1}x - \int (n-1)\cosh^{n-2}x \sinh^2 x \, dx$ | M1A1 | M1: Integration by parts in correct direction. Must be of form $\pm\sinh x\cosh^{n-1}x \pm k\int \cosh^{n-2}x\sinh^2 x\,dx$. A1: Correct expression |
| $= \sinh x\cosh^{n-1}x - \int(n-1)\cosh^{n-2}x(\cosh^2 x - 1)\,dx$ | dM1 | Replaces $\sinh^2 x$ with $\pm\cosh^2 x \pm 1$ on the integration part. Dependent on first method mark |
| $= \sinh x\cosh^{n-1}x - (n-1)\int\cosh^n x\,dx + (n-1)\int\cosh^{n-2}x\,dx$ | ddM1 | Introduces $I_n$ and $I_{n-2}$. Dependent on both previous method marks |
| $\left[\sinh x\cosh^{n-1}x\right]_0^{\ln 2} = \sinh(\ln 2)\cosh^{n-1}(\ln 2)(−0) = \left(\frac{3}{4}\right)\left(\frac{5}{4}\right)^{n-1}$ | M1 | Use of given limits on $\sinh x\cosh^{n-1}x$. Note $\cosh(\ln 2)=\frac{5}{4}$, $\sinh(\ln 2)=\frac{3}{4}$ |
| $I_n = \dfrac{3\times 5^{n-1}}{n\times 4^n} + \dfrac{(n-1)}{n}I_{n-2}$ * | A1* | cao |
## Part (a) – Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int\cosh^{n-2}x\cosh^2 x\,dx = \int\cosh^{n-2}x\,dx + \int\cosh^{n-2}x\sinh^2 x\,dx$ | M1 | Writes $\cosh^n x$ as $\cosh^{n-2}x\cosh^2 x$ and uses $\sinh^2 x = \pm\cosh^2 x \pm 1$ |
| $\int\cosh^{n-2}x\sinh^2 x\,dx = \left[\dfrac{\sinh x\cosh^{n-1}x}{n-1}\right] - \dfrac{1}{n-1}\int\cosh^n x\,dx$ | dM1A1 | M1: Integration by parts in correct direction, of form $p\sinh x\cosh^{n-1}x \pm q\int\cosh^n x\,dx$. A1: Correct expression |
| $(n-1)I_n = (n-1)I_{n-2} + \left[\sinh x\cosh^{n-1}x\right] - I_n$ | ddM1 | Introduces $I_n$ and $I_{n-2}$. Dependent on both previous method marks |
| $\left[\sinh x\cosh^{n-1}x\right]_0^{\ln 2} = \left(\frac{3}{4}\right)\left(\frac{5}{4}\right)^{n-1}$ | M1 | Use of given limits. Note $\cosh(\ln 2)=\frac{5}{4}$, $\sinh(\ln 2)=\frac{3}{4}$ |
| $I_n = \dfrac{3\times 5^{n-1}}{n\times 4^n} + \dfrac{(n-1)}{n}I_{n-2}$ * | A1* | cao |
---
## Part (b) – Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_4 = \dfrac{3\times 5^3}{4\times 4^4} + \dfrac{3}{4}I_2$ | M1 | Correct first application of their or the given reduction formula |
| $= \dfrac{3\times 5^3}{4\times 4^4} + \dfrac{3}{4}\left(\dfrac{3\times 5}{2\times 4^2} + \dfrac{1}{2}I_0\right)$ | M1 | Correct second application consistent with first application, to obtain $I_4$ in terms of $I_0$ |
| $I_0 = \ln 2$ | B1 | |
| $I_4 = \dfrac{735}{1024} + \dfrac{3}{8}\ln 2$ | A1 | cao. Allow equivalent exact forms, fractions must be collected |
## Part (b) – Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_4 = \dfrac{3\times 5^3}{4\times 4^4} + \dfrac{3}{4}I_2$ | M1 | Correct application of their reduction formula |
| $I_2 = \int_0^{\ln 2}\cosh^2 x\,dx = \int_0^{\ln 2}\left(\dfrac{1}{2}+\dfrac{1}{2}\cosh 2x\right)dx$ | | |
| $\int\left(\dfrac{1}{2}+\dfrac{1}{2}\cosh 2x\right)dx = \dfrac{x}{2}+\dfrac{1}{4}\sinh 2x$ | B1 | Correct integration |
| $I_2 = \left[\dfrac{x}{2}+\dfrac{1}{4}\sinh 2x\right]_0^{\ln 2} = \dfrac{1}{2}\ln 2 + \dfrac{15}{32}$ | M1 | Correct use of limits on expression of form $\alpha x + \beta\sinh 2x$ |
| $I_4 = \dfrac{735}{1024} + \dfrac{3}{8}\ln 2$ | A1 | cao. Allow equivalent exact forms, fractions must be collected |
## Part (b) – Way 3
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^{\ln 2}\left(\dfrac{1}{4}+\dfrac{1}{2}\cosh 2x+\dfrac{1}{4}\cosh^2 2x\right)dx$ | B1 | $\cosh^4 x = \dfrac{1}{4}+\dfrac{1}{2}\cosh 2x+\dfrac{1}{4}\cosh^2 2x$ |
| $\dfrac{1}{4}\int_0^{\ln 2}\left(1+2\cosh 2x+\dfrac{1}{2}(1+\cosh 4x)\right)dx$ | M1 | $\cosh^2 2x = \pm\dfrac{1}{2}\pm\dfrac{1}{2}\cosh 4x$ and attempt to integrate |
| $\dfrac{1}{4}\left[\dfrac{3x}{2}+\sinh 2x+\dfrac{1}{8}\sinh 4x\right]_0^{\ln 2}$ | M1 | Correct use of correct limits |
| $I_4 = \dfrac{735}{1024}+\dfrac{3}{8}\ln 2$ | A1 | cao |
## Part (b) – Way 4
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_4 = \int_0^{\ln 2}\left(\dfrac{e^x+e^{-x}}{2}\right)^4 dx = \dfrac{1}{16}\int_0^{\ln 2}(e^{4x}+4e^{2x}+6+4e^{-2x}+e^{-4x})\,dx$ | B1 | Correct expansion |
| $= \dfrac{1}{16}\left[\dfrac{e^{4x}}{4}+2e^{2x}+6x-2e^{-2x}-\dfrac{e^{-4x}}{4}\right]_0^{\ln 2}$ | M1 | Attempts to integrate their expansion |
| $= \dfrac{1}{16}\left[\left(4+8+6\ln 2-\dfrac{1}{2}-\dfrac{1}{64}\right)-(0)\right]$ | M1 | Correct use of correct limits |
| $I_4 = \dfrac{735}{1024}+\dfrac{3}{8}\ln 2$ | A1 | cao |7.
$$I _ { n } = \int _ { 0 } ^ { \ln 2 } \cosh ^ { n } x \mathrm {~d} x , \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geqslant 2$,
$$I _ { n } = \frac { 3 a ^ { n - 1 } } { n b ^ { n } } + \frac { n - 1 } { n } I _ { n - 2 }$$
where $a$ and $b$ are integers to be found.
\item Hence, or otherwise, find the exact value of
$$\int _ { 0 } ^ { \ln 2 } \cosh ^ { 4 } x \mathrm {~d} x$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2017 Q7 [10]}}