| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2015 |
| 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 F3 reduction formula question requiring integration by parts to derive the recurrence relation, then recursive application to evaluate a definite integral. While the technique is standard for this module, it demands careful algebraic manipulation with hyperbolic identities, multiple applications of the formula, and exact evaluation at ln 2. The hyperbolic context and multi-step nature elevate it above average A-level difficulty. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \cosh^n x\, dx = \int \cosh x \cosh^{n-1} x\, dx\) | B1 | Writes \(\cosh^n x\) as \(\cosh x \cosh^{n-1} x\) |
| \(\int \cosh x \cosh^{n-1} x\, dx = \sinh x \cosh^{n-1} x - \int (n-1)\cosh^{n-2} x \sinh^2 x\, dx\) | M1A1 | M1: Parts in correct direction (sign errors only allowed). A1: Correct expression |
| \(= \sinh x \cosh^{n-1} x - (n-1)\int \cosh^{n-2} x(\cosh^2 x - 1)\, dx\) | dM1 | Writes \(\sinh^2 x\) as \(\cosh^2 x - 1\) |
| \(= \sinh x \cosh^{n-1} x - (n-1)\int \cosh^n x\, dx + (n-1)\int \cosh^{n-2} x\, dx\) | ||
| \(= \sinh x \cosh^{n-1} x - (n-1)I_n + (n-1)I_{n-2}\) | ddM1 | Substitutes for \(I_n\) and \(I_{n-2}\) |
| \(nI_n = \sinh x \cosh^{n-1} x + (n-1)I_{n-2}\) * | A1* | Correct answer with at least one intermediate line of working and no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \cosh^n x\, dx = \int \cosh^2 x \cosh^{n-2} x\, dx = \int(1+\sinh^2 x)\cosh^{n-2} x\, dx\) | B1 | Writes \(\cosh^n x\) as \((1+\sinh^2 x)\cosh^{n-2} x\) |
| \(\int \sinh x \sinh x \cosh^{n-2} x\, dx = \frac{1}{n-1}\sinh x \cosh^{n-1} x - \frac{1}{n-1}\int \cosh^n x\, dx\) | M1A1 | M1: Parts in correct direction. A1: Correct expression |
| \(\int \cosh^n x\, dx = \int \cosh^{n-2} x\, dx + \frac{1}{n-1}\sinh x \cosh^{n-1} x - \frac{1}{n-1}\int \cosh^n x\, dx\) | dM1 | Adds \(I_{n-2}\) to their integration by parts |
| \(I_n = I_{n-2} + \frac{1}{n-1}\sinh x \cosh^{n-1} x - \frac{1}{n-1}I_n\) | ddM1 | Substitutes for \(I_n\) and \(I_{n-2}\) |
| \(nI_n = \sinh x \cosh^{n-1} x + (n-1)I_{n-2}\) * | A1* | Correct answer with at least one intermediate line and no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_5 = \frac{1}{5}\left[\sinh x \cosh^4 x\right]_0^{\ln 2} + \frac{4}{5}I_3\) or \(I_3 = \frac{1}{3}\left[\sinh x \cosh^2 x\right]_0^{\ln 2} + \frac{2}{3}I_1\) | M1 | One application of reduction formula (\(I_5\) in terms of \(I_3\) or \(I_3\) in terms of \(I_1\)) |
| \(I_5 = \frac{1}{5}\left[\sinh x \cosh^4 x\right]_0^{\ln 2} + \frac{4}{5}I_3\) and \(I_3 = \frac{1}{3}\left[\sinh x \cosh^2 x\right]_0^{\ln 2} + \frac{2}{3}I_1\) | M1 | Second application of reduction formula (\(I_5\) in terms of \(I_3\) and \(I_3\) in terms of \(I_1\)) |
| \(I_1 = \int_0^{\ln 2}\cosh x\, dx = \left[\sinh x\right]_0^{\ln 2} = \frac{3}{4}\) | B1 | \(I_1 = \frac{3}{4}\) |
| \(= \frac{1}{5}\cdot\frac{3}{4}\cdot\left(\frac{5}{4}\right)^4 + \frac{4}{5}\left(\frac{1}{3}\cdot\frac{3}{4}\cdot\left(\frac{5}{4}\right)^2+\frac{2}{3}\cdot\frac{3}{4}\right)\) | ||
| \(\int_0^{\ln 2}\cosh^5 x\, dx = \frac{5523}{5120}\) | A1 | Must be exact |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \cosh^n x\, dx = \int \cosh x \cosh^{n-1} x\, dx$ | B1 | Writes $\cosh^n x$ as $\cosh x \cosh^{n-1} x$ |
| $\int \cosh x \cosh^{n-1} x\, dx = \sinh x \cosh^{n-1} x - \int (n-1)\cosh^{n-2} x \sinh^2 x\, dx$ | M1A1 | M1: Parts in correct direction (sign errors only allowed). A1: Correct expression |
| $= \sinh x \cosh^{n-1} x - (n-1)\int \cosh^{n-2} x(\cosh^2 x - 1)\, dx$ | dM1 | Writes $\sinh^2 x$ as $\cosh^2 x - 1$ |
| $= \sinh x \cosh^{n-1} x - (n-1)\int \cosh^n x\, dx + (n-1)\int \cosh^{n-2} x\, dx$ | | |
| $= \sinh x \cosh^{n-1} x - (n-1)I_n + (n-1)I_{n-2}$ | ddM1 | Substitutes for $I_n$ and $I_{n-2}$ |
| $nI_n = \sinh x \cosh^{n-1} x + (n-1)I_{n-2}$ * | A1* | Correct answer with at least one intermediate line of working and no errors seen |
**Alternative Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \cosh^n x\, dx = \int \cosh^2 x \cosh^{n-2} x\, dx = \int(1+\sinh^2 x)\cosh^{n-2} x\, dx$ | B1 | Writes $\cosh^n x$ as $(1+\sinh^2 x)\cosh^{n-2} x$ |
| $\int \sinh x \sinh x \cosh^{n-2} x\, dx = \frac{1}{n-1}\sinh x \cosh^{n-1} x - \frac{1}{n-1}\int \cosh^n x\, dx$ | M1A1 | M1: Parts in correct direction. A1: Correct expression |
| $\int \cosh^n x\, dx = \int \cosh^{n-2} x\, dx + \frac{1}{n-1}\sinh x \cosh^{n-1} x - \frac{1}{n-1}\int \cosh^n x\, dx$ | dM1 | Adds $I_{n-2}$ to their integration by parts |
| $I_n = I_{n-2} + \frac{1}{n-1}\sinh x \cosh^{n-1} x - \frac{1}{n-1}I_n$ | ddM1 | Substitutes for $I_n$ and $I_{n-2}$ |
| $nI_n = \sinh x \cosh^{n-1} x + (n-1)I_{n-2}$ * | A1* | Correct answer with at least one intermediate line and no errors |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_5 = \frac{1}{5}\left[\sinh x \cosh^4 x\right]_0^{\ln 2} + \frac{4}{5}I_3$ or $I_3 = \frac{1}{3}\left[\sinh x \cosh^2 x\right]_0^{\ln 2} + \frac{2}{3}I_1$ | M1 | One application of reduction formula ($I_5$ in terms of $I_3$ **or** $I_3$ in terms of $I_1$) |
| $I_5 = \frac{1}{5}\left[\sinh x \cosh^4 x\right]_0^{\ln 2} + \frac{4}{5}I_3$ **and** $I_3 = \frac{1}{3}\left[\sinh x \cosh^2 x\right]_0^{\ln 2} + \frac{2}{3}I_1$ | M1 | Second application of reduction formula ($I_5$ in terms of $I_3$ **and** $I_3$ in terms of $I_1$) |
| $I_1 = \int_0^{\ln 2}\cosh x\, dx = \left[\sinh x\right]_0^{\ln 2} = \frac{3}{4}$ | B1 | $I_1 = \frac{3}{4}$ |
| $= \frac{1}{5}\cdot\frac{3}{4}\cdot\left(\frac{5}{4}\right)^4 + \frac{4}{5}\left(\frac{1}{3}\cdot\frac{3}{4}\cdot\left(\frac{5}{4}\right)^2+\frac{2}{3}\cdot\frac{3}{4}\right)$ | | |
| $\int_0^{\ln 2}\cosh^5 x\, dx = \frac{5523}{5120}$ | A1 | Must be exact |
**Total: 10 marks**4.
$$I _ { n } = \int \cosh ^ { n } x \mathrm {~d} x , \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geqslant 2$
$$n I _ { n } = \sinh x \cosh ^ { n - 1 } x + ( n - 1 ) I _ { n - 2 }$$
\item Hence find the exact value of
$$\int _ { 0 } ^ { \ln 2 } \cosh ^ { 5 } x \mathrm {~d} x$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2015 Q4 [10]}}