| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Standard integral of 1/√(a²-x²) |
| Difficulty | Standard +0.8 This FP3 question requires completing the square to recognize the arcsin form, then applying inverse trig integration (part i), followed by manipulating hyperbolic identities and using substitution to integrate a rational function (part ii). While these are standard Further Maths techniques, the multi-step nature, algebraic manipulation required, and combination of inverse trig and hyperbolic topics makes this moderately challenging, above average difficulty but not requiring novel insight. |
| Spec | 1.08h Integration by substitution4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(15 + 2x - x^2 = 16 - (x-1)^2\) | B1 | Correct completion of the square; allow e.g. \(15+2x-x^2 = -[(x-1)^2 - 16]\); allow \(4^2\) for 16 |
| \(\int \frac{1}{\sqrt{16-(x-1)^2}}\,dx = \arcsin\left(\frac{x-1}{4}\right)\) | M1A1 | M1: \(k\arcsin(f(x))\); A1: correct integration |
| \(\left[\arcsin\left(\frac{x-1}{4}\right)\right]_3^5 = \arcsin 1 - \arcsin\frac{1}{2}\) | dM1 | Correct use of correct limits |
| \(= \frac{\pi}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(15 + 2x - x^2 = 16-(x-1)^2\) | B1 | Correct completion of square |
| \(x-1=4\sin\theta \Rightarrow \int\frac{1}{\sqrt{16-(x-1)^2}}\,dx = \int\frac{1}{\sqrt{16-(4\sin\theta)^2}}\,4\cos\theta\,d\theta = \int d\theta = \theta\) | M1A1 | M1: full substitution leading to \(k\theta\) or \(k\times\) their variable; A1: correct integration |
| \([\theta]_{\pi/6}^{\pi/2} = \frac{\pi}{2} - \frac{\pi}{6}\) | dM1 | Correct use of correct limits |
| \(= \frac{\pi}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(5\cosh x - 4\sinh x = 5\left(\frac{e^x+e^{-x}}{2}\right) - 4\left(\frac{e^x-e^{-x}}{2}\right)\) | B1 | Substitutes correct exponential forms |
| \(= \frac{e^x + 9e^{-x}}{2}\) or \(\frac{e^x}{2} + \frac{9e^{-x}}{2}\) | M1 | Expands and collects terms in \(e^x\) and \(e^{-x}\) |
| \(= \frac{e^{2x}+9}{2e^x}\)* | A1* | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = e^x \Rightarrow \frac{du}{dx} = e^x\) | B1 | Correct derivative. Allow equivalents e.g. \(\frac{dx}{du} = \frac{1}{u}\), \(du = e^x dx\) |
| \(\int \frac{2e^x}{e^{2x}+9}\,dx = \int \frac{2u}{u^2+9} \cdot \frac{du}{u}\) | M1 | Complete substitution into \(\int \frac{2e^x}{e^{2x}+9}\,dx\). Condone omission of \(du\) provided substitution otherwise complete. May be implied by \(\int \frac{2}{u^2+9}\,du\) |
| \(= \frac{2}{3}\arctan\!\left(\frac{u}{3}\right)(+c)\) | dM1 | \(k\arctan(\text{f}(u))\) only. Dependent on first method mark |
| \(= \frac{2}{3}\arctan\!\left(\frac{e^x}{3}\right)(+c)\) | A1 | Cao (+c not required) |
# Question 4(i)
| Working | Mark | Notes |
|---------|------|-------|
| $15 + 2x - x^2 = 16 - (x-1)^2$ | B1 | Correct completion of the square; allow e.g. $15+2x-x^2 = -[(x-1)^2 - 16]$; allow $4^2$ for 16 |
| $\int \frac{1}{\sqrt{16-(x-1)^2}}\,dx = \arcsin\left(\frac{x-1}{4}\right)$ | M1A1 | M1: $k\arcsin(f(x))$; A1: correct integration |
| $\left[\arcsin\left(\frac{x-1}{4}\right)\right]_3^5 = \arcsin 1 - \arcsin\frac{1}{2}$ | dM1 | Correct use of correct limits |
| $= \frac{\pi}{3}$ | A1 | |
**By substitution 1** ($x-1 = 4\sin\theta$):
| Working | Mark | Notes |
|---------|------|-------|
| $15 + 2x - x^2 = 16-(x-1)^2$ | B1 | Correct completion of square |
| $x-1=4\sin\theta \Rightarrow \int\frac{1}{\sqrt{16-(x-1)^2}}\,dx = \int\frac{1}{\sqrt{16-(4\sin\theta)^2}}\,4\cos\theta\,d\theta = \int d\theta = \theta$ | M1A1 | M1: full substitution leading to $k\theta$ or $k\times$ their variable; A1: correct integration |
| $[\theta]_{\pi/6}^{\pi/2} = \frac{\pi}{2} - \frac{\pi}{6}$ | dM1 | Correct use of correct limits |
| $= \frac{\pi}{3}$ | A1 | |
---
# Question 4(ii)(a)
| Working | Mark | Notes |
|---------|------|-------|
| $5\cosh x - 4\sinh x = 5\left(\frac{e^x+e^{-x}}{2}\right) - 4\left(\frac{e^x-e^{-x}}{2}\right)$ | B1 | Substitutes correct exponential forms |
| $= \frac{e^x + 9e^{-x}}{2}$ or $\frac{e^x}{2} + \frac{9e^{-x}}{2}$ | M1 | Expands and collects terms in $e^x$ and $e^{-x}$ |
| $= \frac{e^{2x}+9}{2e^x}$* | A1* | Correct completion with no errors |
# Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = e^x \Rightarrow \frac{du}{dx} = e^x$ | B1 | Correct derivative. Allow equivalents e.g. $\frac{dx}{du} = \frac{1}{u}$, $du = e^x dx$ |
| $\int \frac{2e^x}{e^{2x}+9}\,dx = \int \frac{2u}{u^2+9} \cdot \frac{du}{u}$ | M1 | Complete substitution into $\int \frac{2e^x}{e^{2x}+9}\,dx$. Condone omission of $du$ provided substitution otherwise complete. May be implied by $\int \frac{2}{u^2+9}\,du$ |
| $= \frac{2}{3}\arctan\!\left(\frac{u}{3}\right)(+c)$ | dM1 | $k\arctan(\text{f}(u))$ only. **Dependent on first method mark** |
| $= \frac{2}{3}\arctan\!\left(\frac{e^x}{3}\right)(+c)$ | A1 | Cao (+c not required) |
**(4 marks)**
---4. (i) Find, without using a calculator,
$$\int _ { 3 } ^ { 5 } \frac { 1 } { \sqrt { 15 + 2 x - x ^ { 2 } } } d x$$
giving your answer as a multiple of $\pi$.\\
(ii)
\begin{enumerate}[label=(\alph*)]
\item Show that
$$5 \cosh x - 4 \sinh x = \frac { \mathrm { e } ^ { 2 x } + 9 } { 2 \mathrm { e } ^ { x } }$$
\item Hence, using the substitution $u = e ^ { x }$ or otherwise, find
$$\int \frac { 1 } { 5 \cosh x - 4 \sinh x } d x$$
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2016 Q4 [12]}}