| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Completing square then standard inverse trig |
| Difficulty | Challenging +1.2 This is a Further Maths FP3 question requiring completing the square and applying standard inverse trig/hyperbolic integral formulas. While the topic is advanced, the execution is methodical: complete the square to get (x-5)²+9 and (x-5)²+9, then apply arctan and arsinh formulas with substitution. The algebraic manipulation and evaluation at limits requires care but follows a well-defined procedure with no novel problem-solving insight needed. |
| Spec | 1.08h Integration by substitution4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Uses limits 3 and 0 to give \(\frac{\pi}{12}\) or uses limits 8 and 5 to give \(\frac{\pi}{12}\) | B1, M1 A1, DM1 A1 | Mark can be earned in either part (a) or (b). CAO for the initial rewriting. |
| Answer | Marks | Guidance |
|---|---|---|
| Uses limits 5 and 8 to give \(\ln(1+\sqrt{2})\) | M1 A1, DM1 A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Uses limits 3 and 0 and ln expression to give \(\ln(1+\sqrt{2})\) | M1 A1, DM1 A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Uses limits 0 and \(\frac{\pi}{4}\) to get \(\ln(1+\sqrt{2})\) | M1 A1, DM1 A1 | (4 marks) |
## Part (a)
**Answer/Working:** $x^2 - 10x + 34 = (x-5)^2 + 9$ so $\frac{1}{x^2-10x+34} = \frac{1}{(x-5)^2+9} = \frac{1}{u^2+9}$ where $u = x-5$
$I = \int \frac{1}{u^2+9}du = \left[\frac{1}{3}\arctan\left(\frac{u}{3}\right)\right]$ or $I = \int \frac{1}{(x-5)^2+9}du = \left[\frac{1}{3}\arctan\left(\frac{x-5}{3}\right)\right]$
Uses limits 3 and 0 to give $\frac{\pi}{12}$ or uses limits 8 and 5 to give $\frac{\pi}{12}$ | **B1, M1 A1, DM1 A1** | Mark can be earned in either part (a) or (b). CAO for the initial rewriting.
## Part (b) - Alt 1
**Answer/Working:** $I = \ln\left(\frac{(x-5)}{3} + \sqrt{\frac{(x-5)^2}{3}+1}\right)$ or $I = \ln\left(\frac{x-5+\sqrt{(x-5)^2+9}}{3}\right)$
or $I = \ln\left((x-5)+\sqrt{(x-5)^2+9}\right)$
Uses limits 5 and 8 to give $\ln(1+\sqrt{2})$ | **M1 A1, DM1 A1** | (4 marks)
## Part (b) - Alt 2
**Answer/Working:** Uses $u = x-5$ to get $I = \int \frac{1}{\sqrt{u^2+9}}du = \left[\text{ar sinh}\left(\frac{u}{3}\right)\right] = \ln\left(u+\sqrt{u^2+9}\right)$
Uses limits 3 and 0 and ln expression to give $\ln(1+\sqrt{2})$ | **M1 A1, DM1 A1** | (4 marks)
## Part (b) - Alt 3
**Answer/Working:** Use substitution $x-5 = 3\tan\theta$, $\frac{dx}{d\theta} = 3\sec^2\theta$ and so
$I = \int \sec\theta d\theta = \ln(\sec\theta + \tan\theta)$
Uses limits 0 and $\frac{\pi}{4}$ to get $\ln(1+\sqrt{2})$ | **M1 A1, DM1 A1** | (4 marks)
---Show that
\begin{enumerate}[label=(\alph*)]
\item $\int_5^8 \frac{1}{x^2 - 10x + 34} dx = k\pi$, giving the value of the fraction $k$,
[5]
\item $\int_5^8 \frac{1}{\sqrt{x^2 - 10x + 34}} dx = \ln(A + \sqrt{n})$, giving the values of the integers $A$ and $n$.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2011 Q3 [9]}}