| Exam Board | Edexcel |
|---|---|
| Module | F3 (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 | Completing square then standard inverse trig |
| Difficulty | Standard +0.8 This Further Maths question requires completing the square twice, recognizing standard inverse trig/hyperbolic forms, and careful algebraic manipulation to reach exact forms. Part (a) needs arctan integration and part (b) needs arsinh, both with non-trivial substitutions and exact evaluation at bounds. While the techniques are standard for F3, the execution demands precision and multiple steps, placing it moderately above average difficulty. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(x^2 + 4x + 13 \equiv (x+2)^2 + 9\) | B1 | |
| \(\int \frac{1}{(x+2)^2 + 9}\,dx = \frac{1}{3}\arctan\left(\frac{x+2}{3}\right)\) | M1A1 | M1: \(k\arctan f(x)\); A1: Correct expression |
| \(\left[\frac{1}{3}\arctan\left(\frac{x+2}{3}\right)\right]_{-2}^{1} = \frac{1}{3}(\arctan 1 - \arctan 0)\) | M1 | Correct use of limits; \(\arctan 0\) need not be shown |
| \(\frac{\pi}{12}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(x^2 + 4x + 13 \equiv (x+2)^2 + 9\) | B1 | |
| \(\frac{dx}{dt} = 3\sec^2 t\); \(x=-2, \tan t=0, t=0\); \(x=1, \tan t=1, t=\frac{\pi}{4}\) | ||
| \(\int \frac{3\sec^2 t}{9\tan^2 t + 9}\,dt = \frac{1}{3}\int dt = \frac{1}{3}t\) | M1A1 | M1: sub and integrate inc use of \(\tan^2 + 1 = \sec^2\); A1: Correct expression, ignore limits |
| \(\left[\frac{\pi}{12}\right]_0^{\frac{\pi}{4}}\) (with limits applied) | M1 | Either change limits and substitute, or reverse substitution and substitute original limits |
| \(\frac{\pi}{12}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(4x^2 - 12x + 34 = 4\left(x - \frac{3}{2}\right)^2 + 25\) or \((2x-3)^2 + 25\) | M1A1 | M1: \(4(x \pm p)^2 \pm q\), \((p,q \neq 0)\); A1: \(4\left(x-\frac{3}{2}\right)^2 + 25\) |
| \(\int \frac{1}{\sqrt{4\left(x-\frac{3}{2}\right)^2 + 25}}\,dx = \frac{1}{2}\int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^2 + \frac{25}{4}}}\,dx = \frac{1}{2}\text{arsinh}\left(\frac{x - \frac{3}{2}}{\frac{5}{2}}\right)\) | M1A1 | M1: \(k\,\text{arsinh}\, f(x)\); A1: Correct expression |
| \(\left[\frac{1}{2}\text{arsinh}\left(\frac{x-\frac{3}{2}}{\frac{5}{2}}\right)\right]_{-1}^{4} = \frac{1}{2}(\text{arsinh}(1) - \text{arsinh}(-1))\) | M1 | Correct use of limits |
| \(= \frac{1}{2}\left(\ln(1+\sqrt{2}) - \ln(-1+\sqrt{2})\right)\) | M1 | Uses the logarithmic form of arsinh |
| \(= \frac{1}{2}\ln(3 + 2\sqrt{2})\) or \(\ln(1+\sqrt{2})\) | A1 | cao |
## Question 3:
### Part (a):
| Answer/Working | Mark | Notes |
|---|---|---|
| $x^2 + 4x + 13 \equiv (x+2)^2 + 9$ | B1 | |
| $\int \frac{1}{(x+2)^2 + 9}\,dx = \frac{1}{3}\arctan\left(\frac{x+2}{3}\right)$ | M1A1 | M1: $k\arctan f(x)$; A1: Correct expression |
| $\left[\frac{1}{3}\arctan\left(\frac{x+2}{3}\right)\right]_{-2}^{1} = \frac{1}{3}(\arctan 1 - \arctan 0)$ | M1 | Correct use of limits; $\arctan 0$ need not be shown |
| $\frac{\pi}{12}$ | A1 | cao |
**Alternative (substitution $x + 2 = 3\tan t$):**
| Answer/Working | Mark | Notes |
|---|---|---|
| $x^2 + 4x + 13 \equiv (x+2)^2 + 9$ | B1 | |
| $\frac{dx}{dt} = 3\sec^2 t$; $x=-2, \tan t=0, t=0$; $x=1, \tan t=1, t=\frac{\pi}{4}$ | | |
| $\int \frac{3\sec^2 t}{9\tan^2 t + 9}\,dt = \frac{1}{3}\int dt = \frac{1}{3}t$ | M1A1 | M1: sub and integrate inc use of $\tan^2 + 1 = \sec^2$; A1: Correct expression, ignore limits |
| $\left[\frac{\pi}{12}\right]_0^{\frac{\pi}{4}}$ (with limits applied) | M1 | Either change limits and substitute, or reverse substitution and substitute original limits |
| $\frac{\pi}{12}$ | A1 | cao |
### Part (b):
| Answer/Working | Mark | Notes |
|---|---|---|
| $4x^2 - 12x + 34 = 4\left(x - \frac{3}{2}\right)^2 + 25$ or $(2x-3)^2 + 25$ | M1A1 | M1: $4(x \pm p)^2 \pm q$, $(p,q \neq 0)$; A1: $4\left(x-\frac{3}{2}\right)^2 + 25$ |
| $\int \frac{1}{\sqrt{4\left(x-\frac{3}{2}\right)^2 + 25}}\,dx = \frac{1}{2}\int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^2 + \frac{25}{4}}}\,dx = \frac{1}{2}\text{arsinh}\left(\frac{x - \frac{3}{2}}{\frac{5}{2}}\right)$ | M1A1 | M1: $k\,\text{arsinh}\, f(x)$; A1: Correct expression |
| $\left[\frac{1}{2}\text{arsinh}\left(\frac{x-\frac{3}{2}}{\frac{5}{2}}\right)\right]_{-1}^{4} = \frac{1}{2}(\text{arsinh}(1) - \text{arsinh}(-1))$ | M1 | Correct use of limits |
| $= \frac{1}{2}\left(\ln(1+\sqrt{2}) - \ln(-1+\sqrt{2})\right)$ | M1 | Uses the logarithmic form of arsinh |
| $= \frac{1}{2}\ln(3 + 2\sqrt{2})$ or $\ln(1+\sqrt{2})$ | A1 | cao |3. Without using a calculator, find
\begin{enumerate}[label=(\alph*)]
\item $\int _ { - 2 } ^ { 1 } \frac { 1 } { x ^ { 2 } + 4 x + 13 } \mathrm {~d} x$, giving your answer as a multiple of $\pi$,
\item $\int _ { - 1 } ^ { 4 } \frac { 1 } { \sqrt { 4 x ^ { 2 } - 12 x + 34 } } \mathrm {~d} x$, giving your answer in the form $p \ln ( q + r \sqrt { 2 } )$,\\
where $p , q$ and $r$ are rational numbers to be found.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2016 Q3 [12]}}