| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2019 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Standard integral of 1/√(x²-a²) |
| Difficulty | Challenging +1.2 This is a Further Maths integration question with three parts of increasing complexity. Part (a) is a standard inverse hyperbolic/trig integral requiring recognition of a formula. Part (b) involves partial fractions and logarithmic integration with algebraic manipulation. Part (c) requires substantial hyperbolic function manipulation and simplification before integration. While requiring multiple techniques and careful algebra, these are well-practiced Further Maths topics without requiring novel insight—moderately above average difficulty. |
| Spec | 4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_3^4 \frac{1}{\sqrt{x^2-4}} dx = \left[\cosh^{-1}\frac{x}{2}\right]_3^4\) | M1A1 | M0 unsupported answer |
| \(= 0.355\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_1^2 \frac{k}{9-x^2} dx = \left[\frac{k}{6}\ln\left | \frac{3+x}{3-x}\right | \right]_1^2\) |
| \(= \frac{k}{6}(\ln 5 - \ln 2)\) (= \(\frac{k}{6}\ln\frac{5}{2}\)) | A1 | |
| METHOD 1: \(\ln\frac{25}{4} = \ln\left(\frac{5}{2}\right)^2 = 2\ln\frac{5}{2}\) | m1 | |
| Equating, \(\frac{k}{6}\ln\frac{5}{2} = 2\ln\frac{5}{2}\) | A1 | |
| \(k = 12\) | ||
| METHOD 2: Equating: \(\frac{k}{6}(\ln 5 - \ln 2) = \ln\frac{25}{4}\) | (m1) | |
| \(\frac{k}{6} = 2\) | (A1) | |
| \(k = 12\) |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1: | ||
| \(\sinh^2 x + \cosh^2 x - \sinh 2x\) | ||
| \(= \sinh^2 x + \cosh^2 x - 2\sinh x\cosh x\) | M1 A1 | |
| \(= (\cosh x - \sinh x)^2\) | ||
| \(\int\frac{(\cosh x - \sinh x)^2}{(\cosh x - \sinh x)^2} dx = \int(\cosh x - \sinh x)dx\) | A1 | |
| \(= \sinh x - \cosh x + c\) | A1 | |
| \(= \frac{e^x - e^{-x}}{2} - \frac{e^x + e^{-x}}{2} + c\) | A1 | |
| \(= -e^{-x} + c\) | A1 | Convincing |
| METHOD 2: | ||
| \(\sinh^2 x + \cosh^2 x - \sinh 2x\) | ||
| \(= \sinh^2 x + \cosh^2 x - 2\sinh x\cosh x\) | (M1) (A1) | |
| \(= (\cosh x - \sinh x)^2\) | ||
| \(\int\frac{(\cosh x - \sinh x)^2}{(\cosh x - \sinh x)^2} dx = \int\cosh x - \sinh x dx\) | (A1) | |
| \(= \int\left(\frac{e^x + e^{-x}}{2} - \frac{e^x - e^{-x}}{2}\right) dx\) | (A1) | |
| \(= \int e^{-x} dx\) | (A1) | |
| \(= -e^{-x} + c\) | (A1) | Convincing |
## Part a)
$\int_3^4 \frac{1}{\sqrt{x^2-4}} dx = \left[\cosh^{-1}\frac{x}{2}\right]_3^4$ | M1A1 | M0 unsupported answer
$= 0.355$ | A1 |
## Part b)
$\int_1^2 \frac{k}{9-x^2} dx = \left[\frac{k}{6}\ln\left|\frac{3+x}{3-x}\right|\right]_1^2$ | M1A1 |
$= \frac{k}{6}(\ln 5 - \ln 2)$ (= $\frac{k}{6}\ln\frac{5}{2}$) | A1 |
METHOD 1: $\ln\frac{25}{4} = \ln\left(\frac{5}{2}\right)^2 = 2\ln\frac{5}{2}$ | m1 |
Equating, $\frac{k}{6}\ln\frac{5}{2} = 2\ln\frac{5}{2}$ | A1 |
$k = 12$ | |
METHOD 2: Equating: $\frac{k}{6}(\ln 5 - \ln 2) = \ln\frac{25}{4}$ | (m1) |
$\frac{k}{6} = 2$ | (A1) |
$k = 12$ | |
## Part c)
METHOD 1: | |
$\sinh^2 x + \cosh^2 x - \sinh 2x$ | |
$= \sinh^2 x + \cosh^2 x - 2\sinh x\cosh x$ | M1 A1 |
$= (\cosh x - \sinh x)^2$ | |
$\int\frac{(\cosh x - \sinh x)^2}{(\cosh x - \sinh x)^2} dx = \int(\cosh x - \sinh x)dx$ | A1 |
$= \sinh x - \cosh x + c$ | A1 |
$= \frac{e^x - e^{-x}}{2} - \frac{e^x + e^{-x}}{2} + c$ | A1 |
$= -e^{-x} + c$ | A1 | Convincing
METHOD 2: | |
$\sinh^2 x + \cosh^2 x - \sinh 2x$ | |
$= \sinh^2 x + \cosh^2 x - 2\sinh x\cosh x$ | (M1) (A1) |
$= (\cosh x - \sinh x)^2$ | |
$\int\frac{(\cosh x - \sinh x)^2}{(\cosh x - \sinh x)^2} dx = \int\cosh x - \sinh x dx$ | (A1) |
$= \int\left(\frac{e^x + e^{-x}}{2} - \frac{e^x - e^{-x}}{2}\right) dx$ | (A1) |
$= \int e^{-x} dx$ | (A1) |
$= -e^{-x} + c$ | (A1) | Convincing\begin{enumerate}[label=(\alph*)]
\item Evaluate $\int_3^4 \frac{1}{\sqrt{x^2 - 4}} \mathrm{d}x$, giving your answer correct to three decimal places. [3]
\item Given that $\int_1^2 \frac{k}{9 - x^2} \mathrm{d}x = \ln \frac{25}{4}$, find the value of $k$. [5]
\item Show that $\int \frac{(\cosh x - \sinh x)^3}{\cosh^2 x + \sinh^2 x - \sinh 2x} \mathrm{d}x$ can be expressed as $-e^{-x} + c$, where $c$ is a constant. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2019 Q12 [14]}}