| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Integration by parts with inverse trig |
| Difficulty | Challenging +1.8 This Further Pure 3 question requires differentiating an inverse hyperbolic function using the chain rule (part a), then applying integration by parts with the result to evaluate a definite integral involving inverse hyperbolic and trigonometric functions (part b). While the techniques are standard for Further Maths, the combination of inverse hyperbolic functions, integration by parts, and exact value calculation with surds makes this significantly harder than typical A-level questions, though it follows a structured path once the approach is identified. |
| Spec | 1.08i Integration by parts4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = \text{artanh}(\cos x)\) | ||
| \(\frac{dy}{dx} = \frac{1}{1-\cos^2 x} \times -\sin x\) | M1 | Correct use of the chain rule |
| \(= \frac{-\sin x}{\sin^2 x} = \frac{-1}{\sin x} = -\text{cosec}\, x\) \(*\) | A1 | Correct completion with no errors |
| Alternative 1: \(\tanh y = \cos x \Rightarrow \text{sech}^2 y \frac{dy}{dx} = -\sin x\) | ||
| \(\frac{dy}{dx} = \frac{-\sin x}{\text{sech}^2 y} = \frac{-\sin x}{1 - \cos^2 x} = \frac{-\sin x}{\sin^2 x} = -\text{cosec}\, x\) \(*\) | M1A1 | M1: Correct differentiation to obtain a function of \(x\). A1: Correct completion with no errors |
| Alternative 2: \(\text{artanh}(\cos x) = \frac{1}{2}\ln\left(\frac{1+\cos x}{1-\cos x}\right)\) | ||
| \(\frac{dy}{dx} = \frac{1}{2} \times \frac{1-\cos x}{1+\cos x} \times \frac{-\sin x(1-\cos x) - \sin x(1+\cos x)}{(1-\cos x)^2} = \frac{-2\sin x}{2(1-\cos^2 x)} = -\text{cosec}\, x\) \(*\) | M1A1 | M1: Correct differentiation to obtain a function of \(x\). A1: Correct completion with no errors |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int \cos x\,\text{artanh}(\cos x)\,dx = \sin x\,\text{artanh}(\cos x) - \int \sin x \times -\text{cosec}\, x\,dx\) | M1A1 | M1: Parts in the correct direction. A1: Correct expression |
| \(\left[\sin x\,\text{artanh}(\cos x) + x\right]_0^{\frac{\pi}{6}} = \frac{1}{2}\text{artanh}\!\left(\frac{\sqrt{3}}{2}\right) + \frac{\pi}{6} - (0)\) | M1 | Correct use of limits on either part (provided both parts are integrated). Lower limit need not be shown |
| \(= \frac{1}{4}\ln\!\left(\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}\right) + \frac{\pi}{6}\) | M1 | Use of the logarithmic form of artanh |
| \(= \frac{1}{4}\ln(7+4\sqrt{3}) + \frac{\pi}{6}\) or \(\frac{1}{2}\ln(2+\sqrt{3}) + \frac{\pi}{6}\) | A1 | Cao (oe) |
| The last 2 M marks may be gained in reverse order. | (5) | |
| Total 7 |
## Question 5:
**Part (a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = \text{artanh}(\cos x)$ | | |
| $\frac{dy}{dx} = \frac{1}{1-\cos^2 x} \times -\sin x$ | M1 | Correct use of the chain rule |
| $= \frac{-\sin x}{\sin^2 x} = \frac{-1}{\sin x} = -\text{cosec}\, x$ $*$ | A1 | Correct completion with no errors |
| **Alternative 1:** $\tanh y = \cos x \Rightarrow \text{sech}^2 y \frac{dy}{dx} = -\sin x$ | | |
| $\frac{dy}{dx} = \frac{-\sin x}{\text{sech}^2 y} = \frac{-\sin x}{1 - \cos^2 x} = \frac{-\sin x}{\sin^2 x} = -\text{cosec}\, x$ $*$ | M1A1 | M1: Correct differentiation to obtain a function of $x$. A1: Correct completion with no errors |
| **Alternative 2:** $\text{artanh}(\cos x) = \frac{1}{2}\ln\left(\frac{1+\cos x}{1-\cos x}\right)$ | | |
| $\frac{dy}{dx} = \frac{1}{2} \times \frac{1-\cos x}{1+\cos x} \times \frac{-\sin x(1-\cos x) - \sin x(1+\cos x)}{(1-\cos x)^2} = \frac{-2\sin x}{2(1-\cos^2 x)} = -\text{cosec}\, x$ $*$ | M1A1 | M1: Correct differentiation to obtain a function of $x$. A1: Correct completion with no errors |
| | | (2) |
**Part (b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int \cos x\,\text{artanh}(\cos x)\,dx = \sin x\,\text{artanh}(\cos x) - \int \sin x \times -\text{cosec}\, x\,dx$ | M1A1 | M1: Parts in the correct direction. A1: Correct expression |
| $\left[\sin x\,\text{artanh}(\cos x) + x\right]_0^{\frac{\pi}{6}} = \frac{1}{2}\text{artanh}\!\left(\frac{\sqrt{3}}{2}\right) + \frac{\pi}{6} - (0)$ | M1 | Correct use of limits on either part (provided both parts are integrated). Lower limit need not be shown |
| $= \frac{1}{4}\ln\!\left(\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}\right) + \frac{\pi}{6}$ | M1 | Use of the logarithmic form of artanh |
| $= \frac{1}{4}\ln(7+4\sqrt{3}) + \frac{\pi}{6}$ or $\frac{1}{2}\ln(2+\sqrt{3}) + \frac{\pi}{6}$ | A1 | Cao (oe) |
| The last 2 M marks may be gained in reverse order. | | (5) |
| | | **Total 7** |
---5. Given that $y = \operatorname { artanh } ( \cos x )$
\begin{enumerate}[label=(\alph*)]
\item show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } x$$
\item Hence find the exact value of
$$\int _ { 0 } ^ { \frac { \pi } { 6 } } \cos x \operatorname { artanh } ( \cos x ) d x$$
giving your answer in the form $a \ln ( b + c \sqrt { 3 } ) + d \pi$, where $a , b , c$ and $d$ are rational numbers to be found.\\
(5)\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2016 Q5 [7]}}