| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2021 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Integrate using hyperbolic substitution |
| Difficulty | Challenging +1.2 This is a standard Further Maths hyperbolic substitution question with a guided substitution provided. Students must apply dx/dθ = 4sinh θ, use the identity cosh²θ - 1 = sinh²θ to simplify the integrand to a basic form, integrate, then substitute back. While it requires knowledge of hyperbolic identities and careful algebraic manipulation across 6 marks, it follows a well-practiced template for this topic with no novel problem-solving required. |
| Spec | 4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 4\cosh\theta \Rightarrow \frac{dx}{d\theta} = 4\sinh\theta\); substitutes to get \(\int\frac{4\sinh\theta}{(16\cosh^2\theta-16)^{3/2}}d\theta\) | M1 | Full attempt to use given substitution. Award for \(k\int\frac{\sinh\theta}{((4\cosh\theta)^2-16)^{3/2}}d\theta\) |
| Simplifies \((16\cosh^2\theta-16)^{3/2}\) to form \(k\sinh^3\theta\) giving \(\int\frac{1}{16\sinh^2\theta}d\theta\) | M1, A1 | M1: Simplifies to form \(k\sinh^3\theta\). A1: Fully correct simplified integral. Allow equivalents e.g. \(\frac{1}{16}\int\text{cosech}^2\theta\,d\theta\) |
| \(= \frac{1}{16}\int\text{cosech}^2\theta\,d\theta = -\frac{1}{16}\coth\theta + c\) | dM1 | Integrates to obtain \(k\coth\theta\). Depends on both previous method marks |
| Back-substitutes: \(\cosh\theta = \frac{x}{4}\), \(\sinh\theta = \sqrt{\left(\frac{x}{4}\right)^2-1}\) | dM1 | Substitutes back correctly for \(x\). Depends on all previous method marks |
| \(\dfrac{-x}{16\sqrt{x^2-16}} + c\) | A1 | Correct answer. Condone omission of \(+c\) |
# Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 4\cosh\theta \Rightarrow \frac{dx}{d\theta} = 4\sinh\theta$; substitutes to get $\int\frac{4\sinh\theta}{(16\cosh^2\theta-16)^{3/2}}d\theta$ | M1 | Full attempt to use given substitution. Award for $k\int\frac{\sinh\theta}{((4\cosh\theta)^2-16)^{3/2}}d\theta$ |
| Simplifies $(16\cosh^2\theta-16)^{3/2}$ to form $k\sinh^3\theta$ giving $\int\frac{1}{16\sinh^2\theta}d\theta$ | M1, A1 | M1: Simplifies to form $k\sinh^3\theta$. A1: Fully correct simplified integral. Allow equivalents e.g. $\frac{1}{16}\int\text{cosech}^2\theta\,d\theta$ |
| $= \frac{1}{16}\int\text{cosech}^2\theta\,d\theta = -\frac{1}{16}\coth\theta + c$ | dM1 | Integrates to obtain $k\coth\theta$. Depends on both previous method marks |
| Back-substitutes: $\cosh\theta = \frac{x}{4}$, $\sinh\theta = \sqrt{\left(\frac{x}{4}\right)^2-1}$ | dM1 | Substitutes back correctly for $x$. Depends on all previous method marks |
| $\dfrac{-x}{16\sqrt{x^2-16}} + c$ | A1 | Correct answer. Condone omission of $+c$ |4. Using the substitution $x = 4 \cosh \theta$ show that
$$\int \frac { 1 } { \left( x ^ { 2 } - 16 \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x = \frac { a x } { \sqrt { x ^ { 2 } - 16 } } + c \quad | x | > 4$$
where $a$ is a constant to be determined and $c$ is an arbitrary constant.\\
(6)\\
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\hfill \mbox{\textit{Edexcel F3 2021 Q4 [6]}}