Question 3 - A-Level Maths

Edexcel FP3 — Question 3 8 marks

Exam Board	Edexcel
Module	FP3 (Further Pure Mathematics 3)
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration using inverse trig and hyperbolic functions
Type	Hyperbolic substitution to evaluate integral
Difficulty	Challenging +1.8 This is a Further Maths FP3 question requiring hyperbolic substitution with careful algebraic manipulation. While the substitution is given, students must correctly handle the change of limits, simplify the resulting hyperbolic expression using identities (cosh²θ - sinh²θ = 1), and integrate to obtain a logarithmic answer. The multi-step nature, need for hyperbolic function fluency, and exact answer requirement make this significantly harder than standard A-level integration, though the provided substitution reduces the difficulty somewhat.
Spec	1.08h Integration by substitution 4.07e Inverse hyperbolic: definitions, domains, ranges

3. Using the substitution $\mathrm { x } = \frac { 3 } { \sinh \theta }$, or otherwise, find the exact value of $$\int _ { 4 } ^ { 3 \sqrt { } 3 } \frac { 1 } { x \sqrt { } \left( x ^ { 2 } + 9 \right) } d x$$ giving your answer in the form a ln b , where a and b are rational numbers.
(Total 8 marks)

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Answer	Marks
$\frac{dx}{d\theta} = -\frac{3\cosh\theta}{\sinh^2\theta}$	B1
$\int \frac{1}{x\sqrt{x^2+9}} dx = \int \frac{1}{\frac{3}{\sinh\theta}\sqrt{\left(\frac{9}{\sinh^2\theta}+9\right)}} \times \frac{-3\cosh\theta}{\sinh^2\theta} d\theta$	M1 A1
$= -\frac{1}{3}\int 1d\theta = -\frac{1}{3}\theta$	A1
$x = 3\sqrt{3} \Rightarrow \sinh\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = \ln\left(\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}\right) = \ln\sqrt{3}$	M1 A1
$x = 4 \Rightarrow \sinh\theta = \frac{3}{4} \Rightarrow \theta = \ln\left(\frac{3}{4} + \frac{5}{4}\right) = \ln 2$	M1 A1
$\left[-\frac{1}{3}\theta\right]_{\ln 2}^{\ln\sqrt{3}} = \frac{1}{3}(\ln 2 - \ln\sqrt{3}) = \frac{1}{3}\left(\frac{1}{2}\ln 4 - \frac{1}{2}\ln 3\right) = \frac{1}{6}\ln\frac{4}{3}$	M1 A1 (8)

$\frac{dx}{d\theta} = -\frac{3\cosh\theta}{\sinh^2\theta}$ | B1 |

$\int \frac{1}{x\sqrt{x^2+9}} dx = \int \frac{1}{\frac{3}{\sinh\theta}\sqrt{\left(\frac{9}{\sinh^2\theta}+9\right)}} \times \frac{-3\cosh\theta}{\sinh^2\theta} d\theta$ | M1 A1 |

$= -\frac{1}{3}\int 1d\theta = -\frac{1}{3}\theta$ | A1 |

$x = 3\sqrt{3} \Rightarrow \sinh\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = \ln\left(\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}\right) = \ln\sqrt{3}$ | M1 A1 |

$x = 4 \Rightarrow \sinh\theta = \frac{3}{4} \Rightarrow \theta = \ln\left(\frac{3}{4} + \frac{5}{4}\right) = \ln 2$ | M1 A1 |

$\left[-\frac{1}{3}\theta\right]_{\ln 2}^{\ln\sqrt{3}} = \frac{1}{3}(\ln 2 - \ln\sqrt{3}) = \frac{1}{3}\left(\frac{1}{2}\ln 4 - \frac{1}{2}\ln 3\right) = \frac{1}{6}\ln\frac{4}{3}$ | M1 A1 (8) |

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3. Using the substitution $\mathrm { x } = \frac { 3 } { \sinh \theta }$, or otherwise, find the exact value of

$$\int _ { 4 } ^ { 3 \sqrt { } 3 } \frac { 1 } { x \sqrt { } \left( x ^ { 2 } + 9 \right) } d x$$

giving your answer in the form a ln b , where a and b are rational numbers.\\
(Total 8 marks)\\

\hfill \mbox{\textit{Edexcel FP3  Q3 [8]}}

This paper (8 questions)

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Q1 6 Q2 7 Q3 8 Q4 9 Q5 10 Q6 11 Q7 12 Q8 12

Answer	Marks
\(\frac{dx}{d\theta} = -\frac{3\cosh\theta}{\sinh^2\theta}\)	B1
\(\int \frac{1}{x\sqrt{x^2+9}} dx = \int \frac{1}{\frac{3}{\sinh\theta}\sqrt{\left(\frac{9}{\sinh^2\theta}+9\right)}} \times \frac{-3\cosh\theta}{\sinh^2\theta} d\theta\)	M1 A1
\(= -\frac{1}{3}\int 1d\theta = -\frac{1}{3}\theta\)	A1
\(x = 3\sqrt{3} \Rightarrow \sinh\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = \ln\left(\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}\right) = \ln\sqrt{3}\)	M1 A1
\(x = 4 \Rightarrow \sinh\theta = \frac{3}{4} \Rightarrow \theta = \ln\left(\frac{3}{4} + \frac{5}{4}\right) = \ln 2\)	M1 A1
\(\left[-\frac{1}{3}\theta\right]_{\ln 2}^{\ln\sqrt{3}} = \frac{1}{3}(\ln 2 - \ln\sqrt{3}) = \frac{1}{3}\left(\frac{1}{2}\ln 4 - \frac{1}{2}\ln 3\right) = \frac{1}{6}\ln\frac{4}{3}\)	M1 A1 (8)