Hyperbolic substitution to evaluate integral

A question is this type if and only if it requires using a substitution like x = a sinh u or x = a cosh u to transform an integral involving square roots of quadratic expressions.

5 questions · Challenging +1.7

4.07e Inverse hyperbolic: definitions, domains, ranges
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Edexcel FP3 Q3
8 marks Challenging +1.8
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)
Edexcel FP3 Specimen Q4
Challenging +1.8
Find \(\int \sqrt{x^2 + 4} \, dx\). (Total 7 marks)
AQA Further Paper 1 2024 June Q17
7 marks Challenging +1.8
By making a suitable substitution, show that $$\int_{-2}^{1} \sqrt{x^2 + 6x + 8} \, dx = 2\sqrt{15} - \frac{1}{2}\cosh^{-1}(4)$$ [7 marks]
Edexcel CP1 2021 June Q9
11 marks Challenging +1.2
  1. Use a hyperbolic substitution and calculus to show that $$\int \frac{x^2}{\sqrt{x^2 - 1}} dx = \frac{1}{2}\left[x\sqrt{x^2 - 1} + \text{arcosh } x\right] + k$$ where \(k\) is an arbitrary constant. [6]
\includegraphics{figure_1} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = \frac{4}{15}x \text{ arcosh } x \quad x \geq 1$$ The finite region \(R\), shown shaded in Figure 1, is bounded by the curve \(C\), the \(x\)-axis and the line with equation \(x = 3\)
  1. Using algebraic integration and the result from part (a), show that the area of \(R\) is given by $$\frac{1}{15}\left[17\ln\left(3 + 2\sqrt{2}\right) - 6\sqrt{2}\right]$$ [5]
SPS SPS FM Pure 2023 November Q8
Challenging +1.8
  1. Use a hyperbolic substitution and calculus to show that $$\int \frac{x^2}{\sqrt{x^2 - 1}} dx = \frac{1}{2}\left[x\sqrt{x^2 - 1} + \arcosh x\right] + k$$ where \(k\) is an arbitrary constant. (6) \includegraphics{figure_8} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = \frac{4}{15}x \arcosh x \quad x \geqslant 1$$ The finite region \(R\), shown shaded in Figure 1, is bounded by the curve \(C\), the \(x\)-axis and the line with equation \(x = 3\)
  2. Using algebraic integration and the result from part (a), show that the area of \(R\) is given by $$\frac{1}{15}\left[17\ln\left(3 + 2\sqrt{2}\right) - 6\sqrt{2}\right]$$ (5) This is the last question on the paper.