4.07f Inverse hyperbolic: logarithmic forms

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1. (a) Show that \(\tanh x = \frac { \mathrm { e } ^ { 2 x } - 1 } { \mathrm { e } ^ { 2 x } + 1 }\).
   (b) Hence, or otherwise, show that, if \(\tanh x = \frac { 1 } { k }\) for \(k > 1\), then \(x = \frac { 1 } { 2 } \ln \left( \frac { k + 1 } { k - 1 } \right)\) and find an expression in terms of \(k\) for \(\sinh 2 x\).
2. A curve has equation \(y = \frac { 1 } { 2 } \ln ( \tanh x )\) for \(\alpha \leqslant x \leqslant \beta\), where \(\tanh \alpha = \frac { 1 } { 3 }\) and \(\tanh \beta = \frac { 1 } { 2 }\). Find, in its simplest exact form, the arc length of this curve.

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1. (a) Given that \(y = \tanh ^ { - 1 } x , - 1 < x < 1\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\).
   (b) Show that \(y = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\).
2. Show that \(\int _ { 0 } ^ { 1 / \sqrt { 3 } } \frac { 2 } { 1 - x ^ { 4 } } \mathrm {~d} x = \ln ( 1 + \sqrt { 3 } ) - \frac { 1 } { 2 } \ln 2 + \frac { 1 } { 6 } \pi\).

12 \includegraphics[max width=\textwidth, alt={}, center]{0f5edc87-cb14-4583-a54d-badec47741d1-08_414_659_804_744} The diagram shows a sketch of the curve \(C\) with polar equation \(r = 4 \cos ^ { 2 } \theta\) for \(- \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\).

1. Explain briefly how you can tell from this form of the equation that \(C\) is symmetrical about the line \(\theta = 0\) and that the tangent to \(C\) at the pole \(O\) is perpendicular to the line \(\theta = 0\).
2. The equation of \(C\) may be expressed in the form \(r = k ( 1 + \cos 2 \theta )\). State the value of \(k\) and use this form to show that the area of the region enclosed by \(C\) is given by $$\int _ { - \frac { 1 } { 2 } \pi } ^ { \frac { 1 } { 2 } \pi } ( 3 + 4 \cos 2 \theta + \cos 4 \theta ) d \theta ,$$ and hence find this area.
3. The length of \(C\) is denoted by \(L\). Show that $$L = 8 \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos \theta \sqrt { 1 + 3 \sin ^ { 2 } \theta } \mathrm {~d} \theta$$ and use the substitution \(\sinh x = \sqrt { 3 } \sin \theta\) to determine \(L\) in an exact form.

Find the first three terms in the Maclaurin's series for \(\tanh^{-1}\left(\frac{1}{2}e^t\right)\) in the form \(\frac{1}{2}\ln a + bx + cx^2\), giving the exact values of the constants \(a\), \(b\) and \(c\). [6]

Given that \(y = \text{artanh}(\cos x)\)

1. show that $$\frac{dy}{dx} = -\text{cosec } x$$ [2]
2. Hence find the exact value of $$\int_{0}^{\frac{\pi}{4}} \cos x \, \text{artanh}(\cos x) \, dx$$ giving your answer in the form \(a \ln\left(b + c\sqrt{3}\right) + d\pi\), where \(a\), \(b\), \(c\) and \(d\) are rational numbers to be found. [5]

\includegraphics{figure_33} Figure 2 shows a sketch of the curve with equation $$y = x \operatorname{arcosh} x, \quad 1 \leq x \leq 2.$$ The region \(R\), as shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the line \(x = 2\). Show that the area of \(R\) is $$\frac{7}{4} \ln(2 + \sqrt{3}) - \frac{\sqrt{3}}{2}.$$ (Total 10 marks)

Show that the solutions to the equation $$3\tanh^2 x - 2\operatorname{sech} x = 2$$ can be expressed in the form $$x = \pm \ln(a + \sqrt{b})$$ where \(a\) and \(b\) are integers to be found. You may use without proof the result \(\cosh^{-1} y = \ln(y + \sqrt{y^2 - 1})\) [5 marks]

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]

It is given that \(f(x) = \cosh^{-1}(x - 3)\) Which of the sets listed below is the greatest possible domain of the function \(f\)? Circle your answer. [1 mark] \(\{x : x \geq 4\}\) \quad \(\{x : x \geq 3\}\) \quad \(\{x : x \geq 1\}\) \quad \(\{x : x \geq 0\}\)

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]

In this question you must show detailed reasoning. Show that $$\int_0^{\frac{\pi}{3}} \operatorname{arcsinh} 2x \, dx = \frac{2}{3} \ln 3 - \frac{1}{3}.$$ [8]