4.07f Inverse hyperbolic: logarithmic forms

1 Find the exact value of \(\int _ { 2 } ^ { \frac { 7 } { 2 } } \frac { 1 } { \sqrt { 4 x - x ^ { 2 } - 1 } } \mathrm {~d} x\).

7

1. Show that $$\frac { \mathrm { d } } { \mathrm {~d} x } \left( \frac { x } { 2 } \sqrt { x ^ { 2 } - 9 } - \frac { 9 } { 2 } \cosh ^ { - 1 } \frac { x } { 3 } \right) = \sqrt { x ^ { 2 } - 9 }$$ \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-14_67_1579_413_324} \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-14_77_1581_497_322}
2. Find the solution of the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } - y = x ^ { 2 } \sqrt { x ^ { 2 } - 9 }$$ given that \(y = 1\) when \(x = 3\). Give your answer in the form \(y = \mathrm { f } ( x )\). \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-14_2716_35_143_2012}

3. (a) Given that \(y = \operatorname { arsech } \left( \frac { x } { 2 } \right)\), where \(0 < x \leqslant 2\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { p } { x \sqrt { q - x ^ { 2 } } }$$ where \(p\) and \(q\) are constants to be determined. In part (b) solutions based entirely on calculator technology are not acceptable. $$\mathrm { f } ( x ) = \operatorname { artanh } ( x ) + \operatorname { arsech } \left( \frac { x } { 2 } \right) \quad 0 < x \leqslant 1$$ (b) Determine, in simplest form, the exact value of \(x\) for which \(\mathrm { f } ^ { \prime } ( x ) = 0\)

3. Given that $$y = x - \operatorname { artanh } \left( \frac { 2 x } { 1 + x ^ { 2 } } \right)$$

1. show that $$1 - \frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { 1 - x ^ { 2 } }$$ where \(k\) is a constant to be found.
2. Hence, or otherwise, show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \left( 1 - \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = 0$$

4. $$y = \operatorname { artanh } \left( \frac { \cos x + a } { \cos x - a } \right)$$ where \(a\) is a non-zero constant.
Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = k \tan x$$ where \(k\) is a constant to be determined.

1. The curve \(C\) has equation

$$y = \frac { 1 } { 2 } \operatorname { arcosh } ( 2 x ) \quad \frac { 7 } { 2 } \leqslant x \leqslant 13$$ Using calculus, determine the exact length of the curve \(C\).
Give your answer in the form \(p \sqrt { q }\), where \(p\) and \(q\) are constants to be found.

4. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation $$y = 40 \operatorname { arcosh } x - 9 x , \quad x \geqslant 1$$ Use calculus to find the exact coordinates of the turning point of the curve, giving your answer in the form \(\left( \frac { p } { q } , r \ln 3 + s \right)\), where \(p , q , r\) and \(s\) are integers.

7. The curve \(C\) has equation $$y = \mathrm { e } ^ { - x } , \quad x \in \mathbb { R }$$ The part of the curve \(C\) between \(x = 0\) and \(x = \ln 3\) is rotated through \(2 \pi\) radians about the \(x\)-axis.

1. Show that the area \(S\) of the curved surface generated is given by $$S = 2 \pi \int _ { 0 } ^ { \ln 3 } \mathrm { e } ^ { - x } \sqrt { 1 + \mathrm { e } ^ { - 2 x } } \mathrm {~d} x$$
2. Use the substitution \(\mathrm { e } ^ { - x } = \sinh u\) to show that $$S = 2 \pi \int _ { \operatorname { arsinh } \alpha } ^ { \operatorname { arsinh } \beta } \cosh ^ { 2 } u \mathrm {~d} u$$ where \(\alpha\) and \(\beta\) are constants to be determined.
3. Show that $$2 \int \cosh ^ { 2 } u \mathrm {~d} u = \frac { 1 } { 2 } \sinh 2 u + u + k$$ where \(k\) is an arbitrary constant.
4. Hence find the value of \(S\), giving your answer to 3 decimal places.

1. The curve \(C\) has equation

$$y = \frac { 1 } { \sqrt { x ^ { 2 } + 2 x - 3 } } , \quad x > 1$$

1. Find \(\int y \mathrm {~d} x\) The region \(R\) is bounded by the curve \(C\), the \(x\)-axis and the lines with equations \(x = 2\) and \(x = 3\). The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
2. Find the volume of the solid generated. Give your answer in the form \(p \pi \ln q\), where \(p\) and \(q\) are rational numbers to be found.

8. The curve \(C\) has equation $$y = \ln \left( \frac { \mathrm { e } ^ { x } + 1 } { \mathrm { e } ^ { x } - 1 } \right) , \quad \ln 2 \leqslant x \leqslant \ln 3$$

1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } - 1 }$$
2. Find the length of the curve \(C\), giving your answer in the form \(\ln a\), where \(a\) is a rational number.
   (6)

4. The curve \(C\) has equation $$y = \operatorname { arsinh } x + x \sqrt { x ^ { 2 } + 1 } , \quad 0 \leqslant x \leqslant 1$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \sqrt { x ^ { 2 } + 1 }\)
2. Hence show that the length of the curve \(C\) is given by $$\int _ { 0 } ^ { 1 } \sqrt { 4 x ^ { 2 } + 5 } d x$$
3. Using the substitution \(x = \frac { \sqrt { 5 } } { 2 } \sinh u\), find the exact length of the curve \(C\), giving your answer in the form \(a + b \ln c\), where \(a , b\) and \(c\) are constants to be found.

8

1. Define tanh \(y\) in terms of \(\mathrm { e } ^ { y }\) and \(\mathrm { e } ^ { - y }\).
2. Given that \(y = \tanh ^ { - 1 } x\), where \(- 1 < x < 1\), prove that \(y = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\).
3. Find the exact solution of the equation \(3 \cosh x = 4 \sinh x\), giving the answer in terms of a logarithm.
4. Solve the equation $$\tanh ^ { - 1 } x + \ln ( 1 - x ) = \ln \left( \frac { 4 } { 5 } \right)$$

7 It is given that \(\mathrm { f } ( x ) = \tanh ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right)\), for \(x > - \frac { 1 } { 2 }\).

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = - \frac { 1 } { 1 + 2 x }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\).
2. Show that the first three terms of the Maclaurin series for \(\mathrm { f } ( x )\) can be written as \(\ln a + b x + c x ^ { 2 }\), for constants \(a , b\) and \(c\) to be found.

4

1. Show that \(\operatorname { arcosh } x = \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)\).
2. Find \(\int _ { 2.5 } ^ { 3.9 } \frac { 1 } { \sqrt { 4 x ^ { 2 } - 9 } } \mathrm {~d} x\), giving your answer in the form \(a \ln b\), where \(a\) and \(b\) are rational numbers.
3. There are two points on the curve \(y = \frac { \cosh x } { 2 + \sinh x }\) at which the gradient is \(\frac { 1 } { 9 }\). Show that one of these points is \(\left( \ln ( 1 + \sqrt { 2 } ) , \frac { 1 } { 3 } \sqrt { 2 } \right)\), and find the coordinates of the other point, in a similar form.

4

1. 1. Prove, from definitions involving exponentials, that $$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$$
   2. Given that \(\sinh x = \tan y\), where \(- \frac { 1 } { 2 } \pi < y < \frac { 1 } { 2 } \pi\), show that
      (A) \(\tanh x = \sin y\),
      (B) \(x = \ln ( \tan y + \sec y )\).
2. 1. Given that \(y = \operatorname { artanh } x\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\). Hence show that \(\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \frac { 1 } { 1 - x ^ { 2 } } \mathrm {~d} x = 2 \operatorname { artanh } \frac { 1 } { 2 }\).
   2. Express \(\frac { 1 } { 1 - x ^ { 2 } }\) in partial fractions and hence find an expression for \(\int \frac { 1 } { 1 - x ^ { 2 } } \mathrm {~d} x\) in terms of logarithms.
   3. Use the results in parts (i) and (ii) to show that \(\operatorname { artanh } \frac { 1 } { 2 } = \frac { 1 } { 2 } \ln 3\).

4

1. Define tanh \(t\) in terms of exponential functions. Sketch the graph of \(\tanh t\).
2. Show that \(\operatorname { artanh } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\). State the set of values of \(x\) for which this equation is valid.
3. Differentiate the equation \(\tanh y = x\) with respect to \(x\) and hence show that the derivative of \(\operatorname { artanh } x\) is \(\frac { 1 } { 1 - x ^ { 2 } }\). Show that this result may also be obtained by differentiating the equation in part (ii).
4. By considering \(\operatorname { artanh } x\) as \(1 \times \operatorname { artanh } x\) and using integration by parts, show that $$\int _ { 0 } ^ { \frac { 1 } { 2 } } \operatorname { artanh } x \mathrm {~d} x = \frac { 1 } { 4 } \ln \frac { 27 } { 16 }$$

4

1. Given that \(\sinh y = x\), show that $$y = \ln \left( x + \sqrt { 1 + x ^ { 2 } } \right)$$ Differentiate (*) to show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 1 + x ^ { 2 } } }$$
2. Find \(\int \frac { 1 } { \sqrt { 25 + 4 x ^ { 2 } } } \mathrm {~d} x\), expressing your answer in logarithmic form.
3. Use integration by substitution with \(2 x = 5 \sinh u\) to show that $$\int \sqrt { 25 + 4 x ^ { 2 } } \mathrm {~d} x = \frac { 25 } { 4 } \left( \ln \left( \frac { 2 x } { 5 } + \sqrt { 1 + \frac { 4 x ^ { 2 } } { 25 } } \right) + \frac { 2 x } { 5 } \sqrt { 1 + \frac { 4 x ^ { 2 } } { 25 } } \right) + c$$ where \(c\) is an arbitrary constant. \section*{OCR}

4

1. Starting with the relationship \(\cosh ^ { 2 } t - \sinh ^ { 2 } t = 1\), deduce a relationship between \(\tanh ^ { 2 } t\) and \(\operatorname { sech } ^ { 2 } t\). You are given that \(y = \operatorname { artanh } x\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 - x ^ { 2 } }\).
3. Show, by integrating the result in part (ii), that \(y = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\).
4. Show that \(\int _ { 0 } ^ { \frac { \sqrt { 3 } } { 6 } } \frac { 1 } { 1 - 3 x ^ { 2 } } \mathrm {~d} x = \frac { 1 } { \sqrt { 3 } } \operatorname { artanh } \frac { 1 } { 2 }\). Express this answer in logarithmic form.
5. Use integration by parts to find \(\int \operatorname { artanh } x \mathrm {~d} x\), giving your answer in terms of logarithms. \section*{END OF QUESTION PAPER}

7

1. 1. Show that $$\frac { \mathrm { d } } { \mathrm {~d} u } \left( 2 u \sqrt { 1 + 4 u ^ { 2 } } + \sinh ^ { - 1 } 2 u \right) = k \sqrt { 1 + 4 u ^ { 2 } }$$ where \(k\) is an integer.
   2. Hence show that $$\int _ { 0 } ^ { 1 } \sqrt { 1 + 4 u ^ { 2 } } \mathrm {~d} u = p \sqrt { 5 } + q \sinh ^ { - 1 } 2$$ where \(p\) and \(q\) are rational numbers.
2. The arc of the curve with equation \(y = \frac { 1 } { 2 } \cos 4 x\) between the points where \(x = 0\) and \(x = \frac { \pi } { 8 }\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
   1. Show that the area \(S\) of the curved surface formed is given by $$S = \pi \int _ { 0 } ^ { \frac { \pi } { 8 } } \cos 4 x \sqrt { 1 + 4 \sin ^ { 2 } 4 x } \mathrm {~d} x$$
   2. Use the substitution \(u = \sin 4 x\) to find the exact value of \(S\).

5

1. Using the definition \(\sinh \theta = \frac { 1 } { 2 } \left( \mathrm { e } ^ { \theta } - \mathrm { e } ^ { - \theta } \right)\), prove the identity $$4 \sinh ^ { 3 } \theta + 3 \sinh \theta = \sinh 3 \theta$$
2. Given that \(x = \sinh \theta\) and \(16 x ^ { 3 } + 12 x - 3 = 0\), find the value of \(\theta\) in terms of a natural logarithm.
3. Hence find the real root of the equation \(16 x ^ { 3 } + 12 x - 3 = 0\), giving your answer in the form \(2 ^ { p } - 2 ^ { q }\), where \(p\) and \(q\) are rational numbers.
   [0pt] [2 marks]

8 A curve has equation \(y = 2 \sqrt { x - 1 }\), where \(x > 1\). The length of the arc of the curve between the points on the curve where \(x = 2\) and \(x = 9\) is denoted by \(s\).

1. Show that \(s = \int _ { 2 } ^ { 9 } \sqrt { \frac { x } { x - 1 } } \mathrm {~d} x\).
2. 1. Show that \(\cosh ^ { - 1 } 3 = 2 \ln ( 1 + \sqrt { 2 } )\).
   2. Use the substitution \(x = \cosh ^ { 2 } \theta\) to show that $$s = m \sqrt { 2 } + \ln ( 1 + \sqrt { 2 } )$$ where \(m\) is an integer.
      [0pt] [6 marks]
      \includegraphics[max width=\textwidth, alt={}]{5287255f-5ac4-401a-b850-758257412ff7-20_1638_1709_1069_153}
      \includegraphics[max width=\textwidth, alt={}, center]{5287255f-5ac4-401a-b850-758257412ff7-24_2489_1728_221_141}

13

1. Using the logarithmic form of \(\operatorname { arcosh } x\), prove that the derivative of \(\operatorname { arcosh } x\) is \(\frac { 1 } { \sqrt { x ^ { 2 } - 1 } }\).
2. Hence find \(\int _ { 1 } ^ { 2 } \operatorname { arcosh } x \mathrm {~d} x\), giving your answer in exact logarithmic form.
3. Ali tries to evaluate \(\int _ { 0 } ^ { 1 } \operatorname { arcosh } x \mathrm {~d} x\) using his calculator, and gets an 'error'. Explain why.

2. $$f ( x ) = \tanh ^ { - 1 } \left( \frac { 3 - x } { 6 + x } \right) \quad | x | < \frac { 3 } { 2 }$$

1. Show that $$f ^ { \prime } ( x ) = - \frac { 1 } { 2 x + 3 }$$
2. Hence determine \(\mathrm { f } ^ { \prime \prime } ( x )\)
3. Hence show that the Maclaurin series for \(\mathrm { f } ( x )\), up to and including the term in \(x ^ { 2 }\), is $$\ln p + q x + r x ^ { 2 }$$ where \(p , q\) and \(r\) are constants to be determined.

4. Given that \(y = \operatorname { arsinh } ( \sqrt { } x ) , x > 0\),

1. find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer as a simplified fraction.
2. Hence, or otherwise, find $$\int _ { \frac { 1 } { 4 } } ^ { 4 } \frac { 1 } { \sqrt { [ x ( x + 1 ) ] } } \mathrm { d } x$$ giving your answer in the form \(\ln \left( \frac { a + b \sqrt { } 5 } { 2 } \right)\), where \(a\) and \(b\) are integers.

12

1. Use the definition of the cosh function to prove that $$\cosh ^ { - 1 } \left( \frac { x } { a } \right) = \ln \left( \frac { x + \sqrt { x ^ { 2 } - a ^ { 2 } } } { a } \right) \quad \text { for } a > 0$$ [6 marks]
   12
2. The formulae booklet gives the integral of \(\frac { 1 } { \sqrt { x ^ { 2 } - a ^ { 2 } } }\) as $$\cosh ^ { - 1 } \left( \frac { x } { a } \right) \text { or } \ln \left( x + \sqrt { x ^ { 2 } - a ^ { 2 } } \right) + c$$ Ronald says that this contradicts the result given in part (a).
   Explain why Ronald is wrong.