4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials

4. (i) Find, without using a calculator, $$\int _ { 3 } ^ { 5 } \frac { 1 } { \sqrt { 15 + 2 x - x ^ { 2 } } } d x$$ giving your answer as a multiple of \(\pi\).
(ii)

1. Show that $$5 \cosh x - 4 \sinh x = \frac { \mathrm { e } ^ { 2 x } + 9 } { 2 \mathrm { e } ^ { x } }$$
2. Hence, using the substitution \(u = e ^ { x }\) or otherwise, find $$\int \frac { 1 } { 5 \cosh x - 4 \sinh x } d x$$

1. Given that \(y = \operatorname { arsinh } ( \tanh x )\), show that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \operatorname { sech } ^ { 2 } x } { \sqrt { 1 + \tanh ^ { 2 } x } }$$ \section*{-} \includegraphics[max width=\textwidth, alt={}, center]{64dc962a-1788-49ac-a4db-af1241b552a0-03_51_51_276_2012}
N

3. (a) Using the definition for \(\cosh x\) in terms of exponentials, show that $$\cosh 2 x \equiv 2 \cosh ^ { 2 } x - 1$$ (b) Find the exact values of \(x\) for which $$29 \cosh x - 3 \cosh 2 x = 38$$ giving your answers in terms of natural logarithms.

7. $$I _ { n } = \int _ { 0 } ^ { \ln 2 } \cosh ^ { n } x \mathrm {~d} x , \quad n \geqslant 0$$

1. Show that, for \(n \geqslant 2\), $$I _ { n } = \frac { 3 a ^ { n - 1 } } { n b ^ { n } } + \frac { n - 1 } { n } I _ { n - 2 }$$ where \(a\) and \(b\) are integers to be found.
2. Hence, or otherwise, find the exact value of $$\int _ { 0 } ^ { \ln 2 } \cosh ^ { 4 } x \mathrm {~d} x$$

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)

1. (a) Starting from the definitions of \(\sinh x\) and \(\cosh x\) in terms of exponentials, show that, for \(x \in \mathbb { R }\)

$$\tanh x = \frac { \mathrm { e } ^ { 2 x } - 1 } { \mathrm { e } ^ { 2 x } + 1 }$$ (b) Hence, given that \(- 1 < \theta < 1\), prove that $$\operatorname { artanh } \theta = \frac { 1 } { 2 } \ln \left( \frac { 1 + \theta } { 1 - \theta } \right)$$ uestion 1 continued \(\_\_\_\_\) 7

2. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation $$y = 5 \cosh x - 6 \sinh x$$ The curve crosses the \(x\)-axis at the point \(A\).

1. Find the exact value of the \(x\) coordinate of the point \(A\), giving your answer as a natural logarithm.
2. Show that $$( 5 \cosh x - 6 \sinh x ) ^ { 2 } \equiv a \cosh 2 x + b \sinh 2 x + c$$ where \(a , b\) and \(c\) are constants to be found. The finite region \(R\), bounded by the curve and the coordinate axes, is shown shaded in Figure 1. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
3. Use calculus to find the volume of the solid generated, giving your answer as an exact multiple of \(\pi\).

4

1. Starting from the definitions of \(\sinh x\) and \(\cosh x\) in terms of exponentials, prove that $$1 + 2 \sinh ^ { 2 } x = \cosh 2 x$$
2. Solve the equation $$2 \cosh 2 x + \sinh x = 5 ,$$ giving the answers in an exact logarithmic form.
3. Show that \(\int _ { 0 } ^ { \ln 3 } \sinh ^ { 2 } x \mathrm {~d} x = \frac { 10 } { 9 } - \frac { 1 } { 2 } \ln 3\).
4. Find the exact value of \(\int _ { 3 } ^ { 5 } \sqrt { x ^ { 2 } - 9 } \mathrm {~d} x\).

4

1. Find \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 9 x ^ { 2 } + 16 } } \mathrm {~d} x\), giving your answer in an exact logarithmic form.
2. 1. Starting from the definitions of \(\sinh x\) and \(\cosh x\) in terms of exponentials, prove that \(\sinh 2 x = 2 \sinh x \cosh x\).
   2. Show that one of the stationary points on the curve $$y = 20 \cosh x - 3 \cosh 2 x$$ has coordinates \(\left( \ln 3 , \frac { 59 } { 3 } \right)\), and find the coordinates of the other two stationary points.
   3. Show that \(\int _ { - \ln 3 } ^ { \ln 3 } ( 20 \cosh x - 3 \cosh 2 x ) \mathrm { d } x = 40\).

4

1. Starting from the definitions of \(\sinh x\) and \(\cosh x\) in terms of exponentials, prove that $$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$$
2. Solve the equation \(4 \cosh ^ { 2 } x + 9 \sinh x = 13\), giving the answers in exact logarithmic form.
3. Show that there is only one stationary point on the curve $$y = 4 \cosh ^ { 2 } x + 9 \sinh x$$ and find the \(y\)-coordinate of the stationary point.
4. Show that \(\int _ { 0 } ^ { \ln 2 } \left( 4 \cosh ^ { 2 } x + 9 \sinh x \right) \mathrm { d } x = 2 \ln 2 + \frac { 33 } { 8 }\).

4

1. Prove, using exponential functions, that $$\sinh 2 x = 2 \sinh x \cosh x$$ Differentiate this result to obtain a formula for \(\cosh 2 x\).
2. Sketch the curve with equation \(y = \cosh x - 1\). The region bounded by this curve, the \(x\)-axis, and the line \(x = 2\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Find, correct to 3 decimal places, the volume generated. (You must show your working; numerical integration by calculator will receive no credit.)
3. Show that the curve with equation $$y = \cosh 2 x + \sinh x$$ has exactly one stationary point.
   Determine, in exact logarithmic form, the \(x\)-coordinate of the stationary point.

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 $$\tanh ^ { - 1 } x + \ln ( 1 - x ) = \ln \left( \frac { 4 } { 5 } \right)$$

8

1. By using the definition of \(\sinh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that $$\sinh ^ { 3 } x = \frac { 1 } { 4 } \sinh 3 x - \frac { 3 } { 4 } \sinh x$$
2. Find the range of values of the constant \(k\) for which the equation $$\sinh 3 x = k \sinh x$$ has real solutions other than \(x = 0\).
3. Given that \(k = 4\), solve the equation in part (ii), giving the non-zero answers in logarithmic form.

4

1. Using the definition of \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), prove that $$\cosh 2 x = 2 \cosh ^ { 2 } x - 1$$
2. Hence solve the equation $$\cosh 2 x - 7 \cosh x = 3$$ giving your answer in logarithmic form.

7

1. Using the definitions of hyperbolic functions in terms of exponentials, prove that $$\cosh x \cosh y - \sinh x \sinh y = \cosh ( x - y )$$
2. Given that \(\cosh x \cosh y = 9\) and \(\sinh x \sinh y = 8\), show that \(x = y\).
3. Hence find the values of \(x\) and \(y\) which satisfy the equations given in part (ii), giving the answers in logarithmic form.

8

1. Use the substitution \(x = \cosh ^ { 2 } u\) to find \(\int \sqrt { \frac { x } { x - 1 } } \mathrm {~d} x\), giving your answer in the form \(\mathrm { f } ( x ) + \ln ( \mathrm { g } ( x ) )\). \includegraphics[max width=\textwidth, alt={}, center]{d25d17c4-a87c-4dcf-900c-400086af6610-3_693_1041_927_593}
2. Hence calculate the exact area of the region between the curve \(y = \sqrt { \frac { x } { x - 1 } }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4\) (see diagram).
3. What can you say about the volume of the solid of revolution obtained when the region defined in part (ii) is rotated completely about the \(x\)-axis? Justify your answer.

1

1. By first expanding \(\left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right) ^ { 3 }\), or otherwise, show that \(\cosh 3 x \equiv 4 \cosh ^ { 3 } x - 3 \cosh x\).
2. Solve the equation \(\cosh 3 x = 6 \cosh x\), giving your answers in exact logarithmic form.

1

1. Starting from the definition of \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\), show that \(\cosh 2 x = 2 \cosh ^ { 2 } x - 1\).
2. Given that \(\cosh 2 x = k\), where \(k > 1\), express each of \(\cosh x\) and \(\sinh x\) in terms of \(k\).

7 The curve with equation $$y = \frac { x } { \cosh x }$$ has one stationary point for \(x > 0\).

1. Show that the \(x\)-coordinate of this stationary point satisfies the equation \(x \tanh x - 1 = 0\). The positive root of the equation \(x \tanh x - 1 = 0\) is denoted by \(\alpha\).
2. Draw a sketch showing (for positive values of \(x\) ) the graph of \(y = \tanh x\) and its asymptote, and the graph of \(y = \frac { 1 } { x }\). Explain how you can deduce from your sketch that \(\alpha > 1\).
3. Use the Newton-Raphson method, taking first approximation \(x _ { 1 } = 1\), to find further approximations \(x _ { 2 }\) and \(x _ { 3 }\) for \(\alpha\).
4. By considering the approximate errors in \(x _ { 1 }\) and \(x _ { 2 }\), estimate the error in \(x _ { 3 }\).

4

1. Solve the equation $$\sinh x + 4 \cosh x = 8$$ giving the answers in an exact logarithmic form.
2. Find the exact value of \(\int _ { 0 } ^ { 2 } \mathrm { e } ^ { x } \sinh x \mathrm {~d} x\).
3. 1. Differentiate \(\operatorname { arsinh } \left( \frac { 2 } { 3 } x \right)\) with respect to \(x\).
   2. Use integration by parts to show that \(\int _ { 0 } ^ { 2 } \operatorname { arsinh } \left( \frac { 2 } { 3 } x \right) \mathrm { d } x = 2 \ln 3 - 1\).

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. Given that \(k \geqslant 1\) and \(\cosh x = k\), show that \(x = \pm \ln \left( k + \sqrt { k ^ { 2 } - 1 } \right)\).
2. Find \(\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt { 4 x ^ { 2 } - 1 } } \mathrm {~d} x\), giving the answer in an exact logarithmic form.
3. Solve the equation \(6 \sinh x - \sinh 2 x = 0\), giving the answers in an exact form, using logarithms where appropriate.
4. Show that there is no point on the curve \(y = 6 \sinh x - \sinh 2 x\) at which the gradient is 5 .

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. Prove, using exponential functions, that $$\cosh 2 x = 1 + 2 \sinh ^ { 2 } x$$ Differentiate this result to obtain a formula for \(\sinh 2 x\).
2. Solve the equation $$2 \cosh 2 x + 3 \sinh x = 3$$ expressing your answers in exact logarithmic form.
3. Given that \(\cosh t = \frac { 5 } { 4 }\), show by using exponential functions that \(t = \pm \ln 2\). Find the exact value of the integral $$\int _ { 4 } ^ { 5 } \frac { 1 } { \sqrt { x ^ { 2 } - 16 } } \mathrm {~d} x$$