Prove inverse hyperbolic logarithmic form

2. Given that $$\cosh y = x \quad \text { and } \quad y < 0$$ use the definition of coshy in terms of exponential functions to prove that $$y = \ln \left( x - \sqrt { x ^ { 2 } - 1 } \right)$$

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

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}

4

1. Prove, using exponential functions, that \(\cosh ^ { 2 } u - \sinh ^ { 2 } u = 1\).
2. Given that \(y = \operatorname { arsinh } x\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 1 + x ^ { 2 } } }$$ and that $$y = \ln \left( x + \sqrt { 1 + x ^ { 2 } } \right)$$
3. Show that $$\int _ { 0 } ^ { 2 } \frac { 1 } { \sqrt { 4 + 9 x ^ { 2 } } } \mathrm {~d} x = \frac { 1 } { 3 } \ln ( 3 + \sqrt { 10 } )$$
4. Find, in exact logarithmic form, $$\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 + x ^ { 2 } } } \operatorname { arsinh } x \mathrm {~d} x$$

6

1. Prove that the derivative of \(\cos ^ { - 1 } x\) is \(- \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\). A curve has equation \(y = \cos ^ { - 1 } \left( 1 - x ^ { 2 } \right)\), for \(0 < x < \sqrt { 2 }\).
2. Find and simplify \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), and hence show that $$\left( 2 - x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = x \frac { \mathrm {~d} y } { \mathrm {~d} x }$$
3. Given that \(y = \sinh ^ { - 1 } x\), prove that \(y = \ln \left( x + \sqrt { x ^ { 2 } + 1 } \right)\).
4. It is given that \(x\) satisfies the equation \(\sinh ^ { - 1 } x - \cosh ^ { - 1 } x = \ln 2\). Use the logarithmic forms for \(\sinh ^ { - 1 } x\) and \(\cosh ^ { - 1 } x\) to show that $$\sqrt { x ^ { 2 } + 1 } - 2 \sqrt { x ^ { 2 } - 1 } = x$$ Hence, by squaring this equation, find the exact value of \(x\).

6

1. Given that \(y = \cosh ^ { - 1 } x\), show that \(y = \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)\).
2. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \cosh ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 2 } - 1 } }\).
3. Solve the equation \(\cosh x = 3\), giving your answers in logarithmic form.

1 By first expressing \(\tanh y\) in terms of exponentials, prove that \(\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\).

6

1. Matthew is finding a formula for the inverse function \(\operatorname { arsinh } x\). He writes his steps as follows: $$\begin{gathered} \text { Let } y = \sinh x \\ y = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } \right) \\ 2 y = \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } \\ 0 = \mathrm { e } ^ { x } - 2 y - \mathrm { e } ^ { - x } \\ 0 = \left( \mathrm { e } ^ { x } \right) ^ { 2 } - 2 y \mathrm { e } ^ { x } - 1 \\ 0 = \left( \mathrm { e } ^ { x } - y \right) ^ { 2 } - y ^ { 2 } - 1 \\ y ^ { 2 } + 1 = \left( \mathrm { e } ^ { x } - y \right) ^ { 2 } \\ \pm \sqrt { y ^ { 2 } + 1 } = \mathrm { e } ^ { x } - y \\ y \pm \sqrt { y ^ { 2 } + 1 } = \mathrm { e } ^ { x } \end{gathered}$$ To find the inverse function, swap \(x\) and \(y : x \pm \sqrt { x ^ { 2 } + 1 } = \mathrm { e } ^ { y }\) $$\begin{gathered} \ln \left( x \pm \sqrt { x ^ { 2 } + 1 } \right) = y \\ \operatorname { arsinh } x = \ln \left( x \pm \sqrt { x ^ { 2 } + 1 } \right) \end{gathered}$$ Identify, and explain, the error in Matthew's proof. 6
2. Solve \(\ln \left( x + \sqrt { x ^ { 2 } + 1 } \right) = 3\)

5

1. Using the definition \(\tanh y = \frac { \mathrm { e } ^ { y } - \mathrm { e } ^ { - y } } { \mathrm { e } ^ { y } + \mathrm { e } ^ { - y } }\), show that, for \(| x | < 1\), $$\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)$$
2. Hence, or otherwise, show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \tanh ^ { - 1 } x \right) = \frac { 1 } { 1 - x ^ { 2 } }\).
3. Use integration by parts to show that $$\int _ { 0 } ^ { \frac { 1 } { 2 } } 4 \tanh ^ { - 1 } x \mathrm {~d} x = \ln \left( \frac { 3 ^ { m } } { 2 ^ { n } } \right)$$ where \(m\) and \(n\) are positive integers.

8. By writing \(x = \sinh y\), show that \(\sinh ^ { - 1 } x = \ln \left( x + \sqrt { x ^ { 2 } + 1 } \right)\).

1. (a) Prove that

$$\tanh ^ { - 1 } ( x ) = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right) \quad - k < x < k$$ stating the value of the constant \(k\).
(b) Hence, or otherwise, solve the equation $$2 x = \tanh ( \ln \sqrt { 2 - 3 x } )$$

8

1. Given that \(u = \tanh x\), use the definition of \(\tanh x\) in terms of exponentials to show that $$x = \frac { 1 } { 2 } \ln \left( \frac { 1 + u } { 1 - u } \right)$$
2. Solve the equation \(4 \tanh ^ { 2 } x + \tanh x - 3 = 0\), giving the solution in the form \(a \ln b\) where \(a\) and \(b\) are rational numbers to be determined.
3. Explain why the equation in part (b) has only one root.

8

1. Prove that
   \(\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\)
   8
2. Prove that the graphs of $$y = \sinh x \quad \text { and } \quad y = \cosh x$$ do not intersect.

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.

4

1. Given that \(u = \tanh x\), use the definition of \(\tanh x\) in terms of exponentials to show that $$x = \frac { 1 } { 2 } \ln \left( \frac { 1 + u } { 1 - u } \right)$$
2. Solve the equation \(4 \tanh ^ { 2 } x + \tanh x - 3 = 0\), giving the solution in the form \(a \ln b\) where \(a\) and \(b\) are rational numbers to be determined.
3. Explain why the equation in part (b) has only one root.