4.07e Inverse hyperbolic: definitions, domains, ranges

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.

9

1. Given that \(y = \sinh ^ { - 1 } x\), prove that \(y = \ln \left( x + \sqrt { x ^ { 2 } + 1 } \right)\).
2. It is given that, for non-negative integers \(n\), $$I _ { n } = \int _ { 0 } ^ { \alpha } \sinh ^ { n } \theta \mathrm {~d} \theta$$ where \(\alpha = \sinh ^ { - 1 } 1\). Show that $$n I _ { n } = \sqrt { 2 } - ( n - 1 ) I _ { n - 2 } , \quad \text { for } n \geqslant 2 .$$
3. Evaluate \(I _ { 4 }\), giving your answer in terms of \(\sqrt { 2 }\) and logarithms.

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, from definitions involving exponential functions, that $$\cosh 2 u = 2 \sinh ^ { 2 } u + 1$$
2. Prove that, if \(y \geqslant 0\) and \(\cosh y = u\), then \(y = \ln \left( u + \sqrt { } \left( u ^ { 2 } - 1 \right) \right)\).
3. Using the substitution \(2 x = \cosh u\), show that $$\int \sqrt { 4 x ^ { 2 } - 1 } \mathrm {~d} x = a x \sqrt { 4 x ^ { 2 } - 1 } - b \operatorname { arcosh } 2 x + c$$ where \(a\) and \(b\) are constants to be determined and \(c\) is an arbitrary constant.
4. Find \(\int _ { \frac { 1 } { 2 } } ^ { 1 } \sqrt { 4 x ^ { 2 } - 1 } \mathrm {~d} x\), expressing your answer in an exact form involving logarithms.

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, $$\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 + x ^ { 2 } } } \operatorname { arsinh } x \mathrm {~d} x$$

4

1. Given that \(\cosh y = x\), show that \(y = \pm \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)\) and that \(\operatorname { arcosh } x = \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)\).
2. Find \(\int _ { \frac { 4 } { 5 } } ^ { 1 } \frac { 1 } { \sqrt { 25 x ^ { 2 } - 16 } } \mathrm {~d} x\), expressing your answer in an exact logarithmic form.
3. Solve the equation $$5 \cosh x - \cosh 2 x = 3$$ giving your answers in an exact logarithmic form.

2 It is given that \(\mathrm { f } ( x ) = \tanh ^ { - 1 } x\).

1. Show that \(\mathrm { f } ^ { \prime \prime \prime } ( x ) = \frac { 2 \left( 1 + 3 x ^ { 2 } \right) } { \left( 1 - x ^ { 2 } \right) ^ { 3 } }\).
2. Hence find the Maclaurin series for \(\mathrm { f } ( x )\), up to and including the term in \(x ^ { 3 }\).

8

1. Without using a calculator, show that \(\sinh \left( \cosh ^ { - 1 } 2 \right) = \sqrt { 3 }\).
2. It is given that, for non-negative integers \(n\), $$I _ { n } = \int _ { 0 } ^ { \beta } \cosh ^ { n } x \mathrm {~d} x , \quad \text { where } \beta = \cosh ^ { - 1 } 2$$ Show that \(n I _ { n } = 2 ^ { n - 1 } \sqrt { 3 } + ( n - 1 ) I _ { n - 2 }\), for \(n \geqslant 2\).
3. Evaluate \(I _ { 5 }\), giving your answer in the form \(k \sqrt { 3 }\).

5 A function is defined by \(\mathrm { f } ( x ) = \sinh ^ { - 1 } x + \sinh ^ { - 1 } \left( \frac { 1 } { x } \right)\), for \(x \neq 0\).

1. When \(x > 0\), show that the value of \(\mathrm { f } ( x )\) for which \(\mathrm { f } ^ { \prime } ( x ) = 0\) is \(2 \ln ( 1 + \sqrt { 2 } )\).
2. \includegraphics[max width=\textwidth, alt={}, center]{72a1330a-c6dc-4f3a-9b0e-333b099f4509-3_497_659_520_708} The diagram shows the graph of \(y = \mathrm { f } ( x )\) for \(x > 0\). Sketch the graph of \(y = \mathrm { f } ( x )\) for \(x < 0\) and state the range of values that \(\mathrm { f } ( x )\) can take for \(x \neq 0\).

3 It is given that \(\mathrm { f } ( x ) = \tanh ^ { - 1 } \left( \frac { 1 - x } { 3 + x } \right)\) for \(x > - 1\).

1. Show that \(\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 1 } { 2 ( x + 1 ) ^ { 2 } }\).
2. Hence find the Maclaurin series for \(\mathrm { f } ( x )\) up to and including the term in \(x ^ { 2 }\).

1 Find \(\int _ { 0 } ^ { 2 } \frac { 1 } { \sqrt { 4 + x ^ { 2 } } } \mathrm {~d} x\), giving your answer exactly in logarithmic form.

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 \(\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\).

5

1. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sinh ^ { - 1 } ( 2 x ) \right) = \frac { 2 } { \sqrt { 4 x ^ { 2 } + 1 } }\).
2. Find \(\int \frac { 1 } { \sqrt { 2 - 2 x + x ^ { 2 } } } \mathrm {~d} x\).

3

1. Show that \(\frac { \mathrm { d } } { \mathrm { d } u } \left( \sinh ^ { - 1 } u \right) = \frac { 1 } { \sqrt { u ^ { 2 } + 1 } }\).
2. Find the equation of the normal to the graph of \(\mathrm { y } = \sinh ^ { - 1 } 2 \mathrm { x }\) at the point where \(x = \sqrt { 6 }\). Give your answer in the form \(\mathrm { y } = \mathrm { mx } + \mathrm { c }\) where \(m\) and \(c\) are given in exact, non-hyperbolic form.

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 \(S\).

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 $$2 x = \tanh ( \ln \sqrt { 2 - 3 x } )$$

3. $$f ( x ) = \frac { 1 } { \sqrt { 4 x ^ { 2 } + 9 } }$$

1. Using a substitution, that should be stated clearly, show that $$\int \mathrm { f } ( x ) \mathrm { d } x = A \sinh ^ { - 1 } ( B x ) + c$$ where \(c\) is an arbitrary constant and \(A\) and \(B\) are constants to be found.
2. Hence find, in exact form in terms of natural logarithms, the mean value of \(\mathrm { f } ( x )\) over the interval \([ 0,3 ]\).

1. Solutions based entirely on graphical or numerical methods are not acceptable.

\begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation $$y = \operatorname { arsinh } x \quad x \geqslant 0$$ and the straight line with equation \(y = \beta\) The line and the curve intersect at the point with coordinates \(( \alpha , \beta )\) Given that \(\beta = \frac { 1 } { 2 } \ln 3\)

1. show that \(\alpha = \frac { 1 } { \sqrt { 3 } }\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve with equation \(y = \operatorname { arsinh } x\), the \(y\)-axis and the line with equation \(y = \beta\) The region \(R\) is rotated through \(2 \pi\) radians about the \(y\)-axis.
2. Use calculus to find the exact value of the volume of the solid generated.

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.

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.

5. Given that \(y = ( \operatorname { arcosh } 3 x ) ^ { 2 }\), where \(3 x > 1\), show that

1. \(\left( 9 x ^ { 2 } - 1 \right) \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = 36 y\),
2. \(\left( 9 x ^ { 2 } - 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 9 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 18\).

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.