Integrate using hyperbolic substitution

7

1. Starting from the definitions of cosh and sinh in terms of exponentials, prove that $$2 \sinh ^ { 2 } A = \cosh 2 A - 1$$ \includegraphics[max width=\textwidth, alt={}, center]{1fa404d4-5e14-4356-9b6d-f176d5a9f6db-12_79_1556_358_347} \includegraphics[max width=\textwidth, alt={}]{1fa404d4-5e14-4356-9b6d-f176d5a9f6db-12_69_1575_466_328} ....................................................................................................................................... ........................................................................................................................................
2. A curve has equation \(\mathrm { y } = \mathrm { x } ^ { 2 }\), for \(0 \leqslant x \leqslant \frac { 2 } { 3 }\). The area of the surface generated when the curve is rotated through \(2 \pi\) radians about the \(x\)-axis is denoted by \(S\).
   Use the substitution \(\mathrm { X } = \frac { 1 } { 2 } \operatorname { sinhu }\) to show that \(S = \frac { 1 } { 32 } \pi \left( \frac { 820 } { 81 } - \ln 3 \right)\).

4. Using the substitution \(x = 4 \cosh \theta\) show that $$\int \frac { 1 } { \left( x ^ { 2 } - 16 \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x = \frac { a x } { \sqrt { x ^ { 2 } - 16 } } + c \quad | x | > 4$$ where \(a\) is a constant to be determined and \(c\) is an arbitrary constant.
(6)

1. (a) Show that, under the substitution \(x = \frac { 3 } { 4 } \sinh u\),

$$\int \frac { x ^ { 2 } } { \sqrt { 16 x ^ { 2 } + 9 } } \mathrm {~d} x = k \int ( \cosh 2 u - 1 ) \mathrm { d } u$$ where \(k\) is a constant to be determined.
(b) Hence show that $$\int _ { 0 } ^ { 1 } \frac { 64 x ^ { 2 } } { \sqrt { 16 x ^ { 2 } + 9 } } \mathrm {~d} x = p + q \ln 3$$ where \(p\) and \(q\) are rational numbers to be found.

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.

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.

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.

8

1. Using exponentials, show that \(\cosh 2 u \equiv 2 \sinh ^ { 2 } u + 1\).
2. By differentiating both sides of the identity in part (a) with respect to \(u\), show that \(\sinh 2 u \equiv 2 \sinh u \cosh u\).
3. Use the substitution \(\mathrm { x } = \sinh ^ { 2 } \mathrm { u }\) to find \(\int \sqrt { \frac { x } { x + 1 } } \mathrm {~d} x\). Give your answer in the form asinh \(^ { - 1 } \mathrm {~b} \sqrt { \mathrm { x } } + \mathrm { f } ( \mathrm { x } )\) where \(a\) and \(b\) are integers and \(\mathrm { f } ( x )\) is a function to be determined.
4. Hence determine the exact area of the region between the curve \(\mathrm { y } = \sqrt { \frac { \mathrm { x } } { \mathrm { x } + 1 } }\), the \(x\)-axis, the line \(x = 1\) and the line \(x = 2\). Give your answer in the form \(\mathrm { p } + \mathrm { q } \mid \mathrm { nr }\) where \(p , q\) and \(r\) are numbers to be determined.

6

1. Show that \(\frac { 1 } { 5 \cosh x - 3 \sinh x } = \frac { \mathrm { e } ^ { x } } { m + \mathrm { e } ^ { 2 x } }\), where \(m\) is an integer.
2. Use the substitution \(u = \mathrm { e } ^ { x }\) to show that $$\int _ { 0 } ^ { \ln 2 } \frac { 1 } { 5 \cosh x - 3 \sinh x } \mathrm {~d} x = \frac { \pi } { 8 } - \frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$$

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}

16

1. Show using exponentials that \(\cosh 2 u = 1 + 2 \sinh ^ { 2 } u\).
2. Show that \(\int _ { 0 } ^ { 2 } \frac { x ^ { 2 } } { \sqrt { 4 + x ^ { 2 } } } \mathrm {~d} x = 2 \sqrt { 2 } - 2 \ln ( 1 + \sqrt { 2 } )\).

5

1. Given that \(u = \cosh ^ { 2 } x\), show that \(\frac { \mathrm { d } u } { \mathrm {~d} x } = \sinh 2 x\).
2. Hence show that $$\int _ { 0 } ^ { 1 } \frac { \sinh 2 x } { 1 + \cosh ^ { 4 } x } \mathrm {~d} x = \tan ^ { - 1 } \left( \cosh ^ { 2 } 1 \right) - \frac { \pi } { 4 }$$

7 A curve has equation \(y = 4 \sqrt { x }\).

1. Show that the length of arc \(s\) of the curve between the points where \(x = 0\) and \(x = 1\) is given by $$s = \int _ { 0 } ^ { 1 } \sqrt { \frac { x + 4 } { x } } \mathrm {~d} x$$
2. 1. Use the substitution \(x = 4 \sinh ^ { 2 } \theta\) to show that $$\int \sqrt { \frac { x + 4 } { x } } \mathrm {~d} x = \int 8 \cosh ^ { 2 } \theta \mathrm {~d} \theta$$
   2. Hence show that $$s = 4 \sinh ^ { - 1 } 0.5 + \sqrt { 5 }$$

3

1. Using exponentials, show that \(\cosh 2 u \equiv 2 \sinh ^ { 2 } u + 1\).
2. By differentiating both sides of the identity in part (a) with respect to \(u\), show that \(\sinh 2 u \equiv 2 \sinh u \cosh u\).
3. Use the substitution \(x = \sinh ^ { 2 } u\) to find \(\int \sqrt { \frac { x } { x + 1 } } \mathrm {~d} x\). Give your answer in the form \(a \sinh ^ { - 1 } b \sqrt { x } + \mathrm { f } ( x )\) where \(a\) and \(b\) are integers and \(\mathrm { f } ( x )\) is a function to be determined.
4. Hence determine the exact area of the region between the curve \(y = \sqrt { \frac { x } { x + 1 } }\), the \(x\)-axis, the line \(x = 1\) and the line \(x = 2\). Give your answer in the form \(p + q \ln r\) where \(p , q\) and \(r\) are numbers to be determined.

\includegraphics{figure_6} The curve \(C\) shown in Fig. 1 has equation \(y^2 = 4x\), \(0 \leq x \leq 1\). The part of the curve in the first quadrant is rotated through \(2\pi\) radians about the \(x\)-axis.

1. Show that the surface area of the solid generated is given by $$4\pi \int_0^1 \sqrt{1+x} \, dx.$$ [4]
2. Find the exact value of this surface area. [3]
3. Show also that the length of the curve \(C\), between the points \((1, -2)\) and \((1, 2)\), is given by $$2 \int_0^1 \sqrt{\frac{x+1}{x}} \, dx.$$ [3]
4. Use the substitution \(x = \sinh^2 \theta\) to show that the exact value of this length is $$2[\sqrt{2} + \ln(1 + \sqrt{2})].$$ [6]

\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)