Use the substitution \(\mathrm { u } = \mathrm { x } ^ { 2 } - 1\) to find \(\int \frac { x } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x\).
\includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-12_778_1548_1007_296}
The diagram shows the curve with equation \(\mathrm { y } = \cosh ^ { - 1 } \mathrm { x }\) together with a set of \(( N - 1 )\) rectangles of unit width.
By considering the sum of the areas of these rectangles, show that
$$\sum _ { r = 2 } ^ { N } \ln \left( r + \sqrt { r ^ { 2 } - 1 } \right) > N \ln \left( N + \sqrt { N ^ { 2 } - 1 } \right) - \sqrt { N ^ { 2 } - 1 }$$
Use a similar method to find, in terms of \(N\), an upper bound for \(\sum _ { \mathrm { r } = 2 } ^ { \mathrm { N } } \ln \left( \mathrm { r } + \sqrt { \mathrm { r } ^ { 2 } - 1 } \right)\).
Starting from the definitions of sech and tanh in terms of exponentials, prove that
$$1 - \operatorname { sech } ^ { 2 } t = \tanh ^ { 2 } t$$
\includegraphics[max width=\textwidth, alt={}]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_77_1547_360_347} ......................................................................................................................................... ......................................................................................................................................... . ........................................................................................................................................ ........................................................................................................................................ .......................................................................................................................................
\includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_72_1573_911_324}
\includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_67_1573_1005_324}
The curve \(C\) has parametric equations
$$\mathrm { x } = \frac { 1 } { 2 } \tanh ^ { 2 } \mathrm { t } + \text { Insecht } , \quad \mathrm { y } = 1 + \tanh ^ { 4 } \mathrm { t } , \quad \text { for } t > 0$$
Show that \(\frac { d y } { d x } = - 4 \operatorname { sech } ^ { 2 } t\).
Find the coordinates of the point on \(C\) with \(\frac { d ^ { 2 } y } { d x ^ { 2 } } = - \frac { 9 } { 2 }\), giving your answer in the form \(( a + \ln b , c )\) where \(a , b\) and \(c\) are rational numbers.
If you use the following page to complete the answer to any question, the question number must be clearly shown.