Question 4 - A-Level Maths

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 \(y = \operatorname { artanh } x\).
Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 - x ^ { 2 } }\).
Show, by integrating the result in part (ii), that \(y = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\).
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.
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}