Starting from the definitions of tanh and sech in terms of exponentials, prove that
$$1 - \tanh ^ { 2 } x = \operatorname { sech } ^ { 2 } x$$
Using the substitution \(\mathrm { u } = \tanh \mathrm { x }\), or otherwise, find \(\int \operatorname { sech } ^ { 2 } x \tanh ^ { 2 } x \mathrm {~d} x\).
It is given that, for \(n \geqslant 0 , \mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { \ln 3 } \operatorname { sech } ^ { \mathrm { n } } x \tanh ^ { 2 } x \mathrm { dx }\).
Show that, for \(n \geqslant 2\),
$$( n + 1 ) I _ { n } = \left( \frac { 4 } { 5 } \right) ^ { 3 } \left( \frac { 3 } { 5 } \right) ^ { n - 2 } + ( n - 2 ) I _ { n - 2 }$$
[You may use the result that \(\frac { \mathrm { d } } { \mathrm { d } x } ( \operatorname { sech } x ) = - \tanh x \operatorname { sech } x\).]
Find the value of \(I _ { 4 }\).
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