Question 5 - A-Level Maths

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