Show that
$$\int _ { 0 } ^ { \ln 2 } \mathrm { e } ^ { 1 - 2 x } \mathrm {~d} x = \frac { 3 } { 8 } \mathrm { e }$$
Use the substitution \(u = \tan x\) to find the exact value of
$$\int _ { 0 } ^ { \frac { \pi } { 4 } } \sec ^ { 4 } x \sqrt { \tan x } d x$$
(8 marks)