Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 20

Given to solve
int sec^4(2x)dx
let u= 2x
=> du = 2dx => dx = (1/2)du
so,
int sec^4(2x)dx
= int sec^4(u) (1/2) du
= (1/2) int sec^4(u) du
let us sovle
int sec^4(u) du
as by the formulae
int sec^n (x) dx
= (sec^(n-1) (x) (sinx))/(n-1) +((n-2)/(n-1))*(int sec^(n-2) (x) dx)

so,
int sec^4(u) du
=(sec^(4-1) (u) (sin u))/(4-1) +((4-2)/(4-1))*(int sec^(4-2) (u) du)
=(sec^(3) (u) (sin u))/(3) +(2/3)*(int sec^(2) (u) du)
= (sec^(3) (u) (sin u))/(3) +(2/3)*(tan u)
so,
int sec^4(2x)dx
= (1/2) int sec^4(u) du
=(1/2)[(sec^(3) (u) (sin u))/(3) +(2/3)*(tan u)]
but u= 2x
so,
(1/2)[(sec^(3) (u) (sin u))/(3) +(2/3)*(tan u)]
= (1/2)[(sec^(3) (2x) (sin (2x)))/(3) +(2/3)*(tan (2x))]
so,
int sec^4(2x)dx
=(1/2)[(sec^(3) (2x) (sin (2x)))/(3) +(2/3)*(tan (2x))] +c