Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 18

Given to solve,
int sqrt(4+x^2) dx
using the Trig Substitutions we can solve these type of integrals easily and the solution is as follows

for sqrt(a+bx^2) we can take x= sqrt(a/b) tan(u)
so ,For
int sqrt(4+x^2) dx
the x= sqrt(4/1)tan(u)= 2tan(u)
=> dx= 2sec^(2) (u) du
so,
int sqrt(4+x^2) dx
= int sqrt(4+(2tan(u))^2) (2sec^(2) (u) du)
= int sqrt(4+4(tan(u))^2) (2sec^(2) (u) du)
=int sqrt(4(1+(tan(u))^2)) (2sec^(2) (u) du)
= int 2sqrt(1+tan^2(u))(2sec^(2) (u) du)
= int 2sec(u)(2sec^(2) (u) du)
= int 4sec^(3) (u) du
= 4int sec^(3) (u) du
by applying the Integral Reduction
int sec^(n) (x) dx
= (sec^(n-1) (x) sin(x))/(n-1) + ((n-2)/(n-1)) int sec^(n-2) (x) dx
so ,
4int sec^(3) (u) du
= 4[(sec^(3-1) (u) sin(u))/(3-1) + ((3-2)/(3-1)) int sec^(3-2) (u)du]
= 4[(sec^(2) (u) sin(u))/(2) + ((1)/(2)) int sec (u)du]
=4[(sec^(2) (u) sin(u))/(2) + (1/2) (ln(tan(u)+sec(u)))]
but x= 2tan(u)
=> x/2 = tan(u)
u = tan^(-1) (x/2)
so,
4[(sec^(2) (u) sin(u))/(2) + (1/2) (ln(tan(u)+sec(u)))]
=4[(sec^(2) ( tan^(-1) (x/2)) sin( tan^(-1) (x/2)))/(2) + (1/2) (ln(tan( tan^(-1) (x/2))+sec( tan^(-1) (x/2))))]
=4[(sec^(2) ( tan^(-1) (x/2)) sin( tan^(-1) (x/2)))/(2) + (1/2) (ln((x/2))+sec( tan^(-1) (x/2)))] +c