Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 17

Given to solve,
int sqrt(9+16x^2) dx
by using the trig substitution , we can solve the integral
for sqrt(a +bx^2) dx the x is given as
x= sqrt(a/b) tan(u)
so,
for the integral
int sqrt(9+16x^2) dx
let x=sqrt(9/16) tan(u) = (3/4) tan(u)
=> dx = (3/4) sec^2(u) du
so,
int sqrt(9+16x^2) dx
=int [sqrt(9(1+16/9 x^2))] ((3/4) sec^2(u) du)
= 3 int [sqrt(1+(16/9)x^2)] ((3/4) sec^2(u) du)
= 3 int sqrt(1+(16/9)((3/4) tan(u))^2) ((3/4) sec^2(u) du)
= 3 int [sqrt(1+(16/9)(9/16)(tan^2 u))] ((3/4) sec^2(u) du)
= (9/4) int sqrt(1+tan^2 u) (sec^2(u) du)
= (9/4) int sqrt(sec^2 u) (sec^2(u) du)
= (9/4) int sec u (sec^2(u) du)
= (9/4) int (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 ,
(9/4)int sec^(3) (u) du
= (9/4)[(sec^(3-1) (u) sin(u))/(3-1) + ((3-2)/(3-1)) int sec^(3-2) (u)du]
= (9/4)[(sec^(2) (u) sin(u))/(2) + ((1)/(2)) int sec (u)du]
=(9/4)[(sec^(2) (u) sin(u))/(2) + (1/2) (ln(tan(u)+sec(u)))]
but we know
x= (3/4) tan(u)
= > 4x/3 = tan(u)
=> u =arctan(4x/3)
so,
=(9/4)[(sec^(2) (u) sin(u))/(2) + (1/2) (ln(tan(u)+sec(u)))]
=(9/4)[(sec^(2) (arctan(4x/3)) sin(arctan(4x/3)))/(2) + (1/2) (ln(tan(arctan(4x/3))+sec(arctan(4x/3))))]

=(9/4)[(sec^(2) (arctan(4x/3)) sin(arctan(4x/3)))/(2) + (1/2) (ln((4x/3))+sec(arctan(4x/3)))]+c