Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 21

Given to solve ,
int 1/sqrt(16-x^2) dx
using the Trig substitutions
for sqrt(a-bx^2)
x= sqrt(a/b) sin(u)
so for ,
int 1/sqrt(16-x^2) dx --------(1)
so , x can be
x= sqrt(16/1) sin(u)
= 4sin(u)
=> dx = 4cos(u) du
so,for (1) we get
int 1/sqrt(16-(4sin(u))^2) (4cos(u) du)
=int (4cos(u))/sqrt(16-16(sin(u))^2) du
= int (4cos(u))/(4sqrt(1-sin^2(u))) du
= int (4cos(u))/(4sqrt(cos^2(u))) du
= int (4cos(u))/(4cos(u)) du
= int (1) du
= u+c
but x= 4sin(u)
=> x/4 = sin(u)
=> u = arcsin(x/4)
so ,
=> u+c
= arcsin(x/4)+c
so ,
int 1/sqrt(16-x^2) dx = arcsin(x/4)+c