Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 32

Given,
int sqrt(1-x)/sqrt(x) dx
let us consider u= sqrt(x) ,
we can write it as u^2 = x
Differentiating on both sides we get
=> (2u)du = dx
Now let us solve the integral ,
int sqrt(1-x)/sqrt(x) dx
=int sqrt(1-u^2)/(u) ((2u)du) [as x= u^2 ]
=int 2*sqrt(1-u^2) du
= 2int sqrt(1-u^2) du---(1)
This can be solved by using the Trigonometric substitutions (Trig substitutions)
For sqrt(a-bx^2) we have to take x= sqrt(a/b) sin(v)

so here , For
2 int sqrt(1-u^2) du let us take u = sqrt(1/1) sin(v) = sin(v)
as u= sin(v) => du = cos(v) dv
now substituting in (1) we get
2int sqrt(1-u^2) du
= 2int sqrt(1-(sin(v))^2) (cos(v) dv)
= 2int sqrt((cos(v))^2) (cos(v) dv)
= 2 int cos(v) cos(v) dv
= 2 int cos^2(v) dv
=2 int (1+cos(2v))/2 dv
= (2/2) int (1+cos(2v)) dv
= int (1+cos(2v))dv
= [v+(1/2)(sin(2v))]+c
but ,
u = sin(v)
=> v= sin^(-1) u and u= sqrt(x)
so,
v= sin^(-1) (sqrt(x))

now,
v+1/2sin(2v)+c
=sin^(-1) (sqrt(x))+1/2sin(2sin^(-1) (sqrt(x)))+c
so,
int sqrt(1-x)/sqrt(x) dx
=
=sin^(-1) (sqrt(x))+1/2sin(2sin^(-1) (sqrt(x)))+c