Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 15

Recall that int f(x) dx = F(x) +C where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration..
For the given problem, the integral: int 1/(sqrt(x)sqrt(1-x))dx
does not yet resemble any formula from table of integrals.
To evaluate this, we have to apply u-substitution by letting:
u =sqrt(x)
Square both sides: (u)^2=(sqrt(x))^2 , we get: u^2 =x
Then plug-in u^2 =x in sqrt(1-x) :
sqrt(1-x) = sqrt(1-u^2) .
Apply implicit differentiation on u^2 =x , we get: 2u du = dx .
Plug-in sqrt(x) =u , sqrt(1-x) = sqrt(1-u^2) , and dx= 2u du , we get:
int 1/(sqrt(x)sqrt(1-x))dx =int 1/(u*sqrt(1-u^2))*(2u du)
=int (2u du)/(usqrt(1-u^2))
Cancel out common factor u:
int 1/(usqrt(1-u^2))*(2u du)=int (2 du)/sqrt(1-u^2)
Apply the basic integration property: int c*f(x) dx = c int f(x) dx :
int(2 du)/(sqrt(1-u^2))= 2int(du)/sqrt(1-u^2)
The integral part resembles the basic integration formula for inverse sine function:
int (du)/sqrt((a^2 -u^2)) = arcsin(u/a) +C
Then,
2int(du)/sqrt(1-u^2) =2arcsin(u/1) +C
=2 arcsin(u) +C
Express it in terms of x by plug-in u =sqrt(x) for the final answer :
int 1/(sqrt(x)sqrt(1-x))dx =2 arcsin(sqrt(x)) +C