`int 1 / (sqrt(x)sqrt(1-x)) dx` Find the indefinite integral

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) =...

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`