Question

Integrate the integral:


`\int (1)/(√(x) √(1-x) )`

Answer 1

`\displaystyle \int {(1)/(√(x) √(1-x) ) } \, dx = 2arcsin(√(x)) + C`

General Formulas and Concepts:

Calculus

• U-Substitution
• Integration Property:
  `\int {cf(x)} \, dx = c\int {f(x)} \, dx`
• Arctrig Integration:
  `\displaystyle \int\limits {(1)/(√(a^2-u^2) ) } \, du = arcsin((u)/(a) ) + C`

Explanation:

Step 1: Define

`\displaystyle \int\limits {(1)/(√(x) √(1-x) ) } \, dx`

Step 2: Identify

Set up variables for u-substitution and integral trig.

`\displaystyleu = √(x)\\a = 1\\du = (1)/(2√(x))dx`

Step 3: integrate

1. Rewrite Integral:
   `\displaystyle 2\int {(1)/(2√(x) √(1-x) ) } \, dx`
2. [Integral] U-Substitution:
   `\displaystyle 2\int {\frac{du}{\sqrt{1^2-(√(x))^2 } }`
3. [Integral] U-Sub Arctrig:
   `\displaystyle 2\int {(du)/(√(1^2-u^2 ) )`
4. [Integral] Arctrig:
   `\displaystyle 2arcsin((u)/(1)) + C`
5. Simplify:
   `\displaystyle 2arcsin(u) + C`
6. Back-substitute:
   `\displaystyle 2arcsin(√(x) ) + C`

Answer 2

We are given the expression:

`\int\limits {\frac{1}{\sqrt[]{x} }*\frac{1}{\sqrt[]{ 1-x^(2)}} } \, dx`

U-substitution:

let u =
`x^(1/2)` [So, x = u²]

`(du)/(dx) = \frac{1}{2\sqrt[]{x}}` [differentiating both sides wrt x]

dx = du*2√(x)

dx = 2u(du) [Since u = √x]

Finding the Integral:

Plugging these values in the given expression:

`\int\limits {u^(-1)*(1-u^(2))^(-1/2) } \, 2\sqrt[]{x}*du`

`\int\limits {(1)/(u)*\frac{1}{\sqrt[]{ 1-u^(2)}}*2u } \, du`

The 'u' in the numerator and denominator will cancel out

`\int\limits {\frac{1}{\sqrt[]{ 1-u^(2)}}*2 } \, du`

Since 2 is a constant

`2\int\limits {\frac{1}{\sqrt[]{ 1-u^(2)}} } \, du`

Using the property:
`\int\limits {\frac{1}{\sqrt[]{1-x^(2)} }} \, dx` = ArcSin(x) + C

2*ArcSin(u) + C

Since u = √x :

2*ArcSin(√x) + C