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Integrate the integral:

[tex]\int \frac{1}{\sqrt{x} \sqrt{1-x} }[/tex]

Sagot :

Space

Answer:

[tex]\displaystyle \int {\frac{1}{\sqrt{x} \sqrt{1-x} } } \, dx = 2arcsin(\sqrt{x}) + C[/tex]

General Formulas and Concepts:

Calculus

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

Step-by-step explanation:

Step 1: Define

[tex]\displaystyle \int\limits {\frac{1}{\sqrt{x} \sqrt{1-x} } } \, dx[/tex]

Step 2: Identify

Set up variables for u-substitution and integral trig.

[tex]\displaystyleu = \sqrt{x}\\a = 1\\du = \frac{1}{2\sqrt{x}}dx[/tex]

Step 3: integrate

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

We are given the expression:                                                                              

[tex]\int\limits {\frac{1}{\sqrt[]{x} }*\frac{1}{\sqrt[]{ 1-x^{2}}} } \, dx[/tex]

U-substitution:                                                                                                      

let u = [tex]x^{1/2}[/tex]             [So, x = u²]

[tex]\frac{du}{dx} = \frac{1}{2\sqrt[]{x}}[/tex]                 [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:

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

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

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

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

Since 2 is a constant

[tex]2\int\limits {\frac{1}{\sqrt[]{ 1-u^{2}}} } \, du[/tex]

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

2*ArcSin(u) + C

Since u = √x :

2*ArcSin(√x) + C

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