Mathematics, 08.01.2021 22:00 573589

Integrate the integral:

$\int \frac{1}{\sqrt{x} \sqrt{1-x} }$

Answers: 3

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Answer from: likevlad2014

$\displaystyle \int {\frac{1}{\sqrt{x} \sqrt{1-x} } } \, dx = 2arcsin(\sqrt{x}) + C$

General Formulas and Concepts:

Calculus

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

Step-by-step explanation:

Step 1: Define

$\displaystyle \int\limits {\frac{1}{\sqrt{x} \sqrt{1-x} } } \, dx$

Step 2: Identify

Set up variables for u-substitution and integral trig.

$\displaystyleu = \sqrt{x}\\a = 1\\du = \frac{1}{2\sqrt{x}}dx$

Step 3: integrate

Rewrite Integral: $\displaystyle 2\int {\frac{1}{2\sqrt{x} \sqrt{1-x} } } \, dx$ [Integral] U-Substitution: $\displaystyle 2\int {\frac{du}{\sqrt{1^2-(\sqrt{x})^2 } }$ [Integral] U-Sub Arctrig: $\displaystyle 2\int {\frac{du}{\sqrt{1^2-u^2 } }$ [Integral] Arctrig: $\displaystyle 2arcsin(\frac{u}{1}) + C$ Simplify: $\displaystyle 2arcsin(u) + C$ Back-substitute: $\displaystyle 2arcsin(\sqrt{x} ) + C$

Answer from: jhernandezvaldez142

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]

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

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

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

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

Answer from: Quest

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a). here, that comes to

)

x = = 1. f(1) = 1^2 - 2(1) - 8 = - 9. vertex is at (1, -9).

2(1)

find another point on the curve: let x=0, for which y=-8. then (0, -8) is on the curve. plot the vertex and this point and draw a smooth parabola thru both.

hope this

Use the parabola tool to graph the quadratic function f(x)=(x-5)^2+1 graph the parabola by first plo

Answer from: Quest

Gotta be d the music of previous eras

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