Expert Answer to "Evaluate the following integral: ∫1−x1−x2dx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

lunnatican4

Answered question

2022-01-05

Evaluate the following integral:

\int \frac{1 - x}{\sqrt{1 - x^{2}}} d x

Answer & Explanation

Papilys3q

Beginner2022-01-06Added 34 answers

Step 1: Given that
Evaluate the following integral:

\int \frac{1 - x}{\sqrt{1 - x^{2}}} d x

Step 2: Solve
We have,

\int \frac{1 - x}{\sqrt{1 - x^{2}}} d x = \int \frac{1}{\sqrt{1 - x^{2}}} d x - \int \frac{x}{\sqrt{1 - x^{2}}} d x

= \sin^{- 1} (x) - \int \frac{x}{\sqrt{t}} \times \frac{d t}{- 2 x}

= \sin^{- 1} x + \frac{1}{2} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C

= \sin^{- 1} x + \sqrt{t} + C

= \sin^{- 1} x + \sqrt{1 - x^{2}} + C

Robert Pina

Beginner2022-01-07Added 42 answers

Given:

\int \frac{1 - x}{\sqrt{1 - x^{2}}} d x

\int \frac{1}{\sqrt{1 - x^{2}}} - \frac{x}{\sqrt{1 - x^{2}}} d x

Use properties

\int \frac{1}{\sqrt{1 - x^{2}}} d x - \int \frac{x}{\sqrt{1 - x^{2}}} d x

\arcsin (x) + \sqrt{1 - x^{2}}

Solution:

\arcsin (x) + \sqrt{1 - x^{2}} + C

karton

Expert2022-01-11Added 613 answers

$\int \frac{1 - x}{\sqrt{1 - x^{2}}} d x = \int (\frac{1}{\sqrt{1 - x^{2}}} - \frac{x}{\sqrt{1 - x^{2}}}) d x = \int \frac{1}{\sqrt{1 - x^{2}}} d x - \int \frac{x}{\sqrt{1 - x^{2}}} d x Now calculate \int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin (x) Now calculate \int \frac{x}{\sqrt{1 - x^{2}}} d x = - \frac{1}{2} \int \frac{1}{\sqrt{u}} d u \int \frac{1}{\sqrt{u}} d u = 2 \sqrt{u} - \frac{1}{2} \int \frac{1}{\sqrt{u}} d u = - \sqrt{u} = - \sqrt{1 - x^{2}} \int \frac{1}{\sqrt{1 - x^{2}}} d x - \int \frac{x}{\sqrt{1 - x^{2}}} d x = \arcsin (x) + \sqrt{1 - x^{2}} = \arcsin (x) + \sqrt{1 - x^{2}} + C$

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