We have an answer to "How do you integrate 1−x2"

Ernest Ryland

Answered question

2021-12-17

How do you integrate

\sqrt{1 - x^{2}}

Answer & Explanation

Thomas White

Beginner2021-12-18Added 40 answers

The answer is

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

Explanation:
Let

x = \sin θ, \Rightarrow, d x = \cos θ d θ

\cos θ = \sqrt{1 - x^{2}}

\sin 2 θ = 2 \sin θ \cos θ = 2 x \sqrt{1 - x^{2}}

Therefore, the integral is

I = \int \sqrt{1 - x^{2}} d x = \int \cos θ \cdot \cos θ d θ

= \int \cos^{2} θ d θ

\cos 2 θ = 2 \cos^{2} θ - 1

\cos^{2} θ = \frac{1 + \cos 2 θ}{2}

Therefore,

I = \frac{1}{2} \int (1 + \cos 2 θ) d θ

= \frac{1}{2} (θ + \frac{1}{2} \sin 2 θ)

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

Juan Spiller

Beginner2021-12-19Added 38 answers

Ill

RizerMix

Expert2021-12-29Added 656 answers

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

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