integration (squareroot)

May 2010
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What would be the best/quickest way to integrate

squareroot(1-x^2) ?
 

earboth

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What would be the best/quickest way to integrate

squareroot(1-x^2) ?
1. Use integration by parts:

\(\displaystyle \int(\sqrt{1-x^2}) dx = \int(1 \cdot \sqrt{1-x^2}) dx\)

2. In the second step keep in mind that

\(\displaystyle \int \left(\frac1{\sqrt{1-x^2}}\right) dx = \arcsin(x)\)
 
Oct 2009
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1. Use integration by parts:

\(\displaystyle \int(\sqrt{1-x^2}) dx = \int(1 \cdot \sqrt{1-x^2}) dx\)

2. In the second step keep in mind that

\(\displaystyle \int \left(\frac1{\sqrt{1-x^2}}\right) dx = \arcsin(x)\)

I don't think the above helps: \(\displaystyle u=\sqrt{1-x^2}\Longrightarrow u'=-\frac{x}{\sqrt{1-x^2}}\,,\,\,v'=1\Longrightarrow v=x\) , so we get the integral \(\displaystyle \int\frac{x^2}{\sqrt{1-x^2}}\,dx\) ...which not only is not an arcsine but it doesn't look easy at all...

Unless, of course, I missed something.

Tonio
 

Prove It

MHF Helper
Aug 2008
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What would be the best/quickest way to integrate

squareroot(1-x^2) ?
Trigonometric or Hyperbolic substitution works well...

Let \(\displaystyle x = \sin{\theta}\) so that \(\displaystyle dx = \cos{\theta}\,d\theta\).


The integral becomes

\(\displaystyle \int{\sqrt{1 - x^2}\,dx} = \int{\sqrt{1 - \sin^2{\theta}}\,\cos{\theta}\,d\theta}\)

\(\displaystyle = \int{\cos{\theta}\cos{\theta}\,d\theta}\)

\(\displaystyle = \int{\cos^2{\theta}\,d\theta}\)

\(\displaystyle = \int{\frac{1}{2}\cos{2\theta} + \frac{1}{2}\,d\theta}\)

\(\displaystyle = \frac{1}{4}\sin{2\theta} + \frac{1}{2}\theta + C\)

\(\displaystyle = \frac{2\sin{\theta}\cos{\theta}}{4} + \frac{\theta}{2} + C\)

\(\displaystyle = \frac{\sin{\theta}\cos{\theta} + \theta}{2} + C\).


Remembering that \(\displaystyle x = \sin{\theta}\), that means \(\displaystyle \cos{\theta} = \sqrt{1 - x^2}\) and \(\displaystyle x = \arcsin{\theta}\) and substituting, we find

\(\displaystyle \int{\sqrt{1 - x^2}\,dx} = \frac{x\sqrt{1 - x^2} + \arcsin{x}}{2} + C\).
 
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Oct 2008
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Unless, of course, I missed something.

Tonio
I think earboth is going to re-write

\(\displaystyle \frac{-x^2}{\sqrt{1 - x^2}}\)

as

\(\displaystyle \frac{1 - x^2 - 1}{\sqrt{1 - x^2}}\ =\ \sqrt{1 - x^2}\ -\ \frac{1}{\sqrt{1 - x^2}}\)

and then solve the top row of...



... for I. Which is no longer, really.

Key:


... is lazy integration by parts, doing without u and v.



... is the product rule, straight continuous lines differentiating downwards (integrating up) with respect to x.

Full size:
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Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
 
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May 2010
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Thanks everyone, really appreciate it. :)