What would be the best/quickest way to integrate

squareroot(1-x^2) ?

Printable View

- May 30th 2010, 11:10 PMgomesintegration (squareroot)
What would be the best/quickest way to integrate

squareroot(1-x^2) ? - May 30th 2010, 11:49 PMearboth
- May 30th 2010, 11:51 PMtonio
- May 30th 2010, 11:54 PMtonio

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 - May 31st 2010, 12:08 AMProve It
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

$\displaystyle \int{\sqrt{1 - x^2}\,dx} = \frac{x\sqrt{1 - x^2} + \arcsin{x}}{2} + C$. - May 31st 2010, 09:40 AMtom@ballooncalculus
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...

http://www.ballooncalculus.org/asy/p...tOneMinus1.png

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

Key:__Spoiler__:

__________________________________________________ ___

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - May 31st 2010, 01:43 PMgomes
Thanks everyone, really appreciate it. :)