integration (squareroot)

**gomes** · May 30th 2010, 11:10 PM

What would be the best/quickest way to integrate

squareroot(1-x^2) ?

**earboth** · May 30th 2010, 11:49 PM

Originally Posted by gomes

What would be the best/quickest way to integrate

squareroot(1-x^2) ?

1. Use integration by parts:

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

2. In the second step keep in mind that

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

**tonio** · May 30th 2010, 11:51 PM

Originally Posted by gomes

What would be the best/quickest way to integrate

squareroot(1-x^2) ?

Substitute $\sin t=x$ ...

Tonio

**tonio** · May 30th 2010, 11:54 PM

Originally Posted by earboth

1. Use integration by parts:

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

2. In the second step keep in mind that

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

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

Unless, of course, I missed something.

Tonio

**Prove It** · May 31st 2010, 12:08 AM

Originally Posted by gomes

What would be the best/quickest way to integrate

squareroot(1-x^2) ?

Trigonometric or Hyperbolic substitution works well...

Let $x = \sin{\theta}$ so that $dx = \cos{\theta}\,d\theta$ .

The integral becomes

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

$= \int{\cos{\theta}\cos{\theta}\,d\theta}$

$= \int{\cos^2{\theta}\,d\theta}$

$= \int{\frac{1}{2}\cos{2\theta} + \frac{1}{2}\,d\theta}$

$= \frac{1}{4}\sin{2\theta} + \frac{1}{2}\theta + C$

$= \frac{2\sin{\theta}\cos{\theta}}{4} + \frac{\theta}{2} + C$

$= \frac{\sin{\theta}\cos{\theta} + \theta}{2} + C$ .

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

$\int{\sqrt{1 - x^2}\,dx} = \frac{x\sqrt{1 - x^2} + \arcsin{x}}{2} + C$ .

**tom@ballooncalculus** · May 31st 2010, 09:40 AM

Originally Posted by tonio

Unless, of course, I missed something.

Tonio

I think earboth is going to re-write

$\frac{-x^2}{\sqrt{1 - x^2}}$

as

$\frac{1 - x^2 - 1}{\sqrt{1 - x^2}}\ =\ \sqrt{1 - x^2}\ -\ \frac{1}{\sqrt{1 - x^2}}$

and then solve the top row of...

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

Key:

Spoiler:

__________________________________________________ ___
Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

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**gomes** · May 31st 2010, 01:43 PM

Thanks everyone, really appreciate it.

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