What would be the best/quickest way to integrate
squareroot(1-x^2) ?
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
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
$\displaystyle \int{\sqrt{1 - x^2}\,dx} = \frac{x\sqrt{1 - x^2} + \arcsin{x}}{2} + C$.
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:Spoiler:
__________________________________________________ ___
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!