$\displaystyle \int \sqrt{6-6x^2} dx$

I cant simplify it.

i got $\displaystyle sin\theta = \frac{\sqrt{6}}{6} x$

$\displaystyle cos\theta = \frac{\sqrt{6-6x^2}}{6}$

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- Aug 3rd 2010, 11:37 PMlarryboi7Finding the Integral
$\displaystyle \int \sqrt{6-6x^2} dx$

I cant simplify it.

i got $\displaystyle sin\theta = \frac{\sqrt{6}}{6} x$

$\displaystyle cos\theta = \frac{\sqrt{6-6x^2}}{6}$ - Aug 4th 2010, 02:19 AMtom@ballooncalculus
Hard to see where you might be at here. Just in case a picture helps...

http://www.ballooncalculus.org/asy/internal/sin.png

... (key in spoiler)...

__Spoiler__:

The point is to see that we're going to integrate the original function as though it were the 'outer' derivative in a chain rule differentiation.

So next we fill out a bit more of the chain rule shape, to see the effect of having to multiply by the derivative of the inner function. Then we'll be able to fill the top of the shape by integrating the whole of the bottom level with respect to t...

__Spoiler__:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Aug 4th 2010, 02:20 AMProve It
$\displaystyle \int{\sqrt{6 - 6x^2}\,dx} = \int{\sqrt{6(1-x^2)}\,dx}$

$\displaystyle = \int{\sqrt{6}\sqrt{1-x^2}\,dx}$

Now let $\displaystyle x = \sin{\theta}$ so that $\displaystyle dx= \cos{\theta}\,d\theta$ and the integral becomes

$\displaystyle \int{\sqrt{6}\sqrt{1 - \sin^2{\theta}}\,\cos{\theta}\,d\theta}$

$\displaystyle = \int{\sqrt{6}\sqrt{\cos^2{\theta}}\,\cos{\theta}\, d\theta}$

$\displaystyle = \int{\sqrt{6}\cos^2{\theta}\,d\theta}$

$\displaystyle = \int{\sqrt{6}\left(\frac{1}{2} + \frac{1}{2}\cos{2\theta}\right)\,d\theta}$

$\displaystyle = \frac{\sqrt{6}\theta}{2} + \frac{\sqrt{6}\sin{2\theta}}{4} + C$

$\displaystyle = \frac{\sqrt{6}\theta + \sqrt{6}\sin{\theta}\sqrt{1 - \sin^2{\theta}}}{2} + C$

$\displaystyle = \frac{\sqrt{6}\arcsin{x} + x\sqrt{6 - 6x^2}}{2} + C$. - Aug 11th 2010, 03:47 PMlarryboi7
Thank You that was really helpful couldnt get past that.