$\displaystyle \int \sqrt{6-6x^2} dx$
I cant simplify it.
i got $\displaystyle sin\theta = \frac{\sqrt{6}}{6} x$
$\displaystyle cos\theta = \frac{\sqrt{6-6x^2}}{6}$
Hard to see where you might be at here. Just in case a picture helps...
... (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:
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Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
$\displaystyle \int{\sqrt{6 - 6x^2}\,dx} = \int{\sqrt{6(1-x^2)}\,dx}$
$\displaystyle = \int{\sqrt{6}\sqrt{1-x^2}\,dx}$
Now let $\displaystyle x = \sin{\theta}$ so that $\displaystyle dx= \cos{\theta}\,d\theta$ and the integral becomes
$\displaystyle \int{\sqrt{6}\sqrt{1 - \sin^2{\theta}}\,\cos{\theta}\,d\theta}$
$\displaystyle = \int{\sqrt{6}\sqrt{\cos^2{\theta}}\,\cos{\theta}\, d\theta}$
$\displaystyle = \int{\sqrt{6}\cos^2{\theta}\,d\theta}$
$\displaystyle = \int{\sqrt{6}\left(\frac{1}{2} + \frac{1}{2}\cos{2\theta}\right)\,d\theta}$
$\displaystyle = \frac{\sqrt{6}\theta}{2} + \frac{\sqrt{6}\sin{2\theta}}{4} + C$
$\displaystyle = \frac{\sqrt{6}\theta + \sqrt{6}\sin{\theta}\sqrt{1 - \sin^2{\theta}}}{2} + C$
$\displaystyle = \frac{\sqrt{6}\arcsin{x} + x\sqrt{6 - 6x^2}}{2} + C$.