# How do you find the integral of  3x (sqrt(81-x^2))?

Apr 20, 2015

Have a look: The answer using this method will look like the other, if you use:
$\cos \left(\arcsin \left(\frac{x}{9}\right)\right) = \sqrt{1 - {x}^{2} / 81}$ So

$- 729 {\cos}^{3} \left(\arcsin \left(\frac{x}{9}\right)\right) = - 729 {\left(\sqrt{1 - {x}^{2} / 81}\right)}^{3} = - {\left(9 \sqrt{1 - {x}^{2} / 81}\right)}^{3} = - {\left(\sqrt{81 - {x}^{2}}\right)}^{3}$

Apr 20, 2015  Apr 20, 2015

The answer is: $- \sqrt{{\left(81 - {x}^{2}\right)}^{3}} + c$

Remembering that:

$\int {\left[f \left(x\right)\right]}^{n} \cdot f ' \left(x\right) \mathrm{dx} = {\left[f \left(x\right)\right]}^{n + 1} / \left(n + 1\right) + c$,

than:

$\int 3 x \sqrt{81 - {x}^{2}} \mathrm{dx} = - \frac{3}{2} \int - 2 x {\left(81 - {x}^{2}\right)}^{\frac{1}{2}} \mathrm{dx} =$

$= - \frac{3}{2} {\left(81 - {x}^{2}\right)}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right) + c = - \frac{3}{2} {\left(81 - {x}^{2}\right)}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + c =$

$= - \sqrt{{\left(81 - {x}^{2}\right)}^{3}} + c$

or, if you want

$= - \left(81 - {x}^{2}\right) \sqrt{81 - {x}^{2}} + c$.