# \intx^2\sqrt(9-x^2)dx?

## Symbolab does it a certain way, but is there a different method of approach?

Apr 30, 2018

$\int {x}^{2} \sqrt{9 - {x}^{2}} \mathrm{dx} = \frac{81}{8} \left[\arcsin \left(\frac{x}{3}\right) - \frac{1}{4} \sin \left(4 \arcsin \left(\frac{x}{3}\right)\right)\right]$

#### Explanation:

I also follow similar way but my steps may be slightly more explained and are as follows.

Let $x = 3 \sin u$ then $\mathrm{dx} = 3 \cos u \mathrm{du}$ and this means $u = \arcsin \left(\frac{x}{3}\right)$

and $\int {x}^{2} \sqrt{9 - {x}^{2}} \mathrm{dx} = \int 9 {\sin}^{2} u \sqrt{9 - 9 {\sin}^{2} u} \cdot 3 \cos u \mathrm{du}$

= $81 \int {\sin}^{2} u {\cos}^{2} u \mathrm{du}$

= $\frac{81}{4} \int {\sin}^{2} 2 u \mathrm{du}$

= $\frac{81}{4} \int \frac{1 - \cos 4 u}{2} \mathrm{du}$

= $\frac{81}{8} \left[\int \mathrm{du} - \int \cos 4 u \mathrm{du}\right]$

= $\frac{81}{8} \left[u - \frac{\sin 4 u}{4}\right]$

= $\frac{81}{8} \left[\arcsin \left(\frac{x}{3}\right) - \frac{1}{4} \sin \left(4 \arcsin \left(\frac{x}{3}\right)\right)\right]$