# int sqrt(9-5x^2) dx ?

## I know that x=asintheta. But idk how to factor out so that i wont have $\sqrt{5}$

Feb 19, 2017

The answer is $= \frac{9}{2 \sqrt{5}} \arcsin \left(\frac{\sqrt{5}}{3} x\right) + \frac{1}{2} x \sqrt{9 - 5 {x}^{2}} + C$

#### Explanation:

We perform this integral by substitution

Let $x = \frac{3}{\sqrt{5}} \sin u$

$\frac{\mathrm{dx}}{\mathrm{du}} = \frac{3}{\sqrt{5}} \cos u$

$\mathrm{dx} = 3 \frac{\mathrm{du}}{\sqrt{5}} \cos u$

and

$\sqrt{9 - 5 {x}^{2}} = \sqrt{9 - 5 \cdot \frac{9}{5} {\sin}^{2} u}$

$= 3 \sqrt{1 - {\sin}^{2} u}$

$= 3 \cos u$

So,

$\int \sqrt{9 - 5 {x}^{2}} \mathrm{dx} = \int \frac{3}{\sqrt{5}} \cos u \cdot 3 \cos u \mathrm{du}$

$= \frac{9}{\sqrt{5}} \int {\cos}^{2} u \mathrm{du}$

$\cos 2 u = 2 {\cos}^{2} u - 1$

${\cos}^{2} u = \frac{1 + \cos \left(2 u\right)}{2}$

So,

$\int \sqrt{9 - 5 {x}^{2}} \mathrm{dx} = \frac{9}{\sqrt{5}} \int \frac{1 + \cos \left(2 u\right)}{2} \left(\mathrm{du}\right)$

$= \frac{9}{2 \sqrt{5}} \int \left(1 + \cos \left(2 u\right)\right) \mathrm{du}$

$= \frac{9}{2 \sqrt{5}} \left(u + \sin \frac{2 u}{2}\right)$

$= \frac{9}{2 \sqrt{5}} \left(\arcsin \left(\frac{\sqrt{5}}{3} x\right) + \sin u \cos u\right)$

$= \frac{9}{2 \sqrt{5}} \left(\arcsin \left(\frac{\sqrt{5}}{3} x\right) + \frac{\sqrt{5}}{3} x \frac{\sqrt{9 - 5 {x}^{2}}}{3}\right) + C$

$= \frac{9}{2 \sqrt{5}} \arcsin \left(\frac{\sqrt{5}}{3} x\right) + \frac{1}{2} x \sqrt{9 - 5 {x}^{2}} + C$