# What is the antiderivative of (9-x^2)^(1/2)?

$\setminus \int {\left(9 - {x}^{2}\right)}^{\frac{1}{2}} \setminus \mathrm{dx} = \frac{1}{2} x \setminus \sqrt{9 - {x}^{2}} + \frac{9}{2} \setminus {\sin}^{- 1} \left(\frac{x}{3}\right) + C$

#### Explanation:

Let $x = 3 \setminus \sin \setminus \theta \setminus \implies \mathrm{dx} = 3 \setminus \cos \setminus \theta \setminus d \setminus \theta$

$\setminus \therefore \setminus \int {\left(9 - {x}^{2}\right)}^{\frac{1}{2}} \setminus \mathrm{dx}$

$= \setminus \int {\left(9 - {\left(3 \setminus \sin \setminus \theta\right)}^{2}\right)}^{\frac{1}{2}} \setminus \left(3 \setminus \cos \setminus \theta \setminus d \setminus \theta\right)$

$= 3 \setminus \int \left(3 \setminus \cos \setminus \theta\right) \setminus \cos \setminus \theta \setminus d \setminus \theta$

$= 9 \setminus \int \setminus {\cos}^{2} \setminus \theta \setminus d \setminus \theta$

$= 9 \setminus \int \left(\setminus \frac{1 + \setminus \cos 2 \setminus \theta}{2}\right) \setminus d \setminus \theta$

$= \frac{9}{2} \setminus \int \left(1 + \setminus \cos 2 \setminus \theta\right) \setminus d \setminus \theta$

$= \frac{9}{2} \left(\setminus \theta + \frac{1}{2} \setminus \sin 2 \setminus \theta\right) + C$

$= \frac{9}{2} \left(\setminus {\sin}^{- 1} \left(\frac{x}{3}\right) + \setminus \sin \setminus \theta \setminus \cos \setminus \theta\right) + C$

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

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

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