# How do you find the integral of sqrt(9-x^2)dx?

Aug 17, 2015

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

#### Explanation:

${\sin}^{2} A + {\cos}^{2} A = 1$
$\cos 2 A = 2 {\cos}^{2} A - 1$
$\sin 2 A = 2 \sin A \cos A$

Use substitution.
$x = 3 \sin t$

$\int \sqrt{9 - {x}^{2}} \mathrm{dx} = \int 3 \sqrt{1 - {\sin}^{2} t} \times 3 \cos t \mathrm{dt} = 9 \int {\cos}^{2} t \mathrm{dt}$

$9 \int {\cos}^{2} t \mathrm{dx} = \frac{9}{2} \int \left(1 + \cos 2 t\right) \mathrm{dt} = \frac{9}{2} \left(t + \frac{1}{2} \sin 2 t\right) + C$

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