# How do you integrate int (-8x^3)/sqrt(9-x^2)dx using trigonometric substitution?

Mar 18, 2016

Replace $x$ with either $3 \sin \theta$ or with $3 \cos \theta$

#### Explanation:

$\int \frac{- 8 {x}^{3}}{\sqrt{9 - {x}^{2}}} \mathrm{dx}$

Let $x = 3 \cos \theta$. This makes

$\mathrm{dx} = - 3 \sin \theta d \theta$, and

${x}^{3} = 27 {\cos}^{3} \theta$, and

$\sqrt{9 - {x}^{2}} = 3 \sin \theta$

With these substitutions, we get

$\int \frac{- 8 {x}^{3}}{\sqrt{9 - {x}^{2}}} \mathrm{dx} = \int \frac{- 8 \left(27 {\cos}^{3} \theta\right)}{3 \sin \theta} \left(- 3 \sin \theta\right) d \theta$

$= 8 \cdot 27 \int {\cos}^{3} \theta d \theta$

$= 8 \cdot 27 \int \left(1 - {\sin}^{2} \theta\right) \cos \theta d \theta$

$= 8 \cdot 27 \left(\sin \theta - {\sin}^{3} \frac{\theta}{3}\right) + C$

With $\cos \theta = \frac{x}{3}$, we get $\sin \theta = \frac{\sqrt{9 - {x}^{2}}}{3}$, so our integral becomes:

$\int \frac{- 8 {x}^{3}}{\sqrt{9 - {x}^{2}}} \mathrm{dx} = 8 \cdot 27 \left(\frac{\sqrt{9 - {x}^{2}}}{3} - {\left(\frac{\sqrt{9 - {x}^{2}}}{3}\right)}^{3} / 3\right)$

$= 8 \cdot 27 \left(\frac{\sqrt{9 - {x}^{2}}}{3} - \frac{\left(9 - {x}^{2}\right) \sqrt{9 - {x}^{2}}}{81}\right) + C$

$= 8 \cdot 27 \left(\frac{27 \sqrt{9 - {x}^{2}}}{81} - \frac{\left(9 - {x}^{2}\right) \sqrt{9 - {x}^{2}}}{81}\right) + C$

$= \frac{8}{3} \sqrt{9 - {x}^{2}} \left(27 - \left(9 - {x}^{2}\right)\right) + C$

$= \frac{8}{3} \sqrt{9 - {x}^{2}} \left({x}^{2} + 18\right) + C$