# How do you find the integral intx^3*sqrt(9-x^2)dx ?

Aug 29, 2014

$= {\left(9 - {x}^{2}\right)}^{\frac{5}{2}} / 5 - 3 {\left(9 - {x}^{2}\right)}^{\frac{3}{2}} + c$, where $c$ is a constant

Explanation :

$= \int {x}^{3} \cdot \sqrt{9 - {x}^{2}} \mathrm{dx}$

let's assume $9 - {x}^{2} = {t}^{2}$, then

$- 2 x \mathrm{dx} = 2 t \mathrm{dt}$ $\implies x \mathrm{dx} = - t \mathrm{dt}$

$= \int - \left(9 - {t}^{2}\right) {t}^{2} \mathrm{dt}$

$= \int \left({t}^{4} - 9 {t}^{2}\right) \mathrm{dt}$

$= \int {t}^{4} \mathrm{dt} - 9 \int {t}^{2} \mathrm{dt}$

$= {t}^{5} / 5 - 9 {t}^{3} / 3 + c$, where $c$ is a constant

$= {t}^{5} / 5 - 3 {t}^{3} + c$, where $c$ is a constant

Substituting $t$ back,

$= {\left(9 - {x}^{2}\right)}^{\frac{5}{2}} / 5 - 3 {\left(9 - {x}^{2}\right)}^{\frac{3}{2}} + c$, where $c$ is a constant