How do you find the integral intx^3/(sqrt(16-x^2))dx ?

Aug 27, 2014

$= {\left(16 - {x}^{2}\right)}^{\frac{3}{2}} / 3 - 16 \sqrt{16 - {x}^{2}} + c$, where $c$ is a constant

Solution

$= \int {x}^{3} / \sqrt{16 - {x}^{2}} \mathrm{dx} = \int \frac{{x}^{2} \cdot x}{\sqrt{16 - {x}^{2}}} \mathrm{dx}$

let's assume $\left(16 - {x}^{2}\right) = {t}^{2}$, $\implies - 2 x \mathrm{dx} = 2 t \mathrm{dt}$

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

$= \int \frac{- \left(16 - {t}^{2}\right) t}{t} \mathrm{dt}$

$= \int \left({t}^{2} - 16\right) \mathrm{dt}$

$= \int {t}^{2} \mathrm{dt} - 16 \int \mathrm{dt}$

$= {t}^{3} / 3 - 16 t + c$, where $c$ is a constant

$= {\left(16 - {x}^{2}\right)}^{\frac{3}{2}} / 3 - 16 \sqrt{16 - {x}^{2}} + c$, where $c$ is a constant