# How do you evaluate the integral int x^3/sqrt(1-9x^2)?

Aug 3, 2017

$- \frac{1}{243} \cdot \sqrt{1 - 9 {x}^{2}} \left(2 + 9 {x}^{2}\right) + C .$

#### Explanation:

Let, $I = \int {x}^{3} / \sqrt{1 - 9 {x}^{2}} \mathrm{dx} .$

We substitute ${t}^{2} = 1 - 9 {x}^{2} \Rightarrow 9 {x}^{2} = 1 - {t}^{2.}$

$\therefore {x}^{2} = \frac{1}{9} \left(1 - {t}^{2}\right) , \mathmr{and} , 2 x \mathrm{dx} = \frac{1}{9} \left(- 2 t\right) \mathrm{dt} ,$

" or, xdx=-1/9tdt.

$\therefore I = \int \frac{{x}^{2} \cdot x \mathrm{dx}}{\sqrt{1 - 9 {x}^{2}}} ,$

$= \int \frac{1}{9} \left(1 - {t}^{2}\right) \cdot \left(- \frac{1}{9}\right) t \cdot \frac{1}{t} \cdot \mathrm{dt} ,$

$= \frac{1}{81} \int \left({t}^{2} - 1\right) \mathrm{dt} ,$

$= \frac{1}{81} \left[{t}^{3} / 3 - t\right] ,$

$= \frac{1}{243} \left[{t}^{3} - 3 t\right] ,$

$= \frac{t}{243} \left[{t}^{2} - 3\right] ,$

$= \frac{1}{243} \cdot \sqrt{1 - 9 {x}^{2}} \left[1 - 9 {x}^{2} - 3\right] ,$

$\Rightarrow I = - \frac{1}{243} \cdot \sqrt{1 - 9 {x}^{2}} \left(2 + 9 {x}^{2}\right) + C .$

Aug 3, 2017

$\int \setminus {x}^{3} / \left(\sqrt{1 - 9 {x}^{2}}\right) \setminus \mathrm{dx} = - \frac{1}{243} \left(2 + 9 {x}^{2}\right) \sqrt{1 - 9 {x}^{2}} + c$

#### Explanation:

We want to evaluate:

$I = \int \setminus {x}^{3} / \left(\sqrt{1 - 9 {x}^{2}}\right) \setminus \mathrm{dx}$

Let $u = 1 - 9 {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 18 x$; ${x}^{2} = \frac{1}{9} \left(1 - u\right)$

Substituting into the integral we get:

$I = - \frac{1}{18} \setminus \int \setminus {x}^{2} / \left(\sqrt{1 - 9 {x}^{2}}\right) \setminus \left(- 18 x\right) \setminus \mathrm{dx}$
$\setminus \setminus = - \frac{1}{18} \setminus \int \setminus \frac{\frac{1}{9} \left(1 - u\right)}{\sqrt{u}} \setminus \mathrm{du}$
$\setminus \setminus = - \frac{1}{162} \setminus \int \setminus \frac{1}{\sqrt{u}} - \sqrt{u} \setminus \mathrm{du}$
$\setminus \setminus = - \frac{1}{162} \left({u}^{\frac{1}{2}} / \left(\frac{1}{2}\right) - {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right) + c$
$\setminus \setminus = - \frac{1}{162} \left(2 \sqrt{u} - \frac{2}{3} u \sqrt{u}\right) + c$
$\setminus \setminus = - \frac{1}{162} \left(\frac{2}{3} \left(3 - u\right) \sqrt{u}\right) + c$
$\setminus \setminus = - \frac{1}{243} \left(3 - u\right) \sqrt{u} + c$

And, if we restore the substitution we get:

$I = - \frac{1}{243} \left(3 - 1 + 9 {x}^{2}\right) \sqrt{1 - 9 {x}^{2}} + c$
$\setminus \setminus = - \frac{1}{243} \left(2 + 9 {x}^{2}\right) \sqrt{1 - 9 {x}^{2}} + c$

Aug 3, 2017

$\int \setminus {x}^{3} / \left(\sqrt{1 - 9 {x}^{2}}\right) \setminus \mathrm{dx} = - \frac{1}{243} \left(2 + 9 {x}^{2}\right) \sqrt{1 - 9 {x}^{2}} + c$

#### Explanation:

Alternatively, using a trigonometric substitution:

We want to evaluate:

$I = \int \setminus {x}^{3} / \left(\sqrt{1 - 9 {x}^{2}}\right) \setminus \mathrm{dx}$

Let $\sin \theta = 3 x \implies \cos \theta \frac{d \theta}{\mathrm{dx}} = 3$; $\setminus \setminus x = \frac{1}{3} \sin \theta$

Substituting into the integral we get:

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

$\setminus \setminus = \frac{1}{81} \int \setminus \frac{{\sin}^{3} \theta \cos \theta}{\sqrt{{\cos}^{2} \theta}} \setminus d \theta$

$\setminus \setminus = \frac{1}{81} \int \setminus {\sin}^{3} \theta \setminus d \theta$

$\setminus \setminus = \frac{1}{81} \left(\frac{1}{3} {\cos}^{3} \theta - \cos \theta\right) + c$

$\setminus \setminus = - \frac{1}{243} \left(3 - {\cos}^{2} \theta\right) \cos \theta + c$

Using the identity ${\sin}^{2} \theta + {\cos}^{2} \theta \equiv 1 \implies {\cos}^{2} \theta = 1 - 9 {x}^{2}$

And, if we restore the substitution, using the above relationship, we get:

$I = - \frac{1}{243} \left(3 - 1 + 9 {x}^{2}\right) \sqrt{1 - 9 {x}^{2}} + c$
$\setminus \setminus = - \frac{1}{243} \left(2 + 9 {x}^{2}\right) \sqrt{1 - 9 {x}^{2}} + c$

Aug 3, 2017

$\int {x}^{3} / \sqrt{1 - 9 {x}^{2}} \mathrm{dx} = - \frac{\left(2 + 9 {x}^{2}\right) \sqrt{1 - 9 {x}^{2}}}{243} + C$

#### Explanation:

Evaluate:

$\int {x}^{3} / \sqrt{1 - 9 {x}^{2}} \mathrm{dx}$

Substitute:

$x = \frac{1}{3} \sin t$

as the integrand is defined only for $x \in \left(- \frac{1}{3} , \frac{1}{3}\right)$ $t$ varies in $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$.

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

$\int {x}^{3} / \sqrt{1 - 9 {x}^{2}} \mathrm{dx} = \frac{1}{81} \int \frac{{\sin}^{3} t \cos t}{\sqrt{1 - {\sin}^{2} t}} \mathrm{dt}$

Now:

$\sqrt{1 - {\sin}^{2} t} = \sqrt{{\cos}^{2} t} = \cos t$

as for $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, $\cos t$ is positive.

$\int {x}^{3} / \sqrt{1 - 9 {x}^{2}} \mathrm{dx} = \frac{1}{81} \int \frac{{\sin}^{3} t \cos t}{\cos} t \mathrm{dt} = \frac{1}{81} \int {\sin}^{3} t \mathrm{dt}$

$\int {x}^{3} / \sqrt{1 - 9 {x}^{2}} \mathrm{dx} = \frac{1}{81} \int \left(1 - {\cos}^{2} t\right) \sin t \mathrm{dt} = \frac{1}{81} \int \left({\cos}^{2} t - 1\right) d \left(\cos t\right)$

$\int {x}^{3} / \sqrt{1 - 9 {x}^{2}} \mathrm{dx} = \frac{{\cos}^{3} t - 3 \cos t}{243} + C$

To undo the substitution:

$\cos t = \sqrt{1 - {\sin}^{2} t} = \sqrt{1 - 9 {x}^{2}}$

so:

$\int {x}^{3} / \sqrt{1 - 9 {x}^{2}} \mathrm{dx} = \cos t \frac{{\cos}^{2} t - 3}{243} + C$

$\int {x}^{3} / \sqrt{1 - 9 {x}^{2}} \mathrm{dx} = \cos t \frac{1 - {\sin}^{2} t - 3}{243} + C$

$\int {x}^{3} / \sqrt{1 - 9 {x}^{2}} \mathrm{dx} = - \frac{\left(2 + 9 {x}^{2}\right) \sqrt{1 - 9 {x}^{2}}}{243} + C$