# How do you integrate int x^3 / sqrt[1-x^2] using trigonometric substitution?

Mar 13, 2018

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

#### Explanation:

Substitute $x = \sin t$, $\mathrm{dx} = \cos t \mathrm{dt}$. As the integrand is defined only for $x \in \left(- 1 , 1\right)$ then $t$ will vary accordingly in $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ and we must keep in mind that in this interval $\cos t$ is positive.

So:

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

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

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

Now:

$\int {\sin}^{3} t \mathrm{dt} = \int {\sin}^{2} t \sin t \mathrm{dt}$

$\int {\sin}^{3} t \mathrm{dt} = - \int \left(1 - {\cos}^{2} t\right) d \left(\cos t\right)$

and using the linearity of the integral:

$\int {\sin}^{3} t \mathrm{dt} = \int {\cos}^{2} t d \left(\cos t\right) - \int d \left(\cos t\right)$

$\int {\sin}^{3} t \mathrm{dt} = {\cos}^{3} \frac{t}{3} - \cos t + C$

to undo the substitution note that:

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

So:

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

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

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

It is however probably more straightforward not to use a trigonometric substitution, in fact:

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

and we can integrate by parts:

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

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

and finally:

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

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

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

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