# Solve \int(x^3/\sqrt(1-x^2))dx using two methods?

## a. Integration by parts b. Trigonometric Substitution (I solved part B, but you can answer it if you want. I'm interested in seeing if I might have messed up in steps anywhere.)

Mar 2, 2018

$I = - {x}^{2} \sqrt{1 - {x}^{2}} - \frac{2}{3} {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} + C$

#### Explanation:

We want to solve

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

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

Let $u = {x}^{2} \implies \mathrm{du} = 2 x \mathrm{dx}$

And $\mathrm{dv} = \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} \implies v = - \sqrt{1 - {x}^{2}}$

$I = - {x}^{2} \sqrt{1 - {x}^{2}} + \int \sqrt{1 - {x}^{2}} 2 x \mathrm{dx}$

Make a substitution $s = 1 - {x}^{2} \implies \mathrm{ds} = - 2 x \mathrm{dx}$

$I = - {x}^{2} \sqrt{1 - {x}^{2}} - \int \sqrt{s} \mathrm{ds}$

$= - {x}^{2} \sqrt{1 - {x}^{2}} - \frac{2}{3} {s}^{\frac{3}{2}} + C$

Substitute back $s = 1 - {x}^{2}$

$I = - {x}^{2} \sqrt{1 - {x}^{2}} - \frac{2}{3} {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} + C$

Alternative method

This can also be done by ordinary substitution, we seek

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

Make a substitution $u = 1 - {x}^{2} \implies \mathrm{du} = - 2 x \mathrm{dx}$

$I = - \frac{1}{2} \int {x}^{2} / \left(\sqrt{u}\right) \mathrm{du}$

But $u = 1 - {x}^{2} \implies {x}^{2} = 1 - u$

$I = - \frac{1}{2} \int \frac{1 - u}{\sqrt{u}} \mathrm{du} = - \frac{1}{2} \int {u}^{- \frac{1}{2}} - {u}^{\frac{1}{2}} \mathrm{du}$

Which is

$I = - {u}^{\frac{1}{2}} + \frac{1}{3} {u}^{\frac{3}{2}} + C$

Substitute back $u = 1 - {x}^{2}$

$I = - {\left(1 - {x}^{2}\right)}^{\frac{1}{2}} + \frac{1}{3} {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} + C$