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

May 12, 2018

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

Explanation:

To get rid of the square root in the denominator, we make the trigonometric substitution $x \setminus \mapsto g \left(u\right)$ with

$g \left(u\right) = \sin \frac{u}{3}$,

$g ' \left(u\right) = \cos \frac{u}{3}$, and

${g}^{- 1} \left(x\right) = {\sin}^{- 1} \left(3 x\right)$,

${\left[\setminus \frac{1}{81} \int \setminus \frac{{\sin}^{3} u \cos u}{\setminus \sqrt{1 - {\sin}^{2} u}} \setminus d u\right]}_{u = {\sin}^{- 1} \left(3 x\right)} =$

$= {\left[\setminus \frac{1}{81} \int {\sin}^{3} u \setminus d u\right]}_{u = {\sin}^{- 1} \left(3 x\right)} =$

$= {\left[\setminus \frac{1}{81} \int \left(1 - {\cos}^{2} u\right) \sin u \setminus d u\right]}_{u = {\sin}^{- 1} \left(3 x\right)}$.

Now, we may view $\cos u$ as an inner function, with derivative $- \sin u$, which is why this is the same as

${\left[- \setminus \frac{1}{81} \int 1 - {t}^{2} \setminus d t\right]}_{t = \cos {\sin}^{- 1} \left(3 x\right) = \setminus \sqrt{1 - 9 {x}^{2}}} =$

$= - \setminus \frac{1}{81} {\left[t - {t}^{3} / 3 + C\right]}_{t = \setminus \sqrt{1 - 9 {x}^{2}}} =$

$= - \setminus \frac{1}{81} \left[\setminus \sqrt{1 - 9 {x}^{2}} - {\left(1 - 9 {x}^{2}\right)}^{\frac{3}{2}} / 3\right] + C =$

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