# \int(x^2dx)/(\sqrt(4-9x^2))?

## I know it involves trigonometric substitution, and that $\setminus \sqrt{{a}^{2} - {x}^{2}}$ form with $x = a \setminus \sin \setminus \theta$ ($- \setminus \frac{\pi}{2} \setminus \le \setminus \theta \setminus \le \setminus \frac{\pi}{2}$, $1 - \setminus {\sin}^{2} x = \setminus {\cos}^{2} x$) However, I am stuck at the substitution part here. I know that $x = \frac{2}{3} \setminus \sin \setminus \theta$, $\setminus \therefore a = \frac{2}{3}$, but not what to do afterwards. Would it be $\mathrm{dx} = \frac{2}{3} \setminus \cos \setminus \theta$?

Apr 13, 2018

$\int \frac{{x}^{2} \mathrm{dx}}{\sqrt{4 - 9 {x}^{2}}} = \frac{2}{27} \left(\arcsin \left(\frac{3}{2} x\right) - \frac{3 x \sqrt{4 - 9 {x}^{2}}}{4}\right) + C$

#### Explanation:

For an integral involving the root $\sqrt{{a}^{2} - {b}^{2} {x}^{2}} ,$ we may use the substitution

$x = \frac{a}{b} \sin \theta$

So, here, $a = 2 , b = 3 , x = \frac{2}{3} \sin \theta , \mathrm{dx} = \frac{2}{3} \cos \theta d \theta$

Thus, we have

${\left(\frac{2}{3}\right)}^{2} \left(\frac{2}{3}\right) \int \frac{{\sin}^{2} \theta \cos \theta}{\sqrt{4 - 4 {\sin}^{2} \theta}} d \theta$

$\left(\frac{4}{9}\right) \left(\frac{\cancel{2}}{3}\right) \left(\frac{1}{\cancel{2}}\right) \int \frac{{\sin}^{2} \theta \cos \theta}{\sqrt{1 - {\sin}^{2} \theta}} d \theta$

$1 - {\sin}^{2} \theta = {\cos}^{2} \theta ,$ so

$\frac{4}{27} \int \frac{{\sin}^{2} \theta \cos \theta}{\sqrt{{\cos}^{2} \theta}} d \theta$

$\frac{4}{27} \int \frac{{\sin}^{2} \theta \cancel{\cos \theta}}{\cancel{\cos \theta}} d \theta$

Recalling that ${\sin}^{2} \theta = \frac{1}{2} \left(1 - \cos 2 \theta\right)$,

$\frac{4}{27} \int {\sin}^{2} \theta d \theta = \frac{4}{27} \left(\frac{1}{2}\right) \int \left(1 - \cos 2 \theta\right) d \theta$

$\frac{2}{27} \left(\theta - \frac{1}{2} \sin 2 \theta\right) + C$

We need things in terms of $x .$

Since $x = \frac{2}{3} \sin \theta , \sin \theta = \frac{3}{2} x , \theta = \arcsin \left(\frac{3}{2} x\right)$

Furthermore, recall that $\frac{1}{2} \sin 2 \theta = \sin \theta \cos \theta$. We'll need to find the cosine using a quick sketch. Since $\sin \theta = \frac{3 x}{2} ,$ we label the opposite and hypotenuse as follows, realizing that the adjacent side will be the radical originally found in the integral:

We see $\cos \theta = \frac{\sqrt{4 - 9 {x}^{2}}}{2} , \frac{1}{2} \sin 2 \theta = \left(\frac{3}{2} x\right) \frac{\sqrt{4 - 9 {x}^{2}}}{2} = \frac{3 x \sqrt{4 - 9 {x}^{2}}}{4}$

Thus,

$\int \frac{{x}^{2} \mathrm{dx}}{\sqrt{4 - 9 {x}^{2}}} = \frac{2}{27} \left(\arcsin \left(\frac{3}{2} x\right) - \frac{3 x \sqrt{4 - 9 {x}^{2}}}{4}\right) + C$

Apr 13, 2018

$\text{Sometimes it is better to avoid"color(red)" trigonometric substitution}$ $\text{and think about"color(blue)" another substitution }$that led us to known formula.
$\left(1\right) \int \frac{1}{\sqrt{{a}^{2} - {x}^{2}}} \mathrm{dx} = {\sin}^{-} 1 \left(\frac{x}{a}\right) + c$

#### Explanation:

color(red)((2)intsqrt(a^2-x^2)dx=x/2sqrt(a^2- x^2)+a^2/2sin^-1(x/a)+c

Here,

$I = \int {x}^{2} / \sqrt{4 - 9 {x}^{2}} \mathrm{dx}$

Let, $3 x = t \implies 3 \mathrm{dx} = \mathrm{dt} \implies \mathrm{dx} = \frac{1}{3} \mathrm{dt} , \mathmr{and} 9 {x}^{2} = {t}^{2} \implies {x}^{2} = {t}^{2} / 9$

$\therefore I = \int \frac{\left({t}^{2} / 9\right)}{\sqrt{4 - {t}^{2}}} \frac{1}{3} \mathrm{dt}$

$= \frac{1}{27} \int {t}^{2} / \sqrt{4 - {t}^{2}} \mathrm{dt}$

$= - \frac{1}{27} \left[\int \frac{- {t}^{2}}{\sqrt{4 - {t}^{2}}} \mathrm{dt}\right]$

$= - \frac{1}{27} \left[\int \frac{4 - {t}^{2} - 4}{\sqrt{4 - {t}^{2}}} \mathrm{dt}\right]$

$= - \frac{1}{27} \left[\int \frac{4 - {t}^{2}}{\sqrt{4 - {t}^{2}}} \mathrm{dt} - \int \frac{4}{\sqrt{4 - {t}^{2}}} \mathrm{dt}\right]$

$= - \frac{1}{27} \left[\int \sqrt{4 - {t}^{2}} \mathrm{dt} - 4 \int \frac{1}{\sqrt{4 - {t}^{2}}} \mathrm{dt}\right]$

=-1/27[intsqrt(2^2-t^2)dt-4int1/sqrt(2^2-t^2)dt]toApply(1)& (2)

$= - \frac{1}{27} \left[\frac{t}{2} \sqrt{{2}^{2} - {t}^{2}} + {2}^{2} / 2 {\sin}^{-} 1 \left(\frac{t}{2}\right) - 4 {\sin}^{-} 1 \left(\frac{t}{2}\right)\right] + c$

$= - \frac{1}{27} \left[\frac{t}{2} \sqrt{4 - {t}^{2}} + 2 {\sin}^{-} 1 \left(\frac{t}{2}\right) - 4 {\sin}^{-} 1 \left(\frac{t}{2}\right)\right] + c$

$= - \frac{1}{27} \left[\frac{t}{2} \sqrt{4 - {t}^{2}} - 2 {\sin}^{-} 1 \left(\frac{t}{2}\right)\right] + c$

$= \frac{1}{27} \left[2 {\sin}^{-} 1 \left(\frac{t}{2}\right) - \frac{t}{2} \sqrt{4 - {t}^{2}}\right] + c$

Subst. back, $. . \to t = 3 x , {t}^{2} = 9 {x}^{2}$

$= \frac{1}{27} \left[2 {\sin}^{-} 1 \left(\frac{3 x}{2}\right) - \frac{3 x}{2} \sqrt{4 - 9 {x}^{2}}\right] + c$