# How do you integrate int (x^2 ) / sqrt(9 - x^2) dx using trigonometric substitution?

Apr 1, 2018

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

#### Explanation:

Let

$x = 3 \sin \theta$

${x}^{2} = 9 {\sin}^{2} \theta$

$\mathrm{dx} = 3 \cos \theta d \theta$

We then have

int(27sin^2thetacostheta)/(sqrt(9(1-sin^2theta))d theta

Recall that $1 - {\sin}^{2} \theta = {\cos}^{2} \theta$. Apply the identity:

$9 \int \frac{{\sin}^{2} \theta \cos \theta}{\sqrt{{\cos}^{2} \theta}} d \theta$

$9 \int \frac{{\sin}^{2} \theta \cancel{\cos} \theta}{\cancel{\cos} \theta} d \theta$

$9 \int {\sin}^{2} \theta d \theta$

Recall the identity ${\sin}^{2} \theta = \frac{1}{2} \left(1 - \cos 2 \theta\right)$:

$\frac{9}{2} \int \left(1 - \cos 2 \theta\right) d \theta$

Integrate:

$\frac{9}{2} \int \left(1 - \cos 2 \theta\right) d \theta = \frac{9}{2} \theta - \frac{9}{4} \sin 2 \theta + C$

We need to revert back to $x .$ Recalling that $x = 3 \sin \theta , \sin \theta = \frac{x}{3} , \theta = \arcsin \left(\frac{x}{3}\right)$

$\sin 2 \theta$ is still needed. Recalling that $\sin 2 \theta = 2 \sin \theta \cos \theta , {\sin}^{2} \theta + {\cos}^{2} \theta = 1 :$

${x}^{2} / 9 + {\cos}^{2} \theta = \frac{9}{9}$
${\cos}^{2} \theta = \frac{9 - {x}^{2}}{9}$
$\cos \theta = \frac{\sqrt{9 - {x}^{2}}}{3}$

Then $\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \left(\frac{x}{3}\right) \left(\frac{\sqrt{9 - {x}^{2}}}{3}\right) = \frac{2 x \sqrt{9 - {x}^{2}}}{9}$

Then,

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

Apr 1, 2018

(9/2)(arcsin(x/3)-1/2sin(2arcsin(x/3))+C

#### Explanation:

$\int {x}^{2} / \left(\sqrt{9 - {x}^{2}}\right) \mathrm{dx} = {x}^{2} / \left(\sqrt{{3}^{2} - {x}^{2}}\right) \mathrm{dx}$ using our trig identities for integrals let us substitute $3 \sin \phi = x$ and $\mathrm{dx} = 3 \cos \phi \mathrm{dp} h i$

$\therefore$ 3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9-(9sin^2phi))

3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9(1-(sin^2phi))
where $1 - {\sin}^{2} \phi = {\cos}^{2} \phi$

3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9(cos^2phi))

$3 \int \left(9 {\sin}^{2} \phi\right) \left(\cancel{\cos} \phi\right) \frac{\mathrm{dp} h i}{3 \left(\cancel{\cos} \phi\right)}$

$\cancel{3} \int \left(9 {\sin}^{2} \phi\right) \frac{\mathrm{dp} h i}{\cancel{3}}$

$\int \left(9 {\sin}^{2} \phi\right) \left(\mathrm{dp} h i\right) = 9 \int \left({\sin}^{2} \phi\right) \left(\mathrm{dp} h i\right)$

Use the identity from trig where $\cos \left(2 \phi\right) = 1 - 2 {\sin}^{2} \phi$ rearranging to solve such that $\frac{\cos \left(2 \phi\right) - 1}{- 2} = {\sin}^{2} \phi$ or $\frac{1 - \cos \left(2 \phi\right)}{2} = {\sin}^{2} \phi$

$9 \int \frac{1 - \cos \left(2 \phi\right)}{2} \left(\mathrm{dp} h i\right) = \frac{9}{2} \int \left(1 - \cos \left(2 \phi\right)\right) \left(\mathrm{dp} h i\right)$ use subtraction rules of the integral

$\frac{9}{2} \int \left(\mathrm{dp} h i\right) - \frac{9}{2} \int \cos \left(2 \phi\right) \mathrm{dp} h i$

$\frac{9}{2} \int \left(\mathrm{dp} h i\right) = \left(\frac{9}{2}\right) \phi$

$- \frac{9}{2} \int \cos \left(2 \phi\right) \mathrm{dp} h i = - \frac{9}{4} \sin \left(2 \phi\right)$

Now piece together and substitute in

$\left(\frac{9}{2}\right) \phi - \frac{9}{4} \sin \left(2 \phi\right) + C$

$\phi = \arcsin \left(\frac{x}{3}\right)$

$\left(\frac{9}{2}\right) \phi - \frac{9}{4} \sin \left(2 \phi\right) + C$

(9/2)(arcsin(x/3)-1/2sin(2arcsin(x/3))+C