# How do you integrate x^2/(sqrt(9-x^2))?

Nov 5, 2016

$\frac{9 \arcsin \left(\frac{x}{3}\right) - x \sqrt{9 - {x}^{2}}}{2} + C$

#### Explanation:

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

Rewrite ${x}^{2}$ as ${x}^{2} - 9 + 9$:

$I = \int \frac{{x}^{2} - 9 + 9}{\sqrt{9 - {x}^{2}}} \mathrm{dx} = \int \frac{{x}^{2} - 9}{\sqrt{9 - {x}^{2}}} \mathrm{dx} + \int \frac{9}{\sqrt{9 - {x}^{2}}} \mathrm{dx}$

Rewriting for simplification:

$I = - \int \frac{9 - {x}^{2}}{\sqrt{9 - {x}^{2}}} \mathrm{dx} + \int \frac{9}{\sqrt{9 - {x}^{2}}} \mathrm{dx}$

$I = - \int \sqrt{9 - {x}^{2}} \mathrm{dx} + \int \frac{9}{\sqrt{9 - {x}^{2}}} \mathrm{dx}$

Let $J = - \int \sqrt{9 - {x}^{2}} \mathrm{dx}$ and $K = \int \frac{9}{\sqrt{9 - {x}^{2}}} \mathrm{dx}$.

$J = - \int \sqrt{9 - {x}^{2}} \mathrm{dx}$

Let $x = 3 \sin \theta$ so that $\mathrm{dx} = 3 \cos \theta d \theta$:

$J = - \int \sqrt{9 - 9 {\sin}^{2} \theta} \left(3 \cos \theta d \theta\right)$

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

$J = - 9 \int {\cos}^{2} \theta d \theta$

Using $\cos 2 \theta = 2 {\cos}^{2} \theta - 1$ so solve for ${\cos}^{2} \theta$:

$J = - 9 \int \frac{\cos 2 \theta + 1}{2} d \theta = - \frac{9}{2} \int \cos 2 \theta d \theta - \frac{9}{2} \int d \theta$

Solve the first integral by sight or by substitution:

$J = - \frac{9}{4} \sin 2 \theta - \frac{9}{2} \theta$

Using $\sin 2 \theta = 2 \sin \theta \cos \theta$:

$J = - \frac{9}{2} \sin \theta \cos \theta - \frac{9}{2} \theta$

From our substitution $x = 3 \sin \theta$ we see that $\theta = \arcsin \left(\frac{x}{3}\right)$.

Furthermore, we see that $\sin \theta = \frac{x}{3}$ and $\cos \theta = \sqrt{1 - {\sin}^{2} \theta} = \sqrt{1 - {x}^{2} / 9} = \frac{1}{3} \sqrt{9 - {x}^{2}}$. Thus:

$J = - \frac{9}{2} \left(\frac{x}{3}\right) \left(\frac{1}{3} \sqrt{9 - {x}^{2}}\right) - \frac{9}{2} \arcsin \left(\frac{x}{3}\right)$

$J = - \frac{1}{2} x \sqrt{9 - {x}^{2}} - \frac{9}{2} \arcsin \left(\frac{x}{3}\right)$

Now solving for $K$:

$K = \int \frac{9}{\sqrt{9 - {x}^{2}}} \mathrm{dx}$

We will use the same substitution, $x = 3 \sin \phi$, so $\mathrm{dx} = 3 \cos \phi \mathrm{dp} h i$.

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

$K = \int \frac{27 \cos \phi}{\sqrt{9} \sqrt{1 - {\sin}^{2} \phi}} \mathrm{dp} h i$

$K = \int \frac{9 \cos \phi}{\cos} \phi \mathrm{dp} h i = 9 \int \mathrm{dp} h i = 9 \phi$

From $x = 3 \sin \phi$ we see that $\phi = \arcsin \left(\frac{x}{3}\right)$:

$K = 9 \arcsin \left(\frac{x}{3}\right)$

Now that we've solved for $J$ and $K$, return to $I$:

$I = J + K$

$I = \left[- \frac{1}{2} x \sqrt{9 - {x}^{2}} - \frac{9}{2} \arcsin \left(\frac{x}{3}\right)\right] + 9 \arcsin \left(\frac{x}{3}\right)$

$I = - \frac{1}{2} x \sqrt{9 - {x}^{2}} + \frac{9}{2} \arcsin \left(\frac{x}{3}\right)$

$I = \frac{9 \arcsin \left(\frac{x}{3}\right) - x \sqrt{9 - {x}^{2}}}{2} + C$