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

Jun 6, 2015

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

You can do this with trig substitution. Notice how this is of the form

$\sqrt{{a}^{2} - {x}^{2}} ,$

which looks like

$\sqrt{1 - {\sin}^{2} \theta} ,$

while

${\sin}^{2} \theta + {\cos}^{2} \theta = 1.$

So, let:

$a = \sqrt{9} = 3$
$x = a \sin \theta = 3 \sin \theta$
$\mathrm{dx} = a \cos \theta d \theta = 3 \cos \theta d \theta$

That gives

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

Now we just have

int x^2/(sqrt(9-x^2))dx = int (9sin^2theta)/(cancel(3costheta))(cancel(3costheta)dtheta)

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

There's a useful identity, where ${\sin}^{2} \theta = \frac{1 - \cos \left(2 \theta\right)}{2}$.

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

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

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

Draw out the right triangle if needed, where $\frac{x}{3} = \sin \theta$:

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

Use the identity

$\sin 2 \theta = 2 \sin \theta \cos \theta ,$

to then get

$\implies \frac{9}{4} \left(2 \sin \theta \cos \theta\right) = \left(\frac{\cancel{9}}{\cancel{4}} ^ 2\right) \cdot \cancel{2} \cdot \left(\frac{x}{\cancel{3}} \frac{\sqrt{9 - {x}^{2}}}{\cancel{3}}\right) = \left(\frac{x}{2}\right) \sqrt{9 - {x}^{2}}$

Thus, our final result is

$\implies \textcolor{b l u e}{\frac{9}{2} \arcsin \left(\frac{x}{3}\right) - \frac{x}{2} \sqrt{9 - {x}^{2}} + C}$