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

Question

80936 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-06T22:05:18

I got $9/2arcsin(x/3) - x/2sqrt(9-x^2) + C$ .

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

$sqrt(a^2-x^2),$

which looks like

$sqrt(1 - sin^2theta),$

while

$sin^2theta + cos^2theta = 1.$

So, let:

$a = sqrt9 = 3$
$x = asintheta = 3sintheta$
$dx = acosthetad theta = 3costhetad theta$

That gives

$sqrt(9 - x^2) = sqrt(3^2 - (3sintheta)^2) = sqrt9sqrt(1-sin^2theta) = 3costheta$
$x^2 = 9sin^2theta$

Now we just have

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

$= 9intsin^2thetad theta.$

There's a useful identity, where $sin^2theta = (1-cos(2theta))/2$ .

$=> 9/2int1-cos(2theta)d$ $theta$

$= 9/2intd$ $theta - 9/2intcos(2theta)d$ $theta$

$= 9/2theta - 9/4sin2theta + C$

Draw out the right triangle if needed, where $x/3 = sintheta$ :

$theta = arcsin(x/3)$
$costheta = sqrt(9-x^2)/3$
$sintheta = x/3$

Use the identity

$sin2theta = 2sinthetacostheta,$

to then get

$=> 9/4(2sinthetacostheta) = (cancel(9)/cancel(4)^2)*cancel(2)*(x/cancel(3)sqrt(9-x^2)/cancel(3)) = (x/2)sqrt(9-x^2)$

Thus, our final result is

$=> color(blue)(9/2arcsin(x/3) - x/2sqrt(9-x^2) + C)$

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