Question

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

Answer 1

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

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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)`