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

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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)#

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