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

Question

Charles Arocha · Accepted Answer

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)#

Samantha Adkison · Answer

undefined To integrate $ \frac{x^2}{\sqrt{9-x^2}} $ with respect to $ x $, you can use trigonometric substitution. Let $ x = 3\sin(\theta) $, $ dx = 3\cos(\theta) \, d\theta $. After substitution and simplification, the integral becomes $ \int \frac{9\sin^2(\theta)}{3\cos(\theta)} \cdot 3\cos(\theta) \, d\theta $. This simplifies to $ \int 3\sin^2(\theta) \, d\theta $. Now, use the identity $ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} $ to rewrite the integral, giving $ \frac{1}{2} \int (3 - 3\cos(2\theta)) \, d\theta $. Integrate each term separately to obtain $ \frac{1}{2} (3\theta - \frac{3}{2}\sin(2\theta)) + C $. Finally, substitute $ \theta = \sin^{-1}(\frac{x}{3}) $ and simplify to get the final answer.