How do you integrate x^2/sqrt(9-x^2) dxx2√9−x2dx?
1 Answer
I got
You can do this with trig substitution. Notice how this is of the form
sqrt(a^2-x^2),√a2−x2,
which looks like
sqrt(1 - sin^2theta),√1−sin2θ,
while
sin^2theta + cos^2theta = 1.sin2θ+cos2θ=1.
So, let:
a = sqrt9 = 3a=√9=3
x = asintheta = 3sinthetax=asinθ=3sinθ
dx = acosthetad theta = 3costhetad thetadx=acosθdθ=3cosθdθ
That gives
sqrt(9 - x^2) = sqrt(3^2 - (3sintheta)^2) = sqrt9sqrt(1-sin^2theta) = 3costheta√9−x2=√32−(3sinθ)2=√9√1−sin2θ=3cosθ
x^2 = 9sin^2thetax2=9sin2θ
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
=> 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
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)
