What is the integral of sqrt(9-x^2)?

1 Answer
Dec 20, 2014

Whenever I see these kind of functions, I recognize (by practicing a lot) that you should use a special substitution here:
int sqrt(9-x^2)dx
x = 3sin(u)
This might look like a weird substitution, but you're going to see why we're doing this.
dx = 3cos(u)du
Replace everyhting in the integral:

int sqrt(9-(3sin(u))^2)*3cos(u)du
We can bring the 3 out of the integral:
3*int sqrt(9-(3sin(u))^2)*cos(u)du
3*int sqrt(9-9sin^2(u))*cos(u)du
You can factor the 9 out:
3*int sqrt(9(1-sin^2(u)))*cos(u)du
3*3int sqrt(1-sin^2(u))*cos(u)du

We know the identity: cos^2x + sin^2x = 1
If we solve for cosx, we get:
cos^2x = 1-sin^2x
cosx = sqrt(1-sin^2x)
This is exactly what we see in the integral, so we can replace it:

9 int cos^2(u)du
You might know this one as a basic antiderivative, but if you don't, you can figure it out like so:

We use the identity: cos^2(u) = (1+cos(2u))/2

9 int (1+cos(2u))/2 du
9/2 int 1+cos(2u) du
9/2 (int 1du + int cos(2u)du)
9/2 (u + 1/2sin(2u)) + C (you can work this out by substitution)
9/2 u + 9/4 sin(2u) + C

Now, all we have to do is put u into the function. Let's look back at how we defined it:
x = 3sin(u)
x/3 = sin(u)
To get u out of this, you need to take the inverse function of sin on both sides, this is arcsin:

arcsin(x/3) = arcsin(sin(u))
arcsin(x/3) = u

Now we need to insert it into our solution:

9/2 arcsin(x/3) + 9/4 sin(2arcsin(x/3)) + C

This is the final solution.