Question

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

Answer 1

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.