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.