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

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.

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