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.