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∫√9−x2dx
x = 3sin(u)x=3sin(u)
This might look like a weird substitution, but you're going to see why we're doing this.
dx = 3cos(u)dudx=3cos(u)du
Replace everyhting in the integral:
int sqrt(9-(3sin(u))^2)*3cos(u)du∫√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)du3⋅∫√9−(3sin(u))2⋅cos(u)du
3*int sqrt(9-9sin^2(u))*cos(u)du3⋅∫√9−9sin2(u)⋅cos(u)du
You can factor the 9 out:
3*int sqrt(9(1-sin^2(u)))*cos(u)du3⋅∫√9(1−sin2(u))⋅cos(u)du
3*3int sqrt(1-sin^2(u))*cos(u)du3⋅3∫√1−sin2(u)⋅cos(u)du
We know the identity: cos^2x + sin^2x = 1cos2x+sin2x=1
If we solve for cosxcosx, we get:
cos^2x = 1-sin^2xcos2x=1−sin2x
cosx = sqrt(1-sin^2x)cosx=√1−sin2x
This is exactly what we see in the integral, so we can replace it:
9 int cos^2(u)du9∫cos2(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))/2cos2(u)=1+cos(2u)2
9 int (1+cos(2u))/2 du9∫1+cos(2u)2du
9/2 int 1+cos(2u) du92∫1+cos(2u)du
9/2 (int 1du + int cos(2u)du)92(∫1du+∫cos(2u)du)
9/2 (u + 1/2sin(2u)) + C92(u+12sin(2u))+C (you can work this out by substitution)
9/2 u + 9/4 sin(2u) + C92u+94sin(2u)+C
Now, all we have to do is put uu into the function. Let's look back at how we defined it:
x = 3sin(u)x=3sin(u)
x/3 = sin(u)x3=sin(u)
To get uu out of this, you need to take the inverse function of sinsin on both sides, this is arcsinarcsin:
arcsin(x/3) = arcsin(sin(u))arcsin(x3)=arcsin(sin(u))
arcsin(x/3) = uarcsin(x3)=u
Now we need to insert it into our solution:
9/2 arcsin(x/3) + 9/4 sin(2arcsin(x/3)) + C92arcsin(x3)+94sin(2arcsin(x3))+C
This is the final solution.
