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

Dec 20, 2014

Whenever I see these kind of functions, I recognize (by practicing a lot) that you should use a special substitution here:
$\int \sqrt{9 - {x}^{2}} \mathrm{dx}$
$x = 3 \sin \left(u\right)$
This might look like a weird substitution, but you're going to see why we're doing this.
$\mathrm{dx} = 3 \cos \left(u\right) \mathrm{du}$
Replace everyhting in the integral:

$\int \sqrt{9 - {\left(3 \sin \left(u\right)\right)}^{2}} \cdot 3 \cos \left(u\right) \mathrm{du}$
We can bring the 3 out of the integral:
$3 \cdot \int \sqrt{9 - {\left(3 \sin \left(u\right)\right)}^{2}} \cdot \cos \left(u\right) \mathrm{du}$
$3 \cdot \int \sqrt{9 - 9 {\sin}^{2} \left(u\right)} \cdot \cos \left(u\right) \mathrm{du}$
You can factor the 9 out:
$3 \cdot \int \sqrt{9 \left(1 - {\sin}^{2} \left(u\right)\right)} \cdot \cos \left(u\right) \mathrm{du}$
$3 \cdot 3 \int \sqrt{1 - {\sin}^{2} \left(u\right)} \cdot \cos \left(u\right) \mathrm{du}$

We know the identity: ${\cos}^{2} x + {\sin}^{2} x = 1$
If we solve for $\cos x$, we get:
${\cos}^{2} x = 1 - {\sin}^{2} x$
$\cos x = \sqrt{1 - {\sin}^{2} x}$
This is exactly what we see in the integral, so we can replace it:

$9 \int {\cos}^{2} \left(u\right) \mathrm{du}$
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: ${\cos}^{2} \left(u\right) = \frac{1 + \cos \left(2 u\right)}{2}$

$9 \int \frac{1 + \cos \left(2 u\right)}{2} \mathrm{du}$
$\frac{9}{2} \int 1 + \cos \left(2 u\right) \mathrm{du}$
$\frac{9}{2} \left(\int 1 \mathrm{du} + \int \cos \left(2 u\right) \mathrm{du}\right)$
$\frac{9}{2} \left(u + \frac{1}{2} \sin \left(2 u\right)\right) + C$ (you can work this out by substitution)
$\frac{9}{2} u + \frac{9}{4} \sin \left(2 u\right) + C$

Now, all we have to do is put $u$ into the function. Let's look back at how we defined it:
$x = 3 \sin \left(u\right)$
$\frac{x}{3} = \sin \left(u\right)$
To get $u$ out of this, you need to take the inverse function of $\sin$ on both sides, this is $\arcsin$:

$\arcsin \left(\frac{x}{3}\right) = \arcsin \left(\sin \left(u\right)\right)$
$\arcsin \left(\frac{x}{3}\right) = u$

Now we need to insert it into our solution:

$\frac{9}{2} \arcsin \left(\frac{x}{3}\right) + \frac{9}{4} \sin \left(2 \arcsin \left(\frac{x}{3}\right)\right) + C$

This is the final solution.