Gingan7mhd

2023-03-25

What is the integral of $\sqrt{9-{x}^{2}}$?

Ryker Lloyd

Whenever I see these kind of functions, I recognize (by practicing a lot) that you should use a special substitution here:
$\int \sqrt{9-{x}^{2}}dx$
$x=3\mathrm{sin}\left(u\right)$
You'll see why we're making this odd substitution, despite the fact that it may seem strange at first.
$dx=3\mathrm{cos}\left(u\right)du$
Replace everyhting in the integral:
$\int \sqrt{9-{\left(3\mathrm{sin}\left(u\right)\right)}^{2}}\cdot 3\mathrm{cos}\left(u\right)du$
We can bring the 3 out of the integral:
$3\cdot \int \sqrt{9-{\left(3\mathrm{sin}\left(u\right)\right)}^{2}}\cdot \mathrm{cos}\left(u\right)du$
$3\cdot \int \sqrt{9-9{\mathrm{sin}}^{2}\left(u\right)}\cdot \mathrm{cos}\left(u\right)du$
You can factor the 9 out:
$3\cdot \int \sqrt{9\left(1-{\mathrm{sin}}^{2}\left(u\right)\right)}\cdot \mathrm{cos}\left(u\right)du$
$3\cdot 3\int \sqrt{1-{\mathrm{sin}}^{2}\left(u\right)}\cdot \mathrm{cos}\left(u\right)du$
We know the identity: ${\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x=1$
If we solve for $\mathrm{cos}x$, we get:
${\mathrm{cos}}^{2}x=1-{\mathrm{sin}}^{2}x$
$\mathrm{cos}x=\sqrt{1-{\mathrm{sin}}^{2}x}$
This is exactly what we see in the integral, so we can replace it:
$9\int {\mathrm{cos}}^{2}\left(u\right)du$
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: ${\mathrm{cos}}^{2}\left(u\right)=\frac{1+\mathrm{cos}\left(2u\right)}{2}$
$9\int \frac{1+\mathrm{cos}\left(2u\right)}{2}du$
$\frac{9}{2}\int 1+\mathrm{cos}\left(2u\right)du$
$\frac{9}{2}\left(\int 1du+\int \mathrm{cos}\left(2u\right)du\right)$
$\frac{9}{2}\left(u+\frac{1}{2}\mathrm{sin}\left(2u\right)\right)+C$ (you can work this out by substitution)
$\frac{9}{2}u+\frac{9}{4}\mathrm{sin}\left(2u\right)+C$
Now, all we have to do is put $u$ into the function. Let's look back at how we defined it:
$x=3\mathrm{sin}\left(u\right)$
$\frac{x}{3}=\mathrm{sin}\left(u\right)$
To get $u$ out of this, you need to take the inverse function of $\mathrm{sin}$ on both sides, this is $\mathrm{arcsin}$:
$\mathrm{arcsin}\left(\frac{x}{3}\right)=\mathrm{arcsin}\left(\mathrm{sin}\left(u\right)\right)$
$\mathrm{arcsin}\left(\frac{x}{3}\right)=u$
Now we need to insert it into our solution:
$\frac{9}{2}\mathrm{arcsin}\left(\frac{x}{3}\right)+\frac{9}{4}\mathrm{sin}\left(2\mathrm{arcsin}\left(\frac{x}{3}\right)\right)+C$

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