Gingan7mhd

2023-03-25

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

Ryker Lloyd

Beginner2023-03-26Added 6 answers

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:

$3\cdot \int \sqrt{9(1-{\mathrm{sin}}^{2}\left(u\right))}\cdot \mathrm{cos}\left(u\right)du$

$3\cdot 3\int \sqrt{1-{\mathrm{sin}}^{2}\left(u\right)}\cdot \mathrm{cos}\left(u\right)du$

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

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

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

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

$\frac{9}{2}(u+\frac{1}{2}\mathrm{sin}\left(2u\right))+C$ (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:

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

$\int \sqrt{9-{x}^{2}}dx$

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

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:

$3\cdot \int \sqrt{9(1-{\mathrm{sin}}^{2}\left(u\right))}\cdot \mathrm{cos}\left(u\right)du$

$3\cdot 3\int \sqrt{1-{\mathrm{sin}}^{2}\left(u\right)}\cdot \mathrm{cos}\left(u\right)du$

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

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

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

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

$\frac{9}{2}(u+\frac{1}{2}\mathrm{sin}\left(2u\right))+C$ (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:

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

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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