What is the integral of 9 - x 2 ?

Question

What is the integral of                     9        -                  x          2                    ?

Ryker Lloyd · Accepted Answer

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}} d x

x = 3 \sin (u)

You'll see why we're making this odd substitution, despite the fact that it may seem strange at first.

d x = 3 \cos (u) d u

Replace everyhting in the integral:

\int \sqrt{9 - {(3 \sin (u))}^{2}} \cdot 3 \cos (u) d u

We can bring the 3 out of the integral:

3 \cdot \int \sqrt{9 - {(3 \sin (u))}^{2}} \cdot \cos (u) d u

3 \cdot \int \sqrt{9 - 9 \sin^{2} (u)} \cdot \cos (u) d u

You can factor the 9 out:

3 \cdot \int \sqrt{9 (1 - \sin^{2} (u))} \cdot \cos (u) d u

3 \cdot 3 \int \sqrt{1 - \sin^{2} (u)} \cdot \cos (u) d u

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} (u) d u

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) = \frac{1 + \cos (2 u)}{2}

9 \int \frac{1 + \cos (2 u)}{2} d u

\frac{9}{2} \int 1 + \cos (2 u) d u

\frac{9}{2} (\int 1 d u + \int \cos (2 u) d u)

\frac{9}{2} (u + \frac{1}{2} \sin (2 u)) + C

(you can work this out by substitution)

\frac{9}{2} u + \frac{9}{4} \sin (2 u) + 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 (u)

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

\arcsin (\frac{x}{3}) = u

Now we need to insert it into our solution:

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

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