Let's see if we can evaluate the indefinite integral 1
over plus 9 plus x squared dx.
And we know that if you have the pattern a squared
minus x squared, it could be a good idea
to make the substitution, x is equal to a sine theta.
But we don't see that pattern over here.
Instead, what we see is a squared plus x squared.
And in this context, it tends to be a good idea-- it's not
always going to work, but it never hurts to try out.
This is a little bit of an art here-- to try out,
x is equal to a tangent theta.
Now you might say, Sal, why is that?
Well, let's make that substitution
and see how this thing would simplify.
This thing would become a squared plus a squared tangent
squared theta, which is a squared times 1
plus tangent squared theta.
And this right over here, we could reprove it.
Actually, let me just reprove it for you.
This is going to become a squared times-- this
is 1 could be written as cosine squared theta
over cosine squared theta.
Tangent squared is sine squared theta
over cosine squared theta.
And this is why I picked cosine squared as the denominator,
so that I can add the two.
And this is going to become a squared times cosine
squared theta plus sine squared theta,
all of that over cosine squared theta.
This numerator from the unit circle definition
of trig functions becomes 1.
So this is 1 over cosine squared theta.
And everything simplifies to a squared secant
squared theta, which might simplify things.
So let's see if we have what's going on over here.
We can rewrite this.
So 9 plus x squared you could rewrite
as 3 squared plus x squared.
In this case, a would be equal to 3.
So we want to make the substitution,
x is equal to 3 tangent of theta.
And if we wanted to solve for x, you
can divide both sides by 3, because we're later
going to have to undo the substitution.
x over 3 is equal to tangent theta,
or theta is equal to arctangent or inverse tangent of x over 3.
Now we're also going to have to figure out what dx is.
We're also going to have to figure out what that is.
So let's take the derivative or we'll write it
in differential form.
dx is equal to 3 derivative of tangent theta with respect
to theta is secant squared theta d theta.
So now it looks like we're armed with all
of the things necessary to rewrite this entire integral.
It's going to be equal to the indefinite integral.
You're going to have dx here, which
is equal to 3 secant squared theta d theta.
That's the dx.
And all of that's going to be over this business
right over here, our a squared plus our x squared.
Now we already know what that's going to simplify to.
Our a squared plus our x squared is
going to-- since we made this substitution,
x is equal to 3 tangent theta is going
to simplify to a squared secant squared theta.
So this is going to simplify to 9 secant squared theta.
And you could essentially go through this logic.
You're going to get 9 plus 9 tangent squared theta.
Factor out a 9, you get 9 times 1 plus tan squared theta.
And so that's going to be 9 times secant
squared theta, exactly what we have here.
Lucky for us, we have the secant squared.
It's canceling out.
The secant squared is canceling out.
You have 3 over 9.
This whole thing can be rewritten
as 1/3-- that's just the 3 over 9--
times the indefinite integral of just d theta, which
is equal to 1/3 theta plus c.
And now we just have to put things in terms of x.
And we see theta is equal to arctangent of x over 3.
So this is going to be equal to 1/3 arctangent of x
over 3 plus c.
And we are done.
So now we know how to deal with cases where we see something
like an a squared minus an x squared
and an a squared plus an x squared.
It won't always work, but it might
be a useful-- it'll definitely allow you to do this thing.
It might not always make the integral solvable,
but it's not a bad thing to try.
When it looks like u-substitution isn't working,
then you can look for these patterns
and try some trig substitution.