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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.