Welcome to the first of several videos
on the integration technique of trig substitution.
All the techniques discussed in this video are based upon
these Pythagorean identities in this form.
We have one minus sine squared theta equals
cosine squared theta; one plus tangent squared theta
is equal to secant squared theta, which also can be
expressed as secant squared theta minus one
is equal to tangent squared theta.
We'll be looking at three different types of integrals
that require trig substitution.
The first type will be the integral involving
the square root of a squared minus x squared.
And if we have an integral in this form, we're going to let
x equal a sine theta, and if we do that then the square root
of a squared minus x squared will equal a cosine theta.
And this is based on the identity
one minus sine squared theta is equal to
cosine squared theta.
For example if we had a problem where a was equal to three,
we'd have the square root of nine minus, if a was three,
x would be three sine theta,
so we'd have nine sine squared theta.
Let's go ahead and take a look at how this simplifies.
We know there's a common factor of nine, so we can
factor out the nine and have one minus sine squared theta,
and then one minus sine squared theta is equal to
cosine squared theta, and of course the square root of nine
would be three, and the square root of cosine square
would be cosine theta, so this just shows that when we have
the square root of a squared minus x squared,
where x is a sine theta, it would equal a cosine theta,
and this is the reason why.
The second type of integral we'll take a look at will be
in the form of the square root of a squared plus x squared,
and if this is the case, we'll let x equal a tangent theta.
So if we use the identity one plus tangent squared theta
equals secant squared theta, the square root of
a squared plus x squared in this case
will be equal to a secant theta.
And then the last case will be when we have an integral
involving the square root of x squared minus a squared,
and in this case we'll let x equal a secant theta.
Therefore the square root of x squared minus a squared
will equal plus or minus a tangent theta.
Notice this one has a plus or minus,
and that's based upon the angle theta.
Since tangent theta is positive
between zero and pi over two, we'd use positive a,
in between pi over two and pi, tangent theta is negative,
so we'd use negative a.
And again this equation here is based upon
the fact that secant squared theta minus one
is equal to tangent squared theta.
One last word of caution before we take a look at
our first example, we don't want to assume that every integral
in the forms discussed will require trig substitution.
For example if we compare these two integrals,
only one of them would require a trig substitution.
The other one only requires basic u substitution.
Can you recognize the one
that requires basic u substitution?
It's actually the second one here, because if we let u equal
the ir-rat-ik-and, nine minus x squared, then differential u
is going to equal a negative two x dx, and notice
this in-te-grand does have xdx, so if we divided both sides
by negative two, we'd have negative one-half du is equal to
x dx, so we could rewrite this in terms of u,
what we would have is again x dx would equal
negative one-half du, and this would be one
over the square root of u, or u to the negative one-half.
So we could find the anti-derivative
using basic u substitution, however on this integral,
we would have to use trig substitution.
Let's go ahead and take a look at our first example.
Notice the integral involves
a square root of nine minus x squared.
So this involves the form
the square root of a squared minus x squared.
So a squared is going to equal nine, therefore a would be
equal to three, so we're going to let x equal three sine theta.
Now let's go ahead and find differential x.
Differential x is going to equal three cosine theta, d theta,
and the last thing we're going to do
before we perform this substitution is sketch a triangle
that represents angle theta.
We're going to need that toward the end of the problem.
So let's sketch a right triangle, identifying angle theta,
and looking at this equation here, if we solve this
for sine theta, sine theta will equal x over three,
which means the opposite side would be x
and the hypotenuse would be three.
So using the Pythagorean Theorem, we can state that
this side here would be the square root of
three squared minus x squared, or nine minus x squared.
Now let's go back to the integral
and perform the substitution.
Notice that dx is equal to three cosine theta d theta.
So we'll have three cosine theta, d theta
over the square root of nine minus x squared,
well that'll be nine minus nine sine squared theta.
Now performing this substitution,
this denominator's going to simplify nicely.
Remember the square root of nine
minus nine sine square theta simplifies nicely
to three cosine theta, so this is equal to
three cosine theta over three cosine theta, d theta.
Well of course this simplifies to one,
and the anti-derivative of one in respects to theta
would be theta plus c, so we do have the anti-derivative
here, but it's not in terms of x like divisional integral,
so we can't leave our answer in terms of theta.
And that's why it's so important
that we have this right triangle here.
We know that sine theta is equal to x over three,
so if we take the inverse sine of both sides,
we can solve this equation for theta in terms of x
to express our anti-derivative in terms of x.
Remember the inverse sine of sine theta equals
the inverse sine of x over three.
On the left we have inverse sine of x over three,
therefore the anti-derivative in terms of x
will be inverse sine x over three plus c,
so it is very important that we remember when we're
setting up this integral from the beginning that we
do sketch a right triangle that models our angle theta.
Okay this is going to be the end of part one.
In the next video we'll take a look at
the form of the square root of a squared plus x squared
when we let x equal a tangent theta.
I hope you'll watch part two.