It’s Professor Dave, let’s tackle integration by trig substitution.
We have already learned two handy tricks for integrating.
The first was the substitution rule, which was useful if the integrand could be expressed
as the product of a function and its derivative.
Then there was integration by parts, which was useful for a variety of integrands present
as the product of two functions.
Let’s learn one more trick right now, a technique called trigonometric substitution.
Trigonometric substitution is a technique that works when one of three very specific
terms is present in the integrand.
These are the square root of A squared minus x squared, the square root of A squared plus
x squared, or the square root of x squared minus A squared.
Here, A can be any positive number.
With integrands like these, we will see that the substitution rule will not work, because
there is no term present to act as du.
But there is a different type of substitution we can use, if we realize that we can change
x into something else that will result in the ability to get rid of the radical.
Let’s first see what we can do to simplify these three expressions, and then we will
put that into the greater context of integration.
Take this first example, with A squared minus x squared under the radical.
Let’s allow x to equal A sine theta.
If we plug that in for x, the exponent distributes and we get A squared minus A squared sine
squared theta.
Since we have A squared in both terms, let’s factor that out to get A squared times the
quantity one minus sine squared theta.
We know from our Pythagorean identities that one minus sine squared theta is cosine squared theta.
So let’s make that change, and we are left with a perfect square.
So let’s take the square root, and we are left with A times the absolute value of cosine theta.
If we take the interval of theta as being between negative half pi and half pi, we know
cosine theta will be positive, so we can lose the absolute value, and we are left with A
cosine theta.
This is quite a bit simpler than what we started with, as you can see.
Now let’s look at the next expression, root A squared plus x squared.
For this one, let’s make x equal to A tangent theta.
Just like the first example, we square it, factor out the A squared, and we recognize
that another Pythagorean identity is now applicable.
One plus tangent squared is secant squared, so we apply that, and then get rid of the
square root to get A secant theta.
Again, much nicer looking.
And lastly, if we have root x squared minus A squared, we can change x into A secant theta.
That gives us A squared secant squared theta minus A squared.
Again we factor out the A squared, and then realize that the same identity we used in
the previous example applies here, just the other way around, so we change secant squared
minus one into tangent squared.
Again, the square root goes away and we have A tangent theta.
So this is the first of three things we have to understand about integrating by trigonometric substitution.
It can only be done when one of these three specific expressions is part of the integrand,
and there is a very specific substitution that must be done in each of the three cases,
changing x into A sine theta, A tangent theta, or A secant theta, depending on which expression
is present.
But these conditions are not too hard to identify if we just remember that the goal is to get
rid of the square roots, and the Pythagorean identities relate sums and differences of
squared things to squared things.
However, this substitution and simplification is only the first part of the technique.
To see what else there is to it, let’s look at a specific example to see why this substitution
becomes useful.
Take something like the integral of root (nine minus x squared), over x squared, dx.
We look at this radical term and we see that it is one of the expressions that qualify
this for trig substitution.
In this particular case, we are going to change x into A sine theta, and for this example,
if the term starts with nine, nine must be A squared, so A will be three.
That means x must specifically be substituted with three sine theta, and by the steps we
already discussed, this whole term will become three cosine theta.
But that’s not the only x in the integrand, this x squared term in the denominator will
become three sine theta quantity squared, or nine sine squared theta.
Now here’s the next thing we must keep in mind with this approach.
This integral was in terms of x, so there is still a dx here.
But if we changed everything to be in terms of theta, then this must become d theta, otherwise
we can’t integrate with respect to theta.
So what is dx in terms of theta?
Well if x is three sine theta, then dx over d theta must be its derivative with respect
to theta.
The derivative of three sine theta is three cosine theta.
Then we bring d theta up here, and that allows us to change dx into three cosine theta d theta.
Now everything is in terms of theta and we are ready to integrate.
We can multiply this first three cosine theta by this new three cosine theta to get nine
cosine squared theta.
The nines cancel, and we are left with cosine squared over sine squared which is cotangent squared.
Cotangent squared is not a common integral, so let’s try to change the way it is expressed
using some identity.
The easiest one would be the identity that tells us that cotangent squared is the same
as cosecant squared minus one.
So we make that change, which is advantageous, because cosecant squared can be directly integrated.
Since the derivative of cotangent is negative cosecant squared, cosecant squared theta integrates
to become negative cotangent theta, and the one must become theta, so we have negative
cotangent theta minus theta, and then plus C as usual.
So the integration was successful, but unfortunately we still are not quite done.
We wanted to integrate in terms of x, but we integrated in terms of theta.
Not to worry, we can convert back to a form with x’s in it without too much trouble.
Let’s just make a triangle that looks like it is in the first quadrant, and label the
angle theta.
Now according to our original substitution for x, what will these sides be?
Well if x was equal to three sine theta, then sine theta must be equal to x over three.
Sine theta is opposite over hypotenuse, so according to our own substitution, this vertical
leg must be equal to x, and the hypotenuse must be equal to three.
By the Pythagorean theorem, we can also get this other leg, as this leg squared, plus
x squared, must equal nine, so we subtract to get nine minus x squared, and then take
the square root to get root (nine minus x squared).
Now with all of these sides labeled, we can express terms with theta in terms of x instead.
We have a cotangent theta, and we know that cotangent is the reciprocal of tangent.
If tangent is opposite over adjacent, cotangent is therefore adjacent over opposite, and in
this case that is root (nine minus x squared), over x.
So negative cotangent theta becomes negative this expression we just found, and then for
theta, if sine theta is x over three, then theta must be the inverse sine of x over three.
Then we include plus C, and that will be our answer.
So that may seem quite involved, but remember that we really just did three things.
First we recognized one of the three expressions that allow for trig substitution, and we made
the substitution.
Then we manipulated dx to get an expression with d theta in it, which allowed us to integrate
by whatever means now possible.
Then to get our answer back in terms of x, we constructed a triangle and labeled everything
in terms of x so that we could express our answer in terms of x.
Don’t worry, this will get easier over time.
Let’s try one more to help things sink in.
How about the integral of one over the quantity [x squared times (root x squared plus four)], dx.
The first part is easy, we just recognize that this radical expression is one of the
three that can be simplified by trig substitution, and we recall that with this particular expression,
we allow x to be substituted by A tangent theta.
In this case, if A squared is four, then A is two, so we will allow x to become two tangent theta.
Let’s go ahead and take care of dx while we are at it.
Finding dx over d theta will require taking the derivative of tangent, which is secant
squared, and then we bring d theta up here to get two secant squared theta d theta.
So we go ahead and transform our integral, putting dx up here in the numerator to make
things simpler, and we get two secant squared theta d theta over four tangent squared theta
times two secant theta.
Now substitution is complete and its time to perform the integration.
First we can simplify a bit.
The numbers simplify to one fourth, so let’s take that out of the integral to make things simpler.
One of the secants will cancel, and we are left with secant theta over tangent squared
theta, d theta.
Let’s change secant to one over cosine, and tangent squared to sine squared over cosine squared.
Then we can multiply the numerator by the reciprocal of the denominator.
One cosine cancels, and we are left with cosine theta over sine squared theta, d theta.
Now once we are here, after performing the trig substitution and simplifying as much
as possible, we ought to have something that can be integrated, but it may require another
one of our techniques.
In this case, we can do another substitution to integrate this.
Let’s say that u equals sine theta.
Then du will be cosine theta d theta.
That ends up coming in handy, because this integral then becomes du over u squared.
This is finally something that we can integrate directly.
U to the negative two becomes u to the negative one over negative one, or negative one over u, plus C.
We take the negative out here to join the one fourth, change u back into sine theta,
and we are left with negative one over four sine theta, plus C.
If we prefer, one over sine can become cosecant, so let’s change this to negative cosecant
theta over four, plus C. So we got from u back to theta, but remember, we now have to
go from theta back to x, so let’s make our triangle.
We place theta, and then we consult the original substitution we made, which was x equals two
tangent theta.
Therefore, tangent theta equals x over two.
Tangent is opposite over adjacent, so this opposite leg must be x, and this adjacent
leg must be two.
Then by the Pythagorean theorem, the hypotenuse must be root x squared plus four.
Now we have our triangle set up, so let’s check back with our answer to see what needs
to be done.
We have a cosecant theta term in there, and if sine is opposite over hypotenuse, then
cosecant is hypotenuse over opposite.
In this case that will be root x squared plus four, over x.
So let’s put that in there in the place of cosecant theta.
The denominator of the numerator can just be brought down to the denominator, and there
is our final answer, negative root x squared plus four, over four x, plus C.
So with that, we now have three solid techniques in our bag of tricks, those being the substitution
rule, integration by parts, and trigonometric substitution.
As we learned each of these techniques, it was quite obvious which strategy we would
be applying in each example, but we also have to get used to looking at integrals without
any context, and figuring out which strategy will work best.
In the next tutorial, we will practice exactly that, but first, let’s check comprehension.