Welcome to Calculus.
I'm Professor Ghrist. We're about to begin lecture 28 on
trigonometric integrals. We're nearly done with chapter three.
At this point we've amassed a large collection of techniques for computing
definite and indefinite integrals. In this lesson what we're going to do is
collect all of those methods together and apply them to a particular class of
integrals involving trigonometric integrands.
In this lesson we'll consider three families of integrals involving
trigonometric functions The first, powers of sine and powers of cosine.
The second, powers of tangent and powers of secant.
The third, multiples of sines and cosines.
Throughout, m and n are going to denote positive integers.
For our first class of integrals, consider, as an integrand, sine to the m,
cosine to the n. Solving such a problem is going to break
up into several different cases depending on whether m and n are even or odd.
This will give us a good chance to practice our various integration
techniques. First of all if you're in the setting
where m, the power of sine, is odd. The the following substitution will be
effective. Expand sine squared as 1 minus cosine
squared. And then substitute n u equals cosign.
This will reduce the integral to something that is easily computed.
Likewise if n, the power of cosine, is odd you perform a similar simplification.
Expand out cosine squared as 1 minus sine squared, and then substitute u equals
sine. The one remaining case is when m and n
are both even. In this case, a substitution will not
work. A reduction formula is required.
This tends to be a bit more involved and difficult.
This is the case that is going to cause us a little bit of trouble.
Let's consider a specific example of the form sine to the m, cosine to the n.
Where m is 5 and n is 4. This puts us in a case where the power of
sine is odd, therefore we can split off an even number of powers of sine, and
substitute in 1 minus cosine squared for every sine squared that we have.
Now, you see that there is one power of sine left over.
So that if we perform the substitution, where u equals cosine theta.
We have a copy of du minus sine theta, d theta, sitting in the integrand.
And we wind up getting minus the integral of quantity 1 minus u squared, squared
times u to the 4th, du. This polynomial can be expanded out, and
then easily integrated substituting back in u equals cosine theta.
Gives us a solution. Negative cosine to the 5th over 5 plus 2
cosine to the 7th over 7 minus cosine to the 9th over 9 plus a constant.
Likewise if we had an odd power of cosine.
Then we could apply the same method where we substitute in 1 minus sine squared for
cosine squared and we have left over a single power of cosine.
Another u substitution will easily solve this integral.
Integrals involving powers of tangent and secant follow a similar pattern depending
on the parity of the powers. If one is in the setting where the power
of tangent is odd, then perform the simplification expanding out every
tangent squared as secant squared minus 1, then substitute in u equals secant.
Likewise if n, the power of secant is even then one can simplify every secant
squared is 1 plus tangent squared. The substitution u equals tangent will
then work. The one remaining case where a reduction
formula is required, is when n the power secant is odd, and m, the power of
tangent is even. Let's consider a example, in this case
where the power of tangent is 5, and the power of secant is 6.
Then, because the power of secant is even.
One way to solve this integral is to split off a secant squared, and replace
the remaining even powers of secant with quantity 1 plus tangent squared.
This means that when we do the substitution u equals tangent we have a
copy of du sitting right there to be absorbed.
This yields a polynomial integral of the form u to the 5th times quantity 1 plus u
squared, squared. By expanding that out integrating that
polynomial and substituting back in tangent for u, we easily obtain the
answer, tangent to the 6th, over 6, plus tangent to the 8th over 4, plus tangent
to the 10th over 10, plus a constant. However, there's another way to solve
this integral as well. Exploiting the fact that m, the power of
tangent, is odd. In this case, what we'll want to do is
split off an even power of tangent, substitute n secant squared minus 1 for
tangent squared. And then use the fact that there is a
secant tangent left over. To perform a substitution where u equals
secant theta. There remains, in each class that we've
considered, one case where a simple substitution does not work.
This requires a reduction formula. The first step is to simplify to sums of
powers of cosine or secant. For example if we look at the integral of
sine to the 4th, cosine to the 4th. This is in that one case where both
powers are even. We can replace sine squared with 1 minus
cosine squared and obtain a sum of integrals, each of which is an even power
of cosine. Then we can apply integration by parts
twice to obtain the following general reduction formula.
The integral of cosine to the n is cosine to the n minus 1, times sine, over n,
plus n minus 1 over n, times the integral of cosine to the n minus 2.
This simplifies or reduces the level of complexity of the integral involved.
Likewise, for powers of secant, one can express the integral of secant to the n
as tangent secant to the n minus 2 over n minus 1 plus n minus 2 over n minus 1
times the integral of secant to the n minus 2.
Now you do not have to memorize these formulae, they're the type of thing that
one looks up when you're stuck on a difficult integral.
Let's apply this reduction formula in the specific case of the integral of cosine
to the n. Recalling the reduction formula for
powers of cosine. We're going to have to apply this
iteratively many times until we get down to a low enough power of cosine that we
can proceed. To make things a bit more concrete, let's
do a definite integral, as theta goes from negative pi over 2 to pi over 2.
When we do so one of the things that's nice is that the term cosine to the n
minus 1 sine over n, in the reduction formula, vanishes.
When we perform evaluation. So, for n greater than 1, we get that the
definite integral from negative pi over 2 to pi over 2 of cosine to the n is n
minus 1 over n times the same integral with the power being n minus 2.
This will allow us to come up with a recursive solution.
So let's look at all the different powers.
In the simplest case, where n equals 0, well we can do that integral.
The integral of d theta is theta evaluated from negative pi over 2 to pi
over 2 gives pi. Likewise we can do n equals 1 explicitly,
integrating cosine, getting sine, evaluating at the limits gives us the
value of 2. Now, for higher powers we can use the
reduction formula. All we have to do is multiply by n minus
1 over n. So, to get n equals 2, we multiply 2
minus 1 over 2. That is 1 half times the integral in the
case where n equals 0. To get the value for n equals 3, we
multiply the value for n equals 1 by 3 minus 1, over 3.
We can continue for increasing values of n, always looking back to n minus 2 and
multiplying by n minus 1 over n. Now let's look at this, what do you
notice? Well, first of all, it seems as though
there's a real dependence on whether n is even or odd.
When n is even, there's a factor of pi involved, and when n is odd, there's not.
That's maybe not so surprising seeing that even and odd powers have been
different all throughout this lesson. In general using a little bit of
induction one can show that when n is even the result of this integral is pi.
Times 1 over 2 times 3 over 4 times 5 over 6 et cetera, all the way down to n
minus 1 over n. When n is odd, one obtains a similar
looking result but starting with 2 instead of pi.
And flipping things, the result is 2 3rds times 4 5ths times 6 7ths all the way up
to n minus 1 over n. All of that times 2.
By putting a 1 down in the denominator, we can see a familiar pattern between
these two definite Integrals. Now, there's one last class of integrals
that we're going to look at, and that is something in the form the integral of
sine of m theta times cosine of n theta. The sine wave and the cosine wave have
potentially different periods. Now this integral requires an algebraic
simplification. The following formula is something that
you do not have to know or have memorized, but which is extremely useful.
That is, sine of m theta times cosine of n theta equals 1 half sine of m plus n
theta, plus sine of m minus n theta. At least in the case where m and n are
not the same. If we integrate both sides, with respect
to theta, then we obtain the formula, negative cosine m plus n, theta over 2
times quantity m plus n minus cosine of m minus n theta over twice quantity m minus
n. And you can see here why m and n have to
be different. Likewise, one can do the same thing for
sine of m theta times sine of n theta. A similar reduction gives an integral
that is computable. Likewise, with a pair of cosines of
different periods. The result works out.
Once again, these are not the kind of formulae that you memorize, but they are
useful to look up. Now why would any of these integrals be
useful to us? One important reason is that sines and
cosines, trigonometric functions pervade the physical universe.
If you go far enough in mathematics you will learn about Fourier Analysis, which
is a mathematics built on sines and cosines.
It's extremely useful in analyzing any kind of wave.
Whether acoustic or electromagnetic. In fact one could argue that all of radar
and signal processing is built on a basis of integrals of sines and cosines.
Now you know a bit about how to compute them.
I think you'll agree, some of those integrals were kind of tricky.
What do you do, when faced with harder and harder integration problems?
Do you have to employ ever more clever tricks?
No. Mathematics is not about tricks.
But rather, about principles. However, when having to solve a difficult
problem, we have to use every available method.
In our next lesson, we'll give a brief introduction to those methods that are
computer assisted.