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Welcome to Calculus. I'm Professor Ghrist.

Â We're about to begin lecture 28 on trigonometric integrals.

Â We're nearly done with chapter three. At this point we've amassed a large

Â collection of techniques for computing definite and indefinite integrals.

Â In this lesson what we're going to do is collect all of those methods together and

Â apply them to a particular class of integrals involving trigonometric

Â integrands. In this lesson we'll consider three

Â families of integrals involving trigonometric functions The first, powers

Â of sine and powers of cosine. The second, powers of tangent and powers

Â of secant. The third, multiples of sines and

Â cosines. Throughout, m and n are going to denote

Â positive integers. For our first class of integrals,

Â consider, as an integrand, sine to the m, cosine to the n.

Â Solving such a problem is going to break up into several different cases depending

Â on whether m and n are even or odd. This will give us a good chance to

Â practice our various integration techniques.

Â First of all if you're in the setting where m, the power of sine, is odd.

Â The the following substitution will be effective.

Â Expand sine squared as 1 minus cosine squared.

Â And then substitute n u equals cosign. This will reduce the integral to

Â something that is easily computed. Likewise if n, the power of cosine, is

Â odd you perform a similar simplification. Expand out cosine squared as 1 minus sine

Â squared, and then substitute u equals sine.

Â The one remaining case is when m and n are both even.

Â In this case, a substitution will not work.

Â A reduction formula is required. This tends to be a bit more involved and

Â difficult. This is the case that is going to cause

Â us a little bit of trouble. Let's consider a specific example of the

Â form sine to the m, cosine to the n. Where m is 5 and n is 4.

Â This puts us in a case where the power of sine is odd, therefore we can split off

Â an even number of powers of sine, and substitute in 1 minus cosine squared for

Â every sine squared that we have. Now, you see that there is one power of

Â sine left over. So that if we perform the substitution,

Â where u equals cosine theta. We have a copy of du minus sine theta, d

Â theta, sitting in the integrand. And we wind up getting minus the integral

Â of quantity 1 minus u squared, squared times u to the 4th, du.

Â This polynomial can be expanded out, and then easily integrated substituting back

Â in u equals cosine theta. Gives us a solution.

Â Negative cosine to the 5th over 5 plus 2 cosine to the 7th over 7 minus cosine to

Â the 9th over 9 plus a constant. Likewise if we had an odd power of

Â cosine. Then we could apply the same method where

Â we substitute in 1 minus sine squared for cosine squared and we have left over a

Â single power of cosine. Another u substitution will easily solve

Â this integral. Integrals involving powers of tangent and

Â secant follow a similar pattern depending on the parity of the powers.

Â If one is in the setting where the power of tangent is odd, then perform the

Â simplification expanding out every tangent squared as secant squared minus

Â 1, then substitute in u equals secant. Likewise if n, the power of secant is

Â even then one can simplify every secant squared is 1 plus tangent squared.

Â The substitution u equals tangent will then work.

Â The one remaining case where a reduction formula is required, is when n the power

Â secant is odd, and m, the power of tangent is even.

Â Let's consider a example, in this case where the power of tangent is 5, and the

Â power of secant is 6. Then, because the power of secant is

Â even. One way to solve this integral is to

Â split off a secant squared, and replace the remaining even powers of secant with

Â quantity 1 plus tangent squared. This means that when we do the

Â substitution u equals tangent we have a copy of du sitting right there to be

Â absorbed. This yields a polynomial integral of the

Â form u to the 5th times quantity 1 plus u squared, squared.

Â By expanding that out integrating that polynomial and substituting back in

Â tangent for u, we easily obtain the answer, tangent to the 6th, over 6, plus

Â tangent to the 8th over 4, plus tangent to the 10th over 10, plus a constant.

Â However, there's another way to solve this integral as well.

Â Exploiting the fact that m, the power of tangent, is odd.

Â In this case, what we'll want to do is split off an even power of tangent,

Â substitute n secant squared minus 1 for tangent squared.

Â And then use the fact that there is a secant tangent left over.

Â To perform a substitution where u equals secant theta.

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There remains, in each class that we've considered, one case where a simple

Â substitution does not work. This requires a reduction formula.

Â The first step is to simplify to sums of powers of cosine or secant.

Â For example if we look at the integral of sine to the 4th, cosine to the 4th.

Â This is in that one case where both powers are even.

Â We can replace sine squared with 1 minus cosine squared and obtain a sum of

Â integrals, each of which is an even power of cosine.

Â Then we can apply integration by parts twice to obtain the following general

Â reduction formula. The integral of cosine to the n is cosine

Â to the n minus 1, times sine, over n, plus n minus 1 over n, times the integral

Â of cosine to the n minus 2. This simplifies or reduces the level of

Â complexity of the integral involved. Likewise, for powers of secant, one can

Â express the integral of secant to the n as tangent secant to the n minus 2 over n

Â minus 1 plus n minus 2 over n minus 1 times the integral of secant to the n

Â minus 2. Now you do not have to memorize these

Â formulae, they're the type of thing that one looks up when you're stuck on a

Â difficult integral. Let's apply this reduction formula in the

Â specific case of the integral of cosine to the n.

Â Recalling the reduction formula for powers of cosine.

Â We're going to have to apply this iteratively many times until we get down

Â to a low enough power of cosine that we can proceed.

Â To make things a bit more concrete, let's do a definite integral, as theta goes

Â from negative pi over 2 to pi over 2. When we do so one of the things that's

Â nice is that the term cosine to the n minus 1 sine over n, in the reduction

Â formula, vanishes. When we perform evaluation.

Â So, for n greater than 1, we get that the definite integral from negative pi over 2

Â to pi over 2 of cosine to the n is n minus 1 over n times the same integral

Â with the power being n minus 2. This will allow us to come up with a

Â recursive solution. So let's look at all the different

Â powers. In the simplest case, where n equals 0,

Â well we can do that integral. The integral of d theta is theta

Â evaluated from negative pi over 2 to pi over 2 gives pi.

Â Likewise we can do n equals 1 explicitly, integrating cosine, getting sine,

Â evaluating at the limits gives us the value of 2.

Â Now, for higher powers we can use the reduction formula.

Â All we have to do is multiply by n minus 1 over n.

Â So, to get n equals 2, we multiply 2 minus 1 over 2.

Â That is 1 half times the integral in the case where n equals 0.

Â To get the value for n equals 3, we multiply the value for n equals 1 by 3

Â minus 1, over 3. We can continue for increasing values of

Â n, always looking back to n minus 2 and multiplying by n minus 1 over n.

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Now let's look at this, what do you notice?

Â Well, first of all, it seems as though there's a real dependence on whether n is

Â even or odd. When n is even, there's a factor of pi

Â involved, and when n is odd, there's not. That's maybe not so surprising seeing

Â that even and odd powers have been different all throughout this lesson.

Â In general using a little bit of induction one can show that when n is

Â even the result of this integral is pi. Times 1 over 2 times 3 over 4 times 5

Â over 6 et cetera, all the way down to n minus 1 over n.

Â When n is odd, one obtains a similar looking result but starting with 2

Â instead of pi. And flipping things, the result is 2 3rds

Â times 4 5ths times 6 7ths all the way up to n minus 1 over n.

Â All of that times 2. By putting a 1 down in the denominator,

Â we can see a familiar pattern between these two definite Integrals.

Â Now, there's one last class of integrals that we're going to look at, and that is

Â something in the form the integral of sine of m theta times cosine of n theta.

Â The sine wave and the cosine wave have potentially different periods.

Â Now this integral requires an algebraic simplification.

Â The following formula is something that you do not have to know or have

Â memorized, but which is extremely useful. That is, sine of m theta times cosine of

Â n theta equals 1 half sine of m plus n theta, plus sine of m minus n theta.

Â At least in the case where m and n are not the same.

Â If we integrate both sides, with respect to theta, then we obtain the formula,

Â negative cosine m plus n, theta over 2 times quantity m plus n minus cosine of m

Â minus n theta over twice quantity m minus n.

Â And you can see here why m and n have to be different.

Â Likewise, one can do the same thing for sine of m theta times sine of n theta.

Â A similar reduction gives an integral that is computable.

Â Likewise, with a pair of cosines of different periods.

Â The result works out. Once again, these are not the kind of

Â formulae that you memorize, but they are useful to look up.

Â Now why would any of these integrals be useful to us?

Â One important reason is that sines and cosines, trigonometric functions pervade

Â the physical universe. If you go far enough in mathematics you

Â will learn about Fourier Analysis, which is a mathematics built on sines and

Â cosines. It's extremely useful in analyzing any

Â kind of wave. Whether acoustic or electromagnetic.

Â In fact one could argue that all of radar and signal processing is built on a basis

Â of integrals of sines and cosines. Now you know a bit about how to compute

Â them. I think you'll agree, some of those

Â integrals were kind of tricky. What do you do, when faced with harder

Â and harder integration problems? Do you have to employ ever more clever

Â tricks? No.

Â Mathematics is not about tricks. But rather, about principles.

Â However, when having to solve a difficult problem, we have to use every available

Â method. In our next lesson, we'll give a brief

Â introduction to those methods that are computer assisted.

Â