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.