In this section we are going to look at quite a few integrals involving trig functions and some of the techniques we can use to help us evaluate them.  Let’s start off with an integral that we should already be able to do.

This integral is easy to do with a substitution because the presence of the cosine, however, what about the following integral.

Notice that we were able to do the rewrite that we did in the previous example because the exponent on the sine was odd.  In these cases all that we need to do is strip out one of the sines.  The exponent on the remaining sines will then be even and we can easily convert the remaining sines to cosines using the identity,

If the exponent on the sines had been even this would have been difficult to do.  We could strip out a sine, but the remaining sines would then have an odd exponent and while we could convert them to cosines the resulting integral would often be even more difficult than the original integral in most cases.

Let’s take a look at another example.

At this point let’s pause for a second to summarize what we’ve learned so far about integrating powers of sine and cosine. 

In this integral if the exponent on the sines (n) is odd we can strip out one sine, convert the rest to cosines using (1) and then use the substitution .  Likewise, if the exponent on the cosines (m) is odd we can strip out one cosine and convert the rest to sines and the use the substitution .

Of course, if both exponents are odd then we can use either method.  However, in these cases it’s usually easier to convert the term with the smaller exponent.

The one case we haven’t looked at is what happens if both of the exponents are even?  In this case the technique we used in the first couple of examples simply won’t work and in fact there really isn’t any one set method for doing these integrals.  Each integral is different and in some cases there will be more than one way to do the integral.

With that being said most, if not all, of integrals involving products of sines and cosines in which both exponents are even can be done using one or more of the following formulas to rewrite the integrand.

The first two formulas are the standard half angle formula from a trig class written in a form that will be more convenient for us to use.  The last is the standard double angle formula for sine, again with a small rewrite.

Let’s take a look at an example.

In the previous example we saw two different solution methods that gave the same answer.  Note that this will not always happen.  In fact, more often than not we will get different answers.  However, as we discussed in the Integration by Parts section, the two answers will differ by no more than a constant.

In general when we have products of sines and cosines in which both exponents are even we will need to use a series of half angle and/or double angle formulas to reduce the integral into a form that we can integrate.  Also, the larger the exponents the more we’ll need to use these formulas and hence the messier the problem.

Sometimes in the process of reducing integrals in which both exponents are even we will run across products of sine and cosine in which the arguments are different.  These will require one of the following formulas to reduce the products to integrals that we can do.

Let’s take a look at an example of one of these kinds of integrals.

Okay, at this point we’ve covered pretty much all the possible cases involving products of sines and cosines.  It’s now time to look at integrals that involve products of secants and tangents.

This time, let’s do a little analysis of the possibilities before we just jump into examples.  The general integral will be,

The first thing to notice is that we can easily convert even powers of secants to tangents and even powers of tangents to secants by using a formula similar to (1).  In fact, the formula can be derived from (1) so let’s do that.

Now, we’re going to want to deal with (3) similarly to how we dealt with (2).  We’ll want to eventually use one of the following substitutions.

So, if we use the substitution  we will need two secants left for the substitution to work.  This means that if the exponent on the secant (n) is even we can strip two out and then convert the remaining secants to tangents using (4).

Next, if we want to use the substitution  we will need one secant and one tangent left over in order to use the substitution.  This means that if the exponent on the tangent (m) is odd and we have at least one secant in the integrand we can strip out one of the tangents along with one of the secants of course.   The tangent will then have an even exponent and so we can use (4) to convert the rest of the tangents to secants.  Note that this method does require that we have at least one secant in the integral as well.  If there aren’t any secants then we’ll need to do something different.

If the exponent on the secant is even and the exponent on the tangent is odd then we can use either case.  Again, it will be easier to convert the term with the smallest exponent.

Let’s take a look at a couple of examples.

Both of the previous examples fit very nicely into the patterns discussed above and so were not all that difficult to work.  However, there are a couple of exceptions to the patterns above and in these cases there is no single method that will work for every problem.  Each integral will be different and may require different solution methods in order to evaluate the integral.

Let’s first take a look at a couple of integrals that have odd exponents on the tangents, but no secants.  In these cases we can’t use the substitution  since it requires there to be at least one secant in the integral.

Note that all odd powers of tangent (with the exception of the first power) can be integrated using the same method we used in the previous example.  For instance,

So, a quick substitution (  ) will give us the first integral and the second integral will always be the previous odd power.

Now let’s take a look at a couple of examples in which the exponent on the secant is odd and the exponent on the tangent is even.  In these cases the substitutions used above won’t work.

It should also be noted that both of the following two integrals are integrals that we’ll be seeing on occasion in later sections of this chapter and in later chapters.  Because of this it wouldn’t be a bad idea to make a note of these results so you’ll have them ready when you need them later.

The idea used in the above example is a nice idea to keep in mind.  Multiplying the numerator and denominator of a term by the same term above can, on occasion, put the integral into a form that can be integrated.  Note that this method won’t always work and even when it does it won’t always be clear what you need to multiply the numerator and denominator by.  However, when it does work and you can figure out what term you need it can greatly simplify the integral.

Here’s the next example.

Again, note that we’ve again used the idea of integrating the right side until the original integral shows up and then moving this to the left side and dividing by its coefficient to complete the evaluation.  We first saw this in the Integration by Parts section and noted at the time that this was a nice technique to remember.  Here is another example of this technique.

Now that we’ve looked at products of secants and tangents let’s also acknowledge that because we can relate cosecants and cotangents by

all of the work that we did for products of secants and tangents will also work for products of cosecants and cotangents.  We’ll leave it to you to verify that.

There is one final topic to be discussed in this section before moving on.

To this point we’ve looked only at products of sines and cosines and products of secants and tangents.  However, the methods used to do these integrals can also be used on some quotients involving sines and cosines and quotients involving secants and tangents (and hence quotients involving cosecants and cotangents).

Let’s take a quick look at an example of this.

So, under the right circumstances, we can use the ideas developed to help us deal with products of trig functions to deal with quotients of trig functions.  The natural question then, is just what are the right circumstances?

First notice that if the quotient had been reversed as in this integral,

we wouldn’t have been able to strip out a sine.

In this case the “stripped out” sine remains in the denominator and it won’t do us any good for the substitution  since this substitution requires a sine in the numerator of the quotient.  Also note that, while we could convert the sines to cosines, the resulting integral would still be a fairly difficult integral.

So, we can use the methods we applied to products of trig functions to quotients of trig functions provided the term that needs parts stripped out in is the numerator of the quotient.