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Online Notes / Calculus II / Integration Techniques / Integrals Involving Trig Functions
Calculus II

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Integration by Parts E-Book   Chapter   Section Trig Substitutions

 Integrals Involving Trig Functions

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.

 

Example 1  Evaluate the following integral.

                                                                 

 

Solution

This integral no longer has the cosine in it that would allow us to use the substitution that we used above.  Therefore, that substitution won’t work and we are going to have to find another way of doing this integral.

 

Let’s first notice that we could write the integral as follows,

                                   

 

Now recall the trig identity,

                                  

 

With this identity the integral can be written as,

                                               

 

and we can now use the substitution .  Doing this gives us,

                                       

 

So, with a little rewriting on the integrand we were able to reduce this to a fairly simple substitution.

 

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,

(1)

 

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.

 

Example 2  Evaluate the following integral.

                                                            

Solution

So, in this case we’ve got both sines and cosines in the problem and in this case the exponent on the sine is even while the exponent on the cosine is odd.  So, we can use a similar technique in this integral.  This time we’ll strip out a cosine and convert the rest to sines.

                       

 

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

(2)

 

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.

 

Example 3  Evaluate the following integral.

                                                            

Solution

As noted above there are often more than one way to do integrals in which both of the exponents are even.  This integral is an example of that.  There are at least two solution techniques for this problem.  We will do both solutions starting with what is probably the harder of the two, but it’s also the one that many people see first.

 

Solution 1

In this solution we will use the two half angle formulas above and just substitute them into the integral.

                            

 

So, we still have an integral that can’t be completely done, however notice that we have managed to reduce the integral down to just one term causing problems (a cosine with an even power) rather than two terms causing problems.

 

In fact to eliminate the remaining problem term all that we need to do is reuse the first half angle formula given above.

                                      

So, this solution required a total of three trig identities to complete.

 

Solution 2

In this solution we will use the half angle formula to help simplify the integral as follows.

                                            

Now, we use the double angle formula for sine to reduce to an integral that we can do.

                                           

This method required only two trig identities to complete.

 

Notice that the difference between these two methods is more one of “messiness”.  The second method is not appreciably easier (other than needing one less trig identity) it is just not as messy and that will often translate into an “easier” process.

 

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.

 

Example 4  Evaluate the following integral.

                                                        

Solution

This integral requires the last formula listed above.

                           

 

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,

(3)

 

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.

 

 

 

 

 

(4)

 

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 to 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.

 

Example 5  Evaluate the following integral.

                                                            

Solution

First note that since the exponent on the secant isn’t even we can’t use the substitution .  However, the exponent on the tangent is odd and we’ve got a secant in the integral and so we will be able to use the substitution .  This means striping out a single tangent (along with a secant) and converting the remaining tangents to secants using (4).

 

Here’s the work for this integral.

                 

 

Example 6  Evaluate the following integral.

                                                            

Solution

So, in this example the exponent on the tangent is even so the substitution  won’t work.  The exponent on the secant is even and so we can use the substitution  for this integral.  That means that we need to strip out two secants and convert the rest to tangents.  Here is the work for this integral.

                       

 

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.

 

Example 7  Evaluate the following integral.

                                                                  

Solution

To do this integral all we need to do is recall the definition of tangent in terms of sine and cosine and then this integral is nothing more than a Calculus I substitution.

                                

 

Example 8  Evaluate the following integral.

                                                                 

Solution

The trick to this one is do the following manipulation of the integrand.

                                          

We can now use the substitution  on the first integral and the results from the previous example to on the second integral.

 

The integral is then,

                                              

 

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 t