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,
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.
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,
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.
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 stripping out a single tangent
(along with a secant) and converting the remaining tangents to secants using (4).
Here’s 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 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 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.
Example 9 Evaluate
the following integral.
Solution
This one isn’t too bad once you see what you’ve got to
do. By itself the integral can’t be
done. However, if we manipulate the
integrand as follows we can do it.
In this form we can do the integral using the substitution
. Doing this gives,

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.
Example 10 Evaluate
the following integral.
Solution
This one is different from any of the other integrals that
we’ve done in this section. The first
step to doing this integral is to perform integration by parts using the
following choices for u and dv.
Note that using integration by parts on this problem is
not an obvious choice, but it does work very nicely here. After doing integration by parts we have,
Now the new integral also has an odd exponent on the
secant and an even exponent on the tangent and so the previous examples of
products of secants and tangents still won’t do us any good.
To do this integral we’ll first write the tangents in the
integral in terms of secants. Again,
this is not necessarily an obvious choice but it’s what we need to do in this
case.
Now, we can use the results from the previous example to do
the second integral and notice that the first integral is exactly the
integral we’re being asked to evaluate with a minus sign in front. So, add it to both sides to get,
Finally divide by two and we’re done.

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.
Example 11 Evaluate
the following integral.
Solution
If this were a product of sines and cosines we would know
what to do. We would strip out a sine
(since the exponent on the sine is odd) and convert the rest of the sines to
cosines. The same idea will work in
this case. We’ll strip out a sine from
the numerator and convert the rest to cosines as follows,
At this point all we need to do is use the substitution and we’re done.

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.