Let’s start off with this section with a couple of integrals
that we should already be able to do to get us started. First let’s take a look at the following.
So, that was simple enough.
Now, let’s take a look at,
To do this integral we’ll use the following substitution.
Again, simple enough to do provided you remember how to do substitutions. By the way make sure that you can do these
kinds of substitutions quickly and easily.
From this point on we are going to be doing these kinds of substitutions
in our head. If you have to stop and
write these out with every problem you will find that it will take you
significantly longer to do these problems.
Now, let’s look at the integral that we really want to do.
If we just had an x
by itself or by itself we could do the integral easily
enough. But, we don’t have them by
themselves, they are instead multiplied together.
There is no substitution that we can use on this integral
that will allow us to do the integral.
So, at this point we don’t have the knowledge to do this integral.
To do this integral we will need to use integration by parts
so let’s derive the integration by parts formula. We’ll start with the product rule.
Now, integrate both sides of this.
The left side is easy enough to integrate and we’ll split up
the right side of the integral.
Note that technically we should have had a constant of
integration show up on the left side after doing the integration. We can drop it at this point since other
constants of integration will be showing up down the road and they would just
end up absorbing this one.
Finally, rewrite the formula as follows and we arrive at
the integration by parts formula.
This is not the easiest formula to use however. So, let’s do a couple of substitutions.
Both of these are just the standard Calc I substitutions
that hopefully you are used to by now.
Don’t get excited by the fact that we are using two substitutions
here. They will work the same way.
Using these substitutions gives us the formula that most
people think of as the integration by parts formula.
To use this formula we will need to identify u and dv, compute du and v and then use the formula. Note as well that computing v is very easy. All we need to do is integrate dv.
So, let’s take a look at the integral above that we
mentioned we wanted to do.
Example 1 Evaluate
the following integral.
Solution
So, on some level, the problem here is the x that is in front of the
exponential. If that wasn’t there we
could do the integral. Notice as well
that in doing integration by parts anything that we choose for u will be differentiated. So, it seems that choosing will be a good choice since upon
differentiating the x will drop
out.
Now that we’ve chosen u
we know that dv will be everything
else that remains. So, here are the
choices for u and dv as well as du and v.
The integral is then,
Once we have done the last integral in the problem we will
add in the constant of integration to get our final answer.

Next, let’s take a look at integration by parts for definite
integrals. The integration by parts
formula for definite integrals is,
Integration by Parts,
Definite Integrals
Note that the in the first term is just the standard
integral evaluation notation that you should be familiar with at this
point. All we do is evaluate the term, uv in this case, at b then subtract off the evaluation of the term at a.
At some level we don’t really need a formula here because we
know that when doing definite integrals all we need to do is do the indefinite
integral and then do the evaluation.
Let’s take a quick look at a definite integral using
integration by parts.
Example 2 Evaluate
the following integral.
Solution
This is the same integral that we looked at in the first
example so we’ll use the same u and
dv to get,

Since we need to be able to do the indefinite integral in
order to do the definite integral and doing the definite integral amounts to
nothing more than evaluating the indefinite integral at a couple of points we
will concentrate on doing indefinite integrals in the rest of this
section. In fact, throughout most of
this chapter this will be the case. We
will be doing far more indefinite integrals than definite integrals.
Let’s take a look at some more examples.
Example 3 Evaluate
the following integral.
Solution
There are two ways to proceed with this example. For many, the first thing that they try is
multiplying the cosine through the parenthesis, splitting up the integral and
then doing integration by parts on the first integral.
While that is a perfectly acceptable way of doing the
problem it’s more work than we really need to do. Instead of splitting the integral up let’s
instead use the following choices for u
and dv.
The integral is then,
Notice that we pulled any constants out of the integral
when we used the integration by parts formula. We will usually do this in order to
simplify the integral a little.

Example 4 Evaluate
the following integral.
Solution
For this example we’ll use the following choices for u and dv.
The integral is then,
In this example, unlike the previous examples, the new
integral will also require integration by parts. For this second integral we will use the
following choices.
So, the integral becomes,
Be careful with the coefficient on the integral for the
second application of integration by parts.
Since the integral is multiplied by we need to make sure that the results of
actually doing the integral are also multiplied by . Forgetting to do this is one of the more
common mistakes with integration by parts problems.

As this last example has shown us, we will sometimes need
more than one application of integration by parts to completely evaluate an
integral. This is something that will
happen so don’t get excited about it when it does.
In this next example we need to acknowledge an important
point about integration techniques. Some
integrals can be done in using several different techniques. That is the case with the integral in the
next example.
Example 5 Evaluate
the following integral
(a) Using
Integration by Parts. [Solution]
(b) Using
a standard Calculus I substitution. [Solution]
Solution
(a) Evaluate using
Integration by Parts.
First notice that there are no trig functions or
exponentials in this integral. While a
good many integration by parts integrals will involve trig functions and/or
exponentials not all of them will so don’t get too locked into the idea of
expecting them to show up.
In this case we’ll use the following choices for u and dv.
The integral is then,
[Return to Problems]
(b) Evaluate Using a
standard Calculus I substitution.
Now let’s do the integral with a substitution. We can use the following substitution.
Notice that we’ll actually use the substitution twice,
once for the quantity under the square root and once for the x in front of the square root. The integral is then,
[Return to Problems]

So, we used two different integration techniques in this
example and we got two different answers.
The obvious question then should be : Did we do something wrong?
Actually, we didn’t do anything wrong. We need to remember the following fact from
Calculus I.
In other words, if two functions have the same derivative
then they will differ by no more than a constant. So, how does this apply to the above
problem? First define the following,
Then we can compute and by integrating as follows,
We’ll use integration by parts for the first integral and
the substitution for the second integral.
Then according to the fact and should differ by no more than a constant. Let’s verify this and see if this is the
case. We can verify that they differ by no more than a constant if we take a
look at the difference of the two and do a little algebraic manipulation and
simplification.
So, in this case it turns out the two functions are exactly
the same function since the difference is zero.
Note that this won’t always happen.
Sometimes the difference will yield a nonzero constant. For an example of this check out the Constant of Integration section
in my Calculus I notes.
So just what have we learned? First, there will, on occasion, be more than
one method for evaluating an integral.
Secondly, we saw that different methods will often lead to different
answers. Last, even though the answers
are different it can be shown, sometimes with a lot of work, that they differ
by no more than a constant.
When we are faced with an integral the first thing that
we’ll need to decide is if there is more than one way to do the integral. If there is more than one way we’ll then need
to determine which method we should use.
The general rule of thumb that I use in my classes is that you should
use the method that you find
easiest. This may not be the method that
others find easiest, but that doesn’t make it the wrong method.
One of the more common mistakes with integration by parts is
for people to get too locked into perceived patterns. For instance, all of the previous examples
used the basic pattern of taking u to
be the polynomial that sat in front of another function and then letting dv be the other function. This will not always happen so we need to be
careful and not get locked into any patterns that we think we see.
Let’s take a look at some integrals that don’t fit into the
above pattern.
Example 6 Evaluate
the following integral.
Solution
So, unlike any of the other integral we’ve done to this
point there is only a single function in the integral and no polynomial sitting
in front of the logarithm.
The first choice of many people here is to try and fit
this into the pattern from above and make the following choices for u and dv.
This leads to a real problem however since that means v must be,
In other words, we would need to know the answer ahead of
time in order to actually do the problem.
So, this choice simply won’t work.
Also notice that with this choice we’d get that which also causes problems and is another
reason why this choice will not work.
Therefore, if the logarithm doesn’t belong in the dv it must belong instead in the u.
So, let’s use the following choices instead
The integral is then,

Example 7 Evaluate
the following integral.
Solution
So, if we again try to use the pattern from the first few
examples for this integral our choices for u and dv would probably
be the following.
However, as with the previous example this won’t work
since we can’t easily compute v.
This is not an easy integral to do. However, notice that if we had an x^{2} in the integral along
with the root we could very easily do the integral with a substitution. Also notice that we do have a lot of x’s floating around in the original
integral. So instead of putting all
the x’s (outside of the root) in
the u let’s split them up as
follows.
We can now easily compute v and after using integration by parts we get,

So, in the previous two examples we saw cases that didn’t
quite fit into any perceived pattern that we might have gotten from the first
couple of examples. This is always
something that we need to be on the lookout for with integration by parts.
Let’s take a look at another example that also illustrates
another integration technique that sometimes arises out of integration by parts
problems.
Example 8 Evaluate
the following integral.
Solution
Okay, to this point we’ve always picked u in such a way that upon
differentiating it would make that portion go away or at the very least put
it the integral into a form that would make it easier to deal with. In this case no matter which part we make u it will never go away in the
differentiation process.
It doesn’t much matter which we choose to be u so we’ll choose in the following
way. Note however that we could choose
the other way as well and we’ll get the same result in the end.
The integral is then,
So, it looks like we’ll do integration by parts
again. Here are our choices this time.
The integral is now,
Now, at this point it looks like we’re just running in
circles. However, notice that we now
have the same integral on both sides and on the right side it’s got a minus
sign in front of it. This means that
we can add the integral to both sides to get,
All we need to do now is divide by 2 and we’re done. The integral is,
Notice that after dividing by the two we add in the
constant of integration at that point.

This idea of integrating until you get the same integral on
both sides of the equal sign and then simply solving for the integral is kind
of nice to remember. It doesn’t show up
all that often, but when it does it may be the only way to actually do the
integral.
We’ve got one more example to do. As we will see some problems could require us
to do integration by parts numerous times and there is a short hand method that
will allow us to do multiple applications of integration by parts quickly and
easily.
Example 9 Evaluate
the following integral.
Solution
We start off by choosing u and dv as we always
would. However, instead of computing du and v we put these into the following table. We then differentiate down the column
corresponding to u until we hit
zero. In the column corresponding to dv we integrate once for each entry in
the first column. There is also a
third column which we will explain in a bit and it always starts with a “+”
and then alternates signs as shown.
Now, multiply along the diagonals shown in the table. In front of each product put the sign in
the third column that corresponds to the “u”
term for that product. In this case
this would give,
We’ve got the integral.
This is much easier than writing down all the various u’s and dv’s that we’d have to do otherwise.

So, in this section we’ve seen how to do integration by
parts. In your later math classes this
is liable to be one of the more frequent integration techniques that you’ll
encounter.
It is important to not get too locked into patterns that you
may think you’ve seen. In most cases any
pattern that you think you’ve seen can (and will be) violated at some point in
time. Be careful!
Also, don’t forget the shorthand method for multiple
applications of integration by parts problems.
It can save you a fair amount of work on occasion.