After the last section we now know how to do the following
integrals.
However, we can’t do the following integrals.
All of these look considerably more difficult than the first
set. However, they aren’t too bad once
you see how to do them. Let’s start with
the first one.
In this case let’s notice that if we let
and we compute the differential
(you remember how to compute these right?) for this we get,
Now, let’s go back to our integral and notice that we can
eliminate every x that exists in the
integral and write the integral completely in terms of u using both the definition of u
and its differential.
In the process of doing this we’ve taken an integral that
looked very difficult and with a quick substitution we were able to rewrite the
integral into a very simple integral that we can do.
Evaluating the integral gives,
As always we can check our answer with a quick derivative if
we’d like to and don’t forget to “back substitute” and get the integral back
into terms of the original variable.
What we’ve done in the work above is called the Substitution Rule. Here is the substitution rule in general.
Substitution Rule
A natural question at this stage is how to identify the
correct substitution. Unfortunately, the
answer is it depends on the integral.
However, there is a general rule of thumb that will work for many of the
integrals that we’re going to be running across.
When faced with an integral we’ll ask ourselves what we know
how to integrate. With the integral
above we can quickly recognize that we know how to integrate
However, we didn’t have just the root we also had stuff in
front of the root and (more importantly in this case) stuff under the
root. Since we can only integrate roots
if there is just an x under the root
a good first guess for the substitution is then to make u be the stuff under the root.
Another way to think of this is to ask yourself what portion
of the integrand has an inside function and can you do the integral with that
inside function present. If you can’t
then there is a pretty good chance that the inside function will be the
substitution.
We will have to be careful however. There are times when using this general rule
can get us in trouble or overly complicate the problem. We’ll eventually see problems where there are
more than one “inside function” and/or integrals that will look very similar
and yet use completely different substitutions.
The reality is that the only way to really learn how to do substitutions
is to just work lots of problems and eventually you’ll start to get a feel for
how these work and you’ll find it easier and easier to identify the proper
substitutions.
Now, with that out of the way we should ask the following
question. How, do we know if we got the
correct substitution? Well, upon
computing the differential and actually performing the substitution every x in the integral (including the x in the dx) must disappear in the substitution process and the only letters
left should be u’s (including a du).
If there are x’s left over
then there is a pretty good chance that we chose the wrong substitution. Unfortunately, however there is at least one
case (we’ll be seeing an example of this in the next section) where the correct
substitution will actually leave some x’s
and we’ll need to do a little more work to get it to work.
Again, it cannot be stressed enough at this point that the
only way to really learn how to do substitutions is to just work lots of
problems. There are lots of different kinds of problems and after working many
problems you’ll start to get a real feel for these problems and after you work
enough of them you’ll reach the point where you’ll be able to do simple
substitutions in your head without having to actually write anything down.
As a final note we should point out that often (in fact in
almost every case) the differential will not appear exactly in the integrand as
it did in the example above and sometimes we’ll need to do some manipulation of
the integrand and/or the differential to get all the x’s to disappear in the substitution.
Let’s work some examples so we can get a better idea on how
the substitution rule works.
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Example 1 Evaluate
each of the following integrals.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
(a)
In this case we know how to integrate just a cosine so
let’s make the substitution the stuff that is inside the cosine.
So, as with the first example we worked the stuff in front
of the cosine appears exactly in the differential. The integral is then,
Don’t forget to go back to the original variable in the
problem.
[Return to Problems]
(b)
Again, we know how to integrate an exponential by itself
so it looks like the substitution for this problem should be,
Now, with the exception of the 3 the stuff in front of the
exponential appears exactly in the differential. Recall however that we can factor the 3 out
of the integral and so it won’t cause any problems. The integral is then,
[Return to Problems]
(c)
In this case it looks like the following should be the
substitution.
Okay, now we have a small problem. We’ve got an x2 out in front of the parenthesis but we don’t have a
“-30”. This is not really the problem
it might appear to be at first. We
will simply rewrite the differential as follows.
With this we can now substitute the “x2 dx” away. In
the process we will pick up a constant, but that isn’t a problem since it can
always be factored out of the integral.
We can now do the integral.
Note that in most problems when we pick up a constant as
we did in this example we will generally factor it out of the integral in the
same step that we substitute it in.
[Return to Problems]
(d)
In this example don’t forget to bring the root up to the
numerator and change it into fractional exponent form. Upon doing this we can see that the
substitution is,
The integral is then,
[Return to Problems]
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In the previous set of examples the substitution was
generally pretty clear. There was
exactly one term that had an “inside function” that we also couldn’t
integrate. Let’s take a look at some
more complicated problems to make sure we don’t come to expect all
substitutions are like those in the previous set of examples.
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Example 2 Evaluate
each of the following integrals.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
(a)
In this problem there are two “inside functions”. There is the that is inside the two trig functions and
there is also the term that is raised to the 4th power.
There are two ways to proceed with this problem. The first idea that many students have is
substitute the away.
There is nothing wrong with doing this but it doesn’t eliminate the
problem of the term to the 4th power. That’s still there and if we used this idea
we would then need to do a second substitution to deal with that.
The second (and much easier) way of doing this problem is
to just deal with the stuff raised to the 4th power and see what
we get. The substitution in this case
would be,
Two things to
note here. First, don’t forget to
correctly deal with the “-”. A common
mistake at this point is to lose it.
Secondly, notice that the turns out to not really be a problem after
all. Because the was “buried” in the substitution that we
actually used it was also taken care of at the same time. The integral is then,
As seen in this
example sometimes there will seem to be two substitutions that will need to
be done however, if one of them is buried inside of another substitution then
we’ll only really need to do one.
Recognizing this can save a lot of time in working some of these
problems.
[Return to Problems]
(b)
This one is a little tricky at first. We can see the correct substitution by
recalling that,
Using this it looks like the correct substitution is,
Notice that we again had two apparent substitutions in
this integral but again the 3z is
buried in the substitution we’re using and so we didn’t need to worry about
it.
Here is the integral.
Note that the one third in front of the integral came
about from the substitution on the differential and we just factored it out
to the front of the integral. This is
what we will usually do with these constants.
[Return to Problems]
(c)
In this case we’ve got a 4t, a secant squared as well as a term cubed. However, it looks like if we use the
following substitution the first two issues are going to be taken care of for
us.
The integral is
now,
[Return to Problems]
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The most important thing to remember in substitution
problems is that after the substitution all the original variables need to
disappear from the integral. After the
substitution the only variables that should be present in the integral should
be the new variable from the substitution (usually u). Note as well that this
includes the variables in the differential!
This next set of examples, while not particular difficult,
can cause trouble if we aren’t paying attention to what we’re doing.
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Example 3 Evaluate
each of the following integrals.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
(a)
We haven’t seen a problem quite like this one yet. Let’s notice that if we take the denominator
and differentiate it we get just a constant and the only thing that we have
in the numerator is also a constant.
This is a pretty good indication that we can use the denominator for
our substitution so,
The integral is
now,
Remember that
we can just factor the 3 in the numerator out of the integral and that makes
the integral a little clearer in this case.
[Return to Problems]
(b)
The integral is very similar to the previous one with a
couple of minor differences but notice that again if we differentiate the
denominator we get something that is different from the numerator by only a
multiplicative constant. Therefore
we’ll again take the denominator as our substitution.
The integral
is,
[Return to Problems]
(c)
Now, this one is almost identical to the previous part
except we added a power onto the denominator.
Notice however that if we ignore the power and differentiate what’s
left we get the same thing as the previous example so we’ll use the same
substitution here.
The integral in
this case is,
Be careful in this case to not turn this into a
logarithm. After working problems like
the first two in this set a common error is to turn every rational expression
into a logarithm. If there is a power
on the whole denominator then there is a good chance that it isn’t a
logarithm.
The idea that we used in the last three parts to determine
the substitution is not a bad idea to remember. If we’ve got a rational expression try
differentiating the denominator (ignoring any powers that are on the whole
denominator) and if the result is the numerator or only differs from the
numerator by a multiplicative constant then we can usually use that as our
substitution.
[Return to Problems]
(d)
Now, this part is completely different from the first
three and yet seems similar to them as well.
In this case if we differentiate the denominator we get a y that is not in the numerator and so
we can’t use the denominator as our substitution.
In fact, because we have y2 in the denominator and no y in the numerator is an indication of how to work this problem.
This integral is going to be an inverse tangent when we are done. To key to seeing this is to recall the
following formula,
We clearly don’t have exactly this but we do have
something that is similar. The
denominator has a squared term plus a constant and the numerator is just a
constant. So, with a little work and
the proper substitution we should be able to get our integral into a form
that will allow us to use this formula.
There is one part of this formula that is really important
and that is the “1+” in the denominator.
That must be there and we’ve got a “4+” but it is easy enough to take
care of that. We’ll just factor a 4
out of the denominator and at the same time we’ll factor the 3 in the
numerator out of the integral as well.
Doing this gives,
Notice that in the last step we rewrote things a little in
the denominator. This will help us to
see what the substitution needs to be.
In order to get this integral into the formula above we need to end up
with a u2 in the
denominator. Our substitution will
then need to be something that upon squaring gives us . With the rewrite we can see what that we’ll
need to use the following substitution.
Don’t get excited about the root in the substitution, these
will show up on occasion. Upon
plugging our substitution in we get,
After doing the
substitution, and factoring any constants out, we get exactly the integral
that gives an inverse tangent and so we know that we did the correct
substitution for this integral. The
integral is then,
[Return to Problems]
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In this last set of integrals we had four integrals that
were similar to each other in many ways and yet all either yielded different
answer using the same substitution or used a completely different substitution
than one that was similar to it.
This is a fairly common occurrence and so you will need to
be able to deal with these kinds of issues.
There are many integrals that on the surface look very similar and yet
will use a completely different substitution or will yield a completely
different answer when using the same substitution.
Let’s take a look at another set of examples to give us more
practice in recognizing these kinds of issues.
Note however that we won’t be putting as much detail into these as we
did with the previous examples.
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Example 4 Evaluate
each of the following integrals.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
(a)
Clearly the derivative of the denominator, ignoring the
exponent, differs from the numerator only by a multiplicative constant and so
the substitution is,
After a little manipulation of the differential we get the
following integral.
[Return to Problems]
(b)
The only difference between this problem and the previous
one is the denominator. In the
previous problem the whole denominator is cubed and in this problem the
denominator has no power on it. The
same substitution will work in this problem but because we no longer have the
power the problem will be different.
So, using the substitution from the previous example the
integral is,
So, in this case we get a logarithm from the integral.
[Return to Problems]
(c)
Here, if we ignore the root we can again see that the
derivative of the stuff under the radical differs from the numerator by only
a multiplicative constant and so we’ll use that as the substitution.
The integral is then,
[Return to Problems]
(d)
In this case we are missing the x in the numerator and so the substitution from the last part
will do us no good here. This integral
is another inverse trig function integral that is similar to the last part of
the previous set of problems. In this
case we need to following formula.
The integral in this problem is nearly this. The only difference is the presence of the
coefficient of 4 on the x2. With the correct substitution this can be
dealt with however. To see what this
substitution should be let’s rewrite the integral a little. We need to figure out what we squared to
get 4x2 and that will be
our substitution.
With this rewrite it looks like we can use the following
substitution.
The integral is then,
[Return to Problems]
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Since this document is also being presented on the web we’re
going to put the rest of the substitution rule examples in the next
section. With all the examples in one
section the section was becoming too large for web presentation.