Example 4 Evaluate
each of the following integrals.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
Solution
(a)
The first question about this problem is probably why is
it here? Substitution rule problems
generally require more than a single function. The key to this problem is to realize that
there really are two functions here.
All we need to do is remember the definition of tangent and we can
write the integral as,
Written in this way we can see that the following
substitution will work for us,
The integral is then,
Now, while this is a perfectly serviceable answer that
minus sign in front is liable to cause problems if we aren’t careful. So, let’s rewrite things a little. Recalling a property of logarithms we can move the
minus sign in front to an exponent on the cosine and then do a little
simplification.
This is the formula that is typically given for the
integral of tangent.
Note that we could integrate cotangent in a similar
manner.
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(b)
This problem also at first appears to not belong in the
substitution rule problems. This is
even more of a problem upon noticing that we can’t just use the definition of
the secant function to write this in a form that will allow the use of the
substitution rule.
This problem is going to require a technique that isn’t
used terribly often at this level, but is a useful technique to be aware
of. Sometimes we can make an integral
doable by multiplying the top and bottom by a common term. This will not always work and even when it
does it is not always clear what we should multiply by but when it works it
is very useful.
Here is how we’ll use this idea for this problem.
First, we will think of the secant as a fraction and then
multiply the top and bottom of the fraction by the same term. It is probably not clear why one would want
to do this here but doing this will actually allow us to use the substitution
rule. To see how this will work let’s
simplify the integrand somewhat.
We can now use the following substitution.
The integral is then,
Sometimes multiplying the top and bottom of a fraction by
a carefully chosen term will allow us to work a problem. It does however take some thought sometimes
to determine just what the term should be.
We can use a similar process for integrating cosecant.
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(c)
This next problem has a subtlety to it that can get us in
trouble if we aren’t paying attention.
Because of the root in the cosine it makes some sense to use the
following substitution.
This is where we need to be careful. Upon rewriting the differential we get,
The root that is in the denominator will not become a u as we might have been tempted to
do. Instead it will get taken care of
in the differential.
The integral is,
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(d)
With this problem we need to very carefully pick our
substitution. As the problem is
written we might be tempted to use the following substitution,
However, this won’t work as you can probably see. The differential doesn’t show up anywhere
in the integrand and we just wouldn’t be able to eliminate all the t’s with this substitution.
In order to work this problem we will need to rewrite the
integrand as follows,
We will now use the substitution,
The integral is,
Some substitutions can be really tricky to see and it’s
not unusual that you’ll need to do some simplification and/or rewriting to
get a substitution to work.
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(e)
This last problem in this set is different from all the
other substitution problems that we’ve worked to this point. Given the fact that we’ve got more than an x under the root it makes sense that
the substitution pretty much has to be,
At first glance it looks like this might not work for the substitution because we have an x^{3} in front of the root. However, if we first rewrite 2x^{3} = x^{2}(2x) we could then move the 2x to the end of the integral so at least the du will show up explicitly in the integral. Doing this gives the following,
This is a real problem.
Our integrals can’t have two variables in them. Normally this would mean that we chose our
substitution incorrectly. However, in
this case we can rewrite the substitution as follows,
and now, we can eliminate the remaining x’s from our integral. Doing this gives,
Sometimes, we will need to use a substitution more than
once.
This kind of problem doesn’t arise all that often and when
it does there will sometimes be alternate methods of doing the integral. However, it will often work out that the
easiest method of doing the integral is to do what we just did here.
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