In the previous section we started looking at indefinite
integrals and in that section we concentrated almost exclusively on notation,
concepts and properties of the indefinite integral. In this section we need to start thinking
about how we actually compute indefinite integrals. We’ll start off with some of the basic
indefinite integrals.
The first integral that we’ll look at is the integral of a
power of x.
The general rule when integrating a power of x we add one onto the exponent and then
divide by the new exponent. It is clear
(hopefully) that we will need to avoid 
in this formula. If we allow in this formula we will end up with division
by zero. We will take care of this case
in a bit.
Next is one of the easier integrals but always seems to
cause problems for people.
If you remember that all we’re asking is what did we
differentiate to get the integrand this is pretty simple, but it does seem to
cause problems on occasion.
Let’s now take a look at the trig functions.
Notice that we only integrated two of the six trig functions
here. The remaining four integrals are
really integrals that give the remaining four trig functions. Also, be careful with signs here. It is easy to get the signs for derivatives
and integrals mixed up. Again, remember
that we’re asking what function we differentiated to get the integrand.
We will be able to integrate the remaining four trig
functions in a couple of sections, but they all require the Substitution Rule.
Now, let’s take care of exponential and logarithm functions.
Integrating logarithms requires a topic that is usually
taught in Calculus II and so we won’t be integrating a logarithm in this
class. Also note the third integrand can
be written in a couple of ways and don’t forget the absolute value bars in the x in the answer to the third integral.
Finally, let’s take care of the inverse trig and hyperbolic functions.
As with logarithms integrating inverse trig functions
requires a topic usually taught in Calculus II and so we won’t be integrating
them in this class. There is also a
different answer for the second integral above.
Recalling that since all we are asking here is what function did we
differentiate to get the integrand the second integral could also be,
Traditionally we use the first form of this integral.
Okay, now that we’ve got most of the basic integrals out of
the way let’s do some indefinite integrals.
In all of these problems remember that we can always check our answer by
differentiating and making sure that we get the integrand.
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Example 1 Evaluate
each of the following indefinite integrals.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
(f) [Solution]
Solution
Okay, in all of these remember the basic rules of
indefinite integrals. First, to
integrate sums and differences all we really do is integrate the individual
terms and then put the terms back together with the appropriate signs. Next, we can ignore any coefficients until
we are done with integrating that particular term and then put the
coefficient back in. Also, do not
forget the “+c” at the end it is
important and must be there.
So, let’s evaluate some integrals.
(a)
There’s not really a whole lot to do here other than use
the first two formulas from the beginning of this section. Remember that when integrating powers (that
aren’t -1 of course) we just add one onto the exponent and then divide by the
new exponent.
Be careful when integrating negative exponents. Remember to add one onto the exponent. One of the more common mistakes that
students make when integrating negative exponents is to “add one” and end up
with an exponent of “-7” instead of the correct exponent of “-5”.
[Return to Problems]
(b)
This is here just to make sure we get the point about
integrating negative exponents.
[Return to Problems]
(c)
In this case there isn’t a formula for explicitly dealing
with radicals or rational expressions.
However, just like with derivatives we can write all these terms so
they are in the numerator and they all have an exponent. This should always be your first step when
faced with this kind of integral just as it was when differentiating.
When dealing with fractional exponents we usually don’t
“divide by the new exponent”. Doing
this is equivalent to multiplying by the reciprocal of the new exponent and
so that is what we will usually do.
[Return to Problems]
(d)
Don’t make this one harder than it is…
In this case we are really just integrating a one!
[Return to Problems]
(e)
We’ve got a product here and as we noted in the previous
section there is no rule for dealing with products. However, in this case we don’t need a
rule. All that we need to do is
multiply things out (taking care of the radicals at the same time of course)
and then we will be able to integrate.
[Return to Problems]
(f)
As with the previous part it’s not really a problem that
we don’t have a rule for quotients for this integral. In this case all we need to do is break up
the quotient and then integrate the individual terms.
Be careful to not think of the third term as x to a power for the purposes of
integration. Using that rule on the
third term will NOT work. The third
term is simply a logarithm. Also,
don’t get excited about the 15. The 15
is just a constant and so it can be factored out of the integral. In other words, here is what we did to
integrate the third term.
[Return to Problems]
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Always remember that you can’t integrate products and
quotients in the same way that we integrate sums and differences. At this point the only way to integrate
products and quotients is to multiply the product out or break up the
quotient. Eventually we’ll see some
other products and quotients that can be dealt with in other ways. However, there will never be a single rule
that will work for all products and there will never be a single rule that will
work for all quotients. Every product
and quotient is different and will need to be worked on a case by case basis.
The first set of examples focused almost exclusively on
powers of x (or whatever variable we
used in each example). It’s time to do
some examples that involve other functions.
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Example 2 Evaluate
each of the following integrals.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
Solution
Most of the problems in this example will simply use the
formulas from the beginning of this section.
More complicated problems involving most of these functions will need
to wait until we reach the Substitution Rule.
(a)
There isn’t anything to this one other than using the
formulas.
[Return to Problems]
(b)
Let’s be a little careful with this one. First break it up into two integrals and
note the rewritten integrand on the second integral.
Rewriting the second integrand will help a little with the
integration at this early stage. We
can think of the 6 in the denominator as a 1/6 out in front of the term and
then since this is a constant it can be factored out of the integral. The answer is then,
Note that we didn’t factor the 2 out of the first integral
as we factored the 1/6 out of the second.
In fact, we will generally not factor the 1/6 out either in later
problems. It was only done here to
make sure that you could follow what we were doing.
[Return to Problems]
(c)
In this one we’ll just use the formulas from above and
don’t get excited about the coefficients.
They are just multiplicative constants and so can be ignored while we
integrate each term and then once we’re done integrating a given term we
simply put the coefficients back in.
[Return to Problems]
(d)
Again, there really isn’t a whole lot to do with this one
other than to use the appropriate formula from above.
[Return to Problems]
(e)
This one can be a little tricky if you aren’t ready for
it. As discussed previously, at this
point the only way we have of dealing with quotients is to break it up.
Notice that upon breaking the integral up we further
simplified the integrand by recalling the definition of cosecant. With this simplification we can do the
integral.
[Return to Problems]
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As shown in the last part of this example we can do some
fairly complicated looking quotients at this point if we remember to do
simplifications when we see them. In
fact, this is something that you should always keep in mind. In almost any problem that we’re doing here
don’t forget to simplify where possible.
In almost every case this can only help the problem and will rarely
complicate the problem.
In the next problem we’re going to take a look at a product
and this time we’re not going to be able to just multiply the product out. However, if we recall the comment about
simplifying a little this problem becomes fairly simple.
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Example 3 Integrate
.
Solution
There are several ways to do this integral and most of
them require the next section.
However, there is a way to do this integral using only the material
from this section. All that is
required is to remember the trig formula that we can use to simplify the
integrand up a little. Recall the
following double angle formula.
A small rewrite of this formula gives,
If we now replace all the t’s with we get,
Using this formula the integral becomes something we can
do.
As noted earlier there is another method for doing this
integral. In fact there are two
alternate methods. To see all three
check out the section on Constant
of Integration in the Extras chapter but be aware that the other two do
require the material covered in the next section.
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The formula/simplification in the previous problem is a nice
“trick” to remember. It can be used on
occasion to greatly simplify some problems.
There is one more set of examples that we should do before
moving out of this section.
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Example 4 Given
the following information determine the function .
(a) [Solution]
(b) [Solution]
Solution
In both of these we will need to remember that
Also note that because we are giving values of the
function at specific points we are also going to be determining what the constant
of integration will be in these problems.
(a)
The first step here is to integrate to determine the most
general possible .
Now we have a value of the function so let’s plug in and determine the value of the constant of
integration c.
So, from this it looks like . This means that the function is,
[Return to Problems]
(b)
This one is a little different from the first one. In order to get the function we will need
the first derivative and we have the second derivative. We can however, use an integral to get the
first derivative from the second derivative, just as we used an integral to
get the function from the first derivative.
So, let’s first get the most general possible first
derivative.
Don’t forget the constant of integration!
We can now find the most general possible function.
Do not get excited about integrating the c.
It’s just a constant and we know how to integrate constants. Also, there will be no reason to think the constants
of integration from the integration in each step will be the same and so
we’ll need to call each constant of integration something different.
Now, plug in the two values of the function that we’ve
got.
This gives us a system of two equations in two unknowns
that we can solve.
The function is then,
Don’t remember how to solve systems? Check out the Solving Systems portion of
my Algebra/Trig Review.
[Return to Problems]
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In this section we’ve started the process of
integration. We’ve seen how to do quite
a few basic integrals and we also saw a quick application of integrals in the
last example.
There are many new formulas in this section that we’ll now
have to know. However, if you think
about it, they aren’t really new formulas.
They are really nothing more than derivative formulas that we should
already know written in terms of integrals.
If you remember that you should find it easier to remember the formulas
in this section.
Always remember that integration is asking nothing more than
what function did we differentiate to get the integrand. If you can remember that many of the basic
integrals that we saw in this section and many of the integrals in the coming
sections aren’t too bad.