Example 1 Evaluate
the following integral.
In this case the substitution will not work and so we’re going to have to
do something different for this integral.
It would be nice if we could get rid of the square root
somehow. The following substitution
will do that for us.
Do not worry about where this came from at this
point. As we work the problem you will
see that it works and that if we have a similar type of square root in the
problem we can always use a similar substitution.
Before we actually do the substitution however let’s
verify the claim that this will allow us to get rid of the square root.
To get rid of the square root all we need to do is recall
Using this fact the square root becomes,
Note the presence of the absolute value bars there. These are important. Recall that
There should always be absolute value bars at this
stage. If we knew that was always positive or always negative we
could eliminate the absolute value bars using,
Without limits we won’t be able to determine if is positive or negative, however, we will
need to eliminate them in order to do the integral. Therefore, since we are doing an indefinite
integral we will assume that will be positive and so we can drop the
absolute value bars. This gives,
So, we were able to eliminate the square root using this
substitution. Let’s now do the
substitution and see what we get. In
doing the substitution don’t forget that we'll also need to substitute for the
This is easy enough to get from the substitution.
Using this substitution the integral becomes,
With this substitution we were able to reduce the given
integral to an integral involving trig functions and we saw how to do these
problems in the previous section. Let’s finish the integral.
So, we’ve got an answer for the integral. Unfortunately the answer isn’t given in x’s as it should be. So, we need to write our answer in terms of
We can do this with some right triangle trig. From our original substitution we have,
This gives the following right triangle.
From this we can see that,
We can deal with the in one of any variety of ways. From our substitution we can see that,
While this is a perfectly acceptable method of dealing
with the we can use any of the possible six inverse
trig functions and since sine and cosine are the two trig functions most
people are familiar with we will usually use the inverse sine or inverse
cosine. In this case we’ll use the
So, with all of this the integral becomes,
We now have the answer back in terms of x.