Example 1  Evaluate the following integral.

Solution

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 the relationship,

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 of negative, however, we will need to eliminate them in order to do the integral.  Therefore, since we 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 dx.  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 x.  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 inverse cosine.

So, with all of this the integral becomes,

We now have the answer back in terms of x.

Example 2  Evaluate the following integral.

Solution

The limits here won’t change the substitution so that will remain the same.

Using this substitution the square root still reduces down to,

However, unlike the previous example we can’t just drop the absolute value bars.  In this case we’ve got limits on the integral and so we can use the limits as well as the substitution to determine the range of  that we’re in.  Once we’ve got that we can determine how to drop the absolute value bars.

Here’s the limits of .

So, if we are in the range  then  is in the range of  and in this range of  ’s tangent is positive and so we can just drop the absolute value bars. 

Let’s do the substitution.  Note that the work is identical to the previous example and so most of it is left out.  We’ll pick up at the final integral and then do the substitution.

Note that because of the limits we didn’t need to resort to a right triangle to complete the problem.

Example 3  Evaluate the following integral.

Solution

Again, the substitution and square root are the same as the first two examples.

Let’s next see the limits  for this problem.

Note that in determining the value of  we used the smallest positive value.  Now for this range of x’s we have  and in this range of  tangent is negative and so in this case we can drop the absolute value bars, but will need to add in a minus sign upon doing so.  In other words,

So, the only change this will make in the integration process is to put a minus sign in front of the integral.  The integral is then,

Example 4  Evaluate the following integral.

Solution

Now, the square root in this problem looks to be (almost) the same as the previous ones so let’s try the same type of substitution and see if it will work here as well.

Using this substitution the square root becomes,

So using this substitution we will end up with a negative quantity (the tangent squared is always positive of course) under the square root and this will be trouble.  Using this substitution will give complex values and we don’t want that.  So, using secant for the substitution won’t work.

However, the following substitution (and differential) will work.

With this substitution the square root is,

We were able to drop the absolute value bars because we are doing an indefinite integral and so we’ll assume that everything is positive.

The integral is now,

In the previous section we saw how to deal with integrals in which the exponent on the secant was even and since cosecants behave an awful lot like secants we should be able to do something similar with this.

Here is the integral.

Now we need to go back to x’s using a right triangle.  Here is the right triangle for this problem and trig functions for this problem.

The integral is then,