Calculus for Engineer: Integration of Trigonometric Substitution

Trig Substitutions

As we have done in the last couple of sections, let’s start off with a couple of integrals that we should already be able to do with a standard substitution.

Both of these used the substitution

and at this point should be pretty easy for you to do. However, let’s take a look at the following integral.

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 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 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.

$TrigSubs_Ex1$

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.

Wow! That was a lot of work. Most of these won’t take as long to work however. This first one needed lot’s of explanation since it was the first one. The remaining examples won’t need quite as much explanation and so won’t take as long to work.

However, before we move onto more problems let’s first address the issue of definite integrals and how the process differs in these cases.

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.

Let’s take a look at a different set of limits for this integral.

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,

In the last two examples we saw that we have to be very careful with definite integrals. We need to make sure that we determine the limits on

and whether or not this will mean that we can drop the absolute value bars or if we need to add in a minus sign when we drop them.

Before moving on to the next example let’s get the general form for the substitution that we used in the previous set of examples.

Let’s work a new and different type of example.

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.

$TrigSubs_Ex4$

The integral is then,

Here’s the general form for this type of square root.

There is one final case that we need to look at. The next integral will also contain something that we need to make sure we can deal with.

Example 5 Evaluate the following integral.

Solution

First, notice that there really is a square root in this problem even though it isn’t explicitly written out. To see the root let’s rewrite things a little.

This square root is not in the form we saw in the previous examples. Here we will use the substitution for this root.

With this substitution the denominator becomes,

Now, because we have limits we’ll need to convert them to

so we can determine how to drop the absolute value bars.

In this range of

secant is positive and so we can drop the absolute value bars.

Here is the integral,

There are several ways to proceed from this point. Normally with an odd exponent on the tangent we would strip one of them out and convert to secants. However, that would require that we also have a secant in the numerator which we don’t have. Therefore, it seems like the best way to do this one would be to convert the integrand to sines and cosines.

We can now use the substitution

and we might as well convert the limits as well.

The integral is then,

The general form for this final type of square root is

We have a couple of final examples to work in this section. Not all trig substitutions will just jump right out at us. Sometimes we need to do a little work on the integrand first to get it into the correct form and that is the point of the remaining examples.

Example 6 Evaluate the following integral.

Solution

In this case the quantity under the root doesn’t obviously fit into any of the cases we looked at above and in fact isn’t in the any of the forms we saw in the previous examples. Note however that if we complete the square on the quadratic we can make it look somewhat like the above integrals.

Remember that completing the square requires a coefficient of one in front of the x². Once we have that we take half the coefficient of the x, square it, and then add and subtract it to the quantity. Here is the completing the square for this problem.

So, the root becomes,

This looks like a secant substitution except we don’t just have an x that is squared. That is okay, it will work the same way.

Using this substitution the root reduces to,

Note we could drop the absolute value bars since we are doing an indefinite integral. Here is the integral.

And here is the right triangle for this problem.

$TrigSubs_Ex6$

The integral is then,

Example 7 Evaluate the following integral.

Solution

This doesn’t look to be anything like the other problems in this section. However it is. To see this we first need to notice that,

With this we can use the following substitution.

Remember that to compute the differential all we do is differentiate both sides and then tack on dx or

onto the appropriate side.

With this substitution the square root becomes,

Again, we can drop the absolute value bars because we are doing an indefinite integral. Here’s the integral.

Here is the right triangle for this integral.

$TrigSubs_Ex7$

The integral is then,

So, as we’ve seen in the final two examples in this section some integrals that look nothing like the first few examples can in fact be turned into a trig substitution problem with a little work.

Before leaving this section let’s summarize all three cases in one place.