We integrate by substitution with the appropriate trigonometric function.

This integral can be interpreted as area of the following region:

We now know a number of integration techniques, yet we are still unable to evaluate
the above integral without resorting to a geometric interpretation! This
section introduces **trigonometric substitution**, a method of integration that
will give us a new tool in our quest to compute more antiderivatives. This
technique works on the same principle as substitution. Recall the substitution
formula.

The idea behind substitution is that by changing the variable of integration from to , then in many cases we can simplify the integrand to a form that we can integrate directly. In previous sections, we used this result to go from the left integral to the right integral. But what if we tried going from right to left?

Let’s rewrite the substitution formula in a more suggestive manner for our purposes:

This is just the substitution formula again but now we are thinking of transforming from left-to-right. Initially this may seem to complicate matters. We go from a simple integral to one that looks more complicated. However in certain cases we may not know how to integrate , but we may be able to integrate .

We start by applying this technique to compute the area that we previously used geometry to compute.

The difficulty with this integral is that we have a square root of a difference of two terms. We can’t directly simplify such an expression. (Recall: .) However, consider the Pythagorean identity: From this we see that .

If our expression was then we could simplify. We could write

This suggests that it would be useful to use a substitution .

Just as in the case of substitutions that we have worked with previously, we must convert the entire integral to be in terms of the new variable . We must also convert as well as the bounds on the integral.

Now we change our limits of integration. We have

and

Now we may transform our integral via

Now we must do something about the absolute value. Remember that since we introduced , this means lies in . Since on , the function is always positivenegativezero , we can drop the absolute value, and then employ a power-reduction formula

Now we have

This matches the answer that we obtained previously using geometry.

As you may recall, there are two ways to use substitution. Above we transformed the limits of integration, and worked with the variable . However, we could have solved the problem by finishing with an antiderivative in and then evaluating from the original -bounds to . We should familiarize ourselves with this method as well since it will be our only option if we need to do an indefinite integral using trig substitution.

Above we found the antiderivative

However, now we must express our answer in terms of . Now since , we know that . However what about the term? You may be tempted to just stick in this term but we can do better and express the antiderivative in a simpler form.

To convert our answer to a function in , we construct a right triangle.

Recall that

Since our substitution was , we can think of as the length of the side opposite the angle and we can think of as the length of the hypotenuse of the right triangle.

We can then use the Pythagorean Theorem to find the adjacent leg of the triangle. Using and solving for we obtain that the length of the adjacent side is .

Now we still need to express in terms of using this triangle. However the angle in our right triangle is , not . Recall the trig identity

We can then write our above antiderivative as

We can then replace and using our triangle and the fact that we can write .

Hence our antiderivative can now be written in terms of :

The key idea with this example was that the expression was difficult to work with directly and using the substitution allowed us to simplify the expression and turn the integral into a trig integral that was easier to calculate.

There are other situations where this kind of strategy is useful. Let’s look at a few more examples.

Recall one of the alternate forms of the Pythagorean identity and note that multiplying the entire equation by yields

This suggests that if we made the substitution then we could use this trig identity to simplify the denominator of our integrand.

Hence we’ll make the substitution

Now we convert our integral to be in terms of the new variable :

We’ll use our expertise with trigonometric integrals

Now we do another substitution

to find

. Since we have

Now that we have an answer in terms of , we must convert it back to being a function in . To convert our answer to a function in , use a reference right triangle, noting that , and so . Since we can label the triangle with as the length of the opposite side and as the length of the adjacent side. Note we have already used the Pythagorean Theorem to find the length of the hypotenuse.

and we may write

This suggests that we should use

and

Now we convert our integral to be in terms of the new variable :

The fact that guarantees that .

Now we can evaluate the integral:

So far we have seen that we can use trig substitutions to simplify expressions like , ,
and . However, sometimes the integral may not be in a form where it is clear that a
trig substitution will help. One very useful tool is to **complete the square**, and
then choose a trigonometric function which allows the use of a Pythagorean identity
to simplify the expression.

Let’s recall how to complete the square for a general quadratic . First we divide through by .

Now, consider the following example:

Our integral becomes

We want to let equal some trigonometric function that will simplify after the application of a Pythagorean identity. Again

will be helpful. We multiple this equation by and rearrange to find

Thus we use the substitution

So making this substitution we have

Since , this guarantees that and thus that :

We’ll again use the Pythagorean identity to write

Now that we have an answer in , we must convert it back to being a function in . Since we have and using a reference triangle and using the Pythagorean Theorem to find the length of the opposite side we have:

Thus using the technique of completing the square, we can always rewrite any quadratic in a form where a trig substitution can be used to simplify the expression.

### Summary

In general, when doing a trig substitution:

- If the quadratic has the form after completing the square, make the substitution .
- If the quadratic has the form after completing the square, make the substitution
- If the quadratic has the form after completing the square, make the substitution

You should always remember the other integration techniques you have learned previously; they may allow you to compute the integral more easily than trig substitutions in some cases. In particular, go back and look at example 2 in this section and see if you can compute this integral using a more efficient integration technique! In some cases, though, trig substitution will be your only option.