In Section 5.2 we defined the definite integral as the “signed area under the curve.” In that section we had not yet learned the Fundamental Theorem of Calculus, so we evaluated special definite integrals which described nice, geometric shapes. For instance, we were able to evaluate
(8.3.1) |
as we recognized that described the upper half of a circle with radius 3.
We have since learned a number of integration techniques, including Substitution and Integration by Parts, 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 fills this gap in our integration skill. This technique works on the same principle as Substitution as found in Section 5.5, though it can feel “backward.” In Section 5.5, we set , for some function , and replaced with . In this section, we will set , where is a trigonometric function, then replace with .
We start by demonstrating this method in evaluating the integral in Equation (8.3.1). After the example, we will generalize the method and give more examples.
Evaluate .
SolutionWe begin by noting that , and hence . If we let , then .
Setting gives . We are almost ready to substitute. We also change our bounds of integration. The bound corresponds to (for when , ). Likewise, the bound of is replaced by the bound . Thus
On , is always positive, so we can drop the absolute value bars, then employ a half-angle formula: | ||||
This matches our answer from before.
We now describe in detail Trigonometric Substitution. This method excels when dealing with integrands that contain , and . The following Key Idea outlines the procedure for each case, followed by more examples.
For integrands containing :
Let , for and .
On this interval, , so
For integrands containing :
Let , for and .
On this interval, , so
For integrands containing :
Let , restricting our work to where ,
so , and .
On this interval, , so
Evaluate .
SolutionUsing Key Idea 8.3.1(b), we recognize and set . This makes . We will use the fact that Substituting, we have:
While the integration steps are over, we are not yet done. The original problem was stated in terms of , whereas our answer is given in terms of . We must convert back to .
The lengths of the sides of the reference triangle in Figure 8.3.1 are determined by the Pythagorean Theorem. With , we have
This gives
We can leave this answer as is, or we can use a logarithmic identity to simplify it. Note:
where the term is absorbed into the constant . (In Section 7.4 we learned another way of approaching this problem.)
Evaluate .
SolutionWe start by rewriting the integrand so that it has the form for some value of :
So we have , and following Key Idea 8.3.1(c), we set , and hence . We now rewrite the integral with these substitutions:
We integrated in Example 8.2.6, finding its antiderivatives to be
Thus
We are not yet done. Our original integral is given in terms of , whereas our final answer, as given, is in terms of . We need to rewrite our answer in terms of . With , and , we use the Pythagorean Theorem to determine the lengths of the sides of the reference triangle in Figure 8.3.2. ††margin: Λ
Therefore,
Evaluate .
SolutionWe use Key Idea 8.3.1(a) with , , and hence . This gives
We need to rewrite our answer in terms of . Using the Pythagorean Theorem we determine the lengths of the sides of the reference triangle in Figure 8.3.3. We have and . Thus
Trigonometric Substitution can be applied in many situations, even those not of the form , or . In the following example, we apply it to an integral we already know how to handle.
Evaluate .
SolutionWe know the answer already as . We apply Trigonometric Substitution here to show that we get the same answer without inherently relying on knowledge of the derivative of the arctangent function.
The next example is similar to the previous one in that it does not involve a square-root. It shows how several techniques and identities can be combined to obtain a solution.
Evaluate .
SolutionWe start by completing the square, then make the substitution , followed by the trigonometric substitution of : ††margin: Note: Remember the sine and cosine double angle identities: They are often needed for writing your final answer in terms of . Λ
Now make the substitution , : | ||||
Applying a half-angle formula, we have | ||||
(8.3.2) |
We need to return to the variable . As , . Using the identity and using the reference triangle found in Key Idea 8.3.1(b), we have
Finally, we return to with the substitution . We start with the expression in Equation (8.3.2):
Stating our final result in one line,
Our last example returns us to definite integrals, as seen in our first example. Given a definite integral that can be evaluated using Trigonometric Substitution, we could first evaluate the corresponding indefinite integral (by changing from an integral in terms of to one in terms of , then converting back to ) and then evaluate using the original bounds. It is much more straightforward, though, to change the bounds as we substitute.
Evaluate .
SolutionUsing Key Idea 8.3.1(b), we set , , and note that . As we substitute, we change the bounds of integration.
The lower bound of the original integral is . As , we solve for and find . Thus the new lower bound is . The original upper bound is , thus the new upper bound is .
The next section introduces Partial Fraction Decomposition, which is an algebraic technique that turns “complicated” fractions into sums of “simpler” fractions, making integration easier.
Trigonometric Substitution works on the same principles as Integration by Substitution, though it can feel “ ”.
If one uses Trigonometric Substitution on an integrand containing , then one should set .
Why does Key Idea 8.3.1(a) state that , and not ?
In Exercises 5–26., apply Trigonometric Substitution to evaluate the indefinite integrals.
In Exercises 27–34., evaluate the indefinite integrals. Some may be evaluated without Trigonometric Substitution.
In Exercises 35–40., evaluate the definite integrals by making the proper trigonometric substitution and changing the bounds of integration.