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 only evaluated special definite integrals which described nice, geometric shapes. For instance, we were able to evaluate
(6.4.1)
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 6.1, though it can feel “backward.” In Section 6.1, 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 (6.4.1). After the example, we will generalize the method and give more examples.
Setting gives . We are almost ready to substitute. We also wish to 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 power-reducing formula:
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. Each right triangle acts as a reference to help us understand the relationships between and .
The diagram is of a right angled triangle. The perpendicular is marked the base as . The hypotenuse is marked . The angle opposite to the perpendicular is marked as .
The diagram is of a right angled triangle. The perpendicular is marked the base as . The hypotenuse is marked . The angle opposite to the perpendicular is marked as .
The diagram is of a right angled triangle. The perpendicular is marked the base as . The hypotenuse is marked . The angle opposite to the perpendicular is marked as .
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 .
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 , the reference triangle in Figure 6.4.6 shows that
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.
We 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.
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.
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.
Evaluate the definite integral by making the proper trigonometric substitution and changing the bounds of integration. (Note: the corresponding indefinite integrals appeared previously in the Section 6.4 exercises.)