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Learning Objectives
- Integrate functions resulting in inverse trigonometric functions
In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions before. Recall from Functions and Graphs that trigonometric functions are not one-to-one unless the domains are restricted. When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also in Derivatives , we developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions.
Integrals that Result in Inverse Sine Functions
Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.
Rule: Integration Formulas Resulting in Inverse Trigonometric Functions
The following integration formulas yield inverse trigonometric functions:
Proof
Let Then Now let’s use implicit differentiation. We obtain
For Thus, applying the Pythagorean identity we have This gives
Then for we have
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Evaluating a Definite Integral Using Inverse Trigonometric Functions
Evaluate the definite integral
We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral. We have
Find the antiderivative of
Solution
Finding an Antiderivative Involving an Inverse Trigonometric Function
Evaluate the integral
Substitute Then and we have
Applying the formula with we obtain
Find the indefinite integral using an inverse trigonometric function and substitution for
Solution
Hint
Use the formula in the rule on integration formulas resulting in inverse trigonometric functions.
Evaluating a Definite Integral
Evaluate the definite integral
Solution
The format of the problem matches the inverse sine formula. Thus,
Integrals Resulting in Other Inverse Trigonometric Functions
There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the integrand is negative, simply factor out -1 and evaluate the integral using one of the formulas already provided. To close this section, we examine one more formula: the integral resulting in the inverse tangent function.
Finding an Antiderivative Involving the Inverse Tangent Function
Find an antiderivative of
Solution
Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for So we use substitution, letting then and Then, we have
Use substitution to find the antiderivative of
Solution
Hint
Use the solving strategy from (Figure) and the rule on integration formulas resulting in inverse trigonometric functions.
Applying the Integration Formulas
Find the antiderivative of
Apply the formula with Then,
Evaluating a Definite Integral
Evaluate the definite integral
Solution
Use the formula for the inverse tangent. We have
Evaluate the definite integral
Solution
Hint
Follow the procedures from (Figure) to solve the problem.
Key Concepts
- Formulas for derivatives of inverse trigonometric functions developed in Derivatives of Exponential and Logarithmic Functions lead directly to integration formulas involving inverse trigonometric functions.
- Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match up the correct format and make alterations as necessary to solve the problem.
- Substitution is often required to put the integrand in the correct form.
Key Equations
- Integrals That Produce Inverse Trigonometric Functions
In the following exercises, evaluate each integral in terms of an inverse trigonometric function.
1.
Solution
2.
3.
Solution
4.
5.
Solution
6.
In the following exercises, find each indefinite integral, using appropriate substitutions.
7.
8.
9.
Solution
10.
11.
Solution
12.
13. Explain the relationship Is it true, in general, that
Solution
So, They differ by a constant.
14. Explain the relationship Is it true, in general, that
15. Explain what is wrong with the following integral:
Solution
is not defined as a real number when
16. Explain what is wrong with the following integral:
In the following exercises, solve for the antiderivative of with then use a calculator to graph and the antiderivative over the given interval Identify a value of C such that adding C to the antiderivative recovers the definite integral
17. [T] over
Solution
The antiderivative is Taking recovers the definite integral.
18. [T] over
19. [T] over
Solution
The antiderivative is Taking recovers the definite integral.
20. [T] over
In the following exercises, compute the antiderivative using appropriate substitutions.
21.
Solution
22.
23.
Solution
24.
25.
Solution
26.
In the following exercises, use a calculator to graph the antiderivative with over the given interval Approximate a value of C , if possible, such that adding C to the antiderivative gives the same value as the definite integral
27. [T] over
Solution
The antiderivative is Taking recovers the definite integral over
28. [T] over
29. [T] over
The general antiderivative is Taking recovers the definite integral.
30. [T] over
31. [T] over
The general antiderivative is Taking recovers the definite integral.
32. [T] over
In the following exercises, compute each integral using appropriate substitutions.
33.
Solution
34.
35.
Solution
36.
37.
Solution
38.
In the following exercises, compute each definite integral.
39.
40.
41.
Solution
42.
43. For compute and evaluate the area under the graph of on
Solution
as
44. For compute and evaluate the area under the graph of over
45. Use the substitution and the identity to evaluate ( Hint: Multiply the top and bottom of the integrand by )
Solution
Using the hint, one has Set Then, and the integral is If one uses the identity then this can also be written
46. [T] Approximate the points at which the graphs of and intersect, and approximate the area between their graphs accurate to three decimal places.
47. [T] Approximate the points at which the graphs of and intersect, and approximate the area between their graphs accurate to three decimal places.
Solution
The left endpoint estimate with is 2.796 and these decimals persist for
48. Use the following graph to prove that
Hint
Substitute