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5.7 Inverse Trigonometric Functions: Integration 375 5.7 Inverse Trigonometric Functions: Integration 375 5.7 Inverse Trigonometric Functions: Integration Integrate functions whose antiderivatives involve inverse trigonometric functions. Use the method of completing the square to integrate a function. Review the basic integration rules involving elementary functions. Integrals Involving Inverse Trigonometric Functions The derivatives of the six inverse trigonometric functions fall into three pairs. In each pair, the derivative of one function is the negative of the other. For example, d 1 ͓arcsin x͔ ϭ dx Ί1 Ϫ x 2 and d 1 ͓arccos x͔ ϭϪ . dx Ί1 Ϫ x 2 When listing the antiderivative that corresponds to each of the inverse trigonometric functions, you need to use only one member from each pair. It is conventional to use arcsin x as the antiderivative of 1͞Ί1 Ϫ x2, rather than Ϫarccos x. The next theorem gives one antiderivative formula for each of the three pairs. The proofs of these integration rules are left to you (see Exercises 75–77). FOR FURTHER INFORMATION THEOREM 5.17 Integrals Involving Inverse Trigonometric For a detailed proof of rule 2 of Functions Theorem 5.17, see the article Let u be a differentiable function of x, and let a > 0. “A Direct Proof of the Integral du u du 1 u Formula for Arctangent” by 1.͵ ϭ arcsin ϩ C 2. ͵ ϭ arctan ϩ C Arnold J. Insel in The College Ίa2 Ϫ u2 a a2 ϩ u2 a a Mathematics Journal. To view this du 1 ԽuԽ 3. ͵ ϭ arcsec ϩ C article, go to MathArticles.com. uΊu2 Ϫ a2 a a Integration with Inverse Trigonometric Functions dx x a. ͵ ϭ arcsin ϩ C Ί4 Ϫ x2 2 dx 1 3 dx b. ͵ ϭ ͵ u ϭ 3x, a ϭ Ί2 2 ϩ 9x2 3 ͑Ί2͒2 ϩ ͑3x͒2 1 3x ϭ arctan ϩ C 3Ί2 Ί2 dx 2 dx c. ͵ ϭ ͵ u ϭ 2x, a ϭ 3 xΊ4x2 Ϫ 9 2xΊ͑2x͒2 Ϫ 32 1 Խ2xԽ ϭ arcsec ϩ C 3 3 The integrals in Example 1 are fairly straightforward applications of integration formulas. Unfortunately, this is not typical. The integration formulas for inverse trigonometric functions can be disguised in many ways. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 376 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions Integration by Substitution dx Find ͵ . Ίe2x Ϫ 1 Solution As it stands, this integral doesn’t fit any of the three inverse trigonometric formulas. Using the substitution u ϭ ex, however, produces du du u ϭ ex du ϭ ex dx dx ϭ ϭ . ex u With this substitution, you can integrate as shown. dx dx ͵ ϭ ͵ Write e2x as ͑e x͒2. Ίe2x Ϫ 1 Ί͑ex͒2 Ϫ 1 du͞u ϭ ͵ Substitute. Ίu2 Ϫ 1 du ϭ ͵ Rewrite to fit Arcsecant Rule. uΊu2 Ϫ 1 ԽuԽ ϭ arcsec ϩ C Apply Arcsecant Rule. 1 ϭ arcsec ex ϩ C Back-substitute. TECHNOLOGY PITFALL A symbolic integration utility can be useful for integrating functions such as the one in Example 2. In some cases, however, the utility may fail to find an antiderivative for two reasons. First, some elementary functions do not have antiderivatives that are elementary functions. Second, every utility has limitations—you might have entered a function that the utility was not programmed to handle. You should also remember that antiderivatives involving trigonometric functions or logarithmic functions can be written in many different forms. For instance, one utility found the integral in Example 2 to be dx ͵ ϭ arctan Ίe2x Ϫ 1 ϩ C. Ίe2x Ϫ 1 Try showing that this antiderivative is equivalent to the one found in Example 2. Rewriting as the Sum of Two Quotients x ϩ 2 Find ͵ dx. Ί4 Ϫ x2 Solution This integral does not appear to fit any of the basic integration formulas. By splitting the integrand into two parts, however, you can see that the first part can be found with the Power Rule and the second part yields an inverse sine function. x ϩ 2 x 2 ͵ dx ϭ ͵ dx ϩ ͵ dx Ί4 Ϫ x2 Ί4 Ϫ x2 Ί4 Ϫ x2 1 1 ϭϪ ͵͑4 Ϫ x2͒Ϫ1͞2͑Ϫ2x͒ dx ϩ 2 ͵ dx 2 Ί4 Ϫ x2 1 ͑4 Ϫ x2͒1͞2 x ϭϪ ΄ ΅ ϩ 2 arcsin ϩ C 2 1͞2 2 x ϭϪΊ4 Ϫ x2 ϩ 2 arcsin ϩ C 2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.7 Inverse Trigonometric Functions: Integration 377 Completing the Square Completing the square helps when quadratic functions are involved in the integrand. For example, the quadratic x2 ϩ bx ϩ c can be written as the difference of two squares by adding and subtracting ͑b͞2͒2. b 2 b 2 b 2 b 2 x2 ϩ bx ϩ c ϭ x2 ϩ bx ϩ ΂ ΃ Ϫ ΂ ΃ ϩ c ϭ ΂x ϩ ΃ Ϫ ΂ ΃ ϩ c 2 2 2 2 Completing the Square See LarsonCalculus.com for an interactive version of this type of example. dx Find ͵ . x2 Ϫ 4x ϩ 7 Solution You can write the denominator as the sum of two squares, as shown. x2 Ϫ 4x ϩ 7 ϭ ͑x2 Ϫ 4x ϩ 4͒ Ϫ 4 ϩ 7 ϭ ͑x Ϫ 2͒2 ϩ 3 ϭ u2 ϩ a2 Now, in this completed square form, let u ϭ x Ϫ 2 and a ϭ Ί3. dx dx 1 x Ϫ 2 ͵ ϭ ͵ ϭ arctan ϩ C x2 Ϫ 4x ϩ 7 ͑x Ϫ 2͒2 ϩ 3 Ί3 Ί3 y When the leading coefficient is not 1, it helps to factor before completing the square. For instance, you can complete the square of 2x2 Ϫ 8x ϩ 10 by factoring first. 1 f(x) = 2 2 3 3x − x2 2 x Ϫ 8x ϩ 10 ϭ 2͑x Ϫ 4x ϩ 5͒ 2 ϭ 2͑x Ϫ 4x ϩ 4 Ϫ 4 ϩ 5͒ 2 ϭ 2͓͑x Ϫ 2͒ ϩ 1͔ 2 To complete the square when the coefficient of x2 is negative, use the same factoring process shown above. For instance, you can complete the square for 3x Ϫ x2 as shown. 1 Ϫ 2 ϭϪ͑ 2 Ϫ ͒ ϭϪ͓ 2 Ϫ ϩ ͑3͒2 Ϫ ͑3͒2͔ ϭ ͑3͒2 Ϫ ͑ Ϫ 3͒2 3x x x 3x x 3x 2 2 2 x 2 x 1 x = 3 23x = 9 2 4 Completing the Square The area of the region bounded by the ϭ 3 ϭ 9 Find the area of the region bounded by the graph of graph of f, the x-axis,x 2, and x 4 is ␲͞6. 1 f͑x͒ ϭ Figure 5.28 Ί3x Ϫ x2 ϭ 3 ϭ 9 the x-axis, and the lines x 2 and x 4. TECHNOLOGY With definite integrals such as the one Solution In Figure 5.28, you can see that the area is ͞ given in Example 5, remember 9 4 1 Area ϭ ͵ dx that you can resort to a numerical Ί 2 3͞2 3x Ϫ x solution. For instance, applying 9͞4 ϭ ) dx Simpson’s Rule (with n 12 ϭ ͵ Use completed square form derived above. Ί 2 2 to the integral in the example, 3͞2 ͑3͞2͒ Ϫ ͓x Ϫ ͑3͞2͔͒ you obtain x Ϫ ͑3͞2͒ 9͞4 ϭ arcsin 9͞4 ͞ ΅ 1 3 2 3͞2 ͵ dx Ϸ 0.523599. Ί Ϫ 2 1 3͞2 3x x ϭ arcsin Ϫ arcsin 0 2 This differs from the exact ␲ value of the integral ϭ ͑␲͞6 Ϸ 0.5235988͒ by less 6 than one-millionth. Ϸ 0.524. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 378 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions Review of Basic Integration Rules You have now completed the introduction of the basic integration rules. To be efficient at applying these rules, you should have practiced enough so that each rule is committed to memory. BASIC INTEGRATION RULES ͧa > 0ͨ 1.͵k f͑u͒ du ϭ k͵f͑u͒ du 2. ͵͓ f͑u͒ ± g͑u͔͒ du ϭ ͵f͑u͒ du ± ͵g͑u͒ du unϩ1 3.͵du ϭ u ϩ C 4. ͵un du ϭ ϩ C, n Ϫ1 n ϩ 1 du 5.͵ ϭ lnԽuԽ ϩ C 6. ͵eu du ϭ eu ϩ C u 1 7.͵au du ϭ ΂ ΃au ϩ C 8. ͵sin u du ϭϪcos u ϩ C ln a 9.͵cos u du ϭ sin u ϩ C 10. ͵tan u du ϭϪlnԽcos uԽ ϩ C 11.͵cot u du ϭ lnԽsin uԽ ϩ C 12. ͵sec u du ϭ lnԽsec u ϩ tan uԽ ϩ C 13.͵csc u du ϭϪlnԽcsc u ϩ cot uԽ ϩ C 14.
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