SalasSV_08_06 - 482 CHAPTER 8 TECHNIQUES OF INTEGRATION 48....

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Unformatted text preview: 482 CHAPTER 8 TECHNIQUES OF INTEGRATION 48. Show that if y = x2 1 , then −1 c 54. Use a CAS to calculate the integrals x2 1 dx where n = 0, 1, 2. + 2x + n d ny ( − 1)n n! 1 1 = − . n n+1 dx 2 (x − 1) (x + 1)n+1 49. Find the volume of the solid generated by revolving the √ region bounded by the graph of y = 1/ 4 − x2 and the xaxis from x = 0 to x = 3/2 about : (a) the x-axis; (b) the y-axis. 50. Calculate x3 tan−1 x dx. 51. Find the centroid of the region under the curve y = (x2 + 1)−1 , x ∈ [0, 1]. 52. Find the centroid of the solid generated by revolving the region of Exercise 51 about: (a) the x-axis; (b) the y-axis. Verify your results by differentiating. c 55. Let f (x ) = x . x2 + 5x + 6 (a) Use a graphing utility to sketch the graph of f . (b) Calculate the area of the region bounded by the graph of f and the x-axis between x = 0 and x = 4. 56. (a) The region in Exercise 55 is revolved about the y-axis. Find the volume of the solid that is generated. (b) Find the centroid of the solid in part (a). c 57. Let f (x ) = 9−x . (x + 3)2 c 53. Use a CAS to find the partial fraction decomposition of f . 6x4 + 11x3 − 2x2 − 5x − 2 (a) f (x) = . x2 (x + 1)3 (b) f (x) = − (c) f (x) = x3 + 20x2 + 4x + 93 . (x2 + 4)(x2 − 9) x2 + 7x + 12 . x(x2 + 2x + 4) ∗ 8.6 (a) Use a graphing utility to sketch the graph of f . (b) Find the area of the region bounded by the graph of f and the x-axis between x = −2 and x = 9. 58. (a) The region in Exercise 57 is revolved about the x-axis. Find the volume of the solid that is generated. (b) Find the centroid of the solid in part (a). SOME RATIONALIZING SUBSTITUTIONS These are integrands which are not rational functions but can be transformed into rational functions by a suitable substitution. Such substitutions are known as rationalizing substitutions. First we consider integrals in which the integrand involves an expression of the form n f (x) for some function f . In such cases, the substitution u = n f (x), which is equivalent to setting un = f (x), is sometimes effective. The idea is to replace fractional exponents by integer exponents; integer exponents are usually easier to handle. Example 1 SOLUTION Find dx √. 1+ x u2 = x , 2u du = dx. To rationalize the integrand, we set √ Then u = x and dx √= 1+ x divide 2u du = 1+u 2− 2 du 1+u ✱ = 2u − 2 ln |1 + u| + C √ √ = 2 x − 2 ln |1 + x| + C . Example 2 Find dx √ √. 3 x+ x ∗ 8.6 SOME RATIONALIZING SUBSTITUTIONS 483 Here the integrand contains two distinct roots, x1/3 and x1/2 . If we set u = x, then both terms will be rationalized simultaneously: SOLUTION 6 u6 = x , Thus, we have dx √ √= 3 x+ x 6u5 du = dx, and x1/3 = u2 , x1/2 = u3 . 6u 5 du = 6 u2 + u3 =6 u3 du 1+u u2 − u + 1 − 1 1+u du divide ✱ √ 1 = 6( 1 u3 − 2 u2 + u − ln |1 + u|) + C 3 √ √ √ √ = 2 x − 3 3 x + 6 6 x − 6 ln |1 + 6 x| + C . Example 3 Find 1 − ex dx SOLUTION To rationalize the integrand, we set u2 = 1 − e x . To find dx in terms of u and du, we solve the equation for x: 1 − u2 = e x , The rest is straightforward: √ 1 − ex dx = = divide, then partial fractions ln (1 − u2 ) = x, − 2u du = dx. 1 − u2 u− 2u 1 − u2 du 2+ 1 1 − u−1 u+1 du 2u2 du = u2 − 1 ✱ = 2u + ln |u − 1| − ln |u + 1| + C = 2u + ln u−1 + C, u+1 √ √ 1 − ex − 1 = 2 1 − ex + ln √ + C. 1 − ex + 1 We now consider rational expressions in sine and cosine. Suppose, for example, that we want to calculate 1 dx. 3 sin x − 4 cos x To convert the integrand to a rational function in u, we set u = tan x , 2 −π < u < π . 484 CHAPTER 8 TECHNIQUES OF INTEGRATION Then u2 u cos x 1 = = 2 sec (x/2) sin 1 1+ tan2 1 =√ , 1 + u2 (x/2) + √1 x 2 and x x x u = cos tan . =√ 2 2 2 1 + u2 The right triangle in Figure 8.6.1 illustrates these relationships. Observe that sin x = 2 sin x and cos x = cos2 x x 1 − u2 − sin2 = . 2 2 1 + u2 x 2u x cos = . 2 2 1 + u2 1 Figure 8.6.1 Since u = tan (x/2), x = 2 tan−1 u, and so dx = 2 du. 1 + u2 To summarize, if an integrand is a rational expression in sine and cosine, then the substitution sin x = 2u , 1 + u2 cos x = 1 − u2 , 1 + u2 dx = 2 du, 1 + u2 where u = tan (x/2), will convert the integrand into a rational function in u. The resulting integral can then be calculated by the methods of Section 8.5. 1 dx. 3 sin x − 4 cos x x SOLUTION Set u = tan . Then 2 Example 4 Find 1 1 + u2 1 = =2 3 sin x − 4 cos x [6u/(1 + u2 )] − [4(1 − u2 )/(1 + u2 )] 4u + 6u − 4 and 1 dx = 3 sin x − 4 cos x Since 2u 2 we have 1 du = − 1 5 2u 2 + 3u − 2 1 du + u+2 1 5 2 5 4u 2 2 1 + u2 · du = + 6u − 4 1 + u 2 1 du. 2u2 + 3u − 2 1 1 −1/5 2/5 = = + , + 3u − 2 (u + 2)(2u − 1) u + 2 2u − 1 1 du 2u − 1 (partial fractions) = − 1 ln |u + 2| + 5 = = 1 5 1 5 ln |2u − 1| + C ln ln 2u − 1 +C u+2 2 tan (x/2) − 1 + C. tan (x/2) + 2 ∗ 8.6 SOME RATIONALIZING SUBSTITUTIONS 485 EXERCISES *8.6 Calculate the integral. 1. 3. 5. 6. 7. 8. 9. 11. 13. 15. 17. 19. 20. 21. 23. 24. 25. 27. 28. 29. dx √. 1− x √ 1 + ex dx. x 1 + x dx. x √ 2 1 + x dx. √ 2 2. 4. √ x dx. 1+x dx . 1/3 − 1) x (x π /3 35. 0 1 1 dx. sin x − cos x − 1 √ x √ dx. x 36. 0 1+ [(a) set u = 1 + x; (b) set u = 1 + x] [(a) set u2 = 1 + x; (b) set u = 1 + x] 37. Use the method of this section to show that sec x dx = 1 1 + tan (x/2) dx = ln + C. cos x 1 − tan (x/2) sec x dx is to write cos x dx. 1 − sin2 x √ (x + 2) x − 1 dx. √ (x − 1) x + 2 dx. x3 dx. (1 + x2 )3 √ x dx. √ x−1 √ x−1+1 dx. √ x−1−1 dx . √ 1 + ex x dx. √ x+4 2x2 (4x + 1)−5/2 dx. √ x2 x − 1 dx. x dx. (ax + b)3/2 1 dx. 1 + cos x − sin x 1 dx. 2 + cos x 1 dx. 2 + sin x 1 dx. sin x + tan x 1 dx. 1 + sin x + cos x 1 − cos x dx. 1 + sin x 4 38. (a) Another way to calculate sec x dx = 10. 12. 14. 16. 18. x(1 + x)1/3 dx. x+1 1 − ex dx. 1 + ex dx . 1 + e−x √ x dx. cos x dx = cos2 x Use the method of this section to show that sec x dx = ln 1 + sin x + C. 1 − sin x (b) Show that the result of part (a) is equivalent to the familiar formula sec x dx = ln | sec x + tan x| + C . 39. (a) Use the approach given in Exercise 38(a) to show that csc x dx = ln 1 − cos x + C. 1 + cos x x+1 dx. √ x x−2 22. √ x ax + b dx. (b) Show that the result of part (a) is equivalent to the formula csc x dx = ln | csc x − cot x| + C . 26. sin x dx. 1 + sin2 x 40. The integral of a rational function of sinh x and cosh x can be transformed into a rational function of u by means of the substitution u = tanh (x/2). Show that this substitution, gives sinh x = 2u , 1 − u2 cosh x = 1 + u2 , 1 − u2 dx = 2 du. 1 − u2 30. 1 dx. 5 + 3 sin x 8 In Exercises 41–44, set u = tanh (x/2) and carry out the integration 41. sech x dx. 1 dx. sinh x + cosh x 42. 1 dx. 1 + cosh x 1 − ex dx. 1 + ex Evaluate the definite integral. 31. 0 x3/2 dx. x+1 sin 2x dx. 2 + cos x 32. 0 1 √ dx. 1+ 3 x 1 dx. 1 + sin x π /2 π /2 33. 0 34. 0 43. 44. ...
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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