Unformatted text preview: 468 CHAPTER 8 TECHNIQUES OF INTEGRATION
π −π (b) Evaluate cos mx cos nx dx. 63. Derive the reduction formula: for n > 1, cscn x dx = −cscn−2 x cot x n − 2 + n−1 n−1 cscn−2 x dx. NOTE: The formulas in Exercises 55 and 56 are important in applied mathematics. They arise in Fourier analysis, which involves the approximation of functions by sums of sines and cosines. 57. Verify reduction formula (8.3.2). 58. (a) Use reduction formula (8.3.1) to show that
π /2 0 64. Use Exercise 63 to calculate the integrals: (a) csc3 x dx; (b) csc4 x dx; (c) csc5 3x dx. sinn x dx = n−1 n π /2 0 sinn−2 x dx. 65. Calculate a. b. c. d. sin x cos x dx in three ways. (b) Show that
π /2 0 sinn x dx = n even, n ≥ 2 n odd, n ≥ 3. Let u = sin x. Let u = cos x. Set sin x cos x = 1 sin 2x. 2 Reconcile the results in parts (a), (b), and (c). (n − 1) · · · 5 · 3 · 1 · π , n···6 · 4 · 2 2 (n − 1) · · · 4 · 2 , n···5 · 3 These formulas are known as the Wallis sine formulas. (c) Show that
π /2 0 c 66. Use a graphing utility to draw the graph of f (x) = x + sin 2x, x ∈ [0, π ]. The region bounded by the graph of f and the xaxis is revolved about the xaxis. (a) Use a CAS to calculate the volume of the resulting solid. (b) Use the methods of this section to calculate the volume of the solid. c 67. Use a graphing utility to draw the graph of g (x) = √ sin2 (x2 ), x ∈ [0, π ]. The region bounded by the graph of g and the xaxis is revolved about the yaxis. (a) Use a CAS to calculate the volume of the resulting solid. (b) Use the methods of this section to calculate the volume of the solid. c 68. Use a graphing utility to draw the graphs of f (x) = 1+cos x and g (x) = sin (x/2), x ∈ [0, 2π ], together. (a) Use a CAS to ﬁnd the points of intersection of the two curves and then ﬁnd the area of the region between the curves. (c) The region between the curves is rotated about the xaxis. Use a CAS to ﬁnd the volume of the solid of revolution. cosn x dx =
0 π /2 sinn x dx. Use the results in Exercise 58 to evaluate the integrals in Exercises 59 and 60.
π /2 59.
0 π /2 sin7 x dx. cos6 x dx. 60.
0 61. Derive the reduction formula: for n > 1, cotn x dx = − cotn−1 x − n−1 cotn−2 x dx. 62. Use Exercise 61 to calculate the integrals: (a) cot3 x dx; (b) cot4 x dx; (c) cot5 2x dx. 8.4 INTEGRALS INVOLVING √ √ √ a2 − x2 , a2 + x2 , x2 − a2 Such integrals can often by calculated by a trigonometric substitution: √ for a2 − x2 , set a sin u = x; √ for a2 + x2 , set a tan u = x; √ for x2 − a2 , set a sec u = x. In each case, take a > 0. 8.4 INTEGRALS INVOLVING a2 − x2 , a2 + x2 , x2 − a2 469 Example 1
SOLUTION Find ( a2 dx . − x2 )3/2 First note that ( a2 dx = − x2 )3/2 dx √ a2 − x 2
3 ;
a x u √a2 – x2 the integral involves √ a2 − x2 . Therefore, we set a sin u = x, a cos u du = dx. Then Now dx = − x2 )3/2 = = =
Example 2 a2 − x2 = a cos u. a cos u du (a cos u)3 1 a2 1 a2 cos u du ( cos u)3 sec2 u du ( a2 x 1 tan u + C = √ + C. 2 2 a2 − x 2 a a Find a2 + x2 dx. a sec2 u du = dx. Then
√a
u
2 SOLUTION Set a tan u = x, a2 + x2 = a sec u. +x 2 x Now a2 + x2 dx = = a2 (a sec u) a sec u du sec3 u du 2 a Example 9, Section 8.3 a2 ( sec u tan u + ln  sec u + tan u) + C = ↑2 √ √ a2 + x 2 x x a2 a2 + x 2 + ln + = +C 2 a a a a √ √ = 1 x a2 + x2 + 1 a2 ln (x + a2 + x2 ) − 1 a2 ln a + C . 2 2 2 We can absorb the constant − 1 a2 ln a in C and write 2
(8.4.1) √ √ a2 + x2 dx = 1 x a2 + x2 + 1 a2 ln (x + a2 + x2 ) + C . 2 2 This is a standard integration formula. 470 CHAPTER 8 TECHNIQUES OF INTEGRATION Example 3
SOLUTION
x √x 2 – 4 Find x2 dx . √ x2 − 4 2 sec u tan u du = dx. x2 − 4 = 2 tan u. Then Set 2 sec u = x, u 2 Now dx = √ x2 x2 − 4 = = =
1 4 1 4 1 4 2 sec u tan u du 4 sec2 u · 2 tan u 1 du sec u cos u du √ x2 − 4 sin u + C = + C. 4x Remark Sometimes there is more than one way to calculate an integral. For example, to calculate x dx √ 2 − x2 a you do not need to set a sin u = x. You can carry out the integration more directly by substituting u = a2 − x2 . (Try both substitutions and decide which you like better.)
Example 4
SOLUTION
√ 4x
u
2 Find x 4x2 + 9 √ dx . Then Set 3 tan u = 2x, 3 sec2 u du = 2 dx. 4x2 + 9 = 3 sec u +9
2x 3 and x 4x 2 + 9 √ dx = = = =
1 3 1 3 1 3 3 2 3 2 sec2 u tan u · 3 sec u du sec u du tan u csc u du ln  csc u − cot u + C √ 1 4x2 + 9 − 3 = ln + C. 3 2x The next example requires that we ﬁrst complete the square under the radical.
Example 5 Find x dx. √ 2 + 2x − 3 x 8.4 INTEGRALS INVOLVING
SOLUTION a2 − x2 , a2 + x2 , x2 − a2 471 First note that x dx = √ 2 + 2x − 3 x x (x + 1)2 − 4 Then dx. Now set 2 sec u = x + 1, 2 sec u tan u du = dx. (x + 1)2 − 4 = 2 tan u and x (x + 1)2 −4 dx = = (2 sec u − 1) 2 sec u tan u du 2 tan u (2 sec2 u − sec u) du = 2 tan u − ln  sec u + tan u + C √ x + 1 + x 2 + 2x − 3 2 + 2x − 3 − ln =x + C. 2 For our last example, we calculate a deﬁnite integral.
r Example 6
SOLUTION Calculate
−r r 2 − x2 dx, where r is a positive constant. r cos u du = dx. Then Set r sin u = x, r 2 − x2 = r cos u. Also, u = − π at x = −r and u = 2
r −r π 2 at x = r . Thus
π /2 −π/2 r 2 − x2 dx = π /2 −π/2 (r cos u) · r cos u du = r 2
1 2 cos2 u du = r2 = √
2 π /2 −π/2 + 1 cos 2u du 2
π /2 −π/2 r u + 1 sin 2u 2 2 = π r2 . 2
y Note that the graph of y = r 2 − x2 , −r ≤ x ≤ r , is the semicircle of radius r shown in Figure 8.4.1. Thus we have found that the area enclosed by a semicircle of radius r is π r 2 /2. The familiar formula for the area enclosed by a circle of radius r , A = π r 2 , follows immediately from this result.
–r r x A trigonometric substitution may be effective even in cases where the quadratic in the integrand is not under a radical. For example, the reduction formula
(8.4.2) Figure 8.4.1 (x 2 1 dx = 2n−1 2 )n +a a cos2(n−1) u du can be obtained by setting a tan u = x, a sec2 u du = dx. The proof is left to you as an exercise. 472 CHAPTER 8 TECHNIQUES OF INTEGRATION EXERCISES 8.4
Calculate the integral. 1. 3. 5. 7. 9.
0 dx . √ a2 − x 2 x2 − 1 dx. x2 dx √ 4 − x2 x dx. (1 − x2 )3/2
1/2 4 2.
5/ 2 x dx. √ x2 − 4 x 4 − x2 x2 x2 − 4 x2 4 + x2 dx. dx. dx. (b) A trigonometric substitution. (c) Reconcile your two results. 37. Verify the reduction formula (8.4.2). In Exercises 38 and 39, use reduction formula (8.4.2) to calculate the integral. 38. 1 dx. (x2 + 1)2 39. 1 dx. (x2 + 1)3 4. 6. 8. 10. 12.
0 √ √ √ In Exercises 40 and 41, use integration by parts and then a trigonometric substitution to calculate the integral. 40. x tan−1 x dx. 41. x sin−1 x dx. x2 dx. (1 − x2 )3/2 x dx. a2 + x 2
2 11. 13.
0 x 4 − x2 dx.
5 x2 25 − x2 dx. x2 dx. + 8)3/2 dx x3 9+ x2 . 14. 16.
0 x3 dx. √ 16 − x2 √ 1 − x2 dx. x4 a2 − x2 dx. a 15. 17. 19.
0 (x 2 √ x a2 − x 2
3 18. 20. 22. 24. 26. 28. 30. 32. 34. √ x2 − 1 dx. x x2 √ dx a2 − x 2 . √ 42. Find the area under the graph of y = ( x2 − 9)/x from x = 3 to x = 5. 43. The region bounded by the graph of f (x) = 1/(1 + x2 ) and the xaxis between x = 0 and x = 1 is revolved about the xaxis. Find the volume of the solid that is generated. 44. In a disc of radius r a chord h units from the center generates a region of the disc called a segment (see the ﬁgure). Find a formula for the area of the segment. √ dx. h r 21. 23.
0 x2
1 dx . √ a2 + x 2 (x 2 ex dx . + 2)3/2 dx 4 + e 2x . dx . (5 − x2 )3/2 √ 25. 27. 29. 31. 33. dx . √ 2 x 2 − a2 x dx . √ ex e2x − 9 dx . (x2 − 4x + 4)3/2 x 6x − x2 − 8 dx. x dx. (x2 + 2x + 5)2 √ √ ex 9 − e 2x dx dx. . x 2 − 2x − 3 x 6x − x2 x+2 x2 dx. 45. Derive the formula A = 1 r 2 θ for the area of a sector of a 2 circle of radius r and central angle θ (measured in radians). HINT: Assume ﬁrst that 0 < θ < 1 π and subdivide the 2 region as indicated in the ﬁgure. Then verify that the formula holds for any sector.
y √ √ + 4x + 13 x dx. √ x2 − 2x + 3 dx.
θ
(r, 0) x 35. Verify the integration formula (8.2.7): sec−1 x dx = x sec−1 x − ln x + 36. Calculate a2 − x 2 dx by using: x √ (a) The substitution u = a2 − x2 . √ x2 − 1 + C . 46. Find the area of the region bounded on the left and right by the two branches of the hyperbola (x2 /a2 ) − (y2 /b2 ) = 1, and above and below by the lines y = ±b. 47. Find the area of the region bounded by the right branch of the hyperbola (x2 /9) − (y2 /16) = 1 and the line x = 5. 8.5 RATIONAL FUNCTIONS; PARTIAL FRACTIONS 473 48. If the circle (x − b)2 + y2 = a2 , b > a > 0, is rotated around the yaxis, the resulting “doughnutshaped” solid is called a torus. Use the shell method to ﬁnd the formula for the volume of the torus. See Exercise 43 in Section 6.3 and Example 5 in Section 6.4. 49. Calculate the mass and the center of mass of a rod that extends from x = 0 to x = a > 0 and has mass density λ(x) = (x2 + a2 )−1/2 . 50. Calculate the mass and the center of mass of the rod in Exercise 49 if the mass density is given by λ(x) = (x2 + a2 )−3/2 . For √ Exercises 51–53, √ let be the region under the curve y = x2 − a2 , x ∈ [a, 2a]. 51. Sketch , ﬁnd its area, and locate the centroid. 52. Find the volume of the solid generated by revolving about the xaxis and determine the centroid of that solid. 53. Find the volume of the solid generated by revolving about the yaxis and determine the centroid of that solid. 54. (a) Use a trigonometric substitution to show that, for a > 0, √ 1 a2 + x 2 dx = ln x + a2 + x 2 + C . 55. (a) Use a trigonometric substitution to show that, for a > 0, √ 1 x2 − a2 dx = ln x + x 2 − a2 + C . (b) Use the hyperbolic substitution x = a cosh u with x > a to show that √ 1 x2 − a2 dx = cosh−1 x + C. a See Exercise 42, Section 7.9. c 56. Let x2 . f (x ) = √ 1 − x2 (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 xaxis from x = 0 to x = 1 . 2 (c) Find the volume of the solid generated by revolving the region in part (b) about the yaxis. c 57. Let
√ f (x ) = x2 − 9 , x2 x ≥ 3. (b) Use the hyperbolic substitution x = a sinh u to show that √ 1 a2 + x 2 dx = sinh−1 x + C. a (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 xaxis from x = 3 to x = 6. (c) Find the centroid of the region. See Exercise 41, Section 7.9. 8.5 RATIONAL FUNCTIONS; PARTIAL FRACTIONS
In this section we present a method for integrating rational functions. Recall that a rational function is, by deﬁnition, the quotient of two polynomials. For example, 1 , 2−4 x are rational functions, but 1 √, x x2 + 1 , ln x  x − 2 x2 + 1 2x2 + 3 , x(x − 1)2 3x4 − 20x2 + 17 x3 + 2x2 − 7 are not rational functions. A rational function R(x) = P (x)/Q(x) is said to be proper if the degree of the numerator is less than the degree of the denominator. If the degree of the numerator is greater than or equal to the degree of the denominator, then the rational function is improper.† We will focus our attention on proper rational functions because any
† These terms are taken from the familiar terms used to described rational numbers P . q ...
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 Spring '10
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 Applied Mathematics, Formulas, dx

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