# Integrales

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510

SSSS

CHAPTER 7 TECHNIQUES OF INTEGRATION

then we know from Part 1 of the Fundamental Theorem of Calculus that F x
2

ex

2

Thus, f x e x has an antiderivative F , but it has been proved that F is not an elementary function. This means that no matter how hard we try, we will never succeed in evalu2 ating x e x dx in terms of the functions we know. (In Chapter 11, however, wewill see how 2 to express x e x dx as an inﬁnite series.) The same can be said of the following integrals:

y

ex dx x
3

y sin x
1 dx 1

2

dx

y cos e y

x

dx

y sx

y ln x dx

sin x dx x

In fact, the majority of elementary functions don’t have elementary antiderivatives. You may be assured, though, that the integrals in the following exercises are all elementaryfunctions.

|||| 7.5
1–80
||||

Exercises
25. 2.

Evaluate the integral.

1.

y y0
2

sin x sec x dx tan x 2t t e 1 3
2 arctan y

y tan 3 y s3
x

y x2

3x 2 2 dx 2x 8 dx

26.

y x3

3x 2 2 dx 2x 8

d
27.

y cot x ln sin x y0 y y s3
1 5

28.

y sin sat dt y2
2

3.

dt dy

4.

x4

dx
29.

3w
w

5.

y1
1

y2

6.

y x csc x cot x dx y0
41 dw 2 x dx x 2x x 2 dx

30.

x2

4x dx 1 dx 3 4 cot x dx cot x

7.

y1

3

r 4 ln r dr x 4x
3

8.

x x2 4x x x2

1 5 1

31.

dx
33.

1 1

32.

y y

s2x 2x
2 4

9.

y x2 y sin y y0

1 5
5

dx

10.

y x4

34.

dx
35.

1 4

11.

cos

d

12.

y sin x cos cos x y y0
s1 ln x dx x ln x
s2 2

dx
37.

y 1 x 8 sin x dx y0 y y
436.

y sin 4x cos 3x dx y0
4

13.

dx 1 x2
1 2

3 2

14.

cos2 tan2 d x dx

38.

tan 5 1

sec 3 d dy

15.

x s1 x2

dx

16.

x2 s1 x2 e 2t dt e 4t

dx

39.

1

x2 tan2 d

s1

x2

40.

y s4y 2 y x 2 tan y s1 y1 y
x x4 1
1

4y x dx

3

17.

y x sin 2 x dx y ex y t 3e y0 (1
1 ex

18.

y1
3

41.

42.

19.

dx dt sx )8 dx

20.

yesx dx y x sin y ln x 2
1

43.

y e x s1 y x 5e y x2
x
x3

e x dx dx

44.

e x dx ex dx ex a4 dx

21.

2t

22.

x dx 1 dx

45.

46.

23.

24.

47.

a dx a2

48.

SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS

SSSS

511

49.

y x s4x

1 1 1

dx dx

50.

y x 2 s4x yx y
x dx x
4

1

1 1

dx

67.

y1

3

arctanstdt st e 2x dx ex x 4x 2 1 2 x2 3 dx dx

68.

y1 y

1 2e x

e

x

dx

51.

y x s4x 2

52.

1

69.

y1

70.

ln x 1 dx x2 st dt 3 st dx e
x

53.

y x 2 sinh mx dx yx y x sx
3

54.

sin x 2 dx x ln x 1 dx x dx dx dx

71.

y x4 y
x

72.

y1 y ex y

55.

4

1 4 sx c dx dx dx

1

dx

56.

y sx 2

73.

74.

4

57.

58.

y x 2 ln 1yx y y y2
3

75.

y sin x sin 2x sin 3x dx y1
sx dx x3

76.

x2

bx sin 2x dx

59.

y e 3x y x 10

1 ex x4 16

60.

1
3 sx

77.

78.

y sin x

sec x cos 2x dx sec x sin x cos x dx cos 4 x
I I I I

61.

62.

x3 x 1
3 4

10

79.
I

y x sin 2 x cos x dx
I I I I I

80.
I

y sin 4 x
I

63.

y sxe sx dx y sx
1 1 sx dx

64.

ln tan x dx sin xcos x 1 du u2

81. The functions y

65.

66.

u3 u3

e x and y antiderivatives, but y 2x 2 2 x 2x 2 1 e x dx.

2

x 2e x don’t have elementary 2 1 e x does. Evaluate

2

|||| 7.6

Integration Using Tables and Computer Algebra Systems
In this section we describe how to use tables and computer algebra systems to integrate functions that have elementary antiderivatives. You shouldbear in mind, though, that even the most powerful computer algebra systems can’t ﬁnd explicit formulas for the antideriv2 atives of functions like e x or the other functions described at the end of Section 7.5.

Tables of Integrals
Tables of indeﬁnite integrals are very useful when we are confronted by an integral that is difﬁcult to evaluate by hand and we don’t have access to a computer...