October 28, 2011 16:20 c05 Sheet number 17 Page number 332 cyan magenta yellow black

332 Chapter 5 / Integration

66. Critique the following “proof” that an arbitrary constant 72. Recall that
must be zero:     d 1
0= x dx − x dx [sec−1 x] = √
dx |x| x 2 − 1
 
= (x − x) dx = 0 dx = C Use this to verify Formula 14 in Table 5.2.1.
73. The speed of sound in air at 0 ◦ C (or 273 K on the Kelvin
67. (a) Show that
scale) is 1087 ft/s, but the speed v increases as the temper-
F (x) = tan−1 x and G(x) = − tan−1 (1/x) ature T rises. Experimentation has shown that the rate of
differ by a constant on the interval (0, +⬁) by show- change of v with respect to T is
ing that they are antiderivatives of the same func- dv 1087 −1/2
tion. = √ T
dT 2 273
(b) Find the constant C such that F (x) − G(x) = C by
evaluating the functions F (x) and G(x) at a partic- where v is in feet per second and T is in kelvins (K). Find
ular value of x. a formula that expresses v as a function of T .
(c) Check your answer to part (b) by using trigonomet-
74. Suppose that a uniform metal rod 50 cm long is insulated
ric identities.
laterally, and the temperatures at the exposed ends are main-
68. Let F and G be the functions defined by
 tained at 25 ◦ C and 85 ◦ C, respectively. Assume that an x-
x 2 + 3x x + 3, x > 0 axis is chosen as in the accompanying figure and that the
F (x) = and G(x) =
x x, x<0 temperature T (x) satisfies the equation
(a) Show that F and G have the same derivative.
(b) Show that G(x)  = F (x) + C for any constant C. d 2T
=0
(c) Do parts (a) and (b) contradict Theorem 5.2.2? Ex- dx 2
plain.
Find T (x) for 0 ≤ x ≤ 50.

69–70 Use a trigonometric identity to evaluate the integral. ■ 25°C 85°C x

 
0 50 Figure Ex-74
69. tan2 x dx 70. cot2 x dx
75. Writing What is an initial-value problem? Describe the
71. Use the identities cos 2θ = 1 − 2 sin2 θ = 2 cos2 θ − 1 to
sequence of steps for solving an initial-value problem.
helpevaluate the integrals 
76. Writing What is a slope field? How are slope fields and
(a) sin2 (x /2) dx (b) cos2 (x /2) dx
integral curves related?

✔QUICK CHECK ANSWERS 5.2

 
1 √
1. F  (x) = f(x) 2. (a) √ dx = x + C (b) 4e4x dx = e4x + C
2 x
3. (a) 41 x 4 + 21 x 2 + 5x + C (b) tan x + csc x + C 4. 2x + 1; x 2 + x + 3 5. 0; − 23

5.3 INTEGRATION BY SUBSTITUTION

In this section we will study a technique, called substitution, that can often be used to
transform complicated integration problems into simpler ones.

u-SUBSTITUTION
The method of substitution can be motivated by examining the chain rule from the viewpoint
of antidifferentiation. For this purpose, suppose that F is an antiderivative of f and that
g is a differentiable function. The chain rule implies that the derivative of F (g(x)) can be
expressed as d
[F (g(x))] = F  (g(x))g  (x)
dx
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5.3 Integration by Substitution 333

which we can write in integral form as


F  (g(x))g  (x) dx = F (g(x)) + C (1)

or since F is an antiderivative of f ,

f(g(x))g  (x) dx = F (g(x)) + C (2)

For our purposes it will be useful to let u = g(x) and to write du/dx = g  (x) in the differ-
ential form du = g  (x) dx. With this notation (2) can be expressed as

f(u) du = F (u) + C (3)

The process of evaluating an integral of form (2) by converting it into form (3) with the
substitution
u = g(x) and du = g  (x) dx
is called the method of u-substitution. Here our emphasis is not on the interpretation of
the expression du = g  (x) dx. Rather, the differential notation serves primarily as a useful
“bookkeeping” device for the method of u-substitution. The following example illustrates
how the method works.

Example 1 Evaluate (x 2 + 1)50 · 2x dx.

Solution. If we let u = x 2 + 1, then du/dx = 2x, which implies that du = 2x dx. Thus,
the given integral can be written as
 
u51 (x 2 + 1)51
(x 2 + 1)50 · 2x dx = u50 du = +C = +C
51 51

It is important to realize that in the method of u-substitution you have control over the
choice of u, but once you make that choice you have no control over the resulting expres-
sion for du. Thus, in the last example we chose u = x 2 + 1 but du = 2x dx was computed.
Fortunately, our choice of u, combined with the computed du, worked out perfectly to pro-
duce an integral involving u that was easy to evaluate. However, in general, the method of
u-substitution will fail if the chosen u and the computed du cannot be used to produce an inte-
grand in which no expressions involving x remain, or if you cannot evaluate the resulting in-
tegral. Thus, for example, the substitution u = x 2 , du = 2x dx will not work for the integral

2x sin x 4 dx

because this substitution results in the integral


sin u2 du

which still cannot be evaluated in terms of familiar functions.

In general, there are no hard and fast rules for choosing u, and in some problems no
choice of u will work. In such cases other methods need to be used, some of which will be
discussed later. Making appropriate choices for u will come with experience, but you may
find the following guidelines, combined with a mastery of the basic integrals in Table 5.2.1,
helpful.
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334 Chapter 5 / Integration

Guidelines for u-Substitution

Step 1. Look for some composition f(g(x)) within the integrand for which the substi-
tution
u = g(x), du = g  (x) dx
produces an integral that is expressed entirely in terms of u and its differential
du. This may or may not be possible.
Step 2. If you are successful in Step 1, then try to evaluate the resulting integral in terms
of u. Again, this may or may not be possible.
Step 3. If you are successful in Step 2, then replace u by g(x) to express your final
answer in terms of x.

EASY TO RECOGNIZE SUBSTITUTIONS

The easiest substitutions occur when the integrand is the derivative of a known function,
except for a constant added to or subtracted from the independent variable.

Example 2
 
sin(x + 9) dx = sin u du = − cos u + C = − cos(x + 9) + C

u=x+9
du = 1 · dx = dx
 
u24 (x − 8)24
(x − 8) dx =
23
u23 du = +C = +C
24 24
u=x−8
du = 1 · dx = dx

Another easy u-substitution occurs when the integrand is the derivative of a known
function, except for a constant that multiplies or divides the independent variable. The
following example illustrates two ways to evaluate such integrals.

Example 3 Evaluate cos 5x dx.

Solution.
  
1 1 1 1
cos 5x dx = (cos u) · du = cos u du = sin u + C = sin 5x + C
5 5 5 5
u = 5x
du = 5 dx or dx = 1
5 du

Alternative Solution. There is a variation of the preceding method that some people
prefer. The substitution u = 5x requires du = 5 dx. If there were a factor of 5 in the inte-
grand, then we could group the 5 and dx together to form the du required by the substitution.
Since there is no factor of 5, we will insert one and compensate by putting a factor of 15 in
front of the integral. The computations are as follows:
  
1 1 1 1
cos 5x dx = cos 5x · 5 dx = cos u du = sin u + C = sin 5x + C
5 5 5 5
u = 5x
du = 5 dx
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5.3 Integration by Substitution 335

More generally, if the integrand is a composition of the form f(ax + b), where f(x) is
an easy to integrate function, then the substitution u = ax + b, du = a dx will work.

Example 4
    −4
dx 3 du −5 3 −4 3 1
1 5 = = 3 u du = − u + C = − x − 8 +C
x−8 u5 4 4 3
3
u = 13 x − 8
du = 1
3dx or dx = 3 du


dx
Example 5 Evaluate .
1 + 3x 2

Solution. Substituting √ √
u= 3x, du = 3 dx
yields
  √
dx 1 du 1 1
=√ = √ tan−1 u + C = √ tan−1 ( 3x) + C
1 + 3x 2
3 1+u 2
3 3

With the help of Theorem 5.2.3, a complicated integral can sometimes be computed by
expressing it as a sum of simpler integrals.

Example 6
    
1 dx
+ sec2 πx dx = + sec2 πx dx
x x

= ln |x| + sec2 πx dx


1
= ln |x| + sec2 u du
π u = πx
1
du = π dx or dx = du
π

1 1
= ln |x| + tan u + C = ln |x| + tan πx + C
π π

The next four examples illustrate a substitution u = g(x) where g(x) is a nonlinear
function.

Example 7 Evaluate sin2 x cos x dx.

Solution. If we let u = sin x, then

du
= cos x, so du = cos x dx
dx
Thus,
 
u3 sin3 x
sin2 x cos x dx = u2 du = +C = +C
3 3
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336 Chapter 5 / Integration

 √
e x
Example 8 Evaluate √ dx.
x
√
Solution. If we let u = x, then
du 1 1 1
= √ , so du = √ dx or 2 du = √ dx
dx 2 x 2 x x
Thus,
 √   √
e x
√ dx = 2eu du = 2 eu du = 2eu + C = 2e x
+C
x
 
3
Example 9 Evaluate t 4 3 − 5t 5 dt.

Solution.
  

3 1 √ 1
t 4 3 − 5t 5 dt = − 3
u du = − u1/3 du
25 25 u = 3 − 5t 5
du = −25t 4 dt or − 25
1
du = t 4 dt

1 u4/3 3  4/3
=− +C =− 3 − 5t 5 +C
/
25 4 3 100


ex
Example 10 Evaluate √ dx.
1 − e2x

Solution. Substituting
u = ex , du = ex dx
yields  
ex du
√ dx = √ = sin−1 u + C = sin−1 (ex ) + C
1− e2x 1 − u2

LESS APPARENT SUBSTITUTIONS

The method of substitution is relatively straightforward, provided the integrand contains
an easily recognized composition f(g(x)) and the remainder of the integrand is a constant
multiple of g  (x). If this is not the case, the method may still apply but may require more
computation.

√
Example 11 Evaluate x 2 x − 1 dx.

√
Solution. The composition x − 1 suggests the substitution
u=x−1 so that du = dx (4)
From the first equality in (4)
x 2 = (u + 1)2 = u2 + 2u + 1
so that
  
√ √
x 2
x − 1 dx = (u + 2u + 1) u du =
2
(u5/2 + 2u3/2 + u1/2 ) du

= 27 u7/2 + 45 u5/2 + 23 u3/2 + C

= 27 (x − 1)7/2 + 45 (x − 1)5/2 + 23 (x − 1)3/2 + C

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5.3 Integration by Substitution 337


Example 12 Evaluate cos3 x dx.

Solution. The only compositions in the integrand that suggest themselves are
cos3 x = (cos x)3 and cos2 x = (cos x)2
However, neither the substitution u = cos x nor the substitution u = cos2 x work (verify).
In this case, an appropriate substitution is not suggested by the composition contained in
the integrand. On the other hand, note from Equation (2) that the derivative g  (x) appears
as a factor in the integrand. This suggests that we write
 
cos x dx = cos2 x cos x dx
3

and solve the equation du = cos x dx for u = sin x. Since sin2 x + cos2 x = 1, we then
have
   
cos3 x dx = cos2 x cos x dx = (1 − sin2 x) cos x dx = (1 − u2 ) du

u3 1
=u− + C = sin x − sin3 x + C
3 3


dx
Example 13 Evaluate dx, where a  = 0 is a constant.
a2 + x2

Solution. Some simple algebra and an appropriate u-substitution will allow us to use
Formula 12 in Table 5.2.1.
  
dx a(dx /a) 1 dx /a u = x /a
= =
a2 + x 2 a 2 (1 + (x /a)2 ) a 1 + (x /a)2 du = dx /a

1 du 1 1 x
= = tan−1 u + C = tan−1 + C
a 1 + u2 a a a

The method of Example 13 leads to the following generalizations of Formulas 12, 13,
and 14 in Table 5.2.1 for a > 0:

du 1 u
= tan−1 + C (5)
a 2 + u2 a a

du u
 = sin−1 + C (6)
a 2 − u2 a
 u
du 1  
 = sec−1   + C (7)
u u −a 2 2 a a


dx
Example 14 Evaluate  .
2 − x2
√
Solution. Applying (6) with u = x and a = 2 yields

dx x
 = sin−1 √ + C
2 − x2 2
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338 Chapter 5 / Integration

T E C H N O LO GY M A ST E R Y INTEGRATION USING COMPUTER ALGEBRA SYSTEMS

The advent of computer algebra systems has made it possible to evaluate many kinds of
If you have a CAS, use it to calculate the
integrals that would be laborious to evaluate by hand. For example, a handheld calculator
integrals in the examples in this sec-
tion. If your CAS produces an answer evaluated the integral
that is different from the one in the text, 
5x 2 3(x + 1)2/3 (5x 2 − 6x + 9)
then confirm algebraically that the two dx = +C
answers agree. Also, explore the effect (1 + x)1/3 8
of using the CAS to simplify the expres-
sions it produces for the integrals.
in about a second. The computer algebra system Mathematica, running on a personal
computer, required even less time to evaluate this same integral. However, just as one
would not want to rely on a calculator to compute 2 + 2, so one would not want to use
a CAS to integrate a simple function such as f(x) = x 2 . Thus, even if you have a CAS,
you will want to develop a reasonable level of competence in evaluating basic integrals.
Moreover, the mathematical techniques that we will introduce for evaluating basic integrals
are precisely the techniques that computer algebra systems use to evaluate more complicated
integrals.

✔QUICK CHECK EXERCISES 5.3 (See page 340 for answers.)

1. Indicate the u-substitution. 2. Supply the missing integrand corresponding to the indicated
 
u-substitution.
(a) 3x (1 + x ) dx = u25 du
2 3 25
if u =  
−1/3
(a) 5(5x − 3) dx = du; u = 5x − 3
and du = .
   
(b) 2x sin x 2 dx = sin u du if u = and (b) (3 − tan x) sec2 x dx = du;
du = . u = 3 − tan x
    √ 
18x
=
1
du if u =
3
8+ x √
(c)
1 + 9x 2
dx
u
and (c) √ dx = du; u = 8 + x
x
du = .  
  (d) e3x dx = du; u = 3x
3 1
(d) dx = du if u = and
1 + 9x 2 1 + u2
du = .

EXERCISE SET 5.3 Graphing Utility C CAS


1–12 Evaluate the integrals using the indicated substitutions. 5. (a) cot x csc2 x dx; u = cot x
 ■ 
1. (a) 2x(x + 1) dx; u = x + 1
2 23 2 (b) (1 + sin t)9 cos t dt; u = 1 + sin t
  
(b) cos3 x sin x dx; u = cos x 6. (a) cos 2x dx; u = 2x (b) x sec2 x 2 dx; u = x 2
 
1 √ √ √
2. (a) √ sin x dx; u = x 7. (a) x 2 1 + x dx; u = 1 + x
 x 
3x dx
(b)  ; u = 4x 2 + 5 (b) [csc(sin x)]2 cos x dx; u = sin x
4x + 5
2


3. (a) sec2 (4x + 1) dx; u = 4x + 1 8. (a) sin(x − π) dx; u = x − π
  
5x 4
(b) y 1 + 2y 2 dy; u = 1 + 2y 2 (b) dx; u = x 5 + 1
(x 5 + 1)2
 √ 
dx
4. (a) sin πθ cos πθ dθ ; u = sin πθ 9. (a) ; u = ln x
  ln x
x
(b) (2x + 7)(x 2 + 7x + 3)4/5 dx; u = x 2 + 7x + 3 (b) e−5x dx; u = −5x
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5.3 Integration by Substitution 339

  
sin 3θ √
10. (a) dθ; u = 1 + cos 3θ 43. cos 4θ 2 − sin 4θ dθ 44. tan3 5x sec2 5x dx
 1 + cos 3θ
ex  
(b) dx; u = 1 + ex sec2 x dx sin θ
1 + ex 45. √ 46. 2θ +1
dθ
  1 − tan x
2
 cos
x 2 dx
11. (a) ; u = x3 47. sec3 2x tan 2x dx 48. [sin(sin θ)] cos θ dθ
 1 + x6  
dx dx √
(b)  ; u = ln x 49. 50. ex dx
x 1 − (ln x)2 e x
   √
dx dx e 2y+1
12. (a) √ ; u = 3x 51. √ (2√x) 52. √ dy
 x 9x 2 − 1 xe 2y + 1
 
dx √ y √
(b) √ ; u= x 53. √ dy 54. x 4 − x dx
x(1 + x)  2y + 1
F O C U S O N C O N C E P TS 55. sin3 2θ dθ

13. Explain the connection between the chain rule for dif- 56. sec4 3θ dθ [Hint: Apply a trigonometric identity.]
ferentiation and the method of u-substitution for inte-
gration. 57–60 Evaluate each integral by first modifying the form of
14. Explain how the substitution u = ax + b helps to per- the integrand and then making an appropriate substitution, if
form an integration in which the integrand is f(ax + b),  ■
needed. 
where f(x) is an easy to integrate function. t +1
57. dt 58. e2 ln x dx
 t 
15–56 Evaluate the integrals using appropriate substitutions. 59. [ln(ex ) + ln(e−x )] dx 60. cot x dx
■
  
15. (4x − 3)9 dx 16. x 3 5 + x 4 dx 61–62 Evaluate the integrals with the aid of Formulas (5), (6),
  and (7). ■
x   
17. sin 7x dx 18. cos dx dx dx dx
3 61. (a)  (b) (c) 
9−x 5 + x 2
   x x −π
2 2
 
19. sec 4x tan 4x dx 20. sec2 5x dx ex dx dy
62. (a) dx (b)  (c) 
  4 + e2x 9 − 4x 2 y 5y 2 − 3
dx
21. e2x dx 22.
2x 63–65 Evaluate the integrals assuming that n is a positive in-
 
dx dx teger and b = 0. ■
23.  24.  
1 + 16x 2 √
1 − 4x 2 63. (a + bx)n dx 64.
n
a + bx dx
  
x 
25. t 7t 2 + 12 dt 26.  dx
4 − 5x 2 65. sinn (a + bx) cos(a + bx) dx
 
6 x2 + 1 66. Use a CAS to check the answers you obtained in Exercises
27. dx 28.  dx C
(1 − 2x)3 x 3 + 3x 63–65. If the answer produced by the CAS does not match
  yours, show that the two answers are equivalent. [Sugges-
x3 sin(1/x)
29. dx 30. dx tion: Mathematica users may find it helpful to apply the
(5x 4 + 2)3 3x 2
  Simplify command to the answer.]
4
31. esin x cos x dx 32. x 3 ex dx
  x F O C U S O N C O N C E P TS
2 −2x 3 e + e−x 
33. x e dx 34. dx 67. (a) Evaluate the integral sin x cos x dx by two meth-
ex − e−x
  ods: first by letting u = sin x, and then by letting
ex t
35. dx 36. dt u = cos x.
1 + e2x t4 + 1
  √ (b) Explain why the two apparently different answers
sin(5/x) sec2 ( x) obtained in part (a) are really equivalent.
37. dx 38. √ dx 
x2 x
  68. (a) Evaluate the integral (5x − 1)2 dx by two meth-
39. cos4 3t sin 3t dt 40. cos 2t sin5 2t dt ods: first square and integrate, then let u = 5x − 1.
  (b) Explain why the two apparently different answers
cos 4θ obtained in part (a) are really equivalent.
41. x sec2 (x 2 ) dx 42. dθ
(1 + 2 sin 4θ )4