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7.2 Antidifferentiation
by Substitution
What you will learn about . . .
Indefinite Integrals
Indefinite Integrals
• Leibniz Notation and Antiderivatives
• Substitution in Indefinite Integrals
• Substitution in Definite Integrals
and why . . .
If y = f(x) we can denote the derivative of f by either dy>dx or f¿(x). What can we use
to denote the antiderivative of f ? We have seen that the general solution to the differential
equation dy>dx = f(x) actually consists of an infinite family of functions of the form
F(x) + C, where F¿(x) = f(x). Both the name for this family of functions and the symbol we use to denote it are closely related to the definite integral because of the
Fundamental Theorem of Calculus.
•
Antidifferentiation techniques were
historically crucial for applying the
results of calculus.
DEFINITION Indefinite Integral
The family of all antiderivatives of a function f(x) is the indefinite integral of f
with respect to x and is denoted by 1 f(x)dx.
If F is any function such that F¿(x) = f(x), then 1 f(x)dx = F(x) + C, where
C is an arbitrary constant, called the constant of integration.
As in Chapter 6, the symbol 1 is an integral sign, the function f is the integrand of
the integral, and x is the variable of integration.
Notice that an indefinite integral is not at all like a definite integral, despite the similarities in notation and name. A definite integral is a number, the limit of a sequence of
Riemann sums. An indefinite integral is a family of functions having a common derivative.
If the Fundamental Theorem of Calculus had not provided such a dramatic link between
antiderivatives and integration, we would surely be using a different name and symbol for
the general antiderivative today.
EXAMPLE 1 Evaluating an Indefinite Integral
Evaluate 1 (x2 - sin x) dx.
SOLUTION
Evaluating this indefinite integral is just like solving the differential equation dy>dx =
x2 - sin x. Our past experience with derivatives leads us to conclude that
L
(x2 - sin x) dx =
x3
+ cos x + C
3
(as you can check by differentiating).
Now Try Exercise 3.
You have actually been finding antiderivatives since Section 6.3, so Example 1 should
hardly have seemed new. Indeed, each derivative formula in Chapters 3 and 4 could be
turned around to yield a corresponding indefinite integral formula. We list some of the
most useful such indefinite integral formulas on the next page. Be sure to familiarize yourself with these before moving on to the next section, in which function composition
becomes an issue. (Incidentally, it is in anticipation of the next section that we give some
of these formulas in terms of the variable u rather than x.)
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Properties of Indefinite Integrals
L
L
k f(x)dx = k
L
f(x) dx
(f(x) g(x)) dx =
L
for any constant k
f(x) dx L
g(x) dx
Power Formulas
L
un du =
un+1
+ C when n Z - 1
n + 1
1
du = ln ƒ u ƒ + C
Lu
(see Example 2)
L
u-1 du =
L
sin u du = - cos u + C
Trigonometric Formulas
L
L
L
cos u du = sin u + C
sec2 u du = tan u + C
L
sec u tan u du = sec u + C
L
csc2 u du = - cot u + C
csc u cot u du = - csc u + C
Exponential and Logarithmic Formulas
L
A Note on Absolute Value
Since the indefinite integral does not specify
a domain, you should always use the
absolute value when finding 1 1>u du. The
function ln u + C is only defined on positive
u-intervals, while the function ln ƒ u ƒ + C
is defined on both the positive and negative intervals in the domain of 1>u (see
Example 2).
eu du = eu + C
L
ln u du = u ln u - u + C
L
loga u du =
L
au du =
au
+ C
ln a
(See Example 2)
ln u
u ln u - u
du =
+ C
ln a
L ln a
EXAMPLE 2 Verifying Antiderivative Formulas
Verify the antiderivative formulas:
1
(a)
u-1 du =
du = ln ƒ u ƒ + C
u
L
L
(b)
L
ln u du = u ln u - u + C
SOLUTION
We can verify antiderivative formulas by differentiating.
d
1
d
1
(a) For u 7 0, we have
(ln ƒ u ƒ + C) =
(ln u + C) = + 0 = .
u
u
du
du
d
d
1
1
(ln ƒ u ƒ + C) =
(ln (-u) + C) =
(- 1) + 0 = .
For u 6 0, we have
-u
u
du
du
1
d
(ln ƒ u ƒ + C) = in either case, ln ƒ u ƒ + C is the general antiderivative of
Since
u
du
1
the function on its entire domain.
u
(b)
d
1
(u ln u - u + C) = 1 # ln u + u ¢ ≤ - 1 + 0 = ln u + 1 - 1 = ln u.
u
du
Now Try Exercise 11.
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Leibniz Notation and Antiderivatives
b
The appearance of the differential “dx” in the definite integral 1a f(x) dx is easily explained
n
by the fact that it is the limit of a Riemann sum of the form a f(xk) # ¢x (see Section 6.2).
k=1
The same “dx” almost seems unnecessary when we use the indefinite integral 1 f(x)dx to
represent the general antiderivative of f, but in fact it is quite useful for dealing with the
effects of the Chain Rule when function composition is involved. Exploration 1 will show
you why this is an important consideration.
EXPLORATION 1
Are 1 f (u) du and 1 f (u) dx the Same Thing?
Let u = x2 and let f(u) = u3.
1. Find 1 f(u) du as a function of u.
2. Use your answer to question 1 to write 1 f(u) du as a function of x.
3. Show that f(u) = x6 and find 1 f(u) dx as a function of x.
4. Are the answers to questions 2 and 3 the same?
Exploration 1 shows that the notation 1 f(u) is not sufficient to describe an antiderivative when u is a function of another variable. Just as du>du is different from du>dx when
differentiating, 1 f(u) du is different from 1 f(u) dx when antidifferentiating. We will
use this fact to our advantage in the next section, where the importance of “dx” or “du” in
the integral expression will become even more apparent.
EXAMPLE 3 Paying Attention to the Differential
Let f(x) = x3 + 1 and let u = x2. Find each of the following antiderivatives in terms
of x:
(a)
L
f(x) dx
(b)
L
f(u) du
(c)
L
f(u) dx
SOLUTION
(a)
(b)
(c)
L
f(x) dx =
L
f(u) du =
L
f(u) dx =
L
L
L
(x3 + 1) dx =
(u3 + 1) du =
(u3 + 1) dx =
x4
+ x + C
4
(x2)4
u4
x8
+ u + C =
+ x2 + C =
+ x2 + C
4
4
4
L
((x2)3 + 1) dx =
L
(x6 + 1) dx =
x7
+ x + C
7
Now Try Exercise 15.
Substitution in Indefinite Integrals
A change of variables can often turn an unfamiliar integral into one that we can evaluate.
The important point to remember is that it is not sufficient to change an integral of the form
1 f(x) dx into an integral of the form 1 g(u) dx. The differential matters. A complete
substitution changes the integral 1 f(x) dx into an integral of the form 1 g(u) du.
EXAMPLE 4 Using Substitution
Evaluate 1 sin x ecos x dx.
continued
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SOLUTION
Let u = cos x. Then du>dx = - sin x, from which we conclude that du = - sin x dx.
We rewrite the integral and proceed as follows:
L
sin x ecos x dx = = -
L
L
(-sin x)ecos x dx
ecos x # (-sin x) dx
eu du
L
= - eu + C
= -ecos x + C
= -
Substitute u for cos x and du for sin x dx.
Re-substitute cos x for u after antidifferentiating.
Now Try Exercise 19.
If you differentiate - ecos x + C, you will find that a factor of - sin x appears when you
apply the Chain Rule. The technique of antidifferentiation by substitution reverses that
effect by absorbing the -sin x into the differential du when you change 1 sin x ecos x dx
into - 1 eu du. That is why a “u-substitution” always involves a “du-substitution” to convert
the integral into a form ready for antidifferentiation.
EXAMPLE 5 Using Substitution
Evaluate 1 x2 25 + 2x3 dx.
SOLUTION
This integral invites the substitution u = 5 + 2x3, du = 6x2 dx.
L
x2 25 + 2x3 dx. =
L
(5 + 2x3)1>2 # x2 dx
=
1
(5 + 2x3)1>2 # 6x2 dx
6L
Set up the substitution with a factor of 6.
=
1
u1>2 du
6L
Substitute u for 5 + 2x3 and du for 6x2 dx.
=
1 2 3>2
¢ ≤u + C
6 3
=
1
(5 + 2x3)3>2 + C
9
Re-substitute after antidifferentiating.
Now Try Exercise 27.
EXAMPLE 6 Using Substitution
Evaluate 1 cot 7x dx.
SOLUTION
We do not recall a function whose derivative is cot 7x, but a basic trigonometric identity changes the integrand into a form that invites the substitution u = sin 7x, du =
7 cos 7x dx. We rewrite the integrand as shown below.
cos 7x
cot 7x dx =
dx
Trigonometric identity
L
L sin 7x
1 7 cos 7x dx Note that du = 7 cos 7x dx when u = sin 7x.
=
1
We multiply by # 7, or 1.
7 L sin 7x
continued
7
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1 du
7L u
1
= ln ƒ u ƒ + C
7
1
= ln ƒ sin 7x ƒ + C
7
=
Substitute u for 7 sin x and du for 7 cos 7x dx.
Notice the absolute value!
Resubstitute sin 7x for u after antidifferentiating.
Now Try Exercise 29.
EXAMPLE 7 Setting Up a Substitution with a Trigonometric Identity
Find the indefinite integrals. In each case you can use a trigonometric identity to set up
a substitution.
dx
cot2 3x dx (c)
cos3 x dx
(a)
(b)
2
L cos 2x
L
L
SOLUTION
dx
1
=
sec2 2x dx =
sec2 2x # 2dx
(a)
2
2
cos
2x
L
L
L
1
=
sec2 u du
2L
1
= tan u + C
2
1
= tan 2x + C
2
(b)
L
cot2 3x dx =
=
=
=
=
=
(c)
L
cos3 x dx =
(csc2 3x - 1) dx
L
1
(csc2 3x - 1) # 3 dx
3L
1
(csc2 u - 1) # du
3L
1
(-cot u - u) + C
3
1
(-cot 3x - 3x) + C
3
1
- cot 3x - x + C
3
L
=
L
=
L
Let u = 2x and du = 2 dx.
Resubstitute after antidifferentiating.
Let u = 3x and du = 3 dx.
Resubstitute after antidifferentiation.
(cos2 x) cos x dx
(1 - sin2 x) cos x dx
(1 - u2) du
u3
+ C
3
sin3 x
= sin x + C
3
Let u = sin x and du = cos x dx.
= u -
Resubstitute after antidifferentiating.
Now Try Exercise 47.
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Substitution in Definite Integrals
Antiderivatives play an important role when we evaluate a definite integral by the
Fundamental Theorem of Calculus, and so, consequently, does substitution. In fact, if we
make full use of our substitution of variables and change the interval of integration to
match the u-substitution in the integrand, we can avoid the “resubstitution” step in the previous four examples.
EXAMPLE 8 Evaluating a Definite Integral by Substitution
p>3
Evaluate
2
tan x sec x dx.
L0
SOLUTION
Let u = tan x and du = sec2 x dx.
Note also that u(0) = tan 0 = 0 and u(p>3) = tan(p>3) = 23.
So
p>3
L0
13
tan x sec2 x dx =
=
L0
u du
Substitute u-interval for x-interval.
u2 13
3
3
` = -0= .
2 0
2
2
Now Try Exercise 55.
EXAMPLE 9 That Absolute Value Again
1
Evaluate
x
dx.
L0 x - 4
2
SOLUTION
Let u = x2 - 4 and du = 2x dx. Then u(0) = 02 - 4 = - 4 and u(1) = 12 - 4 = -3.
So
1
1
x
2x dx
1
dx =
2
2 L0 x2 - 4
L0 x - 4
-3
=
1
du
2 L-4 u
=
-3
1
ln ƒ u ƒ `
2
-4
=
1
3
1
(ln 3 - ln 4) = ln ¢ ≤ .
2
2
4
Substitute u-interval for x-interval.
Notice that ln u would not have existed over the interval of integration [-4, -3]. The
absolute value in the antiderivative is important.
Now Try Exercise 63.
Finally, consider this historical note. The technique of u-substitution derived its importance from the fact that it was a powerful tool for antidifferentiation. Antidifferentiation
derived its importance from the Fundamental Theorem, which established it as the way to
evaluate definite integrals. Definite integrals derived their importance from real-world
applications. While the applications are no less important today, the fact that the definite
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integrals can be easily evaluated by technology has made the world less reliant on antidifferentiation, and hence less reliant on u-substitution. Consequently, you have seen in this
book only a sampling of the substitution tricks calculus students would have routinely
studied in the past. You may see more of them in a differential equations course.
Quick Review 7.2 (For help, go to Sections 4.1 and 4.4.)
Exercise numbers with a gray background indicate problems that the
authors have designed to be solved without a calculator.
In Exercises 1 and 2, evaluate the definite integral.
2
2.
L1
L0
In Exercises 3–10, find dy>dx.
2x - 1 dx
L2
8. y = ln sin x
9. y = ln (sec x + tan x)
x
3. y =
x
3t dt
6. y = sin2 (4x - 5)
7. y = ln cos x
5
x4 dx
1.
5. y = (x3 - 2x2 + 3)4
4. y =
L0
10. y = ln (csc x + cot x)
3t dt
Section 7.2 Exercises
In Exercises 1–6, find the indefinite integral.
1.
3.
5.
L
L
L
2
2.
(cos x - 3x ) dx
1
¢t 2
t2
≤ dt
4.
L
x
20.
-2
dx
21.
dt
(2ex + sec x tan x - 2x) dx
L
In Exercises 7–12, use differentiation to verify the antiderivative
formula.
L
csc2 u du = - cot u + C
1
9.
e2x dx = e2x + C
2
L
11.
1
L 1 + u2
du = tan
-1
8.
L
L
24.
L
1
L 1 - u2
2
17.
18.
19.
L
sin 3x dx,
L
L
sec 2x tan 2x dx,
L 21 - r3
u = 1 - cos
,
u = 1 - r3
t
2
u = y4 + 4y2 + 1
26.
L
sec2 (x + 2) dx
2tan x sec2 x dx
sec ¢ u +
p
p
≤ tan ¢ u + ≤ du
2
2
L
29.
L
tan(4x + 2) dx
30.
L
31.
L
cos (3z + 4) dz
32.
L
34.
L
36.
L sin2 3x
33.
35.
ln6x
dx
L x
L
s1/3 cos (s4/3 - 8) ds
sin (2t + 1)
37.
L cos (2t + 1)
39.
L x ln x
u = 2x2
u = 2x
9r2 dr
28.
u = 3x
x cos (2x2) dx,
t
dt,
2
dx
L
14. f(u) = u and u = x
In Exercises 17–24, use the indicated substitution to evaluate the
integral. Confirm your answer by differentiation.
2
22.
8(y4 + 4y2 + 1)2(y3 + 2y) dy,
27.
du = sin u + C
16. f(u) = sin u and u = 4x
t
2
L (1 - x)2
5
15. f(u) = eu and u = 7x
x
3
u =
¢ 1 - cos ≤ sin
25.
-1
In Exercises 13–16, verify that 1 f(u) du Z 1 f(u) dx
13. f(u) = 2u and u = x2 (x 7 0)
,
u = 7x - 2
In Exercises 25–46, use substitution to evaluate the integral.
csc u cot u = - csc u + C
1 x
5x dx =
10.
5 + C
ln 5
L
u + C 12.
L x2 + 9
23.
(3x4 - 2x-3 + sec2 x) dx
28(7x - 2)3 dx,
dx
L t2 + 1
6.
7.
L
2
dx
dt
3(sin x)-2 dx
2cot x csc2 x dx
x
x
tan7 ¢ ≤ sec2 ¢ ≤ dx
2
2
dx
6 cos t
38.
L (2 + sin t)2
40.
L
dt
tan2 x sec2 x dx
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Antidifferentiation by Substitution
Section 7.2
x dx
41.
Lx + 1
dx
43.
L cot 3x
45.
L
40 dx
42.
2
1
67.
L x + 25
dx
2
44.
x3
dx
L0 2x4 + 9
p/3
L 25x + 8
68.
(1 - cos 3x) sin 3x dx
Lp/6
69. Show that
sec x dx [Hint: Multiply the integrand by
y = ln `
sec x + tan x
sec x + tan x
L
csc x dx [Hint: Multiply the integrand by
(See the discussion following Example 4, Section 6.4.)
csc x + cot x
csc x + cot x
70. Show that
y = ln `
and then use a substitution to integrate the result.]
In Exercises 47–52, use the given trigonometric identity to set up a
u-substitution and then evaluate the indefinite integral.
3
47.
L
48.
L
2
2
sin 2x dx,
sin 2x = 1 - cos 2x
sec4 x dx,
sec2 x = 1 + tan2 x
49.
L
50.
L
51.
L
52.
L
p/4
71. True or False By u-substitution, 10
p/4 3
10 u du. Justify your answer.
4 cos2 x dx,
cos 2x = 2 cos2 x - 1
72. True or False If f is positive and differentiable on [a, b], then
tan4 x dx,
b f¿(x)dx
tan2 x = sec2 x - 1
(cos4 x - sin4 x)dx,
La
cos 2x = cos2 x - sin2 x
1
2y + 1 dy
L0
54.
L0
102u
du
L0 (1 + u3/2)2
56.
p
58.
2t5 + 2t (5t4 + 2) dt
60.
dx
x
+ 2
L0
dt
L1 t - 3
3
dx
-3
2u sin 2u du
5
62.
2
63.
cos
L0
dx
2x
- 3
L2
3p/4
64.
Lp/4
cot x dx
2
ex dx
x
L-1 x2 + 1
L0 3 + e
Two Routes to the Integral In Exercises 67 and 68, make a
substitution u = Á (an expression in x), du = Á . Then
65.
x dx
b. Justify your answer.
e
2
(C) ln ƒ cos x ƒ + C
(E) - ln ƒ cot x ƒ + C
2 2x
10 e
(B) e4 - 1
dx =
(C) e4 - 2
(D) 2e4 - 2
(E)
e4 - 1
2
5
p>6
7
61.
cos x
f(a)
(D) - ln ƒ cos x ƒ + C
4
dr
L-p 24 + 3 sin x
1
L0
5r
f(b)
73. Multiple Choice 1 tan x dx =
tan2 x
+ C (B) ln ƒ cot x ƒ + C
(A)
2
(A)
L-1 (4 + r2)2
f(x)
= ln a
74. Multiple Choice
r 21 - r2 dr
1
tan x sec2 x dx
L-p>4
1
59.
tan3 x sec2 x dx =
cos 2x = 1 - 2 sin2 x
0
57.
is the solution to the initial value problem
dy
= cot x, f(2) = 6.
dx
2 sin2 x dx,
3
55.
sin x
` +6
sin 2
Standardized Test Questions
In Exercises 53–66, make a u-substitution and integrate from u(a)
to u(b).
53.
cos 3
` +5
cos x
is the solution to the initial value problem
dy
= tan x, f(3) = 5.
dx
and then use a substitution to integrate the result.]
46.
343
66.
(a) integrate with respect to u from u(a) to u(b).
(b) find an antiderivative with respect to u, replace u by the
expression in x, then evaluate from a to b.
75. Multiple Choice If 13 f(x - a) dx = 7 where a is a constant,
5-a
then 13-a f(x) dx =
(A) 7 + a
(B) 7
(C) 7 - a
(D) a - 7
(E) - 7
76. Multiple Choice If the differential equation dy>dx = f(x)
leads to the slope field shown below, which of the following
could be 1 f(x) dx?
(A) sin x + C
(D) - cos x + C
(B) cos x + C
(E)
sin x
+ C
2
(C) - sin x + C
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Explorations
77. Constant of Integration Consider the integral
L
2x + 1 dx.
2
2x + 1 dx = (x + 1)3/2 + C.
(a) Show that
3
L
(b) Writing to Learn Explain why
x
y1 =
L0
x
2t + 1 dt
and
y2 =
L3
2t + 1 dt
are antiderivatives of 2x + 1.
(c) Use a table of values for y1 - y2 to find the value of C for
which y1 = y2 + C.
(c) Writing to Learn Explain why the different-looking
answers in parts (a) and (b) are actually equivalent.
80. Different Solutions? Consider the integral 1 2 sec2 x tan x dx.
(a) Evaluate the integral using the substitution u tan x.
(b) Evaluate the integral using the substitution u sec x.
(c) Writing to Learn Explain why the different-looking
answers in parts (a) and (b) are actually equivalent.
Extending the Ideas
81. Trigonometric Substitution Suppose u = sin-1 x. Then
cos u 7 0.
(a) Use the substitution x = sin u, dx = cos u du to show that
dx
3
C =
L0
2x + 1 dx.
78. Group Activity Making Connections Suppose that
f(x) dx = F(x) + C.
L
(a) Explain how you can use the derivative of F(x) + C to confirm the integration is correct.
(b) Explain how you can use a slope field of f and the graph of
y = F(x) to support your evaluation of the integral.
(c) Explain how you can use the graphs of y1 = F(x) and
x
y2 = 10 f(t) dt to support your evaluation of the integral.
(d) Explain how you can use a table of values for y1 - y2,
y1 and y2 defined as in part (c), to support your evaluation of
the integral.
(e) Explain how you can use graphs of f and NDER of F(x) to
support your evaluation of the integral.
x
(f) Illustrate parts (a)–(e) for f(x) =
.
2x2 + 1
79. Different Solutions? Consider the integral 1 2 sin x cos x dx.
(a) Evaluate the integral using the substitution u sin x.
(b) Evaluate the integral using the substitution u cos x.
=
L 21 - x2
(d) Writing to Learn Give a convincing argument that
L
1 du.
dx
1 du to show that
= sin-1x + C.
L
L 21 - x2
82. Trigonometric Substitution Suppose u = tan-1 x.
(b) Evaluate
(a) Use the substitution x = tan u, dx = sec2 u du to show that
dx
=
L 1 + x2
(b) Evaluate
L
L
1 du to show that
1 du.
dx
L 1 + x2
= tan-1 x + C.
83. Trigonometric Substitution Suppose 2x = sin y.
(a) Use the substitution x = sin2 y, dx = 2 sin y cos y dy to
show that
1/2
L0
p/4
2x dx
21 - x
=
L0
2 sin2 y dy.
(b) Use the identity given in Exercise 49 to evaluate the definite
integral without a calculator.
84. Trigonometric Substitution Suppose u = tan-1 x.
(a) Use the substitution x = tan u, dx = sec2 u du to show that
23
L0
p/3
dx
21 + x2
=
L0
sec u du.
(b) Use the hint in Exercise 45 to evaluate the definite integral
without a calculator.