Download E INTEGRATION STRATEGIES

INTEGRATION
E
STRATEGIES
he set of exercises in this section consists of an assortment of indefinite integrals that
can be evaluated using the techniques developed in Chapter 8. Your task is to decide
which techniques are best suited to the given integral and to carry out the computation.
Let’s review the tools at our disposal. The two most powerful tools are substitution
and integration by parts. In addition, we have special methods for treating specific types
of integrals:
T
(A) Trigonometric integrals: use trigonometric identities and reduction formulas (Section 8.3)
√
√
(B) Square root expressions x 2 ± a 2 and a 2 − x 2 : use trigonometric substitution
(Section 8.4)
(C) Rational functions: use the method of partial fractions (Section 8.5)
When confronted with an integration problem, you may ask yourself the following questions:
1. Can the integral be evaluated directly (perhaps after algebraic manipulation or a substitution)?
2. Is the integral amenable to integration by parts (or perhaps substitution followed by
integration by parts)?
3. Does the integral fall into one of the types (A)–(C) described above? Or can it be
reduced to one of these types by a substitution?
You may also refer to the Table of Integrals at the back of the book. This table
contains a list of standard integration formulas, including reduction formulas for trigonometric, exponential, and inverse trigonometric integrals. Sometimes it is necessary to
transform your integral before applying one of the formulas listed in the table. In all
cases, be on the lookout for the simplest method.
E X A M P L E 1 Look for the Simplest Method
(a)
x7 dx
x4 − 1
and
(b)
x2 dx
x4 − 1
Evaluate the integrals
.
Solution At first glance, both integrals appear to be candidates for the method of partial
fractions. However, the first integral may be treated much more simply using the substitution u = x 4 − 1.
E1
E2
I N T E G R AT I O N S T R AT E G I E S
APPENDIX E
(a) Let u = x 4 − 1 and du = 4x 3 d x. Then x 4 = u + 1 and
x7 d x = x4x3 d x =
1
(u + 1) du
4
(u + 1) du
u
1
1
=
1+
du
4
u
x7 dx
1
=
4
4
x −1
1
(simplify algebraically)
1
1
u + ln |u| + C
4
4
1
1 4
= (x − 1) + ln |x 4 − 1| + C
4
4
=
We may absorb the term − 14 into the constant C and write our result as
x7 dx
1
1
= x 4 + ln |x 4 − 1| + C
4
4
x4 − 1
The substitution u = x 4 − 1 is not effective
in integral (b) of Example 1 because
(b) Direct substitution cannot be used to evaluate the second integral. We must find the
partial fraction decomposition using the factorization x 4 − 1 = (x − 1)(x + 1)(x 2 + 1):
1
x 2 d x = x −1 x 3 d x = (u + 1)−1/4 du
4
1
1
1
x2
4
4
2
=
−
+
x − 1 x + 1 x2 + 1
x4 − 1
dx
dx
dx
x2 dx
1
1
1
=
−
+
4
2
4
x −1 4
x +1 2
x −1
x +1
The substitution does not lead to a simpler
integral:
1
x2 dx
=
4
x4 − 1
(u + 1)−1/4 du
u
=
1
1
1
ln |x − 1| − ln |x + 1| + tan−1 x + C
4
4
2
E X A M P L E 2 Find a Simplifying Substitution Find approaches to evaluate
(a)
ex d x
√
1 − e2x
and (b)
e2x d x
.
√
1 − e2x
Solution Substitution is a good first step for both of these integrals.
(a) Both e x and e2x appear in the first integral, so it makes sense to try
u = ex ,
du = e x d x
Note that u 2 = (e x )2 = e2x . Thus
ex d x
du
=
= sin−1 u + C = sin−1 (e x ) + C
√
√
2x
1−e
1 − u2
(b) Only e2x appears in the second integral, so we may try u = e2x :
u = e2x ,
e2x d x
=
√
1 − e2x
du = 2e2x d x,
1
du
1
=
√2
2
1−u
√
1
= − (1 − e2x )1/2 + C
2
1
du = e2x d x
2
du
1−u
1
= − (1 − u)1/2 + C
2
APPENDIX E
I N T E G R AT I O N S T R AT E G I E S
E3
An even better substitution is u = 1 − e2x , du = −2e2x d x:
1
e2x d x
1
1
2 du
√
=−
√ = − u 1/2 + C = − (1 − e2x )1/2 + C
2
2
u
1 − e2x
E X A M P L E 3 Evaluate
e3x d x
.
e2x + 1
Solution It may not be obvious which substitution to use for
e3x d x
.
e2x + 1
First approach. One possible choice is u = e2x . Then du = 2e2x d x and since u 1/2 =
(e2x )1/2 = e x , we obtain
3x
1 1/2
du
1
e dx
2u
=
e3x d x = e x e2x d x = u 1/2 du ⇒
2x
2
u+1
e +1
Next, apply the substitution w = u 1/2 , dw = 12 u −1/2 du:
3x
1 1/2
du
e dx
w2 dw
2u
=
=
u+1
e2x + 1
w2 + 1
Rather than continue, we consider another approach that requires just one substitution.
A better approach. Let u = e x , du = e x d x. Noting that e2x = (e x )2 = u 2 , we obtain
3x
e dx
u 2 du
=
e3x d x = e2x (e x d x) = u 2 du ⇒
2x
e +1
u2 + 1
du
REMINDER A rational function
f (x) = P(x)/Q(x) is called “improper” if
deg(P(x)) ≥ deg(Q(x)).
The integrand on the right is an improper rational function, so we must rewrite the integrand first using long division (or in this case, simple algebra):
u2
(u 2 + 1) − 1
1
=
=1− 2
2
2
u +1
u +1
u +1
3x
e dx
1
=
du = u − tan−1 u + C = e x − tan−1 (e x ) + C
1− 2
e2x + 1
u +1
Often there is more than one way to evaluate an integral and in some cases, different
methods lead to different but equivalent expressions for the antiderivative. Showing that
the two answers are equivalent usually requires algebra or identities involving trigonometric functions or logarithms. Keep this in mind when comparing your answer to an
answer produced by a computer algebra system.
E X A M P L E 4 Find an approach to evaluate
(cos x − sin 2x)2 d x.
Solution The integrand involves powers of sine and cosine, so we may expect trigonometric identities or formulas from the Table of Integrals to play a role.
First approach. Use the identity sin 2x = 2 sin x cos x:
(cos x − sin 2x)2 d x = (cos x − 2 sin x cos x)2 d x
=
(cos2 x − 4 sin x cos2 x + 4 sin2 x cos2 x) d x
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APPENDIX E
I N T E G R AT I O N S T R AT E G I E S
Each term in the integrand may be computed using the methods of Section 8.3 or the
reduction formulas (35) and (57) in the Table of Integrals at the back of the book. The
result is
1
4
(cos x − sin 2x)2 d x = x + sin 2x + cos3 x − sin x cos3 x + C
2
2
3
Alternate approach. Expand the integrand directly:
(cos x − sin 2x)2 d x = (cos2 x − 2 cos x sin 2x + sin2 2x) d x
=
cos2 x d x − 2
cos x sin 2x d x +
sin2 2x d x
These three integrals may be evaluated using formulas (34), (35), and (52). The result is
1
1
1
(cos x − sin 2x)2 d x = x + cos x + sin 2x − sin 4x + cos 3x + C
3
4
8
3
We can use the addition formulas for sine and cosine to show (2) and (3) are equivalent.
REMINDER Integration by Parts formula
uv d x = uv − u v d x
In many cases, integration by parts may be used to evaluate or simplify integrals
involving x n times a transcendental function. In Example 6 of Section 8.2, we used integration by parts with u = x n , v = e x to obtain the reduction formula
n x
n x
x e d x = x e − n x n−1 e x d x
However, finding the best choice in integration by parts often requires trial and error.
E X A M P L E 5 Trial and Error Find an approach to evaluate x 2 ln2 x d x.
Solution First try. We might be tempted to use integration by parts with u = x 2 and
v = ln2 x. However, this choice does not simplify our
problem because we would have to
find an antiderivative v of ln2 x and then compute
u v d x (which is more complicated
than our original integral).
Second try. The following choice leads to a simpler integral:
u = ln2 x,
v = x 2
1
v = x3
3
1
2
x 2 ln x d x
x 2 ln2 x d x = x 3 ln2 x −
3
3
u = 2(ln x)x −1 ,
Now we may evaluate the integral on the right using integration by parts, or formula (22)
in the Table of Integrals with n = 2:
1
x n ln x d x =
(n + 1)x n+1 ln x − x n+1
(n + 1)2
2
1 3
1
x 2 ln2 x d x = x 3 ln2 x −
3x ln x − x 3 + C
3
3
9
=
2
2
1 3 2
x ln x − x 3 ln x + x 3 + C
3
9
27
APPENDIX E
Standard references such as Table of
Integrals, Series, and Products by
I. Gradshteyn, I. Ryzhik, and A. Jeffrey
(Academic Press, 2000) contain hundreds
of pages of integral formulas. Computer
algebra systems are also a valuable
resource. These systems make use of
tables stored in memory together with
sophisticated algorithms for finding
antiderivatives.
I N T E G R AT I O N S T R AT E G I E S
E5
Finally, keep in mind that many integrals arising in practice cannot be evaluated
in “elementary terms,” that is, in terms of the basic classes of functions (polynomial,
rational, trigonometric, algebraic, exponential, and logarithm functions). However, it is
not always evident whether or not √
a given integral can be evaluated in elementary terms.
For example, the substitution u = x followed by an integration by parts yields
√
√
√
√
cos x d x = 2 cos x + 2 x sin x + C
Compare this with the following “Fresnel integral”:
cos(x 2 ) d x
This integral plays an important role in optics, but it is known that no elementary expression for an antiderivative exists. When faced with an integral that resists all of your efforts,
consult a more extensive table of integrals or use a computer algebra system. If this fails,
the reason may be that the antiderivative does not have an elementary expression.
Exercises
Evaluate the integral using any method or combination of methods
covered in the text.
x +4
dx
1.
√
x −2
x9
2.
dx
5
x −1
3.
sin 2x ln(cos 2x) d x
4.
√
ex ex + 1 d x
5.
x5
x3 − 1
√
6.
√
7.
√
8.
9.
11.
15.
17.
cos7 x d x
dx
x
x +2
x2
19.
x +1
20.
dx
x 2 − 16 d x
(sin x + cos 2x)2 d x
√
1 + x dx
x 11
dx
x4 − 1
x5 dx
x4 − 1
tan x sec5/4 x d x
21.
dx
e x e2x − 1 d x
18.
x
dx
x −1
dx
x(x 2 − 6x − 7)
sin5 x cos2 x d x
16.
(3 sec x − cos x)2 d x
22.
x 3 ln x d x
23.
x 3 ln2 x d x
24.
25.
√
ex + 1 d x
12.
ln(x 2 + 9) d x
14.
10.
x ln(x + 12) d x
13.
sin2 x tan x d x
26.
dx
x 2 − 36
E6
APPENDIX E
I N T E G R AT I O N S T R AT E G I E S
27.
28.
29.
30.
e x ln(1 + e x ) d x
48.
(3 − x 2 ) d x
x(x 2 − 1)
49.
(x 2 + 3x + 1) d x
x3 + x2
51.
32.
33.
34.
35.
36.
e3x d x
1 − e2x
55.
dx
x 1/3 + 1
56.
40.
58.
x ln(x 2 + 4) d x
59.
(x 2 + 2x)e x d x
41.
dx
16 − x 2
43.
1 + x 1/3 d x
60.
61.
63.
45.
46.
47.
64.
65.
66.
x 2 cos−1 x d x
67.
cos(x −1 ) d x
x3
68.
x 4x 2 − 9
69.
(x 1/2 + x −1/2 )
dx
x2 − 1
sin 2x d x
sin2 x + 1
ln(tan x)
dx
sin x cos x
x2 dx
(x + 1)1/3
ln(tan x)
dx
cos2 x
sec3 x d x
dx
x2 dx
√
x +1
tan3/4 x sec2 x d x
62.
dx
x 2 + 2x + 3
dx
ex + 1
sec3 4x d x
42.
44.
sin 9x sin 3x d x
x tan−1 3x d x
x 3 tan−1 x d x
sin 5x cos 2x d x
57.
x 5 ln x d x
x 2 ln(x − 1) d x
cos(tan−1 x + 1)
dx
x2 + 1
39.
tan6 x d x
54.
(9x + 18) d x
(x + 1)(x 2 − 7x + 10)
x sin 4x d x
x2 + x + 1
dx
x +2
53.
38.
dx
√
1+ x
(1 + x 1/3 )20 d x
52.
37.
sin 4x cos 12x d x
50.
1 − e4x
31.
e1/x d x
x2
e2x d x
sin x cos 3x d x
sinh−1
x
dx
3
sec2 x d x
1 + tan x
sin2 x sin 2x d x
APPENDIX E
cot4 x sec2 x d x
70.
86.
cosh2 5x d x
71.
87.
x tanh(x 2 ) d x
72.
73.
tan 3x sec 3x d x
74.
89.
sinh 2x cosh x d x
90.
x sech2 x d x
75.
91.
x tanh2 x d x
76.
√
78.
79.
x +1
dx
x
x2 dx
x2 + 4
81.
82.
83.
84.
dx
x2 x2 − 1
x 2 sin2 x d x
96.
97.
x 2 + 2x + 3
x 5 ln2 x d x
98.
x tan−1 3x d x
99.
ln x d x
(x + 2)2
dx
x2 + x + 1
94.
dx
dx
x(x 2 + 2x + 5)
x sin x cos x d x
95.
x2 dx
(x 2 + 9)2
x 3 − 4x + 2
dx
x +1
93.
x ln3 x d x
80.
dx
x 2 + 2x − 1
xe x sin x d x
92.
tan 3x sec2 3x d x
77.
85.
88.
dx
e x + 4e2x
100.
x5 dx
4 − x2
x4 dx
4 − x2
x cos−1 x d x
(sin−1 x)2 d x
x2 dx
x2 + 5
ln(x + x −1 ) d x
I N T E G R AT I O N S T R AT E G I E S
E7
E8
APPENDIX E
I N T E G R AT I O N S T R AT E G I E S
Solutions to Appendix E Odd Exercises
1.
3
√
2
√
√
2 √
x − 2 + 6 x − 2 + 32 x − 2 + 32 ln x − 2 + C
3
cos 2x
(1 − ln(cos 2x)) + C
2
1
1 5. x 3 + ln x 3 − 1 + C
3
3
3.
2
(x + 2)3/2 − 4(x + 2)1/2 + C
3
x 2
9.
x − 16 − 8 ln x + x 2 − 16 + C
2
√ 3/2 4 √ 5/2
4
+ 1+ x
+C
11. − 1 + x
3
5
7.
13.
1
1 2
x ln(x + 12) − 72 ln(x + 12) − x 2 + 6x + C
2
4
1
1
1
15. − ln |x| + ln |x + 1| +
ln |x − 7| + C
7
8
56
1 e x 2x
e − 1 − ln e x + e2x − 1 + C
17.
2
2
1
1
1 19. x 8 + x 4 + ln x 4 − 1 + C
8
4
4
21.
4
sec5/4 x + C
5
1 4
1 4
x ln x −
x +C
4
16
√
√
ex + 1 − 1
25. 2 e x + 1 + ln √
+C
ex + 1 + 1
23.
27. (1 + e x ) ln(1 + e x ) − (1 + e x ) + C
29. −e1/x + C
31.
1 −1 2x
sin (e ) + C
2
33.
1
7
ln |x + 1| − 4 ln |x − 2| + ln |x − 5| + C
2
2
35. sin(tan−1 x + 1) + C
1
37. 12 (x 2 + 4) ln(x 2 + 4) − (x 2 + 4) + C
2
39.
1 6
1 6
x ln x −
x +C
6
36
41. x 2 e x + C
43.
7/2 12 5/2
3/2
6
−
+ 2 1 + x 1/3
+C
1 + x 1/3
1 + x 1/3
7
5
1
2
1 3
x cos−1 x − x 2 (1 − x 2 )1/2 − (1 − x 2 )1/2 + C
3
9
9
2 −9
1
4x
47. tan−1
+C
3
3
45.
49.
1
1
cos 8x −
cos 16x + C
16
32
51.
1 2
x − x + 3 ln |x + 2| + C
2
53.
1
1
tan5 x − tan3 x + tan x − x + C
5
3
1
1
55. − cos 3x −
cos 7x + C
6
14
x e
57. ln x
+C
e +1
4
2
(x + 1)5/2 − (x + 1)3/2 + 2(x + 1)1/2 + C
5
3
61. 2 sin2 x + 1 + C
59.
63.
1
ln(tan x)2 + C
2
65. tan x ln(tan x) − tan x + C
x x2
−1
−3 1+
67. x sinh
+C
3
9
69.
1 4
sin x + C
2
71.
1
1
cosh 5x sinh 5x + x + C
10
2
73.
1
sec 3x + C
3
75. x tanh x − ln(cosh x) + C
1
sec2 3x + C
6
x 2
x + 4 − 2 ln x + x 2 + 4 + C
79.
2
1 x
1
−1 1 x + C
81. −
tan
+
2 x2 + 9
6
3
1
1 2
1
83. x 6
ln x −
ln x +
+C
6
18
108
77.
1 x
1
ln x − ln(x + 2) + C
2 x +2
2
√ 1
x + 1 − 2
87. √ ln √ +C
x + 1 + 2
2 2
85.
1
1
1
ln |x| −
ln(x 2 + 2x + 5) −
tan−1
5
10
10
x2 − 1
91.
+C
x
1
1
93.
−x cos 2x + sin 2x + C
4
2
89.
x +1
2
+C
1
8
95. − (4 − x 2 )5/2 + (4 − x 2 )3/2 − 16(4 − x 2 )1/2 + C
5
3
1
1
1 2
97. x cos−1 x − x 1 − x 2 + sin−1 x + C
2
4
4
√
x
+C
99. x − 5 tan−1 √
5