Download y (sx + x)dx y y y y y ax dx y ex dx y y xn dx

STRATEGY FOR INTEGRATION
As we have seen, integration is more challenging than differentiation. In finding the derivative of a function it is obvious which differentiation formula we should apply. But it may
not be obvious which technique we should use to integrate a given function.
Until now individual techniques have been applied in each section. For instance, we
usually used substitution in Exercises 4.5, integration by parts in Exercises 6.1, and partial
fractions in Exercises 6.3. But in this section we present a collection of miscellaneous integrals in random order and the main challenge is to recognize which technique or formula
to use. No hard and fast rules can be given as to which method applies in a given situation,
but we give some advice on strategy that you may find useful.
A prerequisite for strategy selection is a knowledge of the basic integration formulas.
In the following table we have collected the integrals that we have learned in Chapters 4,
5, and 6. Most of them should be memorized. It is useful to know them all, but the ones
marked with an asterisk need not be memorized since they are easily derived. Formula 19
can be avoided by using partial fractions, and trigonometric substitutions can be used in
place of Formula 20.
Copyright © 2013, Cengage Learning. All rights reserved.
Table of Integration Formulas Constants of integration have been omitted.
x n1
n1
1.
yx
n
dx 3.
ye
x
dx e x
5.
n 1
1
dx ln x
x
2.
y
4.
ya
y sin x dx cos x
6.
y cos x dx sin x
7.
y sec x dx tan x
8.
y csc x dx cot x
9.
y sec x tan x dx sec x
10.
y csc x cot x dx csc x
11.
y sec x dx ln sec x tan x 12.
y csc x dx ln csc x cot x 13.
y tan x dx ln sec x 14.
y cot x dx ln sin x 15.
y sinh x dx cosh x
16.
y cosh x dx sinh x
17.
y
dx
1
x
tan1
2 x a
a
a
18.
y sa
2
*19.
y
dx
1
xa
ln
x2 a2
2a
xa
*20.
y sx
2
2
2
x
dx ax
ln a
2
dx
x
sin1
2
x
a
dx
ln x sx 2 a 2
a2
Once you are armed with these basic integration formulas, if you don’t immediately see
how to attack a given integral, you might try the following four-step strategy.
1. Simplify the Integrand if Possible Sometimes the use of algebraic manipulation or
trigonometric identities will simplify the integrand and make the method of integration
obvious. Here are some examples:
y sx (1 sx ) dx y (sx x) dx
1
2 ■ STRATEGY FOR INTEGRATION
tan sin y sec d y cos cos d
2
2
y sin cos d 12 y sin 2 d
y sin x cos x dx y sin x 2 sin x cos x cos x dx
2
2
2
y 1 2 sin x cos x dx
2. Look for an Obvious Substitution Try to find some function u tx in the inte-
grand whose differential du tx dx also occurs, apart from a constant factor.
For instance, in the integral
x
y x 2 1 dx
we notice that if u x 2 1, then du 2x dx. Therefore, we use the substitution u x 2 1 instead of the method of partial fractions.
3. Classify the Integrand According to Its Form If Steps 1 and 2 have not led to the
solution, then we take a look at the form of the integrand f x.
(a) Trigonometric functions. If f x is a product of powers of sin x and cos x,
of tan x and sec x, or of cot x and csc x, then we use the substitutions recommended in Section 6.2.
(b) Rational functions. If f is a rational function, we use the procedure involving
partial fractions in Section 6.3.
(c) Integration by parts. If f x is a product of a power of x (or a polynomial)
and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing u and dv according to the advice given in Section 6.1. If you look at the functions in Exercises 6.1, you will see that most of them are the type just described.
(d) Radicals. Particular kinds of substitutions are recommended when certain
radicals appear.
(i) If sx 2 a 2 occurs, we use a trigonometric substitution according to the
table in Section 6.2.
n
n
(ii) If s
ax b occurs, we use the rationalizing substitution u s
ax b.
n
More generally, this sometimes works for stx.
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4. Try Again If the first three steps have not produced the answer, remember that
there are basically only two methods of integration: substitution and parts.
(a) Try substitution. Even if no substitution is obvious (Step 2), some inspiration
or ingenuity (or even desperation) may suggest an appropriate substitution.
(b) Try parts. Although integration by parts is used most of the time on products
of the form described in Step 3(c), it is sometimes effective on single functions. Looking at Section 6.1, we see that it works on tan1x, sin1x, and ln x,
and these are all inverse functions.
(c) Manipulate the integrand. Algebraic manipulations (perhaps rationalizing the
denominator or using trigonometric identities) may be useful in transforming
the integral into an easier form. These manipulations may be more substantial
than in Step 1 and may involve some ingenuity. Here is an example:
dx
y 1 cos x
y
1
1 cos x
1 cos x
dx y
dx
1 cos x 1 cos x
1 cos 2x
y
1 cos x
dx sin 2x
y
csc 2x cos x
sin 2x
dx
(d) Relate the problem to previous problems. When you have built up some experience in integration, you may be able to use a method on a given integral that
STRATEGY FOR INTEGRATION ■ 3
is similar to a method you have already used on a previous integral. Or you
may even be able to express the given integral in terms of a previous one. For
instance, x tan 2x sec x dx is a challenging integral, but if we make use of the
identity tan 2x sec 2x 1, we can write
y tan x sec x dx y sec x dx y sec x dx
2
3
and if x sec 3x dx has previously been evaluated (see Example 8 in Section 6.2),
then that calculation can be used in the present problem.
(e) Use several methods. Sometimes two or three methods are required to evaluate an integral. The evaluation could involve several successive substitutions
of different types, or it might combine integration by parts with one or more
substitutions.
In the following examples we indicate a method of attack but do not fully work out the
integral.
EXAMPLE 1
tan 3x
y cos x dx
3
In Step 1 we rewrite the integral:
y
tan 3x
dx y tan 3x sec 3x dx
cos 3x
HOW TO INTEGRATE POWERS
OF tan x AND sec x
(i) If the power of sec x is even,
save a factor of sec2 x and use
sec2 x 苷 1 tan 2 x to express the
remaining factors in terms of tan x.
Then substitute u 苷 tan x.
The integral is now of the form x tan m x sec n x dx with m odd, so we can use the advice
shown in the margin.
Alternatively, if in Step 1 we had written
(ii) If the power of tan x is odd,
save a factor of sec x tan x and use
tan 2x 苷 sec 2x 1 to express the
remaining factors in terms of sec x.
Then substitute u 苷 sec x.
then we could have continued as follows with the substitution u cos x :
■
y
y
EXAMPLE 2
ye
sx
tan 3x
sin 3x 1
sin 3x
dx
3 dx y
3
3 dx y
cos x
cos x cos x
cos 6x
sin 3x
1 cos 2x
1 u2
sin x dx y
du
6 dx y
6
cos x
cos x
u6
u2 1
y
du y u 4 u 6 du
u6
dx
According to Step 3(d)(ii) we substitute u sx. Then x u 2, so dx 2u du and
ye
sx
dx 2 y ue u du
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The integrand is now a product of u and the transcendental function e u so it can be integrated by parts.
x5 1
dx
3
3x 2 10x
No algebraic simplification or substitution is obvious, so Steps 1 and 2 don’t apply here.
The integrand is a rational function so we apply the procedure of Section 6.3, remembering that the first step is to divide.
EXAMPLE 3
yx
EXAMPLE 4
y xsln x
dx
Here Step 2 is all that is needed. We substitute u ln x because its differential is
du dxx, which occurs in the integral.
4 ■ STRATEGY FOR INTEGRATION
1x
dx
1x
Although the rationalizing substitution
EXAMPLE 5
y
u
1x
1x
works here [Step 3(d)(ii)], it leads to a very complicated rational function. An easier
method is to do some algebraic manipulation [either as Step 1 or as Step 4(c)]. Multiplying numerator and denominator by s1 x, we have
y
1x
1x
dx y
dx
1x
s1 x 2
1
x
dx y
dx
s1 x 2
s1 x 2
y
sin1x s1 x 2 C
EXERCISES
A Click here for answers.
1–80
1.
3.
5.
7.
Copyright © 2013, Cengage Learning. All rights reserved.
9.
Click here for solutions.
S
Evaluate the integral.
sin x sec x
dx
tan x
y
y
2
y
1
y
3
0
2t
dt
t 32
4.
e arctan y
dy
1 y2
6.
1
1
4
8.
r ln r dr
x1
dx
x 2 4x 5
y
2.
x
dx
s3 x 4
y
y x csc x
y
4
0
x
dx
x4 x2 1
10.
y
y sin x coscos x dx
13.
y
dx
1 x 2 32
14.
y
15.
y
16.
y
17.
y x sin x dx
18.
y 1e
19.
y e xe dx
20.
y esx dx
21.
yt e
22.
y x sin
23.
y (1 sx ) dx
24.
y lnx
2
x
3 2t
1
0
dt
8
26.
yx
27.
y cot x lnsin x dx
28.
y sin sat dt
29.
y
3w 1
dw
w2
30.
y x
31.
y
1x
dx
1x
32.
y
33.
y s3 2x x
34.
y
35.
y
36.
y sin 4x cos 3x dx
37.
y
38.
y
39.
y
40.
y s4y
s1 ln x
dx
x ln x
s22
41.
y tan d
42.
yx
0
x2
dx
s1 x 2
e 2t
4t
43.
y e s1 e
44.
y s1 e
45.
yx e
46.
y 1e
47.
yx
48.
y
dt
3
1
2
49.
y x s4x 1 dx
50.
yx
5
0
cot x dx
x1
dx
2
x 4x 5
12.
0
3x 2 2
dx
2x 8
y tan d
y sin 3 cos 5 d
x
dx
s1 x 2
yx
2
3
3x 2 2
dx
2x 8
3
11.
12
25.
x dx
1 dx
1
2
dx
x 8 sin x dx
1
4
0
cos2 tan2 d
x
dx
1 x 2 s1 x 2
2
x
x
dx
5 x 3
dx
xa
dx
2
a2
1
2
2
2
4x dx
s2x 1
dx
2x 3
2
4
4
0
2
1 4 cot x
dx
4 cot x
tan 5 sec 3 d
2
1
dy
4y 3
tan1x dx
1 ex
x
x
dx
dx
x
dx
x4 a4
2
1
dx
s4x 1
STRATEGY FOR INTEGRATION ■ 5
1
51.
y x s4x
53.
yx
55.
57.
59.
2
2
1
dx
sinh mx dx
1
dx
x 4 4 sx 1
y
y x sx c dx
3
1
dx
e 3x e x
y
y x x
54.
y x sin x dx
56.
58.
60.
63.
65.
Copyright © 2013, Cengage Learning. All rights reserved.
67.
yx
10
x
dx
16
y sxe
sx
dx
1
y sx 1 sx dx
y
3
1
arctanst
dt
st
62.
64.
66.
68.
4
x ln x
dx
sx 2 1
y
yx
2
y 1e
71.
yx
73.
y x 2x
75.
y sin x sin 2x sin 3x dx
77.
y 1x
79.
y x sin
ln1 x dx
1
dx
3
xs
x
y
x
3
y x 1
10
3
y
4
y
3
2
y
dx
lntan x
dx
sin x cos x
u 1
du
u3 u2
3
1
dx
1 2e x ex
e 2x
69.
2
4
61.
dx
1
52.
■
4
x
dx
x
dx
4x 2 3
1
sx
■
2
■
3
2
4
dx
dx
x cos x dx
■
✤
■
■
lnx 1
dx
x2
70.
y
72.
y 1 st dt
74.
ye
76.
y x
78.
y sin x sec x dx
80.
y sin
st
3
dx
e x
2
bx sin 2x dx
sec x cos 2x
sin x cos x
dx
4
x cos 4 x
■
2
x
■
■
■
2
■
■
81. The functions y e x and y x 2e x don’t have elementary
antiderivatives, but y 2x 1e
2
x 2x 2 1e x dx.
2
x2
does. Evaluate
■
6 ■ STRATEGY FOR INTEGRATION
ANSWERS
3
45. 3 x 3 1ex C
1
Click here for solutions.
S
47. ln sx 2 a 2 tan1xa C
1. sin x ln csc x cot x C
5. e 4 e4
3. 4 ln 9
7.
11.
ln 3 242
25
243
5
49. ln
9.
lnx 2 4x 5 tan1x 2 C
1
2
cos8 16 cos6 C (or 14 sin4 13 sin6 18 sin8 C)
1
8
(or
1
4
1
x x sin 2x cos 2 x C)
23.
1
25. 3x 4097
45
61.
21. 8 e2t4t 3 6t 2 6t 3 C
19. e e C
23
3
1
20
( x )C
63. 2( x 2sx 2) e sx C
2
71. 4 lnx 2 3 4 lnx 2 1 C
2
2
1
1
33. 2 sin x 12 x 12 s3 2x x C
75.
39. ln(1 s1 x 2 ) C
77.
2
3
2
1
4
41. tan 2 2 ln sec C
43. 3 1 e x 32 C
2
1
1
24
1
69. e x ln1 e x C
ln x 2 161 lnx 2 4 18 tan1x2 C
1
8
1
tan
65. 3 x 132 x 32 C
73.
37. 8 5
29. 15 7 ln 7
31. sin1x s1 x 2 C
35. 0
1 1
4
67. 3 s3 2 ln 2
1
Copyright © 2013, Cengage Learning. All rights reserved.
3
ln x 4 53 ln x 2 C
27. 2 ln sin x2 C
s4x 2 1 1
C
2x
55. 3 ln(sx 1 3) ln(sx 1 1) C
1
1
8
x
59. ex 2 ln e x 1e x 1 C
1
1
4
2
51. ln
53. 1mx 2 coshmx 2m2 x sinhmx 2m3 coshmx C
3
1
17. 4 x 2 2 x sin x cos x 4 sin 2 x C
1
s4x 1 1
C
s4x 1 1
57. 7 x c73 4 cx c43 C
15. 1 2 s3
13. xs1 x 2 C
cos 6x 161 cos 4x 18 cos 2x C
tan1 x 32 C
2
81. xe x C
79. 3 x sin 3 x 3 cos x 9 cos 3x C
1
1
1
STRATEGY FOR INTEGRATION ■ 7
SOLUTIONS
3.
sin x + sec x
dx =
tan x
1.
2
0
2t
dt =
(t − 3)2
−1
−3
sin x
sec x
+
tan x
tan x
dx =
(cos x + csc x) dx = sin x + ln |csc x − cot x| + C
2 (u + 3)
du [u = t − 3, du = dt] =
u2
−1
−3
6
2
+
u u2
−1
6
du = 2 ln |u| −
u −3
= (2 ln 1 + 6) − (2 ln 3 + 2) = 4 − 2 ln 3 or 4 − ln 9
5. Let u = arctan y. Then du =
7.
3
1
9.
⇒
1
−1
earctan y
dy =
1 + y2
π/4
−π/4
π/4
eu du = [eu ]−π/4 = eπ/4 − e−π/4 .


3
3 3
u = ln r, dv = r4 dr,
243
1 4
1 5
1 5
r ln r −
r dr =
ln 3 − 0 −
r
r ln r dr 
1 5 =
dr
5
5
25
v= r
du =
1 5
1
1
5
r
1
= 243
= 243
ln 3 − 243
− 25
ln 3 − 242
5
25
5
25
4
x−1
dx =
x2 − 4x + 5
=
11.
dy
1 + y2
1
2
(x − 2) + 1
1
u
+ 2
dx =
du [u = x − 2, du = dx]
u2 + 1
u +1
(x − 2)2 + 1
2
ln u + 1 + tan−1 u + C = 12 ln x2 − 4x + 5 + tan−1(x − 2) + C
cos5 θ sin2 θ sin θ dθ = − cos5 θ (1 − cos2 θ)(− sin θ) dθ
u = cos θ,
= − u5 (1 − u2 ) du
du = − sin θ dθ
7
5
1 8
= (u − u ) du = 8 u − 16 u6 + C = 18 cos8 θ − 16 cos6 θ + C
sin3 θ cos5 θ dθ =
Another solution:
sin3 θ cos5 θ dθ = sin3 θ (cos2 θ)2 cos θ dθ = sin3 θ (1 − sin2 θ)2 cos θ dθ
3
u = sin θ,
2 2
= u3 (1 − 2u2 + u4 ) du
= u (1 − u ) du
du = cos θ dθ
= (u3 − 2u5 + u7 ) du = 14 u4 − 13 u6 + 18 u8 + C = 14 sin4 θ − 13 sin6 θ +
13. Let x = sin θ, where − π2 ≤ θ ≤
π
.
2
1
8
sin8 θ + C
Then dx = cos θ dθ and
2 1/2
(1 − x )
= cos θ, so
cos θ dθ
dx
=
=
sec2 θ dθ
(cos θ)3
(1 − x2 )3/2
x
= tan θ + C = √
+C
1 − x2
15. Let u = 1 − x2
1/2
0
Copyright © 2013, Cengage Learning. All rights reserved.
17.
⇒ du = −2x dx. Then
1
x
√
dx = −
2
1 − x2
3/4
1
1
1
√ du =
2
u
1
3/4
u−1/2 du =
1
2
2u1/2
1
3/4
√ 1
u 3/4 = 1 −
√
3
2
u = x,
dv = sin2 x dx,
x sin x dx
v = sin2 x dx = 12 (1 − cos 2x) dx = 12 x −
du = dx
1
= 12 x2 − 12 x sin x cos x −
x − 12 sin x cos x dx
2
2
=
1
2
sin x cos x
= 12 x2 − 12 x sin x cos x − 14 x2 + 14 sin2 x + C = 14 x2 − 12 x sin x cos x + 14 sin2 x + C
Note: sin x cos x dx = s ds = 12 s2 + C [where s = sin x, ds = cos x dx].
A slightly different method is to write x sin2 x dx = x · 12 (1 − cos 2x) dx = 12 x dx − 12 x cos 2x dx. If we
evaluate the second integral by parts, we arrive at the equivalent answer 14 x2 − 14 x sin 2x −
19. Let u = ex . Then
x
ex+e dx =
x
ee ex dx =
x
eu du = eu + C = ee + C.
1
8
cos 2x + C.
8 ■ STRATEGY FOR INTEGRATION
21. Integrate by parts three times, first with u = t3 , dv = e−2t dt:
3 −2t
dt = − 12 t3 e−2t + 12 3t2 e−2t dt = − 12 t3 e−2t − 34 t2 e−2t + 12 3te−2t dt
t e
= −e−2t 12 t3 + 34 t2 − 34 te−2t + 34 e−2t dt = −e−2t 12 t3 + 34 t2 + 34 t + 38 + C
= − 18 e−2t 4t3 + 6t2 + 6t + 3 + C
23. Let u = 1 +
√
x. Then x = (u − 1)2 , dx = 2(u − 1) du ⇒
1
0
(1 +
2
2
2
√ 8
x ) dx = 1 u8 · 2(u − 1) du = 2 1 u9 − u8 du = 15 u10 − 2 · 19 u9 1
=
25.
1024
5
−
1024
9
−
1
5
+
2
9
=
4097
45
6x + 22
A
B
3x2 − 2
=3+
=3+
+
− 2x − 8
(x − 4)(x + 2)
x−4
x+2
x2
⇒ 6x + 22 = A(x + 2) + B(x − 4). Setting
5
x = 4 gives 46 = 6A, so A = 23
3 . Setting x = −2 gives 10 = −6B, so B = − 3 . Now
23/3
5/3
3x2 − 2
3+
dx =
−
dx = 3x + 23
ln |x − 4| − 53 ln |x + 2| + C.
3
2
x − 2x − 8
x−4
x+2
27. Let u = ln(sin x). Then du = cot x dx ⇒
cot x ln(sin x) dx = u du = 12 u2 + C = 12 [ln(sin x)]2 + C.
5
5
5
3w − 1
7
dw =
29.
3−
dw = 3w − 7 ln |w + 2|
w+2
w+2
0
0
0
= 15 − 7 ln 7 + 7 ln 2 = 15 + 7(ln 2 − ln 7) = 15 + 7 ln 27
31. As in Example 5,
√
√
1+x
dx
x dx
1+x
1+x
1+x
√
√
√
√
·√
dx =
dx =
dx =
+
2
2
1−x
1−x
1+x
1−x
1−x
1 − x2
= sin−1 x − 1 − x2 + C
Another method: Substitute u = (1 + x)/(1 − x).
33. 3 − 2x − x2 = −(x2 + 2x + 1) + 4 = 4 − (x + 1)2 . Let
x + 1 = 2 sin θ, where − π2 ≤ θ ≤ π2 . Then dx = 2 cos θ dθ and
√
3 − 2x − x2 dx =
4 − (x + 1)2 dx =
4 − 4 sin2 θ 2 cos θ dθ
= 4 cos2 θ dθ = 2 (1 + cos 2θ) dθ
= 2θ + sin 2θ + C = 2θ + 2 sin θ cos θ + C
√
x+1
x+1
3 − 2x − x2
+2·
·
+C
= 2 sin−1
2
2
2
x+1
x + 1√
3 − 2x − x2 + C
= 2 sin−1
+
2
2
35. Because f (x) = x8 sin x is the product of an even function and an odd function, it is odd. Therefore,
1 8
x sin x dx = 0 [by (4.5.6)(b)].
−1
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37.
π/4
0
π/4
sin2 θ dθ = 0 12 (1 − cos 2θ) dθ = 12 θ −
= π8 − 14 − (0 − 0) = π8 − 14
cos2 θ tan2 θ dθ =
π/4
0
1
4
sin 2θ
π/4
0
39. Let u = 1 − x2 . Then du = −2x dx ⇒
√
v dv
du
1
x dx
√ =−
√
[v = u, u = v 2 , du = 2v dv]
=−
2
v2 + v
u+ u
1 − x2 + 1 − x2
dv
= − ln |v + 1| + C = − ln
1 − x2 + 1 + C
=−
v+1
STRATEGY FOR INTEGRATION ■ 9
41. Let u = θ, dv = tan2 θ dθ = sec2 θ − 1 dθ ⇒ du = dθ and v = tan θ − θ. So
θ tan2 θ dθ = θ(tan θ − θ) − (tan θ − θ) dθ = θ tan θ − θ2 − ln |sec θ| + 12 θ2 + C
= θ tan θ − 12 θ2 − ln |sec θ| + C
43. Let u = 1 + ex , so that du = ex dx. Then
x√
e 1 + ex dx = u1/2 du = 23 u3/2 + C = 23 (1 + ex )3/2 + C.
√
Or: Let u = 1 + ex , so that u2 = 1 + ex and 2u du = ex dx. Then
x√
e 1 + ex dx = u · 2u du = 2u2 du = 23 u3 + C = 23 (1 + ex )3/2 + C.
3
45. Let t = x3 . Then dt = 3x2 dx ⇒ I = x5 e−x dx = 13 te−t dt. Now integrate by parts with u = t,
3
dv = e−t dt: I = − 13 te−t + 13 e−t dt = − 13 te−t − 13 e−t + C = − 13 e−x x3 + 1 + C.
x
x+a
1
1
dx
2x dx
47.
+C
dx
=
+
a
= 12 ln x2 + a2 + a · tan−1
2
2
2
2
2
2
x +a
2
x +a
x +a
a
a
√
= ln x2 + a2 + tan−1 (x/a) + C
√
49. Let u = 4x + 1 ⇒ u2 = 4x + 1 ⇒ 2u du = 4 dx ⇒ dx = 12 u du. So
1
1 u − 1
u du
du
1
2
√
dx =
=
2
=
2
ln
1
2
u + 1 + C [by Formula 19]
2
u2 − 1
x 4x + 1
4 (u − 1) u
√
4x + 1 − 1 +C
= ln √
4x + 1 + 1 √
51. Let 2x = tan θ ⇒ x = 12 tan θ, dx = 12 sec2 θ dθ, 4x2 + 1 = sec θ, so
1
2
sec θ
dx
2 sec θ dθ
√
=
=
dθ
=
csc θ dθ
1
tan θ
tan θ sec θ
x 4x2 + 1
2
= − ln |csc θ + cot θ| + C
[or ln |csc θ − cot θ| + C]
√
√
4x2 + 1
4x2 + 1
1 1 +
C
or
ln
+
C
= − ln
+
−
2x
2x 2x
2x dv = sinh(mx) dx,
u = x2 ,
1 2
2
2
x cosh(mx) dx
53.
x sinh(mx)dx = x cosh(mx) −
1
m
m
du = 2x dx
cosh(mx)
v= m
2
1
= x2 cosh(mx) −
m
m
1
m x sinh(mx)
1 sinh(mx) dx
−
m
U = x, dV = cosh(mx) dx,
1
dU = dx
V = m
sinh(mx)
2
2
1 2
x cosh(mx) − 2 x sinh(mx) + 3 cosh(mx) + C
m
m
m
√
55. Let u = x + 1. Then x = u2 − 1 ⇒
dx
2u du
−1
3
√
=
=
+
du
u2 + 3 + 4u
u+1
u+3
x+4+4 x+1
√
√
= 3 ln |u + 3| − ln |u + 1| + C = 3 ln x + 1 + 3 − ln x + 1 + 1 + C
√
57. Let u = 3 x + c. Then x = u3 − c ⇒
√
3
x 3 x + c dx =
u − c u · 3u2 du = 3 u6 − cu3 du = 37 u7 − 34 cu4 + C
=
Copyright © 2013, Cengage Learning. All rights reserved.
= 37 (x + c)7/3 − 34 c(x + c)4/3 + C
59. Let u = ex . Then x = ln u, dx = du/u ⇒
du
1/2
1
du/u
1/2
dx
=
=
−
=
−
du
e3x − ex
u3 − u
(u − 1)u2 (u + 1)
u−1
u2
u+1
1 u − 1 1 ex − 1 1
−x
+
C
=
e
+C
ln
+
= + ln
u
2
u + 1
2 ex + 1 10 ■ STRATEGY FOR INTEGRATION
61. Let u = x5 . Then du = 5x4 dx ⇒
1
x4 dx
5 du
=
= 15 · 14 tan−1 14 u + C =
10
2
x + 16
u + 16
1
20
tan−1
1
4
x5 + C.
√
1
√ dx ⇒ dx = 2 x dy = 2y dy. Then
x
√ √x
u = 2y 2 , dv = ey dy,
x e dx = yey (2y dy) = 2y 2 ey dy
v = ey
du = 4y dy
U = 4y, dV = ey dy,
= 2y 2 ey − 4yey dy
dU = 4 dy
V = ey
= 2y 2 ey − 4yey − 4ey dy = 2y 2 ey − 4yey + 4ey + C
√
√
= 2(y 2 − 2y + 2)ey + C = 2(x − 2 x + 2) e x + C
√
√ √
√ 1
dx
x+1− x
√
√
dx =
x + 1 − x dx
65.
√ =
√ ·√
√
x+1+ x
x+1+ x
x+1− x
3/2
3/2
2
+C
= 3 (x + 1)
−x
63. Let y =
√
x so that dy =
2
√ √
⇒
67. Let u = t. Then du = dt/ 2 t
√
√3
3
√3
arctan t
√
dt =
tan−1 u (2 du) = 2 u tan−1 u − 12 ln(1 + u2 ) 1
[Example 5 in Section 6.1]
t
1
1
√
√
=2
3 tan−1 3 − 12 ln 4 − tan−1 1 − 12 ln 2
√
√ π
=2
3 · 3 − ln 2 − π4 − 12 ln 2 = 23 3 π − 12 π − ln 2
69. Let u = ex . Then x = ln u, dx = du/u ⇒
e2x
dx =
1 + ex
u2 du
=
1+u u
u
du =
1+u
1−
1
1+u
du
= u − ln|1 + u| + C = ex − ln(1 + ex ) + C
71.
x
Ax + B
Cx + D
x
= 2
= 2
+ 2
x4 + 4x2 + 3
(x + 3)(x2 + 1)
x +3
x +1
⇒
x = (Ax + B) x2 + 1 + (Cx + D) x2 + 3 = Ax3 + Bx2 + Ax + B + Cx3 + Dx2 + 3Cx + 3D
= (A + C)x3 + (B + D)x2 + (A + 3C)x + (B + 3D)
⇒
A + C = 0, B + D = 0, A + 3C = 1, B + 3D = 0 ⇒ A = − 12 , C = 12 , B = 0, D = 0. Thus,
x
dx =
x4 + 4x2 + 3
=
Thomson
Brooks-Cole
copyrightLearning.
2007 All rights reserved.
Copyright
© 2013, Cengage
73.
− 14
1
− 12 x
x
+ 22
2
x +3
x +1
ln x2 + 3 +
1
4
dx
ln x2 + 1 + C
2
x +1
1
or ln 2
+C
4
x +3
A
Bx + C
1
=
+ 2
⇒
(x − 2)(x2 + 4)
x−2
x +4
1 = A x2 + 4 + (Bx + C)(x − 2) = (A + B)x2 + (C − 2B)x + (4A − 2C). So 0 = A + B = C − 2B,
1 = 4A − 2C. Setting x = 2 gives A =
1
dx =
(x − 2)(x2 + 4)
=
1
8
1
8
⇒ B = − 18 and C = − 14 . So
− 18 x − 14
1
1
1
dx
2x dx
dx
dx
=
−
−
x−2
x2 + 4
8
x−2
16
x2 + 4
4
x2 + 4
1
ln|x − 2| − 16
ln x2 + 4 − 18 tan−1 (x/2) + C
1
8
+
STRATEGY FOR INTEGRATION ■ 11
75.
U
U
sin x · 12 [cos(2x − 3x) − cos(2x + 3x)] dx
U
= 12 (sin x cos x − sin x cos 5x) dx
sin x sin 2x sin 3x dx =
=
1
4
U
− 18
sin 2x dx −
U
U
1
2
%
&
by Formula 18a
[sin(x + 5x) + sin(x − 5x)] dx
− 18
cos 2x +
(sin 6x − sin 4x) dx =
√
77. Let u = x3/2 so that u2 = x3 and du = 32 x1/2 dx ⇒
x dx = 23 du. Then
] √
]
2
2
2
x
3
dx =
du = tan−1 u + C = tan−1 (x3/2 ) + C.
1 + x3
1 + u2
3
3
=
cos 2x −
1
4
1
2
[by Formula 18b in Appendix A]
in Appendix A
1
24
cos 6x −
1
16
cos 4x + C
79. Let u = x, dv = sin2 x cos x dx ⇒ du = dx, v = 13 sin3 x. Then
U
U
U
x sin2 x cos x dx = 13 x sin3 x − 13 sin3 x dx = 13 x sin3 x − 13 (1 − cos2 x) sin x dx
]
y = cos x,
1
1
(1 − y 2 ) dy
= x sin3 x +
3
3
dy = − sin x dx
= 13 x sin3 x + 13 y − 19 y 3 + C = 13 x sin3 x +
2
1
3
cos x −
1
9
cos3 x + C
Copyright © 2013, Cengage Learning. All rights reserved.
81. The function y = 2xex does have an elementary antiderivative, so we’ll use this fact to help evaluate the integral.
U
U 2
U
U 2
U
2
2
2
(2x2 + 1)ex dx = 2x2 ex dx + ex dx = x(2xex ) dx + ex dx
&
%
2
U 2
U 2
u = x, dv = 2xex dx,
2
2
= xex + C
= xex − ex dx + ex dx
x2
du = dx
v=e