Download Strategy for Integration

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 5.5, integration by parts in Exercises 5.6, and partial
fractions in Exercises 5.7 and Appendix G. 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 from our previous list together with
several additional formulas that we have learned in this chapter. 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.
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
1
2 ■ STRATEGY FOR INTEGRATION
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
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 5.7 and Additional Topics: Trigonometric Integrals.
(b) Rational functions. If f is a rational function, we use the procedure involving
partial fractions in Section 5.7 and Appendix G.
(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 5.6. If you look at the functions in Exercises 5.6, 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 Additional Topics: Trigonometric Substitution.
n
n
ax b occurs, we use the rationalizing substitution u s
ax b.
(ii) If s
n
More generally, this sometimes works for stx.
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 5.6, 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
STRATEGY FOR INTEGRATION ■ 3
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
y
1
1 cos x
1 cos x
dx y
dx
1 cos x 1 cos x
1 cos 2x
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
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 Additional
Topics: Trigonometric Integrals), 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
The integral is now of the form x tan m x sec n x dx with m odd, so we can use the advice in
Additional Topics: Trigonometric Integrals.
Alternatively, if in Step 1 we had written
y
tan 3x
sin 3x 1
sin 3x
dx
dx
dx
y
y
cos 3x
cos 3x cos 3x
cos 6x
then we could have continued as follows with the substitution u cos x :
y
sin 3x
1 cos 2x
1 u2
dx
sin
x
dx
du
y
y
cos 6x
cos 6x
u6
y
EXAMPLE 2
ye
sx
u2 1
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
4 ■ STRATEGY FOR INTEGRATION
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 Appendix G, 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.
EXAMPLE 5
y
1x
dx
1x
Although the rationalizing substitution
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
y
1
x
dx y
dx
s1 x 2
s1 x 2
sin1x s1 x 2 C
Can We Integrate All Continuous Functions?
The question arises: Will our strategy for integration enable us to find the integral of every
2
continuous function? For example, can we use it to evaluate x e x dx ? The answer is no, at
least not in terms of the functions that we are familiar with.
The functions that we have been dealing with in this book are called elementary functions. These are the polynomials, rational functions, power functions x a , exponential
functions a x , logarithmic functions, trigonometric and inverse trigonometric functions,
hyperbolic and inverse hyperbolic functions, and all functions that can be obtained from
these by the five operations of addition, subtraction, multiplication, division, and composition. For instance, the function
f x x2 1
lncosh x xe sin 2x
x 3 2x 1
is an elementary function.
If f is an elementary function, then f is an elementary function but x f x dx need not
2
be an elementary function. Consider f x e x . Since f is continuous, its integral exists,
and if we define the function F by
x
2
Fx y e t dt
0
STRATEGY FOR INTEGRATION ■ 5
then we know from Part 1 of the Fundamental Theorem of Calculus that
Fx e x
2
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 8, however, we will 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
y sx
3
y sinx
1 dx
y
2
dx
1
dx
ln x
y cose
y
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 elementary
functions.
Exercises
A Click here for answers.
1–80
1.
sin x sec x
dx
tan x
y
2
31.
y
33.
y s3 2x x
x1
dx
x 2 4x 5
35.
y
x
dx
x2 1
37.
y
39.
y
41.
y tan d
43.
y e s1 e
45.
yx e
47.
yx
1 dx
49.
y x s4x 1 dx
3x 2 2
dx
2x 8
51.
y x s4x
6.
y x csc x
8.
y
y
9.
yx
1
1
3
1
r 4 ln r dr
2
x1
dx
4x 5
y sin 3
cos 5 d
dx
y 1 x
y
y tan d
e arctan y
dy
1 y2
7.
0
2.
2 32
x
dx
s1 x 2
12
0
2
10.
12.
14.
16.
18.
y x sin x dx
x
dx
s3 x 4
4
0
yx
4
cot x dx
y sin x coscos x dx
s1 ln x
dx
x ln x
y
y
x2
dx
s1 x 2
s22
0
y
e 2t
dt
1 e 4t
19.
ye
xe x
dx
20.
ye
21.
y t 3e2t dt
22.
y x sin1x dx
23.
y (1 sx ) dx
24.
y lnx
25.
yx
26.
yx
1
8
0
2
3x 2 2
dx
2x 8
3
sx
3
3w 1
dw
w2
5
0
3
y
y
17.
y
4.
5.
15.
29.
2t
dt
t 32
y
13.
y cot x lnsin x dx
Evaluate the integral.
3.
11.
27.
Click here for solutions.
S
1
1x
dx
1x
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
28.
y sin sat dt
30.
y x
32.
y
34.
y
36.
y sin 4x cos 3x dx
38.
y
40.
y s4y
42.
yx
44.
y s1 e
46.
y 1e
48.
y
50.
yx
52.
y x x
dx
2
xa
dx
2
a2
1
1
2
1
dx
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
x a4
4
2
1
dx
s4x 1
dx
1
4
6 ■ STRATEGY FOR INTEGRATION
53.
55.
57.
59.
yx
2
sinh mx dx
1
54.
y x 4 4 sx 1 dx
56.
y x sx c dx
58.
1
dx
ex
60.
3
ye
3x
y x sin x dx
2
2
yx
10
x
dx
16
63.
y sxe
65.
y sx 1 sx dx
67.
sx
dx
1
y
3
1
arctanst
dt
st
62.
71.
yx
ln1 x dx
73.
y x 2x
75.
y sin x sin 2x sin 3x dx
77.
y 1x
79.
y x sin
4
x
dx
x
dx
4x 2 3
1
2
4
dx
3
y
72.
y 1 st dt
74.
ye
76.
y x
78.
y sin x sec x dx
80.
y sin
st
3
x
3
y x 1
10
3
y
66.
y
4
3
2
dx
lntan x
dx
sin x cos x
u3 1
du
u3 u2
x
■
2
■
3
dx
x cos x dx
■
■
■
■
ex
x
dx
e x
2
bx sin 2x dx
sec x cos 2x
sin x cos x
dx
4
x cos 4 x
■
2
■
■
■
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
1
y 1 2e
■
sx
lnx 1
dx
x2
70.
1
64.
68.
x ln x
dx
2 1
y x sx dx
4
61.
y 1e
y sx
yx
e 2x
69.
dx
x2
does. Evaluate
■
STRATEGY FOR INTEGRATION ■ 7
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
3. 4 ln 9
5. e 4 e4
243
242
1
7. 5 ln 3 25
9. 2 lnx 2 4x 5 tan1x 2 C
11.
1
8
cos8 16 cos6 C (or 14 sin4 13 sin6 18 sin8 C)
13. xs1 x 2 C
15. 1 2 s3
1
(or
1
4
1
x 2 x sin 2x cos 2 x C)
x
19. e e C
23.
4097
45
1
2
1
1
8
1 2t
8
23
3
21. e
25. 3x 27. ln sin x C
31.
33.
35.
41.
61.
4t 3 6t 2 6t 3 C
ln x 4 ln x 2 C
5
3
29. 15 7 ln
sin x s1 x 2 C
2 sin1x 12 x 12 s3 2x x 2 C
1
0
37. 8 4
39. ln(1 s1 x 2 ) C
1 2
2
tan 2 ln sec C
43. 3 1 e x 32 C
2
1
1
1
4
59. ex 2 ln e x 1e x 1 C
17. 4 x 2 2 x sin x cos x 4 sin 2 x C
1
s4x 1 1
s4x 2 1 1
C
C
51. ln
2x
s4x 1 1
53. 1mx 2 coshmx 2m2 x sinhmx 2m3 coshmx C
55. 3 ln(sx 1 3) ln(sx 1 1) C
3
3
57. 7 x c73 4 cx c43 C
49. ln
2
7
1
20
tan1( 14 x 5) C
63. 2( x 2sx 2) e sx C
65. 3 x 132 x 32 C
2
67. 3 s3 2 ln 2
2
69. e x ln1 e x C
1
71. 4 lnx 2 3 4 lnx 2 1 C
1
1
ln x 2 73.
1
8
75.
1
24
1
16
lnx 2 4 18 tan1x2 C
cos 6x 161 cos 4x 18 cos 2x C
79. 3 x sin 3 x 3 cos x 9 cos 3x C
1
1
1
77.
2
3
2
tan1 x 32 C
81. xe x C
8 ■ STRATEGY FOR INTEGRATION
Solutions: Strategy for Integration
¶
Z µ
Z
sin x
sec x
sin x + sec x
dx =
+
dx = (cos x + csc x) dx = sin x + ln |csc x − cot x| + C
tan x
tan x
tan x
¶
·
¸−1
Z 2
Z −1
Z −1 µ
2
6
2t
2 (u + 3)
6
3.
du
=
2
ln
|u|
−
dt
=
du
[u
=
t
−
3,
du
=
dt]
=
+
2
u2
u
u2
u −3
0 (t − 3)
−3
−3
1.
5.
7.
9.
11.
Z
= (2 ln 1 + 6) − (2 ln 3 + 2) = 4 − 2 ln 3 or 4 − ln 9
Z 1 arctan y
Z π/4
e
dy
π/4
⇒
dy
=
eu du = [eu ]−π/4 = eπ/4 − e−π/4 .
Let u = arctan y. Then du =
2
1 + y2
−1 1 + y
−π/4


·
¸3
¸3 Z 3
·
Z 3
u = ln r, dv = r 4 dr,
1 5
1 4
243
1 5
r 4 ln r dr 
r ln r −
r dr =
ln 3 − 0 −
r
1 5 =
dr
5
5
25
v= r
du =
1
1 5
1
1
5
r
¢
¡ 243
243
1
243
242
= 5 ln 3 − 25 − 25 = 5 ln 3 − 25
¶
Z
Z µ
Z
x−1
u
(x − 2) + 1
1
dx
=
dx
=
+
du [u = x − 2, du = dx]
x2 − 4x + 5
u2 + 1
u2 + 1
(x − 2)2 + 1
¡
¢
¡
¢
= 12 ln u2 + 1 + tan−1 u + C = 12 ln x2 − 4x + 5 + tan−1(x − 2) + C
R
R
R
sin3 θ cos5 θ dθ = cos5 θ sin2 θ sin θ dθ = − cos5 θ (1 − cos2 θ)(− sin θ) dθ
·
¸
R
u = cos θ,
= − u5 (1 − u2 ) du
du = − sin θ dθ
R 7
5
1 8
= (u − u ) du = 8 u − 16 u6 + C = 18 cos8 θ − 16 cos6 θ + C
Another solution:
R
R
R
sin3 θ cos5 θ dθ = sin3 θ (cos2 θ)2 cos θ dθ = sin3 θ (1 − sin2 θ)2 cos θ dθ
·
¸
R
R 3
u = sin θ,
2 2
= u3 (1 − 2u2 + u4 ) du
= u (1 − u ) du
du = cos θ dθ
R
= (u3 − 2u5 + u7 ) du = 14 u4 − 13 u6 + 18 u8 + C = 14 sin4 θ − 13 sin6 θ + 18 sin8 θ + C
13. Let x = sin θ, where − π2 ≤ θ ≤
π
.
2
Then dx = cos θ dθ and
(1 − x2 )1/2 = cos θ, so
Z
Z
Z
dx
cos θ dθ
=
= sec2 θ dθ
2
3/2
3
(cos θ)
(1 − x )
x
+C
= tan θ + C = √
1 − x2
15. Let u = 1 − x2 ⇒ du = −2x dx. Then
Z
Z
Z 1/2
h
i1
£√ ¤1
1 1 −1/2
x
1 3/4 1
√
√
du =
dx = −
u
du = 12 2u1/2
=
u 3/4 = 1 −
2
2
2
u
3/4
1
−
x
0
1
3/4
"
#
u = x,
R
dv = sin2 x dx,
R
R
17. x sin2 x dx
v = sin2 x dx = 12 (1 − cos 2x) dx = 12 x − 12 sin x cos x
du = dx
R ¡1
¢
x − 12 sin x cos x dx
= 12 x2 − 12 x sin x cos x −
2
√
3
2
= 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
R
R
Note: sin x cos x dx = s ds = 12 s2 + C [where s = sin x, ds = cos x dx].
R
R
R
R
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 −
1
8
cos 2x + C.
STRATEGY FOR INTEGRATION ■ 9
19. Let u = ex . Then
R
x
ex+e dx =
R
x
ee ex dx =
R
x
eu du = eu + C = ee + C.
21. Integrate by parts three times, first with u = t3 , dv = e−2t dt:
R 3 −2t
R
R
t e
dt = − 12 t3 e−2t + 12 3t2 e−2t dt = − 12 t3 e−2t − 34 t2 e−2t + 12 3te−2t dt
Z
£
¤
£
¤
= −e−2t 12 t3 + 34 t2 − 34 te−2t + 34 e−2t dt = −e−2t 12 t3 + 34 t2 + 34 t + 38 + C
23. Let u = 1 +
¡
¢
= − 18 e−2t 4t3 + 6t2 + 6t + 3 + C
√
x. Then x = (u − 1)2 , dx = 2(u − 1) du ⇒
R1
0
25.
(1 +
¢
£
¤2
R2
R2¡
√ 8
x ) dx = 1 u8 · 2(u − 1) du = 2 1 u9 − u8 du = 15 u10 − 2 · 19 u9 1
=
1024
5
−
1024
9
−
1
5
+
2
9
=
4097
45
3x2 − 2
6x + 22
A
B
=3+
= 3+
+
− 2x − 8
(x − 4)(x + 2)
x−4
x+2
x2
⇒
6x + 22 = A(x + 2) + B(x − 4). Setting
x = 4 gives 46 = 6A, so A = 23
. Setting x = −2 gives 10 = −6B, so B = − 53 . Now
3
¶
Z
Z µ
3x2 − 2
23/3
5/3
5
dx
=
3
+
−
dx = 3x + 23
3 ln |x − 4| − 3 ln |x + 2| + C.
x2 − 2x − 8
x−4
x+2
R
R
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.
¶
Z 5
Z 5µ
h
i5
3w − 1
7
29.
dw =
dw = 3w − 7 ln |w + 2|
3−
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,
√
Z
Z
Z
Z √
Z r
1+x
1+x
1+x
dx
x dx
1+x
√
√
√
√
dx =
+
·√
dx =
dx =
1−x
1−x
1+x
1 − x2
1 − x2
1 − x2
p
= sin−1 x − 1 − x2 + C
p
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
R√
Rp
Rp
3 − 2x − x2 dx =
4 − (x + 1)2 dx =
4 − 4 sin2 θ 2 cos θ dθ
R
R
= 4 cos2 θ dθ = 2 (1 + cos 2θ) dθ
= 2θ + sin 2θ + C = 2θ + 2 sin θ cos θ + C
√
¶
µ
3 − 2x − x2
x+1
−1 x + 1
+2·
·
+C
= 2 sin
2
2
2
µ
¶
x+1
x + 1√
+
= 2 sin−1
3 − 2x − x2 + C
2
2
35. Because f (x) = x8 sin x is the product of an even function and an odd function, it is odd. Therefore,
R1 8
x sin x dx = 0 [by (5.5.7)(b)].
−1
10 ■ STRATEGY FOR INTEGRATION
37.
R π/4
0
cos2 θ tan2 θ dθ =
=
2
R π/4
0
¡π
8
R π/4
£
sin2 θ dθ = 0 12 (1 − cos 2θ) dθ = 12 θ −
¢
− 14 − (0 − 0) = π8 − 14
1
4
sin 2θ
¤π/4
0
39. Let u = 1 − x . Then du = −2x dx ⇒
Z
Z
Z
√
v dv
x dx
du
1
√ =−
√
[v = u, u = v 2 , du = 2v dv]
=−
2
v2 + v
u+ u
1 − x2 + 1 − x2
Z
³p
´
dv
=−
= − ln |v + 1| + C = − ln
1 − x2 + 1 + C
v+1
¡
¢
41. Let u = θ, dv = tan2 θ dθ = sec2 θ − 1 dθ ⇒ du = dθ and v = tan θ − θ. So
R
R
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
R x√
R
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
R x√
R
R
e 1 + ex dx = u · 2u du = 2u2 du = 23 u3 + C = 23 (1 + ex )3/2 + C.
R
R
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,
¢
R
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.
Z
Z
Z
³x´
¡
¢
1
dx
x+a
1
2x dx
dx
=
+
a
= 12 ln x2 + a2 + a · tan−1
47.
+C
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
Z
1
√
dx =
x 4x + 1
53.
Z
1
u du
2
1
2
(u − 1) u
4
=2
Z
¯√
¯
¯ 4x + 1 − 1 ¯
¯+C
= ln¯¯ √
4x + 1 + 1 ¯
51. Let 2x = tan θ
Z
Z
⇒
dx
√
=
x 4x2 + 1
Z
x=
1
2
tan θ, dx =
1
sec2 θ dθ
2
1
2 tan θ sec θ
=
Z
1
2
¯
¯
¡1¢ ¯u − 1¯
du
¯
¯
=
2
ln
2
¯u + 1¯ + C
u2 − 1
sec2 θ dθ,
sec θ
dθ =
tan θ
[by Formula 19]
√
4x2 + 1 = sec θ, so
Z
csc θ dθ
= − ln |csc θ + cot θ| + C
[or ln |csc θ − cot θ| + C]
¯√
¯
¯√
¯
·
¸
¯ 4x2 + 1
¯ 4x2 + 1
1 ¯¯
1 ¯¯
¯
¯
= − ln¯
+
+C
or ln¯
−
+C
2x
2x ¯
2x
2x ¯
#
"
Z
dv = sinh(mx) dx,
u = x2 ,
1 2
2
x cosh(mx) dx
x sinh(mx)dx = x cosh(mx) −
1
m
m
du = 2x dx
v= m
cosh(mx)
"
#
µ
¶
U = x, dV = cosh(mx) dx,
1 2
2 1
1 R
= x cosh(mx) −
x sinh(mx) −
sinh(mx) dx
1
m
m m
m
dU = dx
sinh(mx)
V = m
2
=
2
2
1 2
x cosh(mx) − 2 x sinh(mx) + 3 cosh(mx) + C
m
m
m
STRATEGY FOR INTEGRATION ■ 11
55. Let u =
Z
√
x + 1. Then x = u2 − 1 ⇒
¸
−1
3
+
du
u+1
u+3
¡√
¢
¡√
¢
= 3 ln |u + 3| − ln |u + 1| + C = 3 ln x + 1 + 3 − ln x + 1 + 1 + C
dx
√
=
x+4+4 x+1
57. Let u =
Z
2u du
=
u2 + 3 + 4u
Z ·
√
3
x + c. Then x = u3 − c ⇒
R
x
R¡ 3
¢
R¡
¢
√
3
x + c dx =
u − c u · 3u2 du = 3 u6 − cu3 du = 37 u7 − 34 cu4 + C
= 37 (x + c)7/3 − 34 c(x + c)4/3 + C
59. Let u = ex . Then x = ln u, dx = du/u ⇒
Z
¸
Z
Z ·
du
1/2
1
du/u
1/2
=
=
−
du
−
u3 − u
(u − 1)u2 (u + 1)
u−1
u2
u+1
¯
¯
¯
¯
1 ¯ u − 1 ¯¯
1
1 ¯¯ ex − 1 ¯¯
−x
+
C
=
e
ln
+C
= + ln¯¯
+
u
2
u + 1¯
2 ¯ ex + 1 ¯
dx
=
e3x − ex
Z
61. Let u = x5 . Then du = 5x4 dx ⇒
Z
Z
1
¡ ¢
du
x4 dx
5
=
= 15 · 14 tan−1 14 u + C =
x10 + 16
u2 + 16
63. Let y =
65.
Z
√
x so that dy =
1
20
tan−1
¡1
4
¢
x5 + C.
√
1
√ dx ⇒ dx = 2 x dy = 2y dy. Then
x
¸
·
Z
Z
Z
√ √x
u = 2y2 , dv = ey dy,
x e dx = yey (2y dy) = 2y2 ey dy
v = ey
du = 4y dy
¸
·
Z
U = 4y, dV = ey dy,
= 2y 2 ey − 4yey dy
dU = 4 dy
V = ey
¡
¢
R
= 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
2
√
√ ¶
Z
¡√
√ ¢
x+1− x
1
√
x + 1 − x dx
dx =
√ ·√
√
x+1+ x
x+1− x
h
i
= 23 (x + 1)3/2 − x3/2 + C
dx
√
√ =
x+1+ x
Z µ
√
¡ √ ¢
⇒
67. Let u = t. Then du = dt/ 2 t
√
√
Z 3
Z 3
£
¤ √3
arctan t
√
dt =
tan−1 u (2 du) = 2 u tan−1 u − 12 ln(1 + u2 ) 1
[Example 5 in Section 7.1]
t
1
1
√
¢
¡
¢¤
£¡√
3 tan−1 3 − 12 ln 4 − tan−1 1 − 12 ln 2
=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 ⇒
Z
e2x
dx =
1 + ex
Z
u2 du
=
1+u u
Z
u
du =
1+u
Z µ
1−
1
1+u
= u − ln|1 + u| + C = ex − ln(1 + ex ) + C
¶
du
12 ■ STRATEGY FOR INTEGRATION
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 = − 12 , C = 12 , B = 0, D = 0. Thus,
A + C = 0, B + D = 0, A + 3C = 1, B + 3D = 0 ⇒
Z
x
dx =
x4 + 4x2 + 3
Z µ
1
− 12 x
2x
+
2
2
x +3
x +1
¡
¢
= − 14 ln x2 + 3 +
73.
¶
dx
¡
¢
ln x2 + 1 + C
or
µ 2
¶
x +1
1
ln 2
+C
4
x +3
1
A
Bx + C
=
+ 2
⇒
(x − 2)(x2 + 4)
x−2
x +4
¡ 2
¢
1 = A x + 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 =
Z
75.
1
4
⇒
R
⇒ B = − 18 and C = − 14 . So
¶
Z
Z
Z
− 18 x − 14
1
dx
1
2x dx
1
dx
dx
=
−
−
x−2
x2 + 4
8
x−2
16
x2 + 4
4
x2 + 4
¡
¢
1
= 18 ln|x − 2| − 16
ln x2 + 4 − 18 tan−1 (x/2) + C
1
dx =
(x − 2)(x2 + 4)
Z µ
1
8
1
8
+
R
sin x · 12 [cos(2x − 3x) − cos(2x + 3x)] dx = 12 (sin x cos x − sin x cos 5x) dx
R
R
= 14 sin 2x dx − 12 12 [sin(x + 5x) + sin(x − 5x)] dx
R
1
1
cos 6x − 16
cos 4x + C
= − 18 cos 2x − 14 (sin 6x − sin 4x) dx = − 18 cos 2x + 24
sin x sin 2x sin 3x dx =
R
√
x dx = 23 du. Then
77. Let u = x3/2 so that u2 = x3 and du = 32 x1/2 dx ⇒
Z
Z √
2
x
2
2
3
dx =
du = tan−1 u + C = tan−1 (x3/2 ) + C.
3
1+x
1 + u2
3
3
79. Let u = x, dv = sin2 x cos x dx ⇒ du = dx, v = 13 sin3 x. Then
R
R
R
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
·
¸
Z
y = cos x,
1
1
2
3
= x sin x +
(1 − y ) dy
dy = − sin x dx
3
3
= 13 x sin3 x + 13 y − 19 y 3 + C = 13 x sin3 x +
2
1
3
cos x −
1
9
cos3 x + C
81. The function y = 2xex does have an elementary antiderivative, so we’ll use this fact to help evaluate the integral.
R
R
R 2
R
R 2
2
2
2
(2x2 + 1)ex dx = 2x2 ex dx + ex dx = x(2xex ) dx + ex dx
"
#
2
R x2
R x2
u = x, dv = 2xex dx,
2
x2
= xe − e dx + e dx
= xex + C
x2
du = dx
v=e