Download 7.3 Solutions

822
CHAPTER 7
TECHNIQUES OF INTEGRATION
(e) Take the limit of both sides of the equation obtained at the conclusion of Exercise 78:
lim
m!1
!
2!2 4!4
2m ! 2m
I2m
D lim
!
!!!
m!1 1 ! 3 3 ! 5
2
.2m " 1/.2m C 1/ I2mC1
!
"!
"
!
2!2 4!4
2m ! 2m
I2m
D
lim
!
!!!
lim
:
m!1 1 ! 3 3 ! 5
m!1 I2mC1
2
.2m " 1/.2m C 1/
Finally, using the result from (d), we have
!
2!2 4!4
2m ! 2m
D lim
!
!!!
:
m!1 1 ! 3 3 ! 5
2
.2m " 1/.2m C 1/
7.3 Trigonometric Substitution
Preliminary Questions
1. State the trigonometric substitution appropriate to the given integral:
Z p
Z
(a)
9 " x 2 dx
(b)
x 2 .x 2 " 16/3=2 dx
Z
Z
(c)
x 2 .x 2 C 16/3=2 dx
(d) .x 2 " 5/!2 dx
SOLUTION
(a) x D 3 sin "
(b) x D 4 sec "
2. Is trigonometric substitution needed to evaluate
Z
(c) x D 4 tan "
p
x 9 " x 2 dx?
(d) x D
p
5 sec "
No. There is a factor of x in the integrand outside the radical and the derivative of 9 " x 2 is "2x, so we may use the
substitution u D 9 " x 2 , du D "2x dx to evaluate this integral.
SOLUTION
3. Express sin 2" in terms of x D sin ".
SOLUTION
p
p
1 " sin2 " D 1 " x 2 . Thus,
p
sin 2" D 2 sin " cos " D 2x 1 " x 2 :
First note that if sin " D x, then cos " D
4. Draw a triangle that would be used together with the substitution x D 3 sec ".
SOLUTION
x
!x 2 − 9
3
Exercises
In Exercises 1–4, evaluate the integral by following the steps given.
Z
dx
1. I D
p
9 " x2
Z
(a) Show that the substitution x D 3 sin " transforms I into d", and evaluate I in terms of ".
(b) Evaluate I in terms of x.
SOLUTION
(a) Let x D 3 sin ". Then dx D 3 cos " d", and
p
p
p
p
9 " x 2 D 9 " 9 sin2 " D 3 1 " sin2 " D 3 cos2 " D 3 cos ":
Thus,
I D
Z
(b) If x D 3 sin ", then " D sin!1 . x3 /. Thus,
p
dx
9 " x2
D
Z
3 cos " d"
D
3 cos "
I D " C C D sin!1
#x $
3
Z
d" D " C C:
C C:
S E C T I O N 7.3
2. I D
Z
Trigonometric Substitution
dx
p
x2 x2 " 2
Z
p
1
2 sec " transforms the integral I into
cos "d", and evaluate I in terms of ".
p2
(b) Use a right triangle to show that with the above substitution, sin " D x 2 " 2=x.
(c) Evaluate I in terms of x.
(a) Show that the substitution x D
SOLUTION
p
p
2 sec ". Then dx D 2 sec " tan " d", and
q
p
p
p
p
x 2 " 2 D 2 sec2 " " 2 D 2.sec2 " " 1/ D 2 tan2 " D 2 tan ":
(a) Let x D
Thus,
I D
(b) Since x D
p
Z
dx
p
D
2
x x2 " 2
2 sec ", sec " D
x
p
,
2
p
Z
2 sec " tan " d"
1
p
D
2
2
.2 sec "/. 2 tan "/
Z
d"
1
D
sec "
2
Z
cos " d" D
1
sin " C C:
2
and we construct the following right triangle:
x
x2 − 2
2
p
From this triangle we see that sin " D x 2 " 2=x.
(c) Combining the results from parts (a) and (b),
3. I D
Z
I D
p
1
sin " C C D
2
p
x2 " 2
C C:
2x
dx
4x 2 C 9
Z
1
sec " d".
2
(b) Evaluate I in terms of " (refer to the table of integrals on page 410 in Section 7.2 if necessary).
(c) Express I in terms of x.
(a) Show that the substitution x D
3
2
tan " transforms I into
SOLUTION
(a) If x D
3
2
3
2
sec2 " d", and
s
!
"2
p
p
p
3
4x 2 C 9 D 4 !
tan "
C 9 D 9 tan2 " C 9 D 3 sec2 " D 3 sec "
2
tan ", then dx D
Thus,
I D
Z
p
(b)
I D
(c) Since x D
3
2
dx
4x 2 C 9
1
2
Z
D
Z
sec " d" D
3
2
sec2 " d"
1
D
3 sec "
2
Z
sec " d"
1
ln j sec " C tan "j C C
2
tan ", we construct a right triangle with tan " D
!4x2 + 9
2x
3 :
2x
3
4x 2 C 9, and therefore
ˇ
ˇ
1 ˇ1p 2
2x ˇˇ
ln j sec " C tan "j C C D ln ˇˇ
4x C 9 C
CC
2
3
3 ˇ
ˇp
ˇ
ˇ 4x 2 C 9 C 2x ˇ
p
p
1
1
1
ˇ
ˇ
ln ˇ
ˇ C C D ln j 4x 2 C 9 C 2xj " ln 3 C C D ln j 4x 2 C 9 C 2xj C C
ˇ
ˇ
3
2
2
2
From this triangle, we see that sec " D
I D
1
2
D
1
2
p
1
3
823
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TECHNIQUES OF INTEGRATION
CHAPTER 7
4. I D
Z
dx
.x 2 C 4/2
Z
1
(a) Show that the substitution x D 2 tan " transforms the integral I into
cos2 " d".
8
Z
1
1
(b) Use the formula cos2 " d" D " C sin " cos " to evaluate I in terms of ".
2
2
x
2
(c) Show that sin " D p
and cos " D p
.
2
2
x C4
x C4
(d) Express I in terms of x.
SOLUTION
(a) If x D 2 tan ", then dx D 2 sec2 " d", and
Z
Z
Z
dx
2 sec2 " d"
2
sec2 " d"
I D
D
D
2
2
2
2
16
.x C 4/
.4 tan " C 4/
.tan2 " C 1/2
Z
Z
Z
1
sec2 " d"
1
d"
1
D
D
D
cos2 " d":
8
8
8
.sec2 "/2
sec2 "
R
(b) Using the formula cos2 d" D 12 " C 12 sin " cos ", we get
Z
1
1
1
I D
cos2 " d" D
"C
sin " cos " C C:
8
16
16
x
2:
(c) Since x D 2 tan ", we construct a right triangle with tan " D
x2 + 4
x
2
From this triangle we see that
sin " D p
x
x2
and
C4
cos " D p
2
x2
C4
:
(d) Since x D 2 tan ", then " D tan!1 . x2 /, and
!
"!
"
# $
#x$
1
1
x
2
1
x
!1 x
I D
tan
C
p
p
CC D
tan!1
C
C C:
16
2
16
16
2
8.x 2 C 4/
x2 C 4
x2 C 4
In Exercises 5–10, use the indicated substitution to evaluate the integral.
Z p
4
5.
16 " 5x 2 dx, x D p sin "
5
SOLUTION
Let x D
p4
5
sin ". Then dx D
Since x D
p4
5
cos " d", and
s
!
"2
Z p
4
4
4
16 " 5 p sin "
! p cos " d" D p
16 " 16 sin2 " ! cos " d"
5
5
5
Z
16
cos " ! cos " d" D p
cos2 " d"
5
"
1
8
C sin " cos " C C D p ." C sin " cos "/ C C
2
5
Z p
Z
I D
16 " 5x 2 dx D
Z
4
D p !4
5
!
16 1
D p
"
5 2
p4
5
sin ", we construct a right triangle with sin " D
x
p
5
4 :
4
!16 − 5x 2
x !5
Trigonometric Substitution
S E C T I O N 7.3
From this triangle we see that cos " D
6.
Z
1
4
p
16 " 5x 2 , so we have
8
I D p ." C sin " cos "/ C C
5
!
p !
p
8
x 5 1p
!1 x 5
2
D p sin
C
!
16 " 5x C C
4
4
4
5
p !
8
1 p
!1 x 5
D p sin
C x 16 " 5x 2 C C
4
2
5
1=2
p
0
x2
1 " x2
dx,
x D sin "
Let x D sin ". Then dx D cos " d", and
SOLUTION
p
1 " x2 D
Converting the limits of integration to ", we find
xD
p
1 " sin2 " D
1
) " D sin!1
2
p
cos2 " D cos ":
! "
1
!
D
2
6
x D 0 ) " D sin!1 .0/ D 0
Therefore
!
"ˇ!=6
Z !=6
ˇ
sin2 "
1
1
2
I D
p
dx D
.cos " d"/ D
sin " d" D
" " sin " cos " ˇˇ
2
cos
"
2
2
0
0
0
1"x
0
"
p
p
! " p !#
!
1 1
3
!
3
2! " 3 3
D
"
" Œ0 " 0# D
"
D
:
12 2 2
2
12
8
24
Z
7.
Z
1=2
Z
x2
!=6
dx
p
, x D 3 sec "
x x2 " 9
Let x D 3 sec ". Then dx D 3 sec " tan " d", and
SOLUTION
p
p
p
p
x 2 " 9 D 9 sec2 " " 9 D 3 sec2 " " 1 D 3 tan2 " D 3 tan ":
Thus,
Z
x
p
dx
x2
"9
Since x D 3 sec ", " D sec!1 . x3 /, and
8.
Z
1
1=2 x 2
dx
p
,
x2 C 4
SOLUTION
Z
D
Z
p
.3 sec " tan " d"/
1
D
.3 sec "/.3 tan "/
3
dx
x x2 " 9
D
Z
d" D
1
" C C:
3
#x$
1
sec!1
C C:
3
3
x D 2 tan "
Let x D 2 tan ". Then dx D 2 sec2 " d", and
p
This gives us
Z
x2 C 4 D
p
dx
D
p
2
x x2 C 4
p
p
4 tan2 " C 4 D 2 tan2 " C 1 D 2 sec2 " D 2 sec ":
Z
2 sec2 " d"
1
D
4 tan2 ".2 sec "/
4
Z
sec " d"
1
D
tan2 "
4
Z
cos "
sin2 "
d":
Now use substitution, with u D sin " and du D cos " d". Then
Z
Z
1 # !1 $
1
1
cos "
1
!2
D
C C:
d"
D
u
du
"u
CC D"
2
"
4
4
4
4
sin
sin "
Since x D 2 tan ", we construct a right triangle with tan " D
x
2:
825
826
TECHNIQUES OF INTEGRATION
CHAPTER 7
x2 + 4
x
2
From this triangle we see that sin " D
Z
1
1=2
9.
Z
.x 2
SOLUTION
dx
,
" 4/3=2
p x
.
x 2 C4
dx
p
D "
x2 x2 C 4
Thus
p
2
3
q
1
C
4
p
p i
16
1 hp
4
7
D" 4 5"
17 " 5 :
5D
1
4
4
2
ˇ1
C 4 ˇˇ
ˇ
ˇ
4x
x2
1=2
x D 2 sec "
Let x D 2 sec ". Then dx D 2 sec " tan " d", and
x 2 " 4 D 4 sec2 " " 4 D 4.sec2 " " 1/ D 4 tan2 ":
This gives
I D
Z
dx
D
2
.x " 4/3=2
Z
2 sec " tan " d"
D
.4 tan2 "/3=2
Z
2 sec " tan " d"
1
D
4
8 tan3 "
Z
sec " d"
1
D
4
tan2 "
Z
cos "
sin2 "
d":
Now use substitution with u D sin " and du D cos " d". Then
Z
1
1
"1
I D
u!2 du D " u!1 C C D
C C:
4
4
4 sin "
Since x D 2 sec ", we construct a right triangle with sec " D
x
2:
x
x2 − 4
2
From this triangle we see that sin " D
p
x 2 " 4=x, so therefore
I D
10.
Z
0
1
dx
,
.4 C 9x 2 /2
SOLUTION
Let x D
xD
2
3
2
3
4.
p
"1
x2
"x
CC D p
C C:
" 4=x/
4 x2 " 4
tan "
tan ". Then dx D
2
3
4 C 9x 2 D 4 C 9
sec2 " d", and
!
2
tan "
3
"2
D 4 C 4 tan2 " D 4.1 C tan2 "/ D 4 sec2 "
This gives
Z
Z
sec2 " d"
1
d"
D
24
16 sec4 "
sec2 "
!
"
Z
1
1 1
1
2
D
cos " d" D
" C sin " cos " C C
24
24 2
2
dx
D
.4 C 9x 2 /2
D
Z
2
3
1
." C sin " cos "/ C C
48
The limits of integration are from x D 0 to x D 1. x D 0 corresponds to " D 0, while x D 1 corresponds to the angle " with
tan " D 32 . So we construct a right triangle with tan " D 23 :
!13
2
3
S E C T I O N 7.3
From this triangle we see that sin " D
Z
1
0
11. Evaluate
SOLUTION
Z
p3
13
and cos " D
p2 ,
13
Trigonometric Substitution
827
so that
ˇtan!1 .3=2/
ˇ
dx
1
D
." C sin " cos "/ˇˇ
48
.4 C 9x 2 /2
0
!
! "
"
! "
1
3
3
2
1
3
1
D
tan!1
C p ! p "0"0 D
tan!1
C
48
2
48
2
104
13
13
x dx
p
in two ways: using the direct substitution u D x 2 " 4 and by trigonometric substitution.
x2 " 4
Let u D x 2 " 4. Then du D 2x dx, and
Z
Z
p
p
x dx
1
du
1 # 1=2 $
I1 D
p
D
p D
2u
C C D u C C D x 2 " 4 C C:
2
2
u
x2 " 4
To use trigonometric substitution, let x D 2 sec ". Then dx D 2 sec " tan " d", x 2 " 4 D 4 sec2 " " 4 D 4 tan2 ", and
Z
Z
Z
x dx
2 sec ".2 sec " tan " d"/
I1 D
p
D
D 2 sec2 " d" D 2 tan " C C:
2 tan "
x2 " 4
Since x D 2 sec ", we construct a right triangle with sec " D
x
2:
x
x2 − 4
2
From this triangle we see that
p
I1 D 2
x2 " 4
2
!
CC D
12. Is the substitution u D x 2 " 4 effective for evaluating the integral
tion.
SOLUTION
p
Z
x 2 " 4 C C:
x 2 dx
p
? If not, evaluate using trigonometric substitux2 " 4
p
If u D x 2 " 4, then du D 2x dx, x 2 D u C 4, dx D du=2x D du=2 u C 4, and
I D
Z
x 2 dx
p
D
x2 " 4
Z
.u C 4/
p
u
!
du
p
2 uC4
"
D
1
2
Z
uC4
p
du
u2 C 4u
This substitution is clearly not effective for evaluating this integral.
Instead, use the trigonometric substitution x D 2 sec ". Then dx D 2 sec " tan ",
p
p
x 2 " 4 D 4 sec2 " " 4 D 2 tan ";
and we have
I D
Z
x 2 dx
p
D
x2 " 4
Z
4 sec2 ".2 sec " tan " d"/
D4
2 tan "
Z
sec3 " d":
R
Now use the reduction formula for secm x dx from Section 8.7.2:
%
&
Z
Z
'
(
tan " sec "
1
4 sec3 " d" D 4
C
sec " d" D 2 tan " sec " C 2 ln j sec " C tan "j C C:
2
2
Since x D 2 sec ", we construct a right triangle with sec " D
x
2:
x
x2 − 4
2
p
From this triangle we see that tan " D 21 x 2 " 4. Therefore
ˇ
ˇ
ˇ #
"# $
! p
$ˇˇ
p
ˇx
ˇ
ˇ1
x
1p 2
1 p
1
x2 " 4
C 2 ln ˇˇ C
x " 4ˇˇ C C D x x 2 " 4 C 2 ln ˇˇ x C x 2 " 4 ˇˇ C C:
I D2
2
2
2
2
2
2
828
CHAPTER 7
TECHNIQUES OF INTEGRATION
Finally, since
ˇ
ˇ
! "
p
p
ˇ1
ˇ
1
2
ˇ
ˇ
ln ˇ .x C x " 4/ˇ D ln
C ln jx C x 2 " 4j;
2
2
and ln. 12 / is a constant, we can “absorb” this constant into the constant of integration, so that
I D
p
1 p 2
x x " 4 C 2 ln jx C x 2 " 4j C C:
2
13. Evaluate using the substitution u D 1 " x 2 or trigonometric substitution.
Z
Z
p
x
(a)
p
dx
(b)
x 2 1 " x 2 dx
1 " x2
Z
Z
p
x4
(c)
x 3 1 " x 2 dx
(d)
p
dx
1 " x2
SOLUTION
(a) Let u D 1 " x 2 . Then du D "2x dx, and we have
Z
Z
Z
x
1
"2x dx
1
du
p
dx D "
p
D"
:
2
2
2
2
u1=2
1"x
1"x
(b) Let x D sin ". Then dx D cos " d", 1 " x 2 D cos2 ", and so
Z
Z
Z
p
x 2 1 " x 2 dx D sin2 ".cos "/ cos " d" D sin2 " cos2 " d":
(c) Use the substitution u D 1 " x 2 . Then du D "2x dx, x 2 D 1 " u, and so
Z
Z
Z
p
p
1
1
x 3 1 " x 2 dx D "
x 2 1 " x 2 ."2x dx/ D "
.1 " u/u1=2 du:
2
2
(d) Let x D sin ". Then dx D cos " d", 1 " x 2 D cos2 ", and so
Z
Z
Z
x4
sin4 "
p
dx D
cos " d" D sin4 " d":
cos "
1 " x2
14. Z
Evaluate:
Z
dt
t dt
(a)
(b)
2
3=2
2
.t C 1/
.t C 1/3=2
SOLUTION
(a) Use the substitution t D tan ", so that dt D sec2 " d". Then
Z
Z
Z
Z
dt
sec2 "
sec2 "
D
d"
D
d"
D
cos " d" D sin " C C
.t 2 C 1/3=2
.tan2 " C 1/3=2
.sec2 "/3=2
Since t D tan ", we construct a right triangle with tan " D t:
!t 2 + 1
t
1
From this we see that sin " D
p t
,
t 2 C1
so that the integral is
Z
.t 2
dt
t
D sin " C C D p
CC
3=2
2
C 1/
t C1
(b) Use the substitution u D t 2 C 1, du D 2t dt; then
Z
Z
1
t dt
1
u!3=2 du D "u!1=2 C C D " p
CC
D
2
2
.t 2 C 1/3=2
t C1
S E C T I O N 7.3
Trigonometric Substitution
In Exercises 15–32, evaluate using trigonometric substitution. Refer to the table of trigonometric integrals as necessary.
Z
x 2 dx
15.
p
9 " x2
SOLUTION Let x D 3 sin ". Then dx D 3 cos " d",
9 " x 2 D 9 " 9 sin2 " D 9.1 " sin2 "/ D 9 cos2 ";
and
I D
Z
x 2 dx
p
D
9 " x2
Z
9 sin2 ".3 cos " d"/
D9
3 cos "
Since x D 3 sin ", we construct a right triangle with sin " D
Z
sin2 " d" D 9
%
&
1
1
" " sin " cos " C C:
2
2
x
3:
3
x
9 − x2
From this we see that cos " D
16.
Z
dt
.16 " t 2 /3=2
SOLUTION
p
9 " x 2 =3, and so
#x$ 9 #x$
9
I D sin!1
"
2
3
2 3
p
9 " x2
3
!
CC D
9 !1 # x $ 1 p
sin
" x 9 " x 2 C C:
2
3
2
Let t D 4 sin ". Then dt D 4 cos " d", and
.16 " t 2 /3=2 D .16 " 16 sin2 "/3=2 D .16 cos2 "/3=2 D .4 cos "/3 D 64 cos3 "
so that
I D
Z
dt
D
.16 " t 2 /3=2
Z
4 cos "
1
d" D
16
64 cos3 "
Z
sec2 " d" C C D
1
tan " C C
16
Since t D 4 sin ", we construct a right triangle with sin " D 4t :
4
t
!16 − t2
From this, we see that tan " D
p t
,
16!t 2
so that
I D
Z
1
t
tan " C C D p
CC
16
16 16 " t 2
dx
p
x x 2 C 16
SOLUTION Use the substitution x D 4 tan ", so that dx D 4 sec2 " d". Then
q
q
p
x x 2 C 16 D 4 tan " .4 tan "/2 C 16 D 4 tan " 16.tan2 " C 1/ D 16 tan " sec "
17.
so that
I D
Z
dx
p
D
x x 2 C 16
Z
4 sec2 "
1
d" D
16 tan " sec "
4
Since x D 4 tan ", we construct a right triangle with tan " D
Z
sec "
1
d" D
tan "
4
x
4:
!16 + x2
4
x
Z
1
csc " d" D " ln j csc x C cot xj C C
4
829
830
TECHNIQUES OF INTEGRATION
CHAPTER 7
From this, we see that csc x D
18.
p
x 2 C16
x
and cot x D
4
x,
so that
ˇp
ˇ
ˇ
ˇ
p
1
1 ˇˇ x 2 C 16
4 ˇˇ
1 ˇˇ 4 C x 2 C 16 ˇˇ
I D " ln j csc x C cot xj C C D " ln ˇ
C ˇ C C D " ln ˇ
ˇCC
ˇ
4
4 ˇ
x
xˇ
4 ˇ
x
Z p
12 C 4t 2 dt
SOLUTION
First simplify the integral:
I D
Now let t D
p
3 tan ". Then dt D
p
Z p
Z p
12 C 4t 2 dt D 2
3 C t 2 dt
3 sec2 " d",
3 C t 2 D 3 C 3 tan2 " D 3.1 C tan2 "/ D 3 sec2 ";
and
I D2
Since t D
p
Z p
3 sec2 "
#p
%
&
Z
Z
$
tan " sec "
1
3 sec2 " d" D 6 sec3 " d" D 6
C
sec " d"
2
2
D 3 tan " sec " C 3 ln j sec " C tan "j C C:
3 tan ", we construct a right triangle with tan " D
pt
3
:
!t 2 + 3
t
!3
p
p
From this we see that sec " D t 2 C 3= 3. Therefore,
ˇp
ˇ
!
!
" p2
!
"
ˇ t2 C 3
ˇp
ˇ
p
t
t C3
t ˇˇ
1
ˇ
ˇ
ˇ
I D3 p
p
C 3 ln ˇ p
C p ˇ C C1 D t t 2 C 3 C 3 ln ˇ t 2 C 3 C t ˇ C 3 ln p
C C1
ˇ
3
3
3
3ˇ
3
ˇp
ˇ
p
ˇ
ˇ
D t t 2 C 3 C 3 ln ˇ t 2 C 3 C t ˇ C C;
where C D 3 ln. p1 / C C1 .
3
Z
dx
19.
p
x2 " 9
SOLUTION
Let x D 3 sec ". Then dx D 3 sec " tan " d",
x 2 " 9 D 9 sec2 " " 9 D 9.sec2 " " 1/ D 9 tan2 ";
and
I D
Z
dx
p
D
x2 " 9
Z
3 sec " tan " d"
D
3 tan "
Since x D 3 sec ", we construct a right triangle with sec " D
Z
sec " d" D ln j sec " C tan "j C C:
x
3:
x
x2 − 9
3
p
From this we see that tan " D x 2 " 9=3, and so
ˇ
ˇ
p
! "
ˇ
ˇ
ˇx
ˇ
ˇ
p
p
1
x 2 " 9 ˇˇ
ˇ
ˇ
ˇ
ˇ
ˇ
I D ln ˇ C
C C1 D ln ˇx C x 2 " 9 ˇ C C;
ˇ C C1 D ln ˇx C x 2 " 9 ˇ C ln
ˇ3
ˇ
3
3
where C D ln
)1*
3
C C1 .
S E C T I O N 7.3
Trigonometric Substitution
Z
dt
p
t 2 t 2 " 25
SOLUTION Let t D 5 sec ". Then dt D 5 sec " tan " d",
20.
t 2 " 25 D 25 sec2 " " 25 D 25.sec2 " " 1/ D 25 tan2 ";
and
I D
Z
Z
dt
p
D
2
t
t 2 " 25
5 sec " tan " d"
1
D
2
25
.25 sec "/.5 tan "/
Z
d"
1
D
sec "
25
Z
cos " d" D
1
sin " C C:
25
Since t D 5 sec ", we construct a right triangle with sec " D 5t :
t
!t 2 − 25
5
From this we see that sin " D
p
t 2 " 25=t, and so
1
I D
25
21.
Z
y2
dy
p
5 " y2
SOLUTION
p
Let y D
5 sin ". Then dy D
p
t 2 " 25
t
!
CC D
p
t 2 " 25
C C:
25t
p
5 cos " d",
5 " y 2 D 5 " 5 sin2 " D 5.1 " sin2 "/ D 5 cos2 ";
and
I D
Since y D
p
Z
dy
p
D
2
y
5 " y2
p
Z
Z
5 cos " d"
1
d"
1
1
p
D
D
csc2 " d" D ." cot "/ C C:
2
2
5
5
5
sin "
.5 sin "/. 5 cos "/
Z
5 sin ", we construct a right triangle with sin " D
y
p
:
5
5
y
5 − y2
From this we see that cot " D
22.
Z
p
x 3 9 " x 2 dx
SOLUTION
p
5 " y 2 =y, which gives us
!
p
p
1 " 5 " y2
5 " y2
I D
CC D "
C C:
5
y
5y
Let x D 3 sin ". Then dx D 3 cos " d",
9 " x 2 D 9 " 9 sin2 " D 9.1 " sin2 "/ D 9 cos2 ";
and
I D
Z
Z
p
x 3 9 " x 2 dx D .27 sin3 "/.3 cos "/.3 cos " d"/
D 243
D 243
Z
%Z
sin3 " cos2 " d" D 243
cos2 " sin " d" "
Z
Z
.1 " cos2 "/ cos2 " sin " d"
&
cos4 " sin " d" :
Now use substitution, with u D cos " and du D " sin " d" for both integrals:
%
&
1
1
I D 243 " cos3 " C cos5 " C C:
3
5
Since x D 3 sin ", we construct a right triangle with sin " D
x
3:
831
832
TECHNIQUES OF INTEGRATION
CHAPTER 7
3
x
θ
9 − x2
From this we see that cos " D
2
1
I D 243 4"
3
p
9 " x 2 =3. Thus
p
9 " x2
3
!3
!5 3
p
9 " x2 5
1
C C D "3.9 " x 2 /3=2 C .9 " x 2 /5=2 C C:
3
5
1
C
5
Alternately, let u D 9 " x 2 . Then
!
"
Z
Z
p
p
1
1
2
I D x 3 9 " x 2 dx D "
.9 " u/ u du D "
6u3=2 " u5=2 C C
2
2
5
23.
Z
1 5=2
1
u
" 3u3=2 C C D .9 " x 2 /5=2 " 3.9 " x 2 /3=2 C C:
5
5
D
p
dx
25x 2 C 2
Let x D
SOLUTION
p
2
5
I D
Since x D
p
2
5
p
2
5
tan ". Then dx D
Z
p
dx
25x 2 C 2
D
Z
sec2 " d", 25x 2 C 2 D 2 tan2 " C 2 D 2 sec2 ", and
p
2
5
sec2 " d"
1
p
D
5
2 sec "
tan ", we construct a right triangle with tan " D
Z
sec " d" D
1
ln j sec " C tan "j C C:
5
5x
p
:
2
!25x 2 + 2
5x
!2
From this we see that sec " D
p1
2
p
25x 2 C 2, so that
ˇp
ˇ
1
1 ˇˇ 25x 2 C 2
5x ˇˇ
I D ln j sec " C tan "j C C D ln ˇ
p
C p ˇCC
5
5 ˇ
2
2ˇ
ˇ
ˇ
p
ˇ 1 p
p
1 ˇˇ 5x C 25x 2 C 2 ˇˇ
1 ˇˇ
ˇ
D ln ˇ
p
ˇ C C D ln ˇ5x C 25x 2 C 2ˇ " ln 2 C C
ˇ
5 ˇ
5
5
2
24.
Z
D
dt
.9t 2 C 4/2
SOLUTION
Now let t D
ˇ
p
1 ˇˇ
ˇ
ln ˇ5x C 25x 2 C 2ˇ C C
5
First factor out the t 2 -coefficient:
Z
Z
Z
dt
dt
1
dt
I D
D
D
'
)
*(
)
*2 :
2
2
2
4
81
.9t C 4/
9 t2 C 9
t 2 C 49
2
3
tan ". Then dt D
2
3
sec2 " d",
t2 C
4
4
4
4
4
D tan2 " C D .tan2 " C 1/ D sec2 ";
9
9
9
9
9
and
I D
Since t D
2
3
1
81
Z
2
3
16
81
sec2 d"
sec4 " d"
D
1
24
Z
tan ", we construct a right triangle with tan " D
cos2 " d" D
3t
2:
&
%
1 1
1
" C sin " cos " C C:
24 2
2
S E C T I O N 7.3
!9t 2 + 4
Trigonometric Substitution
833
3t
2
p
p
From this we see that sin " D 3t= 9t 2 C 4 and cos " D 2= 9t 2 C 4. Thus
! "
!
"!
"
! "
1
3t
1
3t
2
1
3t
t
I D
tan!1
C
p
p
CC D
tan!1
C
C C:
2 C4
2 C4
48
2
48
48
2
8.9t 2 C 4/
9t
9t
Z
dz
25.
p
z3 z2 " 4
SOLUTION Let z D 2 sec ". Then dz D 2 sec " tan " d",
z 2 " 4 D 4 sec2 " " 4 D 4.sec2 " " 1/ D 4 tan2 ";
and
Z
Z
Z
dz
2 sec " tan " d"
1
d"
1
p
D
D
D
cos2 " d"
8
8
.8 sec3 "/.2 tan "/
sec2 "
z3 z2 " 4
%
&
1 1
1
1
1
D
" C sin " cos " C C D
"C
sin " cos " C C:
8 2
2
16
16
I D
Z
As explained in the text, this computation is valid if we choose " in Œ0; !=2/ if z # 2 and in Œ!; 3!=2/ if z $ "2. If z # 2, we may
construct a right triangle with sec " D z2 :
z
z2 − 4
2
p
z 2 " 4=z and cos " D 2=z. Then
!! "
p
p
#z $
#z $
1
1
z2 " 4
2
1
z2 " 4
I D
sec!1
C
CC D
sec!1
C
C C:
16
2
16
z
z
16
2
8z 2
) *
)
(
However, if z $ "2 then sec!1 z2 lies in !2 ; ! according to the definition of sec!1 x used in the text. But since " is the angle
h
$
p
) *
in !; 3!
satisfying sec " D z=2, we find that " D 2! " sec!1 z2 . Similarly, sin " D " z 2 " 4=z and cos " D "2=z: So, for
2
) * pz 2 !4
) *
1
z $ "2, I D " 16
sec!1 z2 C 8z
C C . Note that although " D 2! " sec!1 z2 ; the 2! is not needed in the expression for
2
I because it may be absorbed in the constant C .
Z
dy
26.
p
y2 " 9
SOLUTION Let y D 3 sec ", so that dy D 3 sec " tan " d" and
From this we see that sin " D
y 2 " 9 D .3 sec "/2 " 9 D 9.sec2 " " 1/ D 9 tan2 "
so that
I D
Z
p
dy
y2 " 9
D
Z
3 sec " tan "
d" D
3 tan "
Since y D 3 sec ", we construct a right triangle with sec " D
Z
sec " d" D ln j sec " C tan "j C C
y
3:
y
!y 2 − 9
3
p
From this, we see that tan " D 13 y 2 " 9, so that
ˇ
ˇ
p
ˇy
y 2 " 9 ˇˇ
ˇ
I D ln j sec " C tan "j C C D ln ˇ C
ˇCC
ˇ3
ˇ
3
ˇ
ˇ
p
ˇ
ˇ
ˇ
ˇ
q
q
ˇ y C y2 " 9 ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2
2
ˇ
ˇ
ˇ
D ln ˇ
ˇ C C D ln ˇy C y " 9ˇ " ln 3 C C D ln ˇy C y " 9ˇˇ C C
ˇ
ˇ
3
834
27.
TECHNIQUES OF INTEGRATION
CHAPTER 7
Z
x 2 dx
.6x 2 " 49/1=2
SOLUTION
Let x D
p7
6
sec "; then dx D
p7
6
sec " tan " d", and
!
"2
7
6x 2 " 49 D 6 p sec "
" 49 D 49.sec2 " " 1/ D 49 tan2 "
6
so that
Z 49 sec2 ". p7 sec " tan "/
6
x 2 dx
6
D
d"
2
1=2
7 tan "
.6x " 49/
!
"
Z
Z
49
49
1
1
D p
sec3 " d" D p
tan " sec " C
sec " d"
2
6 6
6 6 2
I D
D
Since x D
p7
6
Z
49
p .tan " sec " C ln j sec " C tan "j/ C C
12 6
sec ", we construct a right triangle with sec " D
x
p
6
7
x !6
:
!6x 2 − 49
7
From this we see that tan " D
1
7
p
6x 2 " 49, so that
49
I D p
12 6
28.
Z
D
49
p
12 6
D
ˇ p
ˇ$
p
1 # p p 2
ˇ
ˇ
p x 6 6x " 49 C 49 ln ˇx 6 C 6x 2 " 49ˇ C C
12 6
dx
.x 2 " 4/2
SOLUTION
ˇ p
ˇ!
p p
ˇ x 6 C p6x 2 " 49 ˇ
x 6 6x 2 " 49
ˇ
ˇ
C ln ˇ
ˇ CC
ˇ
ˇ
49
7
!
p p
ˇ p
ˇ
p
x 6 6x 2 " 49
ˇ
ˇ
2
C ln ˇx 6 C 6x " 49ˇ " ln 7 C C
49
Let x D 2 sec ". Then dx D 2 sec " tan " d",
x 2 " 4 D 4 sec2 " " 4 D 4.sec2 " " 1/ D 4 tan2 ";
and
Z
Z
dx
2 sec " tan " d"
1
sec " d"
D
D
8
.x 2 " 4/2
16 tan4 "
tan3 "
Z
Z
Z
Z
1
cos2 "
1
1 " sin2 "
1
1
3
D
d"
D
d"
D
csc
"
d"
"
csc " d":
8
8
8
8
sin3 "
sin3 "
Z
Now use the reduction formula for csc3 " d":
I D
I D
Z
%
&
Z
Z
Z
1
cot " csc "
1
1
1
1
"
C
csc " d" "
csc " d" D " cot " csc " "
csc " d"
8
2
2
8
16
16
D"
1
1
cot " csc " "
ln j csc " " cot "j C C:
16
16
Since x D 2 sec ", we construct a right triangle with sec " D
x
2:
x
2
x2 − 4
Trigonometric Substitution
S E C T I O N 7.3
p
p
From this we see that cot " D 2= x 2 " 4 and csc " D x= x 2 " 4. Thus
ˇ
ˇ
!
"!
"
ˇ
1
2
x
1 ˇˇ
x
2
ˇCC
I D"
p
p
"
ln ˇ p
"p
ˇ
2
2
2
2
16
16
x "4
x "4
x "4
x "4
ˇ
ˇ
"x
1 ˇˇ x " 2 ˇˇ
D
"
ln p
C C:
8.x 2 " 4/ 16 ˇ x 2 " 4 ˇ
Z
dt
29.
.t 2 C 9/2
SOLUTION
Let t D 3 tan ". Then dt D 3 sec2 " d",
t 2 C 9 D 9 tan2 " C 9 D 9.tan2 " C 1/ D 9 sec2 ";
and
I D
Z
dt
D
.t 2 C 9/2
Z
3 sec2 " d"
1
D
27
81 sec4 "
Z
cos2 " d" D
%
&
1 1
1
" C sin " cos " C C:
27 2
2
Since t D 3 tan ", we construct a right triangle with tan " D 3t :
!t 2 + 9
t
3
p
p
From this we see that sin " D t= t 2 C 9 and cos " D 3= t 2 C 9. Thus
! "
!
"!
"
! "
1
t
1
t
3
1
t
t
I D
tan!1
C
p
p
CC D
tan!1
C
C C:
2 C 9/
2
2
54
3
54
54
3
18.t
t C9
t C9
Z
dx
30.
.x 2 C 1/3
SOLUTION
Let x D tan ". Then dx D sec2 " d", x 2 C 1 D tan2 " C 1 D sec2 ", and
I D
Using the reduction formula for
I D
Z
Z
dx
D
2
.x C 1/3
Z
sec2 " d"
D
sec6 "
Z
cos4 " d":
cos4 " d", we get
cos3 " sin "
3
C
4
4
Z
cos2 " d" D
1
3
cos3 " sin " C
4
4
!
"
1
1
" C sin " cos " C C:
2
2
Since x D tan ", we construct the following right triangle:
x2 + 1
x
1
p
p
From this we see that sin " D x= x 2 C 1 and cos " D 1= x 2 C 1. Thus
I D
D
31.
Z
x 2 dx
.x 2 " 1/3=2
SOLUTION
1
4
!
p
1
x2 C 1
"3 !
p
x
x2 C 1
"
C
3
3
tan!1 x C
8
8
x
3x
3
C
C tan!1 x C C:
8
4.x 2 C 1/2
8.x 2 C 1/
!
p
x
x2 C 1
"!
p
1
x2 C 1
Let x D sec ". Then dx D sec " tan " d", and x 2 " 1 D sec2 " " 1 D tan2 ". Thus
I D
Z
x2
dx D
2
.x " 1/3=2
Z
sec2 "
sec " tan " d"
.tan2 "/3=2
"
CC
835
836
CHAPTER 7
TECHNIQUES OF INTEGRATION
Z
sec2 " sec " tan "
sec3 "
d"
D
d"
tan3 "
tan2 "
Z
Z
Z
sec2 "
2
D
sec
"
d"
D
csc
"
sec
"
d"
D
.1 C cot2 "/ sec " d"
tan2 "
Z
D sec " C cot " csc " d" D ln j sec " C tan "j " csc " C C
D
Z
Since x D sec ", we construct the following right triangle:
x
!x2 − 1
1
From this we see that tan " D
Z
p
x 2 " 1 and that csc " D
p x
,
x 2 !1
so that
ˇ
ˇ
p
x
ˇ
ˇ
I D ln ˇx C x 2 " 1ˇ " p
CC
2
x "1
x 2 dx
C 1/3=2
SOLUTION Let x D tan ". Then dx D sec2 " d", x 2 C 1 D tan2 " C 1 D sec2 ", and
Z
Z
Z
Z
Z
x 2 dx
tan2 ".sec2 " d"/
tan2 "
sin2 "
1 " cos2 "
I D
D
D
d"
D
d"
D
d"
sec "
cos "
cos "
.x 2 C 1/3=2
.sec2 "/3=2
Z
Z
Z
Z
1
cos2 "
D
d" "
d" D sec " d" " cos " d" D ln j sec " C tan "j " sin " C C:
cos "
cos "
32.
.x 2
Since x D tan ", we construct the following right triangle:
x2 + 1
x
1
From this we see that sec " D
33. Prove for a > 0:
p
p
x 2 C 1 and sin " D x= x 2 C 1. Thus
ˇp
ˇ
x
ˇ
ˇ
I D ln ˇ x 2 C 1 C x ˇ " p
C C:
x2 C 1
Z
dx
1
x
D p tan!1 p C C
x2 C a
a
a
p
p
SOLUTION Let x D a u. Then, x 2 D au2 , dx D a du, and
!
"
Z
Z
dx
1
du
1
1
x
!1
!1
D
p
D
p
tan
u
C
C
D
p
tan
p
C C:
x2 C a
u2 C 1
a
a
a
a
34. Prove for a > 0:
!
"
Z
dx
1
x
1
!1 x
D
C
p
tan
p
CC
2a x 2 C a
.x 2 C a/2
a
a
p
p
SOLUTION Let x D a u. Then, x 2 D au2 , dx D a du, and
Z
Z
dx
1
du
D
:
.x 2 C a/2
.u2 C 1/2
a3=2
Now, let u D tan ". Then du D sec2 " d", and
!
"
Z
Z
Z
sec2 "
1
1
1
1
dx
1
2
cos
"
d"
D
D
d"
D
sin
"
cos
"
C
"
CC
2
.x 2 C a/2
.sec2 "/2
a3=2
a3=2
a3=2 2
p
!
""
!
"
!
1
x
u
1
x= a
!1
!1
D
C
tan
p
C
tan
p
CC
u
C
C
D
a
2a3=2 1 C u2
2a3=2 1 C .x= a/2
!
!
""
x
1
x
1
!1
C C:
D
C
p
tan
p
2a x 2 C a
a
a
S E C T I O N 7.3
Z
Trigonometric Substitution
dx
.
x 2 " 4x C 8
(a) Complete the square to show that x 2 " 4x C 8 D .x " 2/2 C 4.
Z
du
(b) Use the substitution u D x " 2 to show that I D
p
. Evaluate the u-integral.
2 C 22
u
ˇq
ˇ
ˇ
ˇ
(c) Show that I D ln ˇ .x " 2/2 C 4 C x " 2ˇ C C .
35. Let I D
p
SOLUTION
(a) Completing the square, we get
x 2 " 4x C 8 D x 2 " 4x C 4 C 4 D .x " 2/2 C 4:
(b) Let u D x " 2. Then du D dx, and
Z
I D
p
dx
x 2 " 4x C 8
D
Z
Now let u D 2 tan ". Then du D 2 sec2 " d",
dx
p
D
.x " 2/2 C 4
Z
p
du
u2 C 4
:
u2 C 4 D 4 tan2 " C 4 D 4.tan2 " C 1/ D 4 sec2 ";
and
I D
Z
2 sec2 " d"
D
2 sec "
Z
sec " d" D ln j sec " C tan "j C C:
Since u D 2 tan ", we construct a right triangle with tan " D
u
2:
u2 + 4
u
2
p
From this we see that sec " D u2 C 4=2. Thus
ˇp
ˇ
"
ˇ u2 C 4
ˇp
ˇ ! 1
ˇp
ˇ
u ˇˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2
I D ln ˇ
C ˇ C C1 D ln ˇ u C 4 C uˇ C ln C C1 D ln ˇ u2 C 4 C uˇ C C:
ˇ
2
2ˇ
2
(c) Substitute back for x in the result of part (b):
Z
ˇq
ˇ
ˇ
ˇ
I D ln ˇˇ .x " 2/2 C 4 C x " 2ˇˇ C C:
dx
. First complete the square to write 12x " x 2 D 36 " .x " 6/2 .
12x " x 2
SOLUTION First complete the square:
#
$
#
$
12x " x 2 D " x 2 " 12x C 36 " 36 D " x 2 " 12x C 36 C 36 D 36 " .x " 6/2 :
36. Evaluate
p
Now let u D x " 6, and du D dx. This gives us
Z
Z
Z
dx
dx
du
I D
p
D
p
D
p
:
2
2
36 " .x " 6/
12x " x
36 " u2
Next, let u D 6 sin ". Then du D 6 cos " d",
36 " u2 D 36 " 36 sin2 " D 36.1 " sin2 "/ D 36 cos2 ";
and
I D
Z
6 cos " d"
D
6 cos "
Z
d" D " C C:
Substituting back, we find
I D sin!1
#u$
6
C C D sin!1
!
x"6
6
"
C C:
837
838
TECHNIQUES OF INTEGRATION
CHAPTER 7
In Exercises 37–42, evaluate the integral by completing the square and using trigonometric substitution.
Z
dx
37.
p
x 2 C 4x C 13
First complete the square:
SOLUTION
x 2 C 4x C 13 D x 2 C 4x C 4 C 9 D .x C 2/2 C 9:
Let u D x C 2. Then du D dx, and
I D
Z
p
dx
x2
Z
D
C 4x C 13
p
Now let u D 3 tan ". Then du D 3 sec2 " d",
dx
.x
C 2/2
C9
D
Z
p
du
u2 C 9
:
u2 C 9 D 9 tan2 " C 9 D 9.tan2 " C 1/ D 9 sec2 ";
and
I D
Z
3 sec2 " d"
D
3 sec "
Z
sec " d" D ln j sec " C tan "j C C:
Since u D 3 tan ", we construct the following right triangle:
u2 + 9
u
3
u2 C 9=3. Thus
ˇp
ˇ
"
ˇ u2 C 9
ˇp
ˇ ! 1
u ˇˇ
ˇ
ˇ
ˇ
I D ln ˇ
C ˇ C C1 D ln ˇ u2 C 9 C uˇ C ln C C1
ˇ
3
3ˇ
3
ˇq
ˇ
ˇ
ˇ
p
ˇ
ˇ
ˇ
ˇ
D ln ˇˇ .x C 2/2 C 9 C x C 2ˇˇ C C D ln ˇ x 2 C 4x C 13 C x C 2ˇ C C:
From this we see that sec " D
38.
Z
p
p
dx
2 C x " x2
First complete the square:
SOLUTION
Let u D x "
1
2
Now let u D
3
2
!
"
!
"
)
*
1
1
9
1 2
2 C x " x2 D " x2 " x C 2 D " x2 " x C
C2C D " x"
:
4
4
4
2
and du D dx. This gives us
Z
I D
p
sin ". Then du D
3
2
dx
2Cx
" x2
D
Z
q
dx
9
4
" .x " 12 /2
D
Z
q
du
9
4
:
" u2
cos " d",
9
9 9
9
9
" u2 D " sin2 " D .1 " sin2 "/ D cos2 ";
4
4 4
4
4
and
I D
39.
Z
p
Z
3
2
cos " d"
3
2
cos "
D
Z
d" D " C C D sin!1
!
2u
3
"
C C D sin!1
2.x " 12 /
3
!
C C D sin!1
dx
x C 6x 2
SOLUTION
First complete the square:
!
"
!
"
p
1
1
1 2
1
6x 2 C x D 6x 2 C x C
"
D
6x C p
"
24
24
24
2 6
!
2x " 1
3
"
C C:
S E C T I O N 7.3
Let u D
p
6x C
1
p
2 6
Z
I D
Now let u D
1
p
2 6
p
so that du D
6 dx. Then
1
p
dx D
x C 6x 2
1
p
2 6
sec ". Then du D
Trigonometric Substitution
Z
r#
p
1
6x C
1
p
2 6
$2
"
1
24
1
dx D p
6
Z
1
q
u2 "
1
24
du
sec " tan ", and
u2 "
1
1
1
D
.sec2 " " 1/ D
tan2 "
24
24
24
so that
1
I D p
6
Since u D
1
p
2 6
Z
1
1
p
2 6
1
1
p sec " tan " d" D p
tan " 2 6
6
Z
1
sec " d" D p ln j sec " C tan "j C C
6
sec ", we construct the following right triangle:
2u!6
!24u2 − 1
1
p
24u2 " 1 and sec " D 2u 6. Thus
ˇ
ˇ
" s !
"
ˇ p
ˇ
ˇ p !p
ˇ
p
1
1
1
1
ˇ
ˇ
ˇ
ˇ
2
2
I D p ln ˇ2u 6 C 24u " 1ˇ C C D p ln ˇ2 6
6x C p
C 24 6x C x C
" 1ˇ C C
ˇ
ˇ
24
6
6
2 6
ˇ
ˇ
p
1
ˇ
ˇ
D p ln ˇ12x C 1 C 144x 2 C 24x ˇ C C
6
from which we see that tan " D
40.
p
Z p
x 2 " 4x C 7 dx
SOLUTION
First complete the square:
x 2 " 4x C 7 D x 2 " 4x C 4 C 3 D .x " 2/2 C 3:
Let u D x " 2. Then du D dx, and
Z p
Z q
Z p
I D
x 2 " 4x C 7 dx D
.x " 2/2 C 3 dx D
u2 C 3 du:
Now let u D
p
3 tan ". Then du D
p
3 sec2 " d",
u2 C 3 D 3 tan2 " C 3 D 3.tan2 " C 1/ D 3 sec2 ";
and
I D
D
Since u D
Z p
%
&
Z
Z
p
tan " sec "
1
3 sec2 " 3 sec2 " d" D 3 sec3 " d" D 3
C
sec " d"
2
2
3
3
tan " sec " C ln j sec " C tan "j C C:
2
2
p
3 tan ", we construct a right triangle with tan " D
u
p
:
3
u2 + 3
u
3
From this we see that sec " D
p
u2 C 3=3. Thus
ˇp
ˇ
!
!
" p 2
3
u
u C3
3 ˇˇ u2 C 3
u ˇˇ
I D
p
p
C ln ˇ p
C p ˇ C C1
2
2 ˇ
3
3
3
3ˇ
839
840
CHAPTER 7
TECHNIQUES OF INTEGRATION
"
ˇ !3
1 p 2
3 ˇˇp
1
ˇ
u u C 3 C ln ˇ u2 C 3 C uˇ C
ln p C C1
2
2
2
3
ˇq
ˇ
q
ˇ
ˇ
1
3
D .x " 2/ .x " 2/2 C 3 C ln ˇˇ .x " 2/2 C 3 C x " 2ˇˇ C C
2
2
ˇ
p
1
3 ˇˇp
ˇ
D .x " 2/ x 2 " 4x C 7 C ln ˇ x 2 " 4x C 7 C x " 2ˇ C C:
2
2
D
41.
Z p
x 2 " 4x C 3 dx
SOLUTION
First complete the square:
x 2 " 4x C 3 D x 2 " 4x C 4 " 1 D .x " 2/2 " 1:
Let u D x " 2. Then du D dx, and
Z p
Z q
Z p
I D
x 2 " 4x C 3 dx D
.x " 2/2 " 1 dx D
u2 " 1 du:
Now let u D sec ". Then du D sec " tan " d", u2 " 1 D sec2 " " 1 D tan2 ", and
Z p
Z
Z #
$
I D
tan2 " .sec " tan " d"/ D tan2 " sec " d" D
sec2 " " 1 sec " d"
D
Z
D
1
1
tan " sec " "
2
2
sec3 " d" "
Z
sec " d" D
Z
!
tan " sec "
1
C
2
2
Z
" Z
sec " d" " sec " d"
1
1
tan " sec " " ln j sec " C tan "j C C:
2
2
sec " d" D
Since u D sec ", we construct the following right triangle:
u
u2 − 1
1
p
From this we see that tan " D u2 " 1. Thus
ˇ
ˇ
q
q
ˇ
p
ˇ
1 p
1 ˇˇ
1
1 ˇ
ˇ
I D u u2 " 1 " ln ˇu C u2 " 1ˇ C C D .x " 2/ .x " 2/2 " 1 " ln ˇˇx " 2 C .x " 2/2 " 1ˇˇ C C
2
2
2
2
ˇ
ˇ
p
p
1
1
ˇ
ˇ
D .x " 2/ x 2 " 4x C 3 " ln ˇx " 2 C x 2 " 4x C 3ˇ C C:
2
2
Z
dx
42.
2
.x C 6x C 6/2
SOLUTION
First complete the square:
x 2 C 6x C 6 D x 2 C 6x C 9 " 3 D .x C 3/2 " 3:
Let u D x C 3. Then du D dx, and
I D
Now let u D
p
3 sec ". Then du D
Z
dx
D
.x 2 C 6x C 6/2
p
3 sec " tan ",
Z
dx
D
..x C 3/2 " 3/2
Z
du
:
.u2 " 3/2
u2 " 3 D 3 sec2 " " 3 D 3.sec2 " " 1/ D 3 tan2 ";
and
p Z
p Z
p Z
Z p
3 sec " tan " d"
3
sec " d"
3
cos2 "
3
.1 " sin2 "/ d"
I D
D
D
d"
D
9
9
9
9 tan4 "
tan3 "
sin3 "
sin3 "
p %Z
p %!
&
" Z
&
Z
Z
3
3
cot " csc "
1
D
csc3 " d" " csc " d" D
"
C
csc " d" " csc " d"
9
9
2
2
p %
p
p
&
Z
3
1
1
3
3
D
" cot " csc " "
csc " d" D "
cot " csc " "
ln j csc " " cot "j C C:
9
2
2
18
18
Since u D
p
3 sec ", we construct a right triangle with sec " D
u
p
:
3
S E C T I O N 7.3
Trigonometric Substitution
841
u
u2 − 3
3
p
p p
From this we see that cot " D 3= u2 " 3 and csc " D u= u2 " 3. Thus
ˇ
ˇ
!!
p
p
p
" p
3
3
u
3 ˇˇ
u
3 ˇˇ
I D"
p
p
"
ln ˇ p
"p
ˇCC
18
18 ˇ u2 " 3
u2 " 3
u2 " 3
u2 " 3 ˇ
ˇ
ˇ
p
p ˇ
p
p ˇ
"u
3 ˇˇ u " 3 ˇˇ
".x C 3/
3 ˇˇ x C 3 " 3 ˇˇ
D
"
ln ˇ p
"
ln ˇ p
ˇCC D
ˇCC
18 ˇ u2 " 3 ˇ
18 ˇ .x C 3/2 " 3 ˇ
6.u2 " 3/
6..x C 3/2 " 3/
ˇ
p
p ˇ
".x C 3/
3 ˇˇ x C 3 " 3 ˇˇ
D
"
ln
p
ˇ
ˇ C C:
18 ˇ x 2 C 6x C 6 ˇ
6.x 2 C 6x C 6/
In Exercises 43–52, indicate a good method for evaluating the integral (but do not evaluate). Your choices are: substitution (specify
u and du), Integration by Parts (specify u and v 0 ), a trigonometric method, or trigonometric substitution (specify). If it appears
that these techniques are not sufficient, state this.
Z
x dx
43.
p
12 " 6x " x 2
p
SOLUTION Complete the square so the the denominator is 15 " .x C 3/2 and then use trigonometric substitution with x C 3 D
sin ".
Z p
44.
4x 2 " 1 dx
SOLUTION
45.
Z
Use trigonometric substitution, with x D
Z
47.
methods: rewrite sin3 x D .1 " cos2 x/ sin x and let u D cos x, or rewrite
x sec2 x dx
SOLUTION
Z
sec ".
sin3 x cos3 x dx
SOLUTION Use one of the following trigonometric
cos3 x D .1 " sin2 x/ cos x and let u D sin x.
46.
1
2
p
Use Integration by Parts, with u D x and v 0 D sec2 x.
dx
9 " x2
SOLUTION
Either use the substitution x D 3u and then recognize the formula for the inverse sine:
Z
du
D sin!1 u C C;
p
1 " u2
or use trigonometric substitution, with x D 3 sin ".
Z p
48.
1 " x 3 dx
SOLUTION
49.
Z
sin3=2 x dx
SOLUTION
50.
Z
51.
Not solvable by any method yet considered.
p
x 2 x C 1 dx
SOLUTION
Z
Not solvable by any method yet considered. (In fact, this has no antiderivative using elementary functions).
Use integration by parts twice, first with u D x 2 and then with u D x.
dx
.x C 1/.x C 2/3
SOLUTION The techniques we have covered thus far are not sufficient to treat this integral. This integral requires a technique
known as partial fractions.
842
52.
TECHNIQUES OF INTEGRATION
CHAPTER 7
Z
dx
.x C 12/4
SOLUTION
Use the substitution u D x C 12, and then recognize the formula
Z
1
u!4 du D " 3 C C:
3u
In Exercises 53–56, evaluate using Integration by Parts as a first step.
Z
53.
sec!1 x dx
SOLUTION
p
Let u D sec!1 x and v 0 D 1. Then v D x, u0 D 1=x x 2 " 1, and
Z
Z
Z
x
dx
I D sec!1 x dx D x sec!1 x "
p
dx D x sec!1 x "
p
:
2
x x "1
x2 " 1
To evaluate the integral on the right, let x D sec ". Then dx D sec " tan " d", x 2 " 1 D sec2 " " 1 D tan2 ", and
Z
Z
Z
ˇ
ˇ
p
dx
sec " tan " d"
ˇ
ˇ
p
D
D sec " d" D ln j sec " C tan "j C C D ln ˇx C x 2 " 1 ˇ C C:
tan "
x2 " 1
Thus, the final answer is
54.
Z
ˇ
ˇ
p
ˇ
ˇ
I D x sec!1 x " ln ˇx C x 2 " 1 ˇ C C:
sin!1 x
dx
x2
SOLUTION
p
Let u D sin!1 x and v 0 D x !2 . Then u0 D 1= 1 " x 2 , v D "x !1 , and
I D
Z
sin!1 x
sin!1 x
dx D "
C
2
x
x
Z
dx
p
:
x 1 " x2
To evaluate the integral on the right, let x D sin ". Then dx D cos " d", 1 " x 2 D 1 " sin2 " D cos2 ", and
Z
Z
Z
dx
cos " d"
p
D
D csc " d" D ln j csc " " cot "j C C:
.sin "/.cos "/
x 1 " x2
Since x D sin ", we construct the following right triangle:
1
x
1 − x2
The final answer is
55.
Z
ˇ
ˇ 1 " p1 " x 2
sin!1 x
ˇ
I D"
C ln ˇ
ˇ
x
x
ln.x 2 C 1/ dx
SOLUTION
p
1 " x 2 =x. Thus
ˇ
ˇ
ˇ
Z
ˇ 1 p1 " x 2 ˇ
ˇ 1 " p1 " x 2
dx
ˇ
ˇ
ˇ
p
D ln ˇ "
ˇ C C D ln ˇ
ˇx
ˇ
ˇ
x
x
x 1 " x2
From this we see that csc " D 1=x and cot " D
ˇ
ˇ
ˇ
ˇ C C:
ˇ
Start by using integration by parts, with u D ln.x 2 C 1/ and v 0 D 1; then u0 D
I D
Z
ln.x 2 C 1/ dx D x ln.x 2 C 1/ " 2
D x ln.x 2 C 1/ " 2x C 2
Z
1
dx
x2 C 1
Z
x2
dx D x ln.x 2 C 1/ " 2
2
x C1
ˇ
ˇ
ˇ
ˇ C C:
ˇ
2x
x 2 C1
and v D x, so that
Z !
1"
1
x2 C 1
"
dx
Trigonometric Substitution
S E C T I O N 7.3
To deal with the remaining integral, use the substitution x D tan ", so that dx D sec2 " d" and
Z
Z
Z
Z
1
sec2 "
sec2 "
dx
D
d"
D
d"
D
1 d" D " D tan!1 x C C
x2 C 1
tan2 " C 1
sec2 "
so that finally
56.
Z
I D x ln.x 2 C 1/ " 2x C 2 tan!1 x C C
x 2 ln.x 2 C 1/ dx
SOLUTION
Start by using integration by parts with u D ln.x 2 C 1/, v 0 D x 2 ; then u0 D
I D
Z
x 2 ln.x 2 C 1/ dx D
1 3
2
x ln.x 2 C 1/ "
3
3
Z
2x
x 2 C1
and v D
1 3
3x ,
so that
x4
dx
C1
x2
To deal with the remaining integral, use the substitution x D tan "; then dx D sec2 " d" and
Z
Z
Z
Z
x4
tan4 "
tan4 "
2
2
dx
D
sec
"
d"
D
sec
"
d"
D
tan4 " d"
x2 C 1
tan2 " C 1
sec2 "
Using the reduction formula for tann gives
Z
Z
1
1
tan4 " d" D tan3 " " tan2 " d" D tan3 " " tan " C " C C
3
3
so that, substituting back for x D tan ", we get
!
"
1
2 1 3
1
2
2
2
I D x 3 ln.x 2 C 1/ "
x " x C tan!1 x C C D x 3 ln.x 2 C 1/ " x 3 C x " tan!1 x C C
3
3 3
3
9
3
3
p
57. Find the average height of a point on the semicircle y D 1 " x 2 for "1 $ x $ 1.
SOLUTION
The average height is given by the formula
yave D
1
1 " ."1/
Z
Z
p
1 1p
1 " x 2 dx D
1 " x 2 dx
2 !1
!1
1
Let x D sin ". Then dx D cos " d", 1 " x 2 D cos2 ", and
Z p
Z
Z
1
1
1 " x 2 dx D .cos "/.cos " d"/ D cos2 " d" D " C sin " cos " C C:
2
2
Since x D sin ", we construct the following right triangle:
1
x
1 − x2
p
1 " x 2 . Therefore,
!
"ˇ1
%!
" !
"&
ˇ
1 1 !1
1 p
1
1
1
!
yave D
sin x C x 1 " x 2 ˇˇ D
! C0 " " ! C0 D :
2 2
2
2
2
2
4
!1
p
58. Find the volume of the solid obtained by revolving the graph of y D x 1 " x 2 over Œ0; 1# about the y-axis.
From this we see that cos " D
SOLUTION
Using the method of cylindrical shells, the volume is given by
V D 2!
Z
0
1
Z
# p
$
x x 1 " x 2 dx D 2!
0
1
p
x 2 1 " x 2 dx:
To evaluate this integral, let x D sin ". Then dx D cos " d",
1 " x 2 D 1 " sin2 " D cos2 ";
and
I D
Z
Z
Z #
Z
Z
$
p
x 2 1 " x 2 dx D sin2 " cos2 " d" D
1 " cos2 " cos2 " d" D cos2 " d" " cos4 " d":
843
844
TECHNIQUES OF INTEGRATION
CHAPTER 7
Now use the reduction formula for
R
cos4 " d":
#
Z
Z
cos3 " sin "
3
1
1
C
cos2 " d" D " cos3 " sin " C
cos2 " d"
4
4
4
4
%
&
1
1 1
1
1
1
1
D " cos3 " sin " C
" C sin " cos " C C D " cos3 " sin " C " C sin " cos " C C:
4
4 2
2
4
8
8
p
Since sin " D x, we know that cos " D 1 " x 2 . Then we have
I D
Z
cos2 " d" "
"
I D"
Now we can complete the volume:
V D 2!
*3=2
1)
1
1 p
1 " x2
x C sin!1 x C x 1 " x 2 C C:
4
8
8
!
"ˇ1
h#
$
i
ˇ
*3=2 1 !1
1 )
1 p
!
!2
" x 1 " x2
C sin x C x 1 " x 2 ˇˇ D 2! 0 C
C 0 " .0/ D
:
4
8
8
16
8
0
59. Find the volume of the solid obtained by revolving the region between the graph of y 2 " x 2 D 1 and the line y D 2 about the
line y D 2.
SOLUTION
First solve the equation y 2 " x 2 D 1 for y:
p
y D ˙ x 2 C 1:
The region in question is bounded in part by the top half of this hyperbola, which is the equation
p
y D x 2 C 1:
The limits of integration are obtained by finding the points of intersection of this equation with y D 2:
p
p
2 D x 2 C 1 ) x D ˙ 3:
p
The radius of each disk is given by 2 " x 2 C 1; the volume is therefore given by
V D
Z
p
3
p !r
! 3
Z p3
D 8!
To evaluate the integral
0
2
dx D 2!
dx " 8!
Z
Z
p
0
p
3#
3p
Z
$2
p
2 " x 2 C 1 dx D 2!
0
x 2 C 1 dx C 2!
0
p
Z
p
0
3
3h
i
p
4 " 4 x 2 C 1 C .x 2 C 1/ dx
.x 2 C 1/ dx:
Z p
x 2 C 1 dx, let x D tan ". Then dx D sec2 " d", x 2 C 1 D sec2 ", and
Z p
Z
Z
1
1
x 2 C 1 dx D sec3 " d" D tan " sec " C
sec " d"
2
2
D
ˇ
1
1
1 p
1 ˇˇp
ˇ
tan " sec " C ln j sec " C tan "j C C D x x 2 C 1 C ln ˇ x 2 C 1 C x ˇ C C:
2
2
2
2
Now we can compute the volume:
%
V D 8!x " 8!
!
&ˇp3
ˇ" 2
ˇ
1 p 2
1 ˇˇp 2
ˇ
x x C 1 C ln ˇ x C 1 C x ˇ C !x 3 C 2!x ˇˇ
2
2
3
0
p
ˇp
ˇ"ˇˇ 3
p
2 3
ˇ
ˇ ˇ
2
2
D 10!x C !x " 4!x x C 1 " 4! ln ˇ x C 1 C x ˇ ˇ
3
0
ˇ
ˇ
#
hp
p
p
p
p ˇˇ$
p ˇˇi
ˇ
ˇ
D 10! 3 C 2! 3 " 8! 3 " 4! ln ˇ2 C 3ˇ " .0/ D 4!
3 " ln ˇ2 C 3ˇ :
!
60. Find the volume of revolution for the region in Exercise 59, but revolve around y D 3.
SOLUTION
Using the washer method, the volume is given by
V D
Z
p
3
#
p ! R
! 3
Z p3 #
D 2!
0
2
"r
2
$
dx D 2!
Z
p
0
3 %#
3"
p
x2 C 1
$2
Z
#
$
$
p
9 " 6 x 2 C 1 C x 2 C 1 " 1 dx D 2!
&
" 12 dx
p
0
3#
$
p
9 " 6 x 2 C 1 C x 2 dx
S E C T I O N 7.3
identity
Z
!
dx
in two ways and verify that the answers agree: first via trigonometric substitution and then using the
"1
x2
!
1
1
D
2
x2 " 1
SOLUTION
845
p
ˇ" 1 &ˇˇ 3
1 p 2
1 ˇˇp 2
ˇ
3 ˇ
D 2! 9x " 6
x x C 1 C ln ˇ x C 1 C x ˇ C x ˇ
2
2
3
0
ˇ
ˇ p $
ˇ
h# p
i
p
p
p
p ˇˇ
ˇ
ˇ
ˇ
D 2! 9 3 " 3 3.2/ " 3 ln ˇ2 C 3ˇ C 3 " .0/ D 8! 3 " 6! ln ˇ2 C 3ˇ :
%
61. Compute
Trigonometric Substitution
1
1
"
x"1 xC1
"
Using trigonometric substitution, let x D sec ". Then dx D sec " tan "d", x 2 " 1 D sec2 " " 1 D tan2 ", and
Z
Z
Z
Z
Z
dx
sec " tan " d"
sec "
d"
I D
D
D
d"
D
D
csc " d" D ln j csc " " cot "j C C:
tan "
sin "
x2 " 1
tan2 "
Since x D sec ", we construct the following right triangle:
x
x2 − 1
1
p
p
From this we see that csc " D x= x 2 " 1 and cot " D 1= x 2 " 1. This gives us
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ x"1 ˇ
x
1
ˇ
ˇ
ˇ
ˇ C C:
I D ln ˇ p
"p
ˇ C C D ln ˇ p 2
ˇ
x2 " 1
x2 " 1
x "1
Using the given identity, we get
"
Z
Z !
Z
Z
dx
1
1
1
1
dx
1
dx
1
1
I D
D
"
dx
D
"
D ln jx " 1j " ln jx C 1j C C:
2
x"1 xC1
2
x"1 2
xC1
2
2
x2 " 1
To confirm that these answers agree, note that
sˇ
ˇp
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ x " 1 px " 1 ˇ
ˇx "1ˇ
ˇ x "1 ˇ
1
1
1 ˇˇ x " 1 ˇˇ
ˇ
ˇ
ˇ
ˇ
ˇp
ˇ:
ln jx " 1j " ln jx C 1j D ln ˇ
D
ln
D
ln
p
!
p
D
ln
ˇ
ˇ
ˇx C 1ˇ
ˇ
ˇ
ˇ xC1
2
2
2
x C 1ˇ
x " 1ˇ
x2 " 1
62.
You want to divide an 18-inch pizza equally among three friends using vertical slices at ˙x as in Figure 1. Find an
equation satisfied by x and find the approximate value of x using a computer algebra system.
y
−9
−x
x
9
x
FIGURE 1 Dividing a pizza into three equal parts.
SOLUTION First find the value of x which divides evenly a pizza with a 1-inch radius. By proportionality, we can then take this
answer and multiply by 9 to get the answer for the 18-inch pizza. The total area of a 1-inch radius pizza is ! ! 12 D ! (in square
inches). The three equal pieces will have an area of !=3. The center piece is further divided into 4 equal pieces, each of area !=12.
From Example 1, we know that
Z xp
1
1 p
1 " x 2 dx D sin!1 x C x 1 " x 2 :
2
2
0
Setting this expression equal to !=12 and solving for x using a computer algebra system, we find x D 0:265. For the 18-inch pizza,
the value of x should be
x D 9.0:265/ D 2:385 inches:
846
CHAPTER 7
TECHNIQUES OF INTEGRATION
63. A charged wire creates an electric field at a point P located at a distance D from the wire (Figure 2). The component E? of
the field perpendicular to the wire (in N/C) is
Z x2
k$D
E? D
dx
x1 .x 2 C D 2 /3=2
where $ is the charge density (coulombs per meter), k D 8:99 % 109 N!m2 /C2 (Coulomb constant), and x1 , x2 are as in the figure.
Suppose that $ D 6 % 10!4 C/m, and D D 3 m. Find E? if (a) x1 D 0 and x2 D 30 m, and (b) x1 D "15 m and x2 D 15 m.
y
P
D
x
x1
x2
FIGURE 2
SOLUTION
Let x D D tan ". Then dx D D sec2 " d",
x 2 C D 2 D D 2 tan2 " C D 2 D D 2 .tan2 " C 1/ D D 2 sec2 ";
and
Z x2
k$D
D sec2 " d"
dx D k$D
x1 .x 2 C D 2 /3=2
x1 .D 2 sec2 "/3=2
ˇx2
Z
Z
ˇ
k$D 2 x2 sec2 " d"
k$ x2
k$
ˇ
D
D
cos
"
d"
D
sin
"
ˇ
D 3 x1
sec3 "
D x1
D
x1
E? D
Z
x2
Since x D D tan ", we construct a right triangle with tan " D x=D:
x 2 + D2
x
D
p
From this we see that sin " D x= x 2 C D 2 . Then
E? D
k$
D
!
"ˇx2
ˇ
x
ˇ
p
ˇ
2
2
x CD
x1
(a) Plugging in the values for the constants k, $, D, and evaluating the antiderivative for x1 D 0 and x2 D 30, we get
%
&
.8:99 % 109 /.6 % 10!4 /
30
V
E? D
p
" 0 & 1:789 % 106
2
2
3
m
30 C 3
(b) If x1 D "15 m and x2 D 15 m, we get
.8:99 % 109 /.6 % 10!4 /
E? D
3
"
p
15
152
C 32
"15
"p
."15/2 C 32
Further Insights and Challenges
64. Let Jn D
Z
dx
. Use Integration by Parts to prove
.x 2 C 1/n
!
"
! "
1
1
x
JnC1 D 1 "
Jn C
2n
2n .x 2 C 1/n
Then use this recursion relation to calculate J2 and J3 .
#
& 3:526 % 106
V
m
Trigonometric Substitution
S E C T I O N 7.3
SOLUTION
Let x D tan ". Then dx D sec2 " d", x 2 C 1 D tan2 " C 1 D sec2 ", and
Z
JnC1 D
Using the reduction formula for
R
dx
D
.x 2 C 1/nC1
Z
sec2 " d"
D
sec2nC2 "
Z
sec!2n " d" D
Z
847
cos2n " d":
cosm " d", we get
JnC1 D
cos2n!1 " sin "
2n " 1
C
2n
2n
Z
cos2n!2 " d":
Since x D tan ", we construct the following right triangle:
x2 + 1
x
1
p
p
From this we see that cos " D 1= x 2 C 1, and sin " D x= x 2 C 1. This gives us
JnC1 D
1
2n
!
p
1
x2 C 1
"2n!1 !
x
p
x2 C 1
"
C
2n " 1
2n
Z !
p
1
x2 C 1
"2n!2 !
p
1
x2 C 1
"2
dx:
Here we’ve used the fact that
dx
d" D
D cos2 " dx D
sec2 "
!
p
1
x2 C 1
"2
dx:
Simplifying, we get
!
1
2n
"
2n " 1
2
2n
2n
. x C 1/
!
"
1
x
1
D
C 1"
Jn :
2
n
2n .x C 1/
2n
JnC1 D
p
x
C
Z
.
p
dx
x2
C 1/2n
D
1
x
2n " 1
C
2n .x 2 C 1/n
2n
Z
dx
.x 2 C 1/n
To use this formula, we first compute J1 :
J1 D
Z
dx
D tan!1 x C C:
x2 C 1
Now use the formula to compute J2 and J3 :
!
"
1 x
1
x
1
J2 D
C
1
"
J1 D
C tan!1 x C C I
2
2
2x C1
2
2
2.x C 1/
!
"
%
&
1
x
1
1
x
3x
3
!1
J3 D
C
1
"
J
D
C
C
tan
x
C C:
2
4 .x 2 C 1/2
4
4 .x 2 C 1/2
8
8.x 2 C 1/
65. Prove the formula
Z p
1
1 p
1 " x 2 dx D sin!1 x C x 1 " x 2 C C
2
2
using geometry by interpreting the integral as the area of part of the unit circle.
Z ap
SOLUTION The integral
1 " x 2 dx is the area bounded by the unit circle, the x-axis, the y-axis, and the line x D a. This
0
area can be divided into two regions as follows:
y
1
II
I
0
a
1
x