CHAPTER
27
Techniques of Integration
27.1
1.
3.
5.
7.
9.
11.
13.
THE GENERAL POWER FORMULA
Ð
sin3 x cos x dx. Let u ¼ sin x and du ¼ cos x dx,
Ð
4
then u3 du ¼ u4 þ C ¼ 14 sin4 x þ C.
ffi
Ð pffiffiffiffiffiffiffiffiffiffiffi
sin3 x cos x dx. Let u ¼ sin x and du ¼
Ð
cos x dx, so we have u3=2 du ¼ 25 u5=2 þ C ¼
5=2
2
x þ C.
5 sin
Ð
2 2
x sec x dx. If you let u ¼ x2 andÐ du ¼ 2 x dx,
or 12 du ¼ x dx, then you get 12 sec2 u du ¼
1
1
2
2 tan u þ C ¼ 2 tan x þ C.
Ð
sec2 x tan x dx. If you let uÐ ¼ tan x and du ¼
sec2 x dx, then you get
u du ¼ 12 u2 þ C ¼
1
2
tan x þ C or rewrite the original integral as
Ð
Ð2
sec2 x tan x dx ¼ sec x ðsec x tan x dxÞ and let
Ð
u ¼ sec x and du ¼ sec x tan x dx. Thus, u du ¼
1 2
1
2
2 u þ C ¼ 2 sec x þ C.
Ð arccos x
1 ffi
pffiffiffiffiffiffiffiffi dx. Let u ¼ arccos x and then du ¼ pffiffiffiffiffiffiffi
1x2
1x2
Ð
2
1
dx, with the result u du ¼ 1
2 u þ C ¼ 2
Ð =4
23.
Ð 2 ½lnðx2 þ1Þ3
25.
27.
ðarccos xÞ2 þ C.
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Ð 1 arccsc x
dx. Let u ¼ arccsc x and du ¼
x
x2 1
Ð
ffiffiffiffiffiffiffi
ffi dx, with the result u1=2 du ¼ 23 u3=2 þ
p1
2
x x 1
C ¼ 23 ðarccsc xÞ3=2 þ C.
Ð ðarcsin 2xÞ3
pffiffiffiffiffiffiffiffiffi dx. Let u ¼ arcsin 2x and du ¼ 2 14x2
1
dx
pffiffiffiffiffiffiffiffiffi
dx, so 12 du ¼ pffiffiffiffiffiffiffiffiffi
. Substitution
14x2
14x2
Ð
4
1
3
1 1 4
1
2 u du ¼ 2 4 u þ C ¼ 8 ðarcsin 2xÞ þ C.
15.
Ð
17.
then
Ð ex dx
19.
21.
yields
4
lnðxþ4Þ
xþ4
Ð 4
dx
dx. Let u ¼ lnðx þ 4Þ and du ¼ xþ4
,
5
1 5
1
u du ¼ 5 u þ C ¼ 5 lnðx þ 4Þ þ C.
. Let u ¼ ex þ 4, and du ¼ ex dx, then have
1
u du ¼ 1u1 þ C ¼ ex þ 4 þ C.
3
Ð 2x
e 1 e2x dx. Let u ¼ e2x 1 and du ¼ 2e2x dx
Ð
so 12 du ¼ e2x dx. Then, 12 u3 du ¼ 12 4
2x
1 4
1
þ C.
4u þ C ¼ 8 e 1
Ð
ðex þ4Þ2
2
29.
sin3 x cos x dx. If you let u ¼ sin x and
Ð 3
du ¼ cos x dx, then you have
u du ¼ 14 u4 .
=4
4
Substituting for u produces 14 sin x0 ¼ 14 sin4 4 h pffiffi i
2 4
1
¼ 04 ¼ 14 14 ¼ 16
.
sin4 0 ¼ 14
2
0
dx. Let u ¼ ln x2 þ 1
1
x2 þ1
2x
1
x
x2 þ1 dx and so 2 du ¼ x2 þ1 dx. Thus,
2
Ð
1
3
1 4
1 4
2
2 u du ¼ 8 u or 8 ln x þ 1 1 ¼
4
4
1
8 ln 5 ln 2 0:8098.
and du ¼
we get
Ð2
e2x dx
2x
2x
1 ðe2x 1Þ2 . Let u ¼ e 1 and du ¼ 2e dx
Ð
so 12 du ¼ e2x dx. Then, 12 u2 du ¼ 12 u1 or
1 2
1
12 e2x 1
¼ 1 ¼ 12 e4 1 1 2
e 1
0:0689.
Ð 2
x dx. Let u ¼ sin x and du ¼ cos x 0 sin x cos
Ð 2
dx, and so u du ¼ 13 u3 . The curve is above the x
axis from 0 to 2 and below from 2 to . Thus, the
=2
area is 13 sin3 x0 13 sin3 x=2 ¼ 13 sin3 2 sin3 0 sin3 þ sin3 2 ¼ 13 ð1 0 0 þ 1Þ ¼ 23.
2
Ð 3 t t
e 2 dt. Let u ¼ et 2 and du ¼
0:7 e
Ð
et dt, and so u2 du ¼ 13 u3 . Substituting
h for
2
Ð 3 t t
1
t
u iwe obtain
h 0:7 e e 2 dt ¼i3 e 3 3
3
3
2
¼ 13 e3 2 e0:7 2
0:7
31.
1:3397 C.
Ð
1020
(a) NðtÞ ¼ 1020
t2 340dt ¼ t 340t þ C.
Since Nð1Þ ¼ 1020 340 þ C ¼ 9750 and so,
C ¼ 9070. Thus, NðtÞ ¼ 1020
t 340t þ 9070.
(b) Nð10Þ ¼ 1020
340ð10Þ
þ 9070 ¼ 102 3400
10
þ 9070 ¼ 5772.
275
276
CHAPTER 27
27.2
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
Ð
BASIC LOGARITHMIC AND EXPONENTIAL INTEGRALS
Let u ¼ 4x þ 1 and du ¼ 4 dx, so 14 du ¼
Ð
1
1
dx. Then, 14 du
u ¼ 4 juj þ C ¼ 4 ln j4x þ 1j þ C.
Ð 3x
e dx. Let u ¼ 3x and du ¼Ð 3 dx, or
13 du ¼ dx, which leads to 13 eu du ¼ 13 eu þ
C ¼ 13 e3x þ C.
Ð sin x
cos x dx. LetÐ u ¼ cos x and du ¼ sin x dx, then
you have du
u ¼ ln jujþ C ¼ ln j cos xj þ C.
Ð 2 sec2 x
2
tan x . If you let u ¼ tan x and du ¼ sec x dx,
2
then 2 du ¼ 2 sec x dx. Substituting we get
Ð
2 du
u ¼ 2 ln juj þ C ¼ 2 ln j tan xj þ C.
Ð x
x
4 dx ¼ ln4 4 þ C.
Ð x
R
Ð
e þ ex dx ¼ ex dx ex dx ¼ ex x
e þ C. (Note: for the second integral let u ¼
x, then du ¼ dx)
Ð e2x
2x
2x
1þe2x dx. Let u ¼ 1 þ e , so duÐ ¼ e 2 dx or
1
1 du
1
2x
2 du ¼ e dx. Substituting we get 2 u ¼ 2 ln jujþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p
C ¼ 12 ln 1 þ e2x þ C ¼ ln 1 þ e2x þ C.
Ð lnð1=xÞ
dx. Let u ¼ ln 1x ¼ lnðxÞ1 ¼ ln x, and
x
Ð
then du ¼ 1x dx. Substituting we get u du ¼
2
12 u2 þ C ¼ 12 ln 1x þ C.
pffi
Ð e p3 ffiffiffi
pffiffiffi
x
dx.pLet
u ¼ 3 x ¼ x1=3 ; du=dt ¼ 13 x2=3 3 du ¼
2=3
x
ffiffiffi
p
ffi
p
ffiffi
Ð 3 x
Ð
3
1
dx ex2=3 dx ¼ 3 eu du ¼ 3eu þ c ¼ 3e x þ c.
x2=3
1 x=2 4
i
2 h 2
Evaluating, we have 5 1 ¼ 1 15 15
ln 5 2 ln 5
¼ 0:1998.
dx
4xþ1.
2
27.
x
2
0 x2 þ1 dx. Let u ¼ x þ 1 and du ¼ 2x dx
Ð
or 12 du ¼ x dx. Substituting we get 12 du
u ¼
2
2
1
1
1
2 ln juj ¼ 2 ln x þ 1 . This leads to 2 ln x þ
4
1 0 ¼ 12 lnð17Þ lnð1Þ ¼ 12 ln 17 1:4166.
3
Ð3 1
Ð3 1
x
1 ln ex dx. Recall that ln e ¼ x so 1 x dx ¼ ln jxj 1
¼ ln :3 ln 1 ¼ ln 3 1:0986.
Ð 1 3 x4
4
3
1
0 x e dx. Let u ¼ x and du ¼ 4x dx or 4 du ¼
Ð
4
x3 dx. Substituting produces 14 eu du ¼ 14 eu ¼ 14 ex
4 1
which leads to 1 ex ¼ 1 e1 e0 ¼ 1 ðe 1Þ 0:4296.
Ð 4 1x=2
0
4
4
dx. Let u ¼ 2x and du ¼ 12 dx. Substi 1 u
x=2
Ð 1 u
2 15
5
¼
.
tuting, we get 2 5 du ¼ 2 ln 15
ln 15
2 5
Since exþ1 > x2 on ½0; 1, the area we want is
Ð1
Ð1
Ð 1 xþ1
x2 dx ¼ 0 exþ1 0 x2 dx ¼ exþ1
0 e
1 x3 1
¼ e2 e1 13 ¼ e2 e 1 4:3374.
0
29.
31.
3 0
3
3
Ð
(a) sðtÞ ¼ 394e0:025t 384 dt ¼
15760e0:025t 384t þ C. At t ¼ 0, the object
is on the ground, so sð0Þ ¼ 0 and we see that,
C ¼ 15;760. Thus, the position function is sðtÞ ¼
15;760e0:025t 384t þ 15;760;
(b) The object reaches its maximum height when
vðtÞ ¼ 0 or when e0:025t ¼ 384
394. Taking the natural
logarithm
of
both
sides
we get 0:025t ¼
ln 384ln 394
ln 384
394 ¼ ln 384 ln 394. So, t ¼
0:025
1:028;
(c) sð1:028Þ ¼ 5:1196.
Ð 2 2
Using the disc method this volume is 0 ex dx
Ð2
¼ 0 e2x dx. Let u ¼ 2x, then du ¼ 2 dx
or 12 du ¼ dx. Thus, the integral becomes Ð
u
2x
. Evaluating, we get
12 eu du ¼ 2 e ¼ 2e
2
2x 4
0
e
¼
e e ¼ 1 e4 1:5420.
2
0
dV
ds
2
2
0:15s
33.
Since ¼ 0:45e
, we see that V ¼
Ð
0:45 0:15s
0:15s
ds ¼ 0:15 e
þ C ¼ 3e0:15s þ
0:45e
0
C. Hence, Vð0Þ ¼ 3 ¼ 3e þ C ¼ 3 þ C so
C ¼ 0 and VðsÞ ¼ 3e0:15s . Also, 3e0:15s ¼ 1:5
yields e0:15s ¼ 12. Taking the natural logarithm of
both sides we get 0:15s ¼ ln 12 or s ¼ lnð0:5Þ ð0:15Þ ¼ 4:6210 mi.
35.
First we find k ¼ PV ¼ 20 0:4
¼ 8. Then,
Ð 1:500
the work done by the gas is W ¼ 0:400 V8 dN ¼
8ðln 1:5 ln 0:4Þ 10:574 ft lb.
Ðx
Ðx
2
2
FðxÞ ¼ 0 12te3t dt ¼ 2 0 ð6tÞe3t dt ¼
2 x
2
2
2
2e3t 0 ¼ 2e3x 2e30 ¼ 2e3x þ
2
2 ¼ 2 2e3x
Ð4
4
25.
TECHNIQUES OF INTEGRATION
37.
SECTION 27.3
27.3
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
BASIC TRIGONOMETRIC AND HYPERBOLIC INTEGRALS
Ð
sin x dx. Let u ¼ 2x and du ¼ 12 dx gives
Ð 2
2 sin u du ¼ 2 cos u þ C ¼ 2 cos 2x þ C.
Ð
tan 5x dx. Let u ¼ 5x, then du ¼ 5 dx and 15 du ¼
Ð
dx. This gives 15 tan u du ¼ 15 ln j cos uj þ C ¼
15 ln j cos 5xj þ C or 15 ln j sec 5xj þ C.
Ð 1
Ð
cos x dx ¼ sec x dx ¼ ln j sec x þ tan xj þ C
Ð 1
Ð
2
cos2 x dx ¼ sec x dx ¼ tan x þ C
Ð tan pffiffix
pffiffiffi
pffiffi dx. Let u ¼
x then du ¼ 2p1 ffiffix dx and 2du ¼
x
Ð
p1ffiffi dx. Substituting produces 2 tan u du ¼
x
pffiffiffi
2 ln j cos uj þ C ¼ 2 ln j cos xj þ C or
pffiffiffi
2 ln j sec xj þ C.
Ð
sinð4x 1Þdx ¼ 14 cosð4x 1Þ þ C
Ð sin x
and du ¼ sin x dx. Subcos2 x dx. Let u ¼ cos x
Ð
stituting, we obtain u2 du ¼ u1 þ C ¼
ðcos xÞ1 þ C ¼ sec x þ C.
Ð sec 3x tan 3x
5þ2 sec 3x dx. Let u ¼ 5 þ 2 sec 3x, and then
du ¼ 6 sec 3x tan 3x dx or 16 du ¼ sec 3x tan 3x dx.
Ð
This yields 16 1u du ¼ 16 ln juj þ C ¼ 16 ln j5 þ
2 sec 3xj þ C.
Ð
Ð
sec4 3x tan 3x dx ¼ sec3 3x sec 3x tan 3x dx. If
you let u ¼ sec
¼ 3 sec 3x tan 3x dx.
Ð 3x, then1 du
1
This leads to 13 u3 du ¼ 12
u4 þC ¼ 12
sec4 3x þ C.
Ð 3 dx
Ð
tan 3x ¼ 3 cotð3xÞdx ¼ ln j sin 3xj þ C
Ð
Ð
ðsec x þ 2Þ2 dx ¼ ð sec2 x þ 4 sec x þ 4Þdx ¼
tan x þ 4 ln j sec x þ tan xj þ 4x þ C
ð =2
23.
0
277
ð =2
ðsec 0:5x þ 5Þ2 dx ¼
0
sec2 0:5x þ 10 sec 0:5x þ 25 dx
¼ ½2 tan 0:5x þ 20 ln j sec 0:5x
=2
þ tan 0:5xj þ 25x 0
h
¼ 2 tan
þ 20 lnsec
4
i 4
þ tan
þ 25
4
2
½2 tan 0 þ 20 lnjsec 0 þ tan 0j þ 25ð0Þ
pffiffiffi
25
¼ 2ð1Þ þ 20 ln 2 þ 1 þ
2
½2ð0Þ þ 20 ln j1 þ 0j þ 25ð0Þ
pffiffiffi
25
58:897
¼ 2 þ 20 ln 2 þ 1 þ
2
25.
Ð
27.
Ð
29.
31.
33.
35.
37.
39.
Ð
Ð
1þcos 4x
1
cos 4x
dx ¼ csc2 4x dx ¼
þ sin
2
4x
sin2 Ð4x
sin2 4x
Ð
cos 4x
dx þ sin
dx. Now, the first integral is csc2 4x
2
4x
2
¼ 1
4 cot 4x þ C. For the second integral, let u ¼
sin 4x then du ¼ 4 cos 4x dx or 14 du ¼ cos 4x dx.
Ð
Substitution yields 14 u2 du ¼ 14 1u1 þ C ¼
14 ðsin 4xÞ1 þ C ¼ 14 csc 4x þ C. Putting these
together we get 14 cot2 4x 14 csc 4x þ C ¼
14 cot2 4x þ csc 4x þ C or 14 cossin4xþ1
þ C.
4x
tan 4x dx ¼ 4 ln j sec 4x j þ C
Ð =2
Ð =2
2
x
Using trig identities, we get =4 1þcot
csc2 x dx ¼ =4
1
Ð =2 2
2
2
csc2 x þ cos x dx ¼ =4 sin x þ cos x dx
Ð =2
=2
¼ =4 dx ¼ xj=4 ¼ 2 4 ¼ 4.
Ð =2 csc pffiffix cot pffiffix
pffiffiffi
pffiffi
dx. Let u ¼ x and du ¼ 2p1 ffiffix. Then,
=4
x
Ð
2 csc u cot u du ¼ 2 csc u. Evaluating, we obtain
pffiffiffi =2
2 csc xj=4 ¼ 0:4765.
Using a Pythagorean trigonometric substitution
Ð =2 2 x
Ð =2 1sin2 x
gives the result =6 cos
sin x dx ¼ =6 sin x dx ¼
Ð =2
Ð =2 1
=6 sin x sin x dx ¼ =6 ðcsc x sin xÞdx ¼
h
=2
ln j csc x cot xj þ cos x =6 ¼ ln j1 0j þ 0 pffiffiffi pffiffi
ln j2 3j 23 ¼ 0:4509.
Ð
x sinh x2 dx. Let u ¼ x2 and du ¼ 2x dx or 12 du ¼
Ð
x dx. Substituting we get 12 sinh u du ¼ 12 cosh uþ
C ¼ 12 cosh x2 þ C
pffiffiffiffiffiffiffiffiffiffiffiffiffi
Ð
3x cosh x2 sinh x2 dx. Let u ¼ sinh x2 , then du ¼
2x cosh x2 dx or 32 du ¼ 3x cosh x2 dx. Substituting
Ð pffiffiffi
3=2
we get 32
u du ¼ 32 23 u3=2 þ C ¼ ðsinh x2 Þ þ C
Ð
3
sinh3 x cosh2 x dx. Recall that sinh
x¼
2
2
sinh x sinh x ¼ sinh x cosh x 1 , and then we
Ð
have sinh x cosh2 x 1 cosh2 x dx ¼
Ð
Ð
cosh4 x sinh x dx cosh2 x sinh x dx ¼
5
3
1
1
5 cosh x 3 cosh x þ C.
41.
u
We begin by rewriting cot u as cos
sin u . Here, if we let
vÐ ¼ sin u, then
Ð dv ¼ cos u du, and the integral is
cot u du ¼ dv
v ¼ ln jvj Ðþ C. Back substitution
gives the desired result: cot u du ¼ ln j sin uj þ C.
43.
For 0 x 2, we see that sin 2x 0, so the area
Ð =2
=2
¼
is the integral 0 sin 2x dx ¼ 1
2 cos 2xj0
ð1
1Þ
¼
1.
12 ðcos cos 0Þ ¼ 1
2
Ð
1 =4
The average value is =4
tan x dx ¼
0
pffiffi
pffiffiffi
=4
4 ln 2
4
4
0:4413.
ln j sec xj0 ¼ ðln 2 ln 1Þ ¼
45.
278
CHAPTER 27
TECHNIQUES OF INTEGRATION
Ð
¼ 1 1 5 sin 4t dt ¼ 5 cos 4tj10 ¼
E
10 0
4
54 ð0:6536 1Þ 2:067 V.
47.
x
x
Here y ¼ 12 cosh 12
and so y0 ¼ sinh 12
. The length
of the cable is the arc length ¼
ffi
Ð 18 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Ð 18
x
x
dx ¼
1 þ sinh2 12
dx ¼ 18 cosh 12
18
49.
51.
The center of the towers is at 0 and the towers are
x
and
at 150 and 150. We are given y ¼ 80 cosh 80
0
x
differentiating, we obtain y ¼ sinh 80. The arc
Ð 150 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffixffi
Ð 150
x
length is 150 1 þ sinh2 80
dx
dx ¼ 150 cosh 80
150
x
¼ 80 sinh 80
¼ 509:397 ft.
150
x 18
12 sinh 12
j18 ¼ 12 sinh 32 12 sinh 3
2 ¼ 51:1027 m.
27.4
1.
3.
7.
9.
11.
MORE TRIGONOMETRIC INTEGRALS
Ð
Ð
1
sin2 3x cos 3x dx. Let u ¼ sin 3x and du ¼ 3
sin5 5x þ C
5.
cos 5x sin4 5x dx ¼ 25
Ð
cos 3x dx. Then, 13 u2 du ¼ 19 u3 þ C ¼ 19 sin3 3x þ C.
Ð
cos x sin3 x dx ¼ 14 sin4 x þ C.
ð
ð
ð
1 cos 2x 1 þ cos 2x 2
1
2
4
sin x cos x dx ¼
dx ¼ ð1 cos 2xÞð1 þ 2 cos 2x þ cos2 2xÞdx
2
2
8
ð
1 ¼
1 þ cos 2x cos2 2x cos3 2x dx
8ð
ð
1
1 1 þ cos 2x cos3 2x cos2 2x dx
¼
8
8
ð
ð
1
1
1 1 1 þ cos 4x
2
dx
¼ x sin 2x 1 sin 2x cos 2x dx 8
16
8
8
2
1
1
1
1
1
1 1
¼ x sin 2x þ sin 2x þ sin3 2x x þ
sin 4x þ C
8
16
16
48
16
16 4
1
1
1
¼ x þ sin3 2x þ sin 4x þ C
16
48
64
ð
ð
ð
1 þ cos 6x 3
1 6
cos 3x dx ¼
1 þ 3 cos 6x þ 3 cos2 6x þ cos3 6x dx
dx ¼
2
8
ð
1
3
2
1 þ 3 cos 6x þ ð1 þ cos 12xÞ þ 1 sin 6x cos 6x dx
¼
8
2
ð
1
3 3
1 þ 3 cos 6x þ þ cos 12x þ cos 6x sin2 6x cos 6x dx
¼
8
2 2
ð
1 5
3
2
þ 4 cos 6x þ cos 12x sin 6x cos 6x dx
¼
8 2
2
1 5x 2
1
1
3
þ sin 6x þ sin 12x sin 6x þ C
¼
8 2 3
8
18
5x 1
1
1
þ sin 6x þ sin 12x sin3 6x þ C
¼
16 12
64
144
ð
ð
ð
1 cos 4 1 þ cos 2 2
1 sin 2 cos 2 d ¼
1 þ cos 4 cos2 4 cos3 4 d
d ¼
2
2
8
ð
1
1
1 þ cos 4 ð1 þ cos 8Þ 1 sin2 4 cos 4 d
¼
8
2
ð
1
1 1
2
¼
1 þ cos 4 cos 8 cos 4 þ sin 4 cos 4 d
8
2 2
ð
1 1 1
1 1
1
2
3
cos 8 þ sin 4 cos 4 d ¼
sin 8 þ sin 4 þ C
¼
8 2 2
8 2 16
12
1
1
sin 8 þ sin3 4 þ C
¼
16 128
96
2
4
SECTION 27.5
Ð
sec2 x tan2 x dx. Let u ¼ tanÐ x and then du ¼
sec2 x dx. Hence, we get u2 du ¼ 13 u3 þ C ¼
1
3
3 tan x þ C.
Ð
Ð
First factor csc4 x cot x dx ¼ csc3 x ðcsc x cot x dxÞ. Then let u ¼ csc
Ð x and du ¼
csc x cot x dx. These produce u3 du ¼ 14 u4 þ
Ð
C ¼ 14 csc4 x þ C1 or csc4 x cot x dx ¼
Ð
Ð
csc2 xð1 þ cot2 xÞ cot x dx ¼ ð cot3 x þ cot xÞ csc2 x dx. Let u ¼ cot x and du ¼ csc2 x dx, and
Ð
then you have ðu3 þ uÞdu ¼ 14 u4 12 u2 þ
C ¼ 14 cot4 x 12 cot2 xþC2 where C2 ¼ C1 0:25.
13.
15.
We first factor the integrand and use a Pythagorean
Ð
Ð
identity: sin1=2 3 cos3 3d ¼ sin1=2 3ð1
sin2 3Þ cos 3 d. Then multiply and integrate,
Ð
with the result sin1=2 3sin5=2 3 cos 3 d ¼
3=2
1 2
3 13 27 sin7=2 3 þ C ¼ 29 sin3=2 3 3 3 sin
7=2
2
3 þ C.
21 sin
Ð
Ð
csc x cot3 x dx ¼ ð csc2 x 1Þ csc x cot x dx ¼
13 csc3 x þ csc x þ C.
Ð
tan3 x sec2 x dx ¼ 14 tan4 x þ C.
Ð
tan6 x sec2 x dx ¼ 17 tan7 x þ C
Ð
Ð
cot6 x dx ¼ cot4 xð csc2 x 1Þdx ¼
Ð
Ð
cot4 x csc2 x dx cot2 xð csc2 x 1Þdx ¼
Ð
Ð
Ð
cot4 x csc2 x dx cot2 x csc2 x dx þ cot2 x dx
Ð
¼ 15 cot5 x þ 13 cot3 x þ ð csc2 x 1Þdx ¼
15 cot5 x þ 13 cot3 xcot xx þ C. Recall: cot2 x ¼
csc2 x 1.
Ð =4 2
Ð =4
tan x dx ¼ 0 ð sec2 x 1Þdx ¼ tan x 0
=4
xj0 ¼ 1 4 0 ¼ 1 4
Ð =2 4
Ð =2 2x 2
sin x dx ¼ 0 1cos
dx by a half-angle
0
2
Ð =2
identity. This expands as 14 0 ð1 2 cos 2xþ
17.
19.
21.
23.
25.
27.
29.
27.5
1.
3.
5.
33.
cos2 2xÞdx which becomes, by another half-angle
Ð =2 4x
dx or
identity, 14 0 1 2 cos 2x þ 1þcos
2
Ð 1 =2 3
cos 4x
1 3x
dx ¼ 4 2 sin 2xþ
4 0
2 2 cos 2x þ 2
=2
1
1 3
¼ 8 4 sin þ 18 sin 2 ¼
8 sin 4x0
3
1 3
4 4 0 þ 0 ¼ 16 .
Ð =3
Using the disc method, you get 0 cos2 x dx ¼
=3
Ð =3
2x
0 1þcos
dx ¼ 2 x þ 12 sin 2x 0 ¼ 2 3 þ
2
pffiffi
3
4 2:3251.
ð
mr 2 2 2
sin d
4 0
ð
mr 2 1 2
ð1 cos 2Þd
¼
4 2 0
2
mr 2
1
sin 2
¼
2
8
0
(a) I ¼
mr 2
mr 2 ½ð2 0Þ ð0 0Þ ¼
4
8
(b) Substituting r ¼ 0:1 m and m ¼ 12:4 kg into
2
2
I ¼ mr4 produces I ¼ ð12:4Þð0:1Þ
0:0974 kg m2 .
4
¼
35.
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð 0:5
pffiffiffiffiffiffiffiffiffi
1
ieff ¼
ð2 sin t cos tÞ2 dt
0:5 0 0
pffiffiffi ð 0:5 2
¼ 8
sin t cos t dt
rffiffiffi 0
8 3 0:5
sin t 0
¼
3
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
8 3
sin 0:5 0:5421
¼
3
The effective current from t ¼ 0 to t ¼ 0:5 s is
about 0.5421 A.
INTEGRALS RELATED TO INVERSE TRIGONOMETRIC AND INVERSE HYPERBOLIC FUNCTIONS
Ð
dx ffi
pffiffiffiffiffiffiffi
¼ arcsin 2x þ C
4x2
Ð dx
pffiffiffiffiffiffiffiffiffi. Let u ¼ 3x and du ¼
49x2
Ð du
ffi
Substituting, we have 13 pffiffiffiffiffiffiffi
4u2
1
3x
3 arcsin 2 þ C.
Ð
31.
279
7.
1
3 du ¼ dx.
1
u
3 arcsin 2 þ C ¼
3 dx so
¼
ffiffiffiffiffiffiffiffiffi.
pdx
3x 9x2 4
Let u ¼ 3x, then du ¼ 3 dx, so 13 du ¼
dx. Thus, we can rewrite the integral as
Ð du
pffiffiffiffiffiffiffiffi ¼ 1 1 arcsec u þ C ¼ 1 arcsec 3x þ C.
3 2
2
6
2
u u2 4
Ð dx
. Here u ¼ 3 x and du ¼ dx and so
1þð3xÞ2 Ð
du
we get 1þu
2 ¼ arctan u þ C ¼
1
3
9.
x
dxffi
p
ffiffiffiffiffiffiffi
.
9x2
Let u ¼ 9 x2 and du ¼ 2x dx so
Ð 1=2
1
1
du ¼ 12 21 u1=2 þ C ¼
2 du ¼ x dx. Then, 2 u
pffiffiffiffiffiffiffiffiffiffiffiffiffi
9 x2 þ C.
Ð
11.
arctan ð3 xÞ þ C.
Ð 4x6
Ð 4x
Ð
4x2 þ25 dx ¼ 4x2 þ25 dx 2
6
4x2 þ25 dx.
In the first
integral let u ¼ 4x Ð þ 25; du ¼ 8x dx and we get
2
by substitution 2 du
u ¼ 2 ln juj ¼ 2 lnð4x þ 25Þ.
280
13.
15.
17.
19.
CHAPTER 27
TECHNIQUES OF INTEGRATION
Ð
The second integral is 6 4x2dxþ25 ¼ 6 1
2x
10 arctan 5 . Putting these together we have 2 ln 2
ð4x þ 25Þ 35 arctan 2x
5 þC
Ð
dx
Completing the square, we obtain x2 þ6xþ10
¼
Ð dx
¼
arctanðx
þ
3Þ
þ
C.
ðxþ3Þ2 þ1
Ð x dx
2
1þx4 ; Let u ¼ xÐ and then du ¼ 2x dx. Substitut1
du
1
ing, we obtain 2 1þu
2 ¼ 2 arctan u þ C ¼
1
2
2 arctan x þ C.
Ð sin x dx
pffiffiffiffiffiffiffiffiffiffiffiffiffi. Let u ¼ cos x and du ¼ sin x dx.
1cos2 x
Ð du
ffi ¼ arcsin u þ
Substituting produces pffiffiffiffiffiffiffi
1u
2
C ¼ arcsinðcos xÞ þ C ¼ 2 x þ C ¼
x 2 þ C.
=4
Ð =4 cos x dx
¼ arctanðsin xÞ0 ¼ arctan sin 4 0
1þsin2 x
pffiffi
arctanðsin 0Þ ¼ arctan 22 arctan 0 ¼ 0:6155
21.
0 ¼ 0:6155.
Ð dx
pffiffiffiffiffiffiffiffiffi ¼ cosh1 x þ C.
5
x2 25
23.
Ð
25.
C ¼ 13 sinh1 3x
5 þ C.
0:5
Ð 0:5 dx
pffiffiffiffiffiffiffiffi ¼ arcsin
¼ arcsin 0:5 arcsin 0 ¼
0
0
1x2
1
1
2x ¼ e arctan x
2
0
0
2
¼ e 1 ðarctan 1 arctan 0Þ
2
2
¼ e 1 0
2
4
2
¼
e 1
7:5685
2
2
35.
dx
pffiffiffiffiffiffiffiffiffiffiffi
.
25þ9x2
Let u ¼ 3x and du ¼ 3 dx and so 13 du ¼
Ð du
ffi ¼ 13 sinh1 u5 þ
dx. Substituting produces pffiffiffiffiffiffiffiffiffi
25þu2
6
27.
29.
31.
33.
0 ¼ 6 0:5236
Ð 5 dx
Ð5
Ð5
dx
dx
1 x2 4xþ13 ¼ 1 ðx2 4xþ4Þþ9 ¼ 1 32 þðx2Þ2 ¼
ðx2Þ 5
1
¼
¼ 13 arctan 1 arctan 1
3 arctan 3
3
1
1 3 4 þ 0:32175 0:36905
Let u ¼ 4 tan x and du ¼ 4 sec2 x dx. Then,
Ð =6 sec2 x dx
Ð
Ð pffiffi
1 =6 4 sec2 x dx
1 4 3=3 du
¼
¼
2
2
0
1þu2 i
1þ16 tan x pffiffi 4 0
1þð4
h tan xÞ pffiffi4 0
4
3
=3
¼ 14 arctan u0
¼ 14 arctan 4 3 3 arctan 0 ¼
pffiffi
4 3
1
0:2905
3
4 arctan
1
Ð1 1
0 1þx2 dx ¼ arctan x 0 ¼ arctan 1 arctan 0 ¼
4 0 ¼ 4.
These two graphs intersect at (0, 1) and, from 0 to
1, ex is always
larger. Using the
washer method we
2 #
ð1"
1
ðex Þ2 pffiffiffiffiffiffiffiffiffiffiffiffiffi
dx
get x2 1
0
ð1
1
2x
e 2
¼
dx
x 1
0
ð1
ð1
1
dx
¼ e2x dx 2
0
0x þ1
37.
pffiffiffiffiffiffiffiffiffiffiffiffiffi
4 x2
1
y0 ¼ ð4 xÞ1=2 ð2xÞ
2
x
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
4 x2
2
x
y02 ¼
4 r
x2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð 1
x2
1 þ ðy 0 Þ2 ¼
1þ
dx
4 x2
1
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r
ð1
4 x2 þ x2
dx
¼
4 x2
1
r
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð1
4
¼
dx
4
x2
1
ð1
2 dx
pffiffiffiffiffiffiffiffiffiffiffiffiffi
¼
1 4 x2
1
x
¼ 2 arcsin 2 1
1
1
¼ 2 arcsin arcsin
2
2
h i 2
¼2 ¼
6
6
3
y¼
1
x
f 0 ðxÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffi
2
1 x2
ð x 1x
ð
1
f ðxÞ ¼ f 0 ðxÞdx ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi dx
xð 1 x2
x
þ pffiffiffiffiffiffiffiffiffiffiffiffiffi dx
1 x2 ð
x dx
¼ sech1 x þ pffiffiffiffiffiffiffiffiffiffiffiffiffi :
1 x2
For the second integral let u ¼ 1 x2 , then du ¼
2x dx or 12 du ¼ x dx. These substitutions
Ð dx
Ð
pffiffiffi
ffi ¼ 12 pduffiffi ¼ 12 21 u1=2 ¼ u ¼
make pxffiffiffiffiffiffiffi
2
u
1x
pffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2 . Putting these together we get f ðxÞ ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffi
sech1 x 1 x2 þ C. Since f ð1Þ ¼ 0 we
pffiffiffiffiffiffiffiffiffiffiffiffiffi
have C ¼ sech1 1 þ 1 12 ¼ sech1 1. Thus,
the desired equation is y ¼ sech1 x
pffiffiffiffiffiffiffiffiffiffiffiffiffi
1 x2 þ sech1 1.
SECTION 27.6
27.6
1.
3.
Ð
TRIGONOMETRIC SUBSTITUTION
x ffi
pffiffiffiffiffiffiffi
dx.
9x2
This can be done with trigonometric
substitutions but it is much easier to do the following. Let u ¼ 9 x2 ; du ¼ 2x dx so x dx ¼ 12 du.
Ð
Substituting yields 12 pduffiffiu ¼ 12 21 u1=2 þ C ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffi
9 x2 þ C.
Ð x2
pffiffiffiffiffiffiffiffi dx. Let x ¼ 3 sin and dx ¼ 3 cos d.
9x2
Substituting produces
ð
ð
ð3 sin Þ2 3 cos d
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 9 sin2 d
9 9 sin2 ð
1 cos 2
¼9
2 ð
ð
9
9
¼
1 d cos 2 d
2
2
9
9 1
¼ sin 2 þ C:
2
2 2
7.
Hence sin 2 ¼ 2 sin cos , which yields sin 2 ¼
pffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2 3x 9x
¼ 29 x 9 x2 . Thus, the final answer is
3
pffiffiffiffiffiffiffiffiffiffiffiffiffi
9
x
1
2
2 arcsin 3 2 x 9 x þ C.
Ð 3 pffiffiffiffiffiffiffiffiffiffiffiffiffi
x 9 x2 dx. Again, this can be done using trig
substitutions but the following method is easier.
Let u ¼ 9 x2 and du ¼ 2x dx, Ðandpffiffiffiffiffiffiffiffiffiffiffiffi
so x2ffi ¼
1
3
9 u and du ¼ x dx. Thus,
x 9 x2 ¼
Ð 2 pffiffiffiffiffiffiffiffiffiffiffiffiffi 2
x 9 x2 x dx. Now substitute
ð
pffiffiffi
1
ð9 uÞ u du
2
ð
1 pffiffiffi
¼
9 u u3=2 du
2
1
2
1 2 5=2
u þC
¼ 9 u3=2 þ
2
3
25
3=2 1 5=2
¼ 3 9 x2
þ 9 x2
þC
5
3=2 1 3=2
¼ 3 9 x2
þ 9 x2 9 x2
þC
5
3=2
15 9 x2
¼
5
3=2
1
þ 9 x2 9 x2
þC
5
3=2 1
¼ 9 x2
6 þ x2 þ C:
5
Ð
dx ffi
pffiffiffiffiffiffiffi
.
x2 9
Let x ¼ 3 sec and then dx ¼ 3 sec tan d. Substitution into the given integral yields
Ð
Ð
Ð 3 sec tan pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d ¼ 3 sec tan d ¼ sec d ¼
2
3 tan 9 sec 9
9.
Since x ¼ 3 sin , we have sin ¼ 3x and ¼ sin1 3x.
5.
281
ln j sec þ tan j þ C. Since sec ¼ 3x and tan ¼
ffi
pffiffiffiffiffiffiffiffi
x pffiffiffiffiffiffiffi
x2 9
x2 9 þ C ¼ ln 13 x þ
3 , the answer is ln 3 þ
3
pffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 9 þ C ¼ ln 13 þ ln x þ x2 9 þ C or
pffiffiffiffiffiffiffiffiffiffiffiffiffi
ln xþ x2 9 þ k where k ¼ C þ ln 13 ¼ C ln 3.
Ð 2
Ð pffiffiffiffiffiffiffiffiffiffiffiffiffi
1=2
ðx 9Þ dx ¼
x2 9 dx. Let x ¼ 3 sec ,
then dx ¼ 3 sec tan d. Substituting yields
Ð
Ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
9 sec2 9 3 sec tan d ¼ 9 tan2 sec
Ð
Ð
d ¼ 9 ðsec2 1Þ sec d ¼ 9 ðsec3 Ð
Ð
sec Þd ¼ 9 sec2 d 9 sec d. The first integral is example 20.50 and the second is a formula.
They yield 9 12 sec tan þ 12 ln j sec þ tan j
ln j sec þ tan j ¼ 92 sec tan 92 ln jpsec
ffi
ffiffiffiffiffiffiffi
x
x2 9
þ tan j. Since sec ¼ 3 and tan ¼ 3 , we get
pffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffi Ð 2
2
2
ðx h 9Þ1=2 dx ¼ 92 3x x39 92 ln i3x þ x39 þ K
pffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 12 x x2 9 9 ln x þ x2 9 þ C where
C ¼ K 92 ln 3.
Ð x
pffiffiffiffiffiffiffiffi dx. Letting u ¼ x2 þ 9 and du ¼ 2x dx, pro11.
x2 þ9 Ð
pffiffiffiffiffiffiffiffiffiffiffiffiffi
duces 12 pduffiffi ¼ 12 21 u1=2 þ C ¼ x2 þ 9 þ C.
u
13.
15.
Ð
dx
x2 þ9
¼ 13 arctan 3x þ C.
Ð 3 pffiffiffiffiffiffiffiffiffiffiffiffiffi
x 9 þ x2 dx. Let u ¼ 9 þ x2 and du ¼ 2x dx or
1
2
these
2 du ¼ x dx. Thus, x ¼ u 9. Substituting
Ð
pffiffiffi
1
in the given integral yields 2 ðu 9Þ u du ¼
Ð 3=2
1
u 9u1=2 du ¼ 12 25 u5=2 92 23 u3=2 þ C ¼
2
1 5=2
3u3=2 þC ¼ 15 uu3=2 3u3=2¼ 15 ð9 þ x2 Þ
5u
2
3=2
3=2
3=2
ð9 þ x2 Þ 3ð9 þ x2 Þ ¼ x 56 ð9 þ x2 Þ .
282
17.
CHAPTER 27
Ð pffiffiffiffiffiffiffiffiffi
43x2
x4
dx. Let
TECHNIQUES OF INTEGRATION
pffiffiffi
3x ¼ 2 sin , so x ¼ p2ffiffi3 sin and
dx ¼ p2ffiffi3 cos d. Substitution produces
ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 4 sin2 2
4 pffiffiffi cos d
3
p2ffiffi sin 3
ð 2 cos p2ffiffi cos d
3
¼
4
16
sin
9
ð
2
4 9 cos ¼ pffiffiffi d
3 16 sin4 pffiffiffi ð
3 3
csc2 cot2 d
¼
4
pffiffiffi
3 3 1 3
¼
cot þ C
3
4
pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi!3
4 3x2
3
pffiffiffi
þC
¼
4
3x
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 ð 4 3x2 Þ3
þC
¼
12
x3
3=2
1 ð4 3x2 Þ
þC
¼
12
x3
19.
21.
dx. Let x ¼ tan and dx ¼ sec2 . Then
ffi
ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð
sec tan2 þ 1 2
sec
sec2 d
d
¼
tan2 tan2 ð
ð
sec3 ð1 þ tan2 Þ sec d
¼
d
¼
2
tan tan2 ð
ð
sec ¼
þ sec d
tan2 ð 1
ð
cos ¼
d þ sec d
2
x2
sin 2
23.
Ð
x3
pffiffiffiffiffiffiffiffiffi
dx; u ¼ 3x; 3x ¼ 2 tan and so x ¼ 23 tan ;
9x2 þ4
and dx ¼ 23 sec2 d. Substituting these values in
the given integral produces
2
3
ð
2 2
3 tan qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
sec d
3
2
ð2 tan Þ þ 4
ð 16 3
tan sec2 d
¼ 81
2 sec ð
8
tan3 sec d
¼
81
ð
8
tan sec2 1 sec d
¼
81
ð
8 tan sec3 tan sec d
¼
81
8 1 3
sec sec þ C
¼
81 3
2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi!
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi!3
3
2
9x þ 4
9x2 þ 4 5
8 1
¼ 4
þC
81 3
2
2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
4 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼
ð9x2 þ 4Þ3 9x2 þ 4 þ C
243
81
ffi
Ð pffiffiffiffiffiffiffi
x2 þ1
25.
ð cos ð
cos ¼
d
þ
sec d
sin2 ¼ ðsin Þ1 þ ln j sec þ tan j þ C
pffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 1
þ ln
x2 þ 1 þ x þ C
¼
x
Ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 ðx 1Þ2 dx. Here we have a ¼ 2 and
u ¼ x 1. Let x 1 ¼ 2 sin , then x ¼
2 sin
þ 1 and dx ¼ 2 cos d. Substitution yields
Ð
Ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 4 sin2 ð2 cos Þ d ¼ ð2 cos Þð2 cos Þ Ð
Ð
Ð
2
d ¼ 2 ð1 þ cos 2Þ
d ¼ 4 cos2 d ¼ 4 1þcos
2
d þ 2 þ sin 2 þ C. Thus ¼ sin1 x1
2 and using
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ffi
4ðx1Þ
the identity sin 2 ¼ 2 sin cos ¼ 2 x1
2
2
p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx1Þ 4ðx1Þ2ffi
1 x1
so the answer is 2 sin
þC
2 þ
2
Ð
dx
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi dx. Let x 1 ¼ 2 sec and dx ¼ 2 sec 2
ðx1Þ 4
tan d. Then, you get
ð
ð
sec tan d
2 sec tan d
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
2 tan 2
ð2 sec Þ 4
ð
¼ sec d
¼ ln j sec þ tan j þ C
x 1
¼ ln 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx 1Þ2 4 þC
þ
2
1
¼ ln x 1
2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
þ ðx 1Þ2 4 þ C
¼ ln x 1
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
þ ðx 1Þ2 4 þ K
where K ¼ C ln 2.
SECTION 27.6
27.
Completing the square produces Ðx2p
2x
3 ¼ x2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
x 2x 3 dx
2xþ1 4 ¼ ðx 1Þ 4. Thus
Ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼
ðx 1Þ2 4 dx. Let x 1 ¼ 2 sec , then
dx ¼ 2 sec tan d. Substitution gives
ð qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð2 sec Þ2 4ð2 sec tan Þ d
ð
¼ 2 tan 2 sec tan d
ð
¼ 4 tan2 sec d
ð
¼ 4 sec2 1 sec d
ð
ð
¼ 4 sec3 d 4 sec d
1
1
¼ 4 sec tan þ ln j sec þ tan j
2
2
4 ln j sec þ tan j þ C
1
¼ 4 sec tan 2
1
ln j sec þ tan j þ C
2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 2x 3
x1
¼2
2
2
x 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 2x 3
þ
2 ln
þC
2
2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
¼ ðx 1Þ x2 2x 3
2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 ln x 1 þ x2 2x 3 þ K
29.
Ð
dx
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðx3Þ x2 6xþ25
¼
Ð
dx
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
. Here u ¼
2
ðx3Þ
ðx3Þ þ16
x 3 and a ¼ 4. Let x 3 ¼ 4 tan , so dx ¼
Ð
2
4p
sec
d
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 sec2 d. Then,
¼
4 tan ð4 tan Þ2 þ16
Ð 4 sec2 d
Ð
Ð
Ð
1 sec 1
1
1
4 tan 4 sec ¼ 4 tan d ¼ 4 sin ¼ 4 csc d ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
1
x2 6xþ25 4 þ
4 ln j csc cot j þ C ¼ 4 ln
x3
x3
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
4 þ C.
C ¼ 14 ln x 6xþ25
x3
31.
Substituting these values, we obtain
ð x¼3
ð 3 pffiffiffiffiffiffiffiffiffiffiffiffi2ffi
3 cos 9x
dx
¼
3 cos d
2
x
ð3
sin Þ2
1
x¼1
ð x¼3
cot2 d
¼
x¼1
ð x¼3
2
csc 1 d
¼
x¼1
¼ ½ cot x¼3
x¼1
" pffiffiffiffiffiffiffiffiffiffiffiffiffi
#
x 3
9 x2
arcsin
¼ 3
x
1
¼ 0 2
ð2:82843 0:33984Þ
1:59747
33.
Using the trigonometric substitution x ¼ 3 sin ,
pffiffiffiffiffiffiffiffiffiffiffiffiffi
produces 9 x2 ¼ 3 cos and dx ¼ 3 cos d.
Let x ¼ 2 sec ; ¼ sec1 2 , and dx ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffi
2 sec tan d so that x2 4 ¼ 2 tan . Then
ð8
ð x¼8
dx
2 sec tan d
pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
4 x x2 4
x¼4 2 sec 2 tan x¼8
ð
1 x¼8
1 ¼
d ¼ 2 x¼4
2 x¼4
x 8
1
¼ sec1
2
2 4
1
¼ sec1 4 sec1 2
2
0:1355
35.
where K ¼ C þ ln 2.
283
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Let x ¼ 6 sin ; dx ¼ 6 cos d; 36 x2 ¼
6 cos , and ¼ arcsin 6 . Thus, when x ¼
0; ¼ 0 and when x ¼ 6; ¼ 2. Then
ð 6 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð =2
2
2
x 36 x dx ¼
ð6 sin Þ2
0
0
xð6 cos Þð6 cos Þ d
ð =2
sin2 cos2 d
¼ 1296
0
¼ 324
¼ 162
ð =2
0
ð =2
sin2 2d
ð1 cos 2Þ d
0
1
¼ 162 sin 2
2
¼ 81 254:4690
=2
0
284
37.
CHAPTER 27
TECHNIQUES OF INTEGRATION
x3
The area under pffiffiffiffiffiffiffiffiffi
from 0 to 3 is
16x2
Ð3
0
x2 dx
pffiffiffiffiffiffiffiffiffi
.
16x2
Let
Substitution gives
ð
4 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
25
625 625 sin2 pffiffiffiffiffi cos d
25
21
ð
4
25
25 cos pffiffiffiffiffi cos d
¼
25
21
ð
ð
100
100 1 þ cos 2
cos2 d ¼ pffiffiffiffiffi
d
¼
25
2
21
50
1
50
¼ pffiffiffiffiffi þ sin 2 ¼ pffiffiffiffiffi ½ þ sin cos 2
21
21
pffiffiffiffiffi
pffiffiffiffiffi
21x
21x
50
þ
¼ pffiffiffiffiffi sin1
25
25
21
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi#5
625 21x2
25
pffiffiffiffiffi5
21
50
¼ pffiffiffiffiffi sin1
5
21
pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi#
21 5 625 21 25
þ
ð2Þ
625
pffiffiffiffiffi pffiffiffiffiffi
21 50
100
1 21
þ
¼ pffiffiffiffiffi sin
5
625
21
pffiffiffiffiffi
100 50 100 1 21
þ pffiffiffiffiffi sin
¼
5
625
21
pffiffiffiffiffi
21
100
104:6073:
¼ 8 þ pffiffiffiffiffi sin1
5
21
x ¼ 4 sin , then dx ¼ 4 cos d. Substituting gives
ð
ð4 sin Þ3 ð4 cos Þ d
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
16 ð4 sin Þ2
ð
64 sin3 4 cos ¼
ð 4 cos ð
3
¼ 64 sin d ¼ 64 1 cos2 sin d
1
¼ 64 cos þ cos3 3
2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi33
3
16 x2
16 x2 5
1
4
¼ 64
3
4
4
0
39.
pffiffiffi 1
1 pffiffiffi3
7 16 7 43 þ 16 4
¼
3
3
pffiffiffi
1 pffiffiffi
¼ ð7 7 48 7 64 þ 192Þ
3
pffiffiffi
1
¼ ð128 41 7Þ 6:5080654
3
Ð 4 pffiffiffiffiffiffiffiffiffiffiffiffiffi
The area described is 0 9 þ x2 dx. Let x ¼ 3 tan Ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
9 þ 9 tan2 and dx ¼ 3 sec2 d. Then,
Ð
R
ð3 sec2 Þ d ¼ ð3 sec Þð3 sec2 Þ d ¼ 9
sec3 d ¼ 9 12 sec tan þ 12 ln j sec þ tan j ¼
hpffiffiffiffiffiffiffiffi i
hpffiffiffiffiffiffiffiffi
i4
9þx2 x
9þx2
9
x 3 þ ln 3 þ 3 or, we get, 92 53 43 þ
2
3
0
ln j 53 þ 43 j 33 03 ln j 33 j ¼ 92 20
9 þ ln j3j
0 0 ¼ 5 þ 92 ln 3 or 5 þ ln 39=2 9:9438.
41.
To find the desired area we first solve
2
2
2
x2
25
2
2
þ y4 ¼ 1
2
x
4x
for y : y4 ¼ 1 25
y2 ¼
or y ¼ 1004x
25
qso
ffiffiffiffiffiffiffiffiffiffiffiffi
ffi 4 p25
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2 . We also
¼
25
x
and we get y ¼ 1004x
25
5
1=2
1=2
need y0 ¼ 25 ð25 x2 Þ
12 2x ¼ 25 xð25x2 Þ
¼
2
2x
02
4x
pffiffiffiffiffiffiffiffiffi
. Thus, y ¼ 25ð25x2 Þ and so
5 25x2
ð5
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2y 1 þ y02 dx
S¼
5
ffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð5
2
2
4x
1þ
dx
¼
2
25 x2
5
25ð25 x2 Þ
5
s
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
ð5
2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 625 21x2
¼
dx
2 25 x
5
25ð25 x2 Þ
5
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð5
2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 1 625 21x2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx
¼
2 25 x 5
5
25 x2
5
ð
4 5 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2
¼
625 21x dx:
25 5
pffiffiffiffiffi
ffi cos d:
If you let 21x ¼ 25 sin , then dx ¼ p25ffiffiffi
21
43.
Let u ¼ t ¼ tan and du ¼ dt ¼ sec2 d. Then
ffi
ð pffiffiffiffiffiffiffiffiffiffiffiffi
ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
t2 þ 1
tan2 þ 1 2
sec d
dt ¼
9t2
9 tan2 ð
ð
1 sec3 1
csc2 sec d
d ¼
¼
9 tan2 9
ð
1 1 þ cot2 sec d
¼
9
ð
1
cos ¼
sec þ 2 d
9
sin i
1h
lnj sec þ tan j ðsin Þ1
¼
9"
#
pffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffi
2
1
þ
t
1
ln
1 þ t2 þ t ¼
t
9
from t ¼ 0:5 to ti¼ 1, we have i ¼ 19 hSo p
ffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffi2 1
ln 1 þ t2 þ t 1þt
0:1829 A
t
0:5
SECTION 27.7
27.7
1.
Ð
INTEGRATION BY PARTS
x ln x dx. Completing the table, we get
u ¼ lnx
v ¼ 12 x2
du ¼ 1x dx
dv ¼ x dx
Combined with the Ðresult from the first table, this
produces 12 x3 e2x 32 x2 e2x dx ¼ 12 x3 e2x Ð 2x 3 1 2 2x
2 2 x e xe dx . A third table produces
Ð
Thus, using the table, we obtain x ln x dx ¼
Ð
Ð
ln x 12 x2 1x dx ¼ 12 x2 ln x 12 x dx ¼
1 2
2x
1 2
2x
3.
5.
Ð
v ¼ 12 cos 2x
du ¼ dx
dv ¼ sin 2x dx
11.
Ð
Using the table,
we see that x sin 2x dx ¼
Ð
12 x cos 2x 12 cos 2x dx ¼ 12 x cos 2x þ
1
4 sin 2x þ C.
Ð 2
x ln x dx
u ¼ ln x
v ¼ 13 x3
du ¼ 1x dx
du ¼ x2 dx
Thus, from the table, we see that x2 ln x dx ¼
Ð
ln x 13 x3 1x dx ¼ 13 x3 ln x 19 x3 þ C.
Ð
x sin1 x2 dx.
u ¼ sin
v¼
x
13.
For the second integral let w ¼ 1 x ; dw ¼
4x3 dx or 14 dw ¼ x3 dx. Substituting, we get
Ð x3
Ð
1 2
1 2
1 2
1 2
x x 14 pdwffiffiwffi ¼ 12
2 x sin
1x2 dx ¼ 2 x sin
Ð
x2 sin1 x2 14 w1=2 dw ¼ 12 x2 sin1 x2 þ 12 w1=2 þ
1
4 1=2
þ C. The
2 ð1 x Þ
1 2
1 2
1
answer
is 2 x sin ix þ 2 ð1 x4 Þ1=2
h
pffiffiffiffiffiffiffiffiffiffiffiffiffi
1 2
1
2
x þ 1 x4 þ C.
2 x sin
final
þC ¼
x3 e2x dx
u ¼ x3
v ¼ 12 e2x
du ¼ 3x2 dx
dv ¼ e2x dx
Ð
Ð
So, x3 e2x dx ¼ 12 x3 e2x 32 x2 e2x dx. Using a second table produces
u ¼ x2
v ¼ 12 e2x
du ¼ 2x dx
dv ¼ e2x dx
Ð
x2 ex=4 dx
u ¼ 8x
du ¼ 8 dx
4
Ð
Ð pffiffiffiffiffiffiffiffiffiffiffiffiffi
x 4x þ 1 dx. Let u ¼ 4x þ 1 and du ¼ 4 dx. So,
u1
¼ x and 1 du ¼ dx. Thus, the given integral is
Ð 4u1 pffiffiffi 1 4
Ð 3=2
1
1 2 5=2
u 4 dx ¼ 16
u u1=2 du ¼ 16
4 5u
h
i
5=2 2
2 3=2
1 2
þC ¼ 16
3 ð4x þ 1Þ3=2 þ C ¼
3u
5 ð4xþ1Þ
h
i
3=2 2
3=2
1 2
ð
Þ
ð
Þ
ð
Þ
4x
þ
1
4x
þ
1
4x
þ
1
þC ¼
16 5
3
h
i
3=2
1
8x
4
1 24x4
þ C ¼ 16
16
5 15 ð4x þ 1Þ
15
v ¼ 4ex=4
du ¼ 2x dx
dv ¼ ex=4 dx
Ð
Ð 2 x=4
Thus, x e dx ¼ 4x2 ex=4 8xex=4 . Using integration by parts a second time, we get the following table:
2x ffi
du ¼ pffiffiffiffiffiffiffi
dx
dv ¼ x dx
1x2
Ð x3
Ð
ffi dx.
Hence,
x sin1 x2 dx ¼ 12 x2 sin1 x2 pffiffiffiffiffiffiffi
1x2
9.
dv ¼ e2x dx
u ¼ x2
1 2
2x
C ¼ 12 x2 sin1 x2 þ
du ¼ dx
1
ð4x þ 1Þ3=2 þ C ¼ 60
ð6x 1Þð4xþ1Þ3=2 þ C.
Ð
1 2
v ¼ 12 e2x
1 3 2x
3 1 2 2x
2x
1 3 2x
3 2 2x
2 x e 2 2 x e xe dx ¼ 2 x e 4 x e
Ð
3
2x
1 2x
1 3 2x
3 2 2x
3
2x
4 xe 2 e dx ¼ 2 x e 4 x e þ 4 xe 1 2x
2x 1 3
3 2
3
1
4e þ C ¼ e
2x 4x þ 4x 4 þ C
x sin 2x dx
u¼x
u¼x
When combined with the previous result, this yields
Ð
þ
ln x 14 x2 þ C or 12 x2 ðln x 12Þ þ C.
1 3
3x
7.
285
15.
v ¼ 4ex=4
dv ¼ ex=4 dx
Ð
Hence, we get 4x2 ex=4 8xex=4 ¼ 4x2 ex=4 Ð
32xex=4 þ 32ex=4 dx ¼ 4x2 ex=4 32xex=4 þ
128ex=4 þ C ¼ ex=4 ð4x2 32x þ 128Þ þ C.
Ð x
e cos x dx
u ¼ ex
du ¼ ex dx
v ¼ sin x
dv ¼ cos x dx
Ð
Ð
This results in ex cos x dx ¼ ex sin x ex sin x
dx. Using integration by parts again produces
u ¼ ex
du ¼ ex dx
v ¼ cos x
dv ¼ sin x dx
Combining thisÐ with the previous result
x
x
ex sin x ex sin x dx
Ðyields
Ð ¼x e sin x þ e x cos x x
e cos x dx.
Hence 2 e cos x dx ¼ e sin x þ
Ð
ex cos x or ex cos x dx ¼ 12 ex ðsin x þ cos xÞ þ C.
286
17.
CHAPTER 27
Ð
Ð
Rewrite the given integral as x3 cos x2 dx ¼
2x cos x2 dx, and then use integration by parts.
2
u ¼ x2
v ¼ sin x2
du ¼ x dx
dv ¼ 2x cos x2 dx
So the integral equals
x2
2
1
2
2 sin x þ 2 cos x þ C
19.
TECHNIQUES OF INTEGRATION
Ð
x2
2
x2
2
27.
x3 ex dx
v ¼ ex
dv ¼ ex dx
Ð
Ð
Thus, we obtain x3 ex dx ¼ x3 ex þ 3x2 ex .
Using integration by parts a second time, we
have the following table.
29.
¼ 14:7781 þ 5:8861 ¼ 20:6642
Ð2
Ð2
2
Mx ¼ 12 1 ðx2 ex Þ dx ¼ 12 1 x4 e2x dx. From Exercise #16, we know
ð
x4 e2x dx
3 2 3
3
2x 1 4
3
x x þ x xþ
¼e
þC
2
2
2
4
CombinedÐ with the earlier result, this produces
Ð
x3 ex þ 3x2 ex ¼ x3 ex 3x2 ex þ 6xex .
We now use integration by parts a third time.
v ¼ ex
dv ¼ ex dx
u ¼ 6x
du ¼ 6 dx
21.
Ð
2 x
Hence, x3 ex 3x
e þ 6xex ¼ x3 ex Ð
3x2 ex 6xex þ 6ex ¼ x3 ex 3x2 ex 6xex 6ex þ C ¼ ex ðx3 þ 3x2 þ 6x þ 6Þ þ C
Ð x
e cos x dx
Thus,
ð
ð
1 2 2 x 2
1 2 4 2x
x e dx ¼
x e dx
2 1
2 1
2
1 ¼ e2x 2x4 4x2 þ 6x2 6x þ 3 1
8
15
21
¼ e4 e2 ¼ 102:0163
8
8
Ð2
The mass is m ¼ 1 x2 ex dx. Once again, we need
to use integration by parts.
Mx ¼
u ¼ ex
v ¼ sin x
x
du ¼ e dx
dv ¼ cos x dx
Ð
Ð x
Thus, e cos x dx ¼ ex sin x þ sin xex dx.
Applying integration by parts again, we have the
following table.
v ¼ ex
u ¼ x2
du ¼ 2x dx
dv ¼ ex dx
Ð
Ð
So, x2 ex dx ¼ x2 ex 2xex dx. A second application of integration by parts produces
u ¼ ex
du ¼ ex dx
23.
v ¼ cos x
dv ¼ sin x dx
Ð
Hence, Ð ex sin x þ sin xex dx ¼ ex sinÐ x ex
cos x ex cos x dx which produces 2 exÐ cos
x dx ¼ ex sin x ex cos x and means that ex
cos x ¼ 12 ex ðsin x cos xÞ þ C.
Ð
x cos 4x dx
u¼x
25.
u ¼ 2x
du ¼ 2 dx
v ¼ ex
dv ¼ ex dx
Ð 2 x
With the result that
x e dx ¼ x2 ex 2xex þ
Ð2
Ð x
2e dx ¼ x2 ex 2xex þ2ex . Thus, m ¼ 1 x2 ex dx ¼
v ¼ 14 sin 4x
du ¼ dx
dv ¼ cos 4x dx
Ð
Ð
Hence, x cos 4x dx ¼ 4x sin 4x 14 sin 4x dx ¼
x
1
4 sin 4x þ 16 cos 4x þ C.
Ð =2
Using the disc method, we see that V ¼ 0 =2
Ð =2
2x
cos2x dx ¼ 0 1þcos
¼ 12 x þ 14 sin 2x 0 ¼
2
2
4 þ 0 0 0 ¼ 4 .
du ¼ 4 dt
dv ¼ sin 2t dt
Ð
Ð
Thus, 4t sin 2t dt ¼ 2t cos 2t þ 2 cos 2t dt ¼
2t cos 2t þ sin 2t. Evaluated from 0 to 4 we have
½8 cos 8 þ sin 8 ¼ 2:1534 C.
Ð2
Ð2
My ¼ 1 x x2 ex dx ¼ 1 x3 ex dx. From Exercise #9,
Ð
we have x3 e2x dx ¼ e2x 12 x3 34 x2 þ 34 x 38 þ C.
Hence, we see that
ð2
2
x x2 ex dx ¼ ex x3 3x2 þ 6x 6 1
My ¼
1
v ¼ ex
dv ¼ ex dx
u ¼ 3x2
du ¼ 6x dx
v ¼ 12 cos 2t
u ¼ 4t
Ð
sin x2 x sin x2 ¼
u ¼ x3
du ¼ 3x2 dx
Charge
Ð is the integral of current. Hence we must
find 4t sin 2t dt. Use the following integration
by parts table.
2
2
½x2 ex 2xex þ 2ex 1 ¼ ½ex ðx2 2x þ 2Þ1 ¼
12:9378. Using the above result, we see that x ¼
My 20:6642
m ¼ 12:9387 ¼ 1:5971
31.
and y ¼ Mmx ¼ 102:0163
12:9387 ¼ 7:8846.
is the integral of force so we get W ¼
ÐWork
x3 cos x dx. Using integration by parts, we have
SECTION 27.7
u ¼ x3
v ¼ sin x
2
du ¼ 3x dx
dv ¼ cos x dx
Ð
Ð 3
Thus, x cos x dx ¼ x3 sin x 3x2 sin x dx.
Using integration by parts again yields
F ¼ 2
v ¼ sin x
dv ¼ cos x dx
¼ x3 sin x þ 3x2 cos x 6x sin x
ð
þ 6 sin x dx
=2
¼ x3 sin x þ 3x2 cos x 6x sin x þ 6 cos xj0
3
¼
þ0 6 0 0 þ 0 0 6
2
2
3
¼
3 þ 6 0:4510
2
33.
As in example 28.54 set up the coordinate axis so
that (0, 0) is at the center of the road between the
towers. Since it is a parabola the equation of the
main cable fits the form y ¼ 4px2 and contains
the point (500, 200). Thus, 200 ¼ 4p ð500Þ or
2
p ¼ 10000
. The equation of the cable is thus
1
1
x. The
y ¼ 1250 x2 . Differentiating we get y0 ¼ 625
q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x 2
Ð 500
length of the cable is 500 1 þ 625 dx. Evalu x 2
Ð qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ 625
¼
ate this integral as follows:
p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
Ð
1
6252 þ x2 . Let x ¼ 625 tan and dx ¼ 625
625
Ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
2
sec d. This leads to 625
6252 þ252 tan2 Ð
1
625 sec d ¼ 625
625 sec 625 sec2 d ¼
Ð
3
we get
625 sec d. Using Example 28.50,
hpffiffiffiffiffiffiffiffiffiffiffi
625þx2
625
625
625 þ
2 ½sec tan þ ln j sec þ tan j ¼ 2
pffiffiffiffiffiffiffiffiffiffiffi
i500
625þx2
x
x þ 625
¼ 625ð1:02445 þ
625 þ ln
625
500
35.
lnð2:08062Þ ¼ 625ð1:02445 þ 0:73267Þ. Thus, the
length of the main cable is 625ð1:75712Þ ¼ 1098 ft.
ÐR
ÐR
(a) F ¼ 2 0 r PðrÞdr ¼ 2 0 P0 ekr dr. Integrating by parts, we have
u ¼ Po r
v ¼ 1k ekr
du ¼ P0 dr
dv ¼ ekr dr
and so,
r P0 ekr dr
0
v ¼ cos x
u ¼ 3x2
du ¼ 6x dx
dv ¼ sin x dx
Ð 3
Hence,
x cos x dx ¼ x3 sin x þ 3x2 cos x
Ð
6x cos x. We need integration by parts one more
time.
u ¼ 6x
du ¼ 6 dx
ð
x3 cos x dx
ðR
287
ðR
R
1
P0 kr
e dr
¼ 2 P0 rekr 0 þ 2
k
0 k
1
P0 kr R
kr
¼ 2 P0 re 2 2 d
k
k
0
2P0 RkekR 2P0 ekR 2P0
þ 2
k2
k
kR
e 2ðkR þ 1ÞP0 2P0
¼
þ 2
k2
k
ÐR 2
ÐR
(b) T ¼ 2 0 r PðrÞdr ¼ 2 0 r 2 ÐR
P ekr dr ¼ 2P0 0 r 2 ekr dr. Integrating
Ð R0 2 kr
dr we will use integration by parts
0 r e
twice. The first time produces
¼
u ¼ r2
v ¼ 1k ekr
du ¼ 2r dr
dv ¼ ekr dr
Ð
Ð
2
and we get r 2 ekr dr ¼ rk ekr þ 2k rekr dr.
The second time we use integration by parts, we
have
u ¼ 2k r
v ¼ 1k ekr
du ¼ 2k dr
dv ¼ ekr dr
and we get
ð
ð
r2
2r
2
r 2 ekr dr ¼ ekr 2 þ 2 ekr dr
k
k
k
r 2 kr 2r 2 kr
¼ e 2 3e þC
k
k
k
Putting the constant multiples and the limits of
integration back in, the result is
ðR
T ¼ 2 r 2 PðrÞdr
0
R
r 2 kr 2r 2 kr
¼ 2 e 2 3 e
k
k
k
0
R
1 kr 2 2
¼ 2 3 e
k r þ 2kr þ 2
k
0
1 kR 2 2
¼ 2 3 e
k R þ 2kR þ 2
k
1
þ 2 3 e0 k2 02 þ 2k 0 þ 2
k
1 kR 2 2
k R þ 2kR þ 2
¼ 2 3 e
k
1
þ 4 3
k
288
37.
CHAPTER 27
C¼
Ð 6 40 lnðtþ2Þ
0
ðtþ2Þ2
TECHNIQUES OF INTEGRATION
dt ¼ 20
3
Ð 6 lnðtþ2Þ
0 ðtþ2Þ2
dt. To determine
the integral, we use integration by parts
u ¼ lnðt þ 2Þ
1
du ¼ tþ2
dt
and
Ð
so,
v ¼ ðt þ 2Þ1
dv ¼ ðt þ 2Þ2 dt
lnðtþ2Þ
dt
ðtþ2Þ2
¼ lnðtþ2Þ
tþ2
Ð
1
ðtþ2Þ2
dt ¼ lnðtþ2Þ
tþ2
1
tþ2
. Combining the pieces, we get C ¼ 20
3 h
i6
lnðtþ2Þþ1
ln 8þ1
ln 2þ1
3:0776 ppm
¼ 20
3
8 2
tþ2
0
39.
(a) As in Example 27.54, we put the origin at the roadbed midway between the towers. This means that the vertex
of the parabolas that describe the cables is at the ð0; 11:86Þ and the cables are attached to the towers at the points
ð1650; 312:34Þ and ð1650; 312:34Þ. By placing the vertex at the midpoint of the roadbed, we know that this
parabola is of the form 4pðy 11:86Þ ¼ x2 . Since ð1650; 312:34Þ is a point on this parabola, we see that
4pð312:34 11:86Þ ¼ 4pð300:48Þ ¼ 1201:92p ¼ 16502 and so p 2265:1258. The parabola for the cables has
x2
þ 11:86. For the arc
the equation 4ð2265:1258Þðy 11:86Þ ¼ 9060:5032ðy 11:86Þ ¼ x2 or y ¼
9060:5032
2x
x
¼
. The length of the cable is
length we need the derivative of y, which is y0 ¼
9060:5032 4530:2516
r
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
ð 1650
2
x
L¼
1þ
dx
4530:2516
1650
ð 1650 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
4530:25162 þ x2 dx
¼
4530:2516 1650
Using the trigonometric substitution of x ¼ 4530:2516 tan , then dx ¼ 4530:2516 sec2 d and the indefinite
integral becomes
ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
L¼
4530:25162 þ x2 dx
4530:2516
ð qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
4530:25162 þ ð4530:2516 tan Þ2 ð4530:2516 sec2 Þ d
¼
4530:2516
ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 4530:2516
1 þ tan2 sec2 d
ð
¼ 4530:2516 sec3 d
Since
ð
1
1
sec3 d ¼ sec tan þ ln j sec þ tan j þ C
2
2
we have
4530:2516
½sec tan þ ln j sec þ tan j þ C
2
¼ 2265:1258½sec tan þ ln j sec þ tan j þ C
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4530:25162 þ x2
x
and so
, we see that sec ¼
Since tan ¼
4530:2516
4530:2516
#1650
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
"pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4530:25162 þ x2
x
x
4530:25162 þ x2
L ¼ 2265:1258
þ ln
þ
4530:2516
4530:2516
4530:2516
4530:2516
L¼
1650
3371:573066
Thus, the length of one of the cables that hold up the center span is about 3371.56 m and the total length of the
four cables is about 13 486.24 m.
SECTION 27.8
289
(b) For this parabola, place the origin at the anchor block so it has an equation of the type 4py ¼ x2 . Since the cable is
fasten to the top of the tower at ð960; 322:5Þ we see that 4pð322:5Þ ¼ 9602 and p 714:4186. Thus,
1
1
1
y ¼ 4714:4186
x2 ¼ 2857:6744
x2 and y0 ¼ 1428:8372
x. The length of one of these cables is
ð 960 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ffi
x
L
1þ
dx
1428:8372
0
Using the procedures in (a) you should find that L 1027:9863 and the total length of the four cables from the top of
the tower on the Sicily side of the bridge to the anchor block is about 4111.95 m.
(c) As in (b), place the origin at the anchor block so it has an equation of the type 4py ¼ x2 . Since the cable is fasten to
1
the top of the tower at ð810; 258Þ we see that 4pð258Þ ¼ 8102 and p 635:7558. Thus, y ¼ 2543:0232
x2 and
0
1
y ¼ 1217:5116 x. The length of one cable is
ð 810 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ffi
x
L¼
1þ
dx
1217:5116
0
866:304592
The total length of the four cables from the top of the tower on the Calabria side of the bridge to the anchor block is
about 3465.22 m.
(d) The total length of the four cables from the anchor block on the Sicily side to the anchor block on the Calabria side
is 13 486:24 þ 4111:95 þ 3465:22 ¼ 21 063:41 m. We can think of this as a cylinder with a length of 21 063:41 m and a
diameter of 1:24 m or a radius of 0:62 m. The volume of a cylinder is V ¼ r 2 h and for these cables we have
V ¼ ð0:62Þ2 ð21 063:41Þ 25 436:77 m3 .
27.8
1.
3.
USING INTEGRATION TABLES
Ð
Ð
ð1 þ tan 3xÞ2 dx ¼ ð1 þ 2 tan 3x þ tan2 3xÞdx by
formulas #12 and #58 we get x þ 2 13 ln j sec 3xjþ
1
2
1
3 ðtan 3x 3xÞ þ C ¼ x þ 3 ln j sec 3xj þ 3 tan 3x x þ C ¼ 13 ð2 ln j sec 3xj þ tan 3xÞ þ C
Ð dx
Ð dx
xð4x3Þ ¼ xð3þ4xÞ Let u ¼ x; a ¼ 3, and du ¼
u
dx. By formula #48 this is 1a ln j aþbu
jþC ¼
þ C.
Ð x2 dx
pffiffiffiffiffiffiffiffiffi. Let u ¼ x3 and du ¼ 3x2 dx or
x6 16þx6
Ð du
1
2
1
pffiffiffiffiffiffiffiffiffi
3 du ¼ 3x dx. Substituting yields 3 u2 42 þu2 . By
pffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffi
2
16þu
16þx6
formula #39 this is 13 16u
þC ¼ 13 16x
þ
3
pffiffiffiffiffiffiffiffiffi
6
16þx
þ C.
C ¼ 48x
3
11.
1
x 3 ln 4x3
5.
7.
9.
Ð
2
x ffi
pffiffiffiffiffiffiffi
dx.
9x2
1 x
9
2 sin
3þ
Ð
pffiffiffiffiffiffiffiffiffiffiffiffiffi
By formula #25 this is 2x 9 x2 þ
13.
15.
C.
cos3 x dx. By formula #65 this is 13 cos2 x sin x þ
Ð
2
¼ 1 cos2 x sin x þ 23 sin x þ C ¼ 13
3 cos x dx
3
2
1sin x sin x þ 23 sin x þ C ¼ sin x 13 sin3 xþ
Ð
Ð
C. Alternate solution: cos3 x dx ¼ 1 sin2 x
Ð
Ð 2
cos x dx ¼ cos x dx sin x cos x dx ¼ sin x
13 sin3 x þ C.
17.
19.
Ð
sin6 3x dx. Let u ¼ 3x, then du ¼ 3 dx or
Ð
1
we get 13 sin6 u du. By for3 du ¼ dx. Substituting
Ð
mula #64 this is 13 16 sin5 u cos u þ 56 sin4 u du .
Using formula #64 once more, you obtain
1 5
1 3
Ð
1
5
u þ 34 sin2 3 6 sin u cos u þ 6 4 sin u cos
u duÞ. By formula #56 we get 13 16 sin5 u cos uþ
3
5
1
3 1
1
6 4 sin u cos u þ 4 2 u 2 sin u cos u þ C ¼
1
5
5
sin5 u cos u 72
sin3 u cos u 48
sin u cos u þ
18
5
3
5
1
5
u
þ
C
¼
sin
3x
cos
3x
sin
3x cos 3x 48
18
72
5
5
48 sin 3x cos 3x þ 48 x þ C.
Ð 10x
e cos 6x dx: By formula #88 we have this is
e10x
1 10x
ð10 cos 6x
102 þ62 ð10 cos 6x þ 6 sin 6xÞ þC ¼ 136 e
þ 6 sin 6xÞ þ C.
Ð 7
x ln x dx. By formula #90 we obtain x8 18 ln x 1
64Þ þ C
Ð
arcsin 4x dx. Letting u ¼
Ð 4x and du ¼ 4 dx and
then substituting we get 14 arcsin u du. By formula
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R
#78, arcsin 4x dx ¼ 14 ð4x sin1 4xþ 1ð4xÞ2 þ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
C ¼ 14 4x sin1 4x þ 1 16x2 þ C.
ffi
pffiffiffiffiffiffiffiffiffiffiffiffiffi
Ð pffiffiffiffiffiffiffi
9þx2
dx. By formula #32 this is
9 þ x2 x
p
ffiffiffiffiffiffiffi
ffi
3þ 9þx2 þ C.
3 ln x
290
21.
CHAPTER 27
Ð
x3 e2x dx. By formula #86 we get 12 x3 e2x Ð
3
2 2x
1 3 2x
2 x e dx. Repeating this process, we get 2 x e
Ð
32 12 x2 e2x þ 32 22 xe2x dx or 12 x3 e2x 34 x2 e2x þ
Ð 2x
3
1 3 2x
2 xe dx. Repeating again produces 2 x e Ð
3 2 2x
3
2x
31
2x
1 3 2x
3 2 2x
4 x e þ 4 xe 2 2 e dx ¼ 2 x e 4 x e þ
3
2x
3 2x
4 xe 8 e
23.
TECHNIQUES OF INTEGRATION
Ð
25.
þ C ¼ 18 e2x ð4x3 6x2 þ 6x 3Þ þ C.
x3 sin 2x dx. Let u ¼ 2x, then u3 ¼ 8x3 or
x ¼ 18 u3 , and also, du ¼ 2dx or dx ¼ 12 du. SubstiÐ 3
1
tution produces 16
u sin u du. By formula #75 this
3
Ð
1
is 16 u cos u þ 3 u2 cos u du . Now using formula #76, on the integral in this result, we obtain
3
2
Ð
1
16 u cos uþ3 u sin u2 u sin u du . Finally,
using formula #75 one more time produces the
Ð
1
½u3 cos uþ3u2 sin uþ6u cos u 6 cos u
result 16
1
½u3 cos uþ3u2 sin uþ6u cos u 6 sin u þ C.
or 16
3
27.
Ð
Back substituting for u produces x3 sin 2x dx ¼
1
3
2
16 ½8x cos 2x þ12x sin 2x þ12x cos 2x6 sin 2x
1 3
þ C ¼ 2 x cos 2x þ 34 x2 sin 2x þ 34 x cos 2x 3
8 sin 2xþ C.
Ð
1 2:5 2 5t
V ¼ 2:5
0 t e dt. Using integration form #85 produces
2:5
1 e5t 2
25t 10t þ 2
V¼
2:5 125
0
1
¼
ð286047:5315Þ 114419:0126
2:5
3xþ7 2
Ð
Ð2 1
1
W ¼ F dx ¼ 0 499x
2 dx ¼ 27 ln 3x7 0 by inte1
ðln 13
gration form #19. This evaluates as 14
1
ln 1Þ ¼ 14 ln 13 0:1832 N m.
CHAPTER 27 REVIEW
1.
3.
5.
7.
9.
Ð
xe3x dx. Using integration by parts with the table
u¼x
v ¼ 13 e3x
du ¼ dx
dv ¼ e3x dx
Ð 3x
Ð
We have
xe dx ¼ 13 xe3x 13 e3x dx ¼ 13 xe3x 1 3x
1 3x
9 e þ C ¼ 9 e ð3x 1Þ þ C.
Ð x
pffiffiffiffiffiffiffiffiffi dx. Let x ¼ 5 sin , then dx ¼ 5 cos d.
25x2
Ð 5 sin 5 cos d
Thus, substitution produces pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
255 sin2 Ð 5 sin 5 cos d
Ð
¼
¼ 5 sin d ¼ 5 cos þ C ¼ 5 5 cos pffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
25x
þ C ¼ 25 x2 þ C. Alternate solution:
5
Let u ¼ 25 x2 , then du ¼ 2x dx or 12 du ¼
x dx. Substituting, we obtain
ð
ð
1 du
1 1=2
1 2
p
ffiffi
ffi
u
du ¼ u1=2 þ C
¼
2
2
2 1
u
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 25 x2 þ C
11.
1
7 ð cos uÞ
þ C ¼ 17 cosð7x þ 2Þ þ C.
Ð
Ð
sin5 3x cos2 3x dx ¼ sin 3x sin4 3x cos2 3x dx ¼
2
sin 3xð1 cos2 3xÞ cos2 3x dx. This expands as
Ð 2
cos 3x2 cos4 3x þ cos6 3x sin 3x dx. Let u ¼
cos 3x and du ¼ 3 sin 3x dx or 13 du ¼ sin 3x dx.
Ð
Substituting we get 13 u2 2u4 þ u6 du ¼ 13
1 3 2 5 1 7 1
2
3
5
3 u 5 u þ 7 u þ C ¼ 9 cos 3x þ 15 cos 3x
1
7
21 cos
13.
3x þ C.
Ð
dx
. Let 2x ¼ 7 tan , or x ¼ 72 tan ð4x2 þ49Þ3=2
dx ¼ 72 sec2 d. Substituting yields
Ð
Ð
and then
Ð
2
sec2 d
d
1
1
¼ 72 sec
73 sec3 ¼ 98 sec d ¼
ðð7 tan Þ2 þ49Þ3=2
Ð
1
1
1 pffiffiffiffiffiffiffiffiffiffiffi
2x
98 cos d ¼ 98 sin þ C ¼ 98 4x2 þ49 þ C
7
2
x
pffiffiffiffiffiffiffiffiffiffiffi
49 4x2 þ49
15.
Ð
sin4 2x cos 2x dx. Let u ¼ sin 2x then du ¼
2 cos 2x dx or 12 du ¼ cos 2x dx. Substitution proÐ
1
sin5 2x þ C.
duces 12 u4 du ¼ 12 15 u5 þ C ¼ 10
Ð dx
1
9xþ5. Let u ¼ 9x þ 5 and du ¼ 9 dx or 9 du ¼ dx.
Ð
1
1
Then 19 du
u ¼ 9 ln juj ¼ 9 ln j9x þ 5j þ C.
Ð
sinð7x þ 2Þdx. Let u ¼ 7x þ 2 and du ¼ 7 dx or
Ð
1
1
7 du ¼ dx. Then, substitution produces 7 sin u du ¼
Ð
17.
19.
Ð
¼
þ C.
3
x2 ex dx. Let u ¼ x3 and du ¼ 3x2 dx or 13 du ¼
Ð
x2 dx. Substituting, we get 13 eu du ¼ 13 eu þ C ¼
1 x3
3 e þ C.
Ð x dx
pffiffiffiffiffiffiffiffiffiffiffi. Let u ¼ 4x2 þ 49 and du ¼ 8x dx or
4x2 þ49
Ð
1
Substituting we get 18 pduffiffiu ¼
8 du ¼ x dx.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Ð 1
1
1 2 1=2
2
þ C ¼ 14 4x2 þ 49 þ C.
8 u du ¼ 8 1 u
Ð
sec 4x tan 4x
9þ2 sec 4x dx.
Let u ¼ 9 þ 2 sec 4x and du ¼
Ð
8 sec 4x tan 4x dx. Substituting we get 18 du
u ¼
1
1
ln
juj
þ
C
¼
ln
j9
þ
2
sec
4xj
þ
C.
8
8
CHAPTER 27 REVIEW
21.
23.
Ð
½lnð2xþ1Þ5
2xþ1 dx. Let u ¼ lnð2x þ 1Þ and then du
Ð 5
2dx
1
1 6
2xþ1. Substituting yields 2 u du ¼ 12 u þ C
6
1
12 ½lnð2x þ 1Þ þ C.
Ð
x2
3
x3 þ4 dx. Let u ¼ x þ 4 and
1
2
3 du ¼ x dx. Now substituting
1
1
3
3 ln juj þ C ¼ 3 ln jx þ 4j þ C.
25.
Ð
27.
Ð
29.
Ð
31.
33.
35.
37.
¼
du ¼ 3x dx or
Ð
we get 13 du
u ¼
tan 5x dx ¼ 5 ln j sec 5x j þ C or 5 ln j cos 5x j þ C.
39.
41.
cos x
dx.
sin2 xþ9
2
x5 ex dx. Using integration by parts, we have the
following table.
u ¼ x4
du ¼ 4x3 dx
v ¼ 12 ex
2
2
dv ¼ xex dx
Ð
Ð 5 x2
2
2
This produces
x e dx ¼ 12 ex x4 2 x3 ex dx.
Using integration by parts again with u ¼ 2x2
2
2
and dv ¼ xex dx, we get v ¼ 12 ex and du ¼
Ð
2
2
2
2
4x dx. Thus, 12 ex x4 2 x3 ex dx ¼ 12 ex x4 ex Ð
ðarctan 2xÞ4
1þ4x2
2
dx. Let u ¼ arctan 2x, then du ¼ 1þ4x
2 dx
Ð
1
4
1 5
and substituting gives 2 u du ¼ 10 u þ C ¼
5
1
10 ðarctan 2xÞ
ecos x sin x dx ¼ ecos x þ C
Let u ¼ sin x and du ¼ cos x dx.
Ð
Substituting we get u2duþ9 ¼ 13 tan1 u3 þ C ¼
1
1 sin x
þ C.
3 tan
3
Ðe
Let u ¼ x2 and du ¼ 2x dx. Then 1 x3 ln x2 dx ¼
Ð
1 e 2
2
2 1 x ln x 2x dx and by integration formula #90,
e
ð
1 e 2
1 2 ln x2 1 x ln x2 2x dx ¼ x2
2 1
2
4 1
2
2
4
e ln e
1
¼
4
2
2
1
1
0
2
4
4
e
1
1
¼
1
þ
4
8
2
4
3e
1
¼
þ 20:5993
8
8
Ð 1=3
Ð
2
sin 4x cos5 4x dx ¼ sin1=3 4xðcos2 4xÞ 2
Ð
cos 4x dx ¼ sin1=3 4x 1 sin2 4x cos 4x dx ¼
Ð 1=3
sin 4x 2 sin7=3 4x þ sin13=3 4x cos 4x dx ¼
4=3
1 3
1
3
4x 2 10
sin10=3 4x þ 16
sin16=3 4x þ
4 4 sin
3
C ¼ 320
sin4=3 4x 20 16 sin2 4x þ 5 sin4 4x þ C.
Ð cos x dx
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi. Let u ¼ 2 sin x and du ¼ 2 cos x dx or
164 sin2 x
Ð du
1
du
¼ cos x dx. Substituting we get 12 pffiffiffiffiffiffiffiffiffi
2
x
42 u2
1 sin x
1
¼ 12 sin u4 þC ¼ 12 sin1 2 sin
sin
þC
¼
4
2
2 þC.
Ð
Ð
2
2
2
2
2 xex ¼ 12 ex x4 ex x2 þ ex þ C ¼
x2 þ
2
4
x2
ex x2 x2 1 þ C ¼ e2 ðx4 2x2 þ 2Þ þ C.
¼
2
291
43.
þ C.
ffi
Ð ln 49 pffiffiffiffiffiffiffiffiffiffiffiffiffi
Ð ln 49 pffiffiffiffiffiffiffi
9þex ex=2
Rewrite 0
9 þ ex dx ¼ 2 0
2 dx. Let
ex=2
1 x=2
x=2
u ¼ e ; du ¼ 2 e dx, and a ¼ 3 and this fits integration formula #32. Remember that ln 49 ¼
ln 72 ¼ 2 ln 7, and so, evaluating ex=2 when x ¼
ln 49 produces eðln 49Þ=2 ¼ eð2 ln 7Þ=2 ¼ eln 7 ¼ 7.
ð ln 49
pffiffiffiffiffiffiffiffiffiffiffiffiffi
9 þ ex dx
0
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼2
9 þ ex =2:
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 þ 9 þ ex=2 ln 49
3 ln ex=2
0
pffiffiffiffiffiffiffiffiffiffiffi
3 þ 9 þ 7
pffiffiffiffiffiffiffiffiffiffiffi
¼ 2 9 þ 7 3 ln
7
pffiffiffiffiffiffiffiffiffiffiffi
3 þ 9 þ 1
pffiffiffiffiffiffiffiffiffiffiffi
9 þ 1 3 ln
1
¼ 2½4 3 ln j1j
hpffiffiffiffiffi
pffiffiffiffiffii
2 10 3 ln 3 þ 10
pffiffiffiffiffi
pffiffiffiffiffi
¼ 8 2 10 þ 6 ln 3 þ 10 12:5861
Using integration by parts, with
pffiffiffiffiffiffiffiffiffiffiffiffiffi
v ¼ 9 þ x2
u ¼ x2
du ¼ 2x dx
x ffi
dv ¼ pffiffiffiffiffiffiffi
dx
9þx2
Then, we obtain
ð3 3
x dx
pffiffiffiffiffiffiffiffiffiffiffiffiffi
9 þ x2
0
ð 3 pffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ x2 9 þ x2 30 2x 9 þ x2 dx
0
pffiffiffiffiffiffiffiffiffiffiffiffiffi
3
2
¼ x2 9 þ x2 ð9 þ x2 Þ3=2
3
0
pffiffiffiffiffi 2
2 3=2
3=2
¼ 9 18 ð18Þ 0 þ ð9Þ
3
3
pffiffiffi 2 pffiffiffi3 2
¼ 27 2 3 2 þ ð27Þ
3
pffiffiffi 3 pffiffiffi
¼ 27 2 36 2 þ 18
pffiffiffi
¼ 18 9 2 5:2721
292
CHAPTER 27
TECHNIQUES OF INTEGRATION
CHAPTER 27 TEST
1.
3.
5.
Ð x dx
ffi¼
Let u ¼ 9 ex , then du ¼ ex dx and peffiffiffiffiffiffiffi
9ex
pffiffiffiffiffiffiffiffiffiffiffiffiffi
Ð 1=2
1=2
x
du ¼ 2u þ C ¼ 2 9 e þ C
u
Ð 5 dx
1
xþC
x2 þ1 ¼ 5 tan
4x
Use integration by parts with u ¼ x and dv ¼ e dx.
Ð
Then du ¼ dx and v ¼ 14 e4x and so xe4x dx ¼ uv
Ð
1 4x
e þC ¼
v du ¼ 14 xe4x 14 e4x dx ¼ 14 xe4x 16
1 4x
e
ð4x
1Þ
þ
C.
16
Ð
7.
Using the disk method, we have V ¼ Ð 1 2x
Ð 1 pffiffiffi x 2
1
2x
0 ð xe Þ dx ¼ 0 xe dx ¼ 4 e ð2x 1Þ0 ¼
2
4 ðe þ 1Þ.