7
TECHNIQUES OF INTEGRATION
7.1 Integration by Parts
Preliminary Questions
1. Which derivative rule is used to derive the Integration by Parts formula?
SOLUTION
The Integration by Parts formula is derived from the Product Rule.
2. For each of the following integrals, state whether substitution or Integration by Parts should be used:
Z
Z
Z
Z
2
2
2 x
x cos.x / dx;
x cos x dx;
x e dx;
xe x dx
SOLUTION
(a)
(b)
(c)
(d)
Z
Z
Z
Z
x cos.x 2 / dx: use the substitution u D x 2 .
x cos x dx: use Integration by Parts.
x 2 e x dx; use Integration by Parts.
2
xe x dx; use the substitution u D x 2 .
3. Why is u D cos x, v 0 D x a poor choice for evaluating
SOLUTION
Z
x cos x dx?
Transforming v 0 D x into v D 12 x 2 increases the power of x and makes the new integral harder than the original.
Exercises
In Exercises 1–6, evaluate the integral using the Integration by Parts formula with the given choice of u and v 0 .
Z
1.
x sin x dx; u D x, v 0 D sin x
SOLUTION
Using the given choice of u and v 0 results in
uDx
u0 D 1
v D ! cos x
v 0 D sin x
Using Integration by Parts,
Z
Z
Z
x sin x dx D x.! cos x/ ! .1/.! cos x/ dx D !x cos x C cos x dx D !x cos x C sin x C C:
Z
2.
xe 2x dx; u D x, v 0 D e 2x
SOLUTION
Using u D x and v 0 D e 2x gives us
1 2x
2e
uDx
vD
u0 D 1
v 0 D e 2x
Integration by Parts gives us
!
" Z
! "
Z
1 2x
1
1
1 1 2x
1
xe 2x dx D x
e
! .1/ e 2x dx D xe 2x !
e C C D e 2x .2x ! 1/ C C:
2
2
2
2 2
4
Z
x
0
x
3. .2x C 9/e dx; u D 2x C 9, v D e
SOLUTION
Using u D 2x C 9 and v 0 D e x gives us
u D 2x C 9
u0 D 2
v D ex
v0 D ex
Integration by Parts gives us
Z
Z
.2x C 9/e x dx D .2x C 9/e x ! 2e x dx D .2x C 9/e x ! 2e x C C D e x .2x C 7/ C C:
782
4.
CHAPTER 7
Z
TECHNIQUES OF INTEGRATION
x cos 4x dx; u D x, v 0 D cos 4x
SOLUTION
Using u D x and v 0 D cos 4x gives us
1
4
sin 4x
uDx
vD
u D1
v D cos 4x
0
0
Integration by Parts gives us
!
"
Z
Z
1
1
1
1
1
x cos 4x dx D x sin 4x ! .1/ sin 4x dx D x sin 4x !
! cos 4x C C
4
4
4
4
4
5.
Z
D
1
1
x sin 4x C
cos 4x C C:
4
16
x 3 ln x dx; u D ln x, v 0 D x 3
SOLUTION
Using u D ln x and v 0 D x 3 gives us
u D ln x
v D 14 x 4
u0 D
v0 D x 3
1
x
Integration by Parts gives us
!
" Z ! "!
"
Z
1 4
1
1 4
x 3 ln x dx D .ln x/
x !
x
dx
4
x
4
Z
1
1
1
1
x4
D x 4 ln x !
x 3 dx D x 4 ln x ! x 4 C C D
.4 ln x ! 1/ C C:
4
4
4
16
16
Z
6.
tan!1 x dx; u D tan!1 x, v 0 D 1
SOLUTION
Using u D tan!1 x and v 0 D 1 gives us
u D tan!1 x
1
u0 D 2
x C1
vDx
v0 D 1
Integration by Parts gives us
Z
tan!1 x dx D x tan!1 x !
Z !
"
1
x dx:
x2 C 1
For the integral on the right we’ll use the substitution w D x 2 C 1, dw D 2x dx. Then we have
"
Z
Z !
Z
1
1
1
dw
!1
tan!1 x dx D x tan!1 x !
2x
dx
D
x
tan
x
!
2
2
2
w
x C1
D x tan!1 x !
1
1
ln jwj C C D x tan!1 x ! ln jx 2 C 1j C C:
2
2
In Exercises 7–36, evaluate using Integration by Parts.
Z
7. .4x ! 3/e !x dx
SOLUTION
Let u D 4x ! 3 and v 0 D e !x . Then we have
u D 4x ! 3
u0 D 4
v D !e !x
v 0 D e !x
Using Integration by Parts, we get
Z
Z
.4x ! 3/e !x dx D .4x ! 3/.!e !x / ! .4/.!e !x / dx
D !e !x .4x ! 3/ C 4
Z
e !x dx D !e !x .4x ! 3/ ! 4e !x C C D !e !x .4x C 1/ C C:
S E C T I O N 7.1
8.
Z
Integration by Parts
.2x C 1/e x dx
SOLUTION
Let u D 2x C 1 and v 0 D e x . Then we have
v D ex
u D 2x C 1
u0 D 2
v0 D ex
Using Integration by Parts, we get
Z
Z
.2x C 1/ e x dx D .2x C 1/.e x / ! .2/.e x / dx
9.
Z
D .2x C 1/e x C 2
Z
e x dx D .2x C 1/e x ! 2e x C C D e x .2x ! 1/ C C:
x e 5xC2 dx
SOLUTION
Let u D x and v 0 D e 5xC2 . Then we have
1 5xC2
e
5
uDx
vD
u0 D 1
v 0 D e 5xC2
Using Integration by Parts, we get
!
" Z
!
"
Z
Z
1 5xC2
1 5xC2
1
1
xe 5xC2 dx D x
e
! .1/
e
dx D xe 5xC2 !
e 5xC2 dx
5
5
5
5
!
"
1 5xC2
1 5xC2
x
1
D xe
! e
CC D
!
e 5xC2 C C
5
25
5 25
Z
10.
x 2 e x dx
SOLUTION
Let u D x 2 and v 0 D e x . Then we have
u D x2
u0 D 2x
v D ex
v0 D ex
Using Integration by Parts, we get
Z
x 2 e x dx D x 2 e x ! 2
We must apply Integration by Parts again to evaluate
Z
Z
Z
xe x dx:
xe x dx. Taking u D x and v 0 D e x , we get
xe x dx D xe x !
Z
.1/e x dx D xe x ! e x C C:
Plugging this into the original equation gives us
Z
#
$
x 2 e x dx D x 2 e x ! 2 xe x ! e x C C D e x .x 2 ! 2x C 2/ C C:
11.
Z
x cos 2x dx
SOLUTION
Let u D x and v 0 D cos 2x. Then we have
uDx
u0 D 1
vD
1
2
sin 2x
v 0 D cos 2x
Using Integration by Parts, we get
" Z
!
"
!
Z
1
1
x cos 2x dx D x
sin 2x ! .1/
sin 2x dx
2
2
Z
1
1
1
1
D x sin 2x !
sin 2x dx D x sin 2x C cos 2x C C:
2
2
2
4
783
784
12.
CHAPTER 7
Z
TECHNIQUES OF INTEGRATION
x sin.3 ! x/ dx
SOLUTION
Let u D x and v 0 D sin.3 ! x/. Then we have
v D cos.3 ! x/
uDx
u0 D 1
v 0 D sin.3 ! x/
Using Integration by Parts, we get
Z
Z
x sin.3 ! x/ dx D x cos.3 ! x/ ! .1/ cos.3 ! x/ dx D x cos.3 ! x/ C sin.3 ! x/ C C
13.
Z
x 2 sin x dx
SOLUTION
Let u D x 2 and v 0 D sin x. Then we have
u D x2
u0 D 2x
v D ! cos x
v 0 D sin x
Using Integration by Parts, we get
Z
Z
Z
x 2 sin x dx D x 2 .! cos x/ ! 2x.! cos x/ dx D !x 2 cos x C 2 x cos x dx:
We must apply Integration by Parts again to evaluate
Z
Z
x cos x dx. Taking u D x and v 0 D cos x, we get
x cos x dx D x sin x !
Z
sin x dx D x sin x C cos x C C:
Plugging this into the original equation gives us
Z
x 2 sin x dx D !x 2 cos x C 2.x sin x C cos x/ C C D !x 2 cos x C 2x sin x C 2 cos x C C:
14.
Z
x 2 cos 3x dx
SOLUTION
Let u D x 2 and v 0 D cos 3x. Then we have
1
sin 3x
3
u D x2
vD
u0 D 2x
v 0 D cos 3x
Using Integration by Parts, we get
Z
Z
Z
1
1
1
2
x 2 cos 3x dx D x 2 sin 3x ! .2x/ sin 3x dx D x 2 sin 3x !
x sin 3x dx
3
3
3
3
Use Integration by Parts again on this integral, with u D x and v 0 D sin 3x to get
!
"
Z
Z
1
2
1
1
x 2 cos 3x dx D x 2 sin 3x !
! x cos 3x C
cos 3x dx
3
3
3
3
15.
Z
D
1 2
2
2
x sin 3x C x cos 3x !
sin 3x C C
3
9
27
e !x sin x dx
SOLUTION
Let u D e !x and v 0 D sin x. Then we have
u D e !x
u0 D !e !x
v D ! cos x
v 0 D sin x
Using Integration by Parts, we get
Z
Z
Z
e !x sin x dx D !e !x cos x ! .!e !x /.! cos x/ dx D !e !x cos x ! e !x cos x dx:
S E C T I O N 7.1
We must apply Integration by Parts again to evaluate
Z
Z
e !x cos x dx D e !x sin x !
Integration by Parts
e !x cos x dx. Using u D e !x and v 0 D cos x, we get
Z
.!e !x /.sin x/ dx D e !x sin x C
Z
e !x sin x dx:
Plugging this into the original equation, we get
%
&
Z
Z
e !x sin x dx D !e !x cos x ! e !x sin x C e !x sin x dx :
Solving this equation for
16.
Z
Z
e !x sin x dx gives us
Z
1
e !x sin x dx D ! e !x .sin x C cos x/ C C:
2
e x sin 2x dx
SOLUTION
Let u D sin 2x and v 0 D e x . Then we have
u D sin 2x
u0 D 2 cos 2x
v D ex
v0 D ex
Using Integration by Parts, we get
Z
e x sin 2x dx D e x sin 2x ! 2
We must apply Integration by Parts again to evaluate
Z
Z
e x cos 2x dx D e x cos 2x !
Z
Z
e x cos 2x dx:
e x cos 2x dx. Using u D cos 2x and v 0 D e x , we get
.!2 sin 2x/e x dx D e x cos 2x C 2
Z
e x sin 2x dx:
Plugging this into the original equation, we get
%
&
Z
Z
Z
x
x
x
x
x
x
e sin 2x dx D e sin 2x ! 2 e cos 2x C 2 e sin 2x dx D e sin 2x ! 2e cos 2x ! 4 e x sin 2x dx:
Solving this equation for
17.
Z
Z
e x sin 2x dx gives us
Z
e x sin 2x dx D
1 x
e .sin 2x ! 2 cos 2x/ C C:
5
e !5x sin x dx
SOLUTION
Let u D sin x and v 0 D e !5x . Then we have
u D sin x
1
v D ! e !5x
5
u0 D cos x
v 0 D e !5x
Using Integration by Parts, we get
!
"
Z
Z
Z
1
1
1
1
e !5x sin x dx D ! e !5x sin x ! cos x ! e !5x dx D ! e !5x sin x C
e !5x cos x dx
5
5
5
5
Apply Integration by Parts again to this integral, with u D cos x and v 0 D e !5x to get
Z
Z
1
1
e !5x cos x dx D ! e !5x cos x !
e !5x sin x dx
5
5
Plugging this into the original equation, we get
"
!
Z
Z
1
1
1
1
e !5x sin x dx D ! e !5x sin x C
! e !5x cos x !
e !5x sin x dx
5
5
5
5
Z
1
1
1
D ! e !5x sin x ! e !5x cos x !
e !5x sin x dx
5
25
25
785
786
TECHNIQUES OF INTEGRATION
CHAPTER 7
Solving this equation for
18.
Z
Z
Z
e !5x sin x dx gives us
e !5x sin x dx D !
5 !5x
1
1
e
sin x ! e !5x cos x C C D ! e !5x .5 sin x C cos x/ C C
26
26
26
e 3x cos 4x dx
SOLUTION
Let u D cos 4x and v 0 D e 3x . Then we have
1 3x
e
3
u D cos 4x
vD
u0 D !4 sin 4x
v 0 D e 3x
Using Integration by Parts, we get
Z
Z
Z
1
1 3x
1
4
e 3x cos 4x dx D e 3x cos 4x !
e .!4 sin 4x/ dx D e 3x cos 4x C
e 3x sin 4x dx
3
3
3
3
Apply Integration by Parts again to this integral, with u D sin 4x and v 0 D e 3x , to get
Z
Z
Z
1
1 3x
1
4
e 3x sin 4x dx D e 3x sin 4x !
e " 4 cos 4x dx D e 3x sin 4x !
e 3x cos 4x dx
3
3
3
3
Plugging this into the original equation, we get
!
"
Z
Z
1
4 1 3x
4
e 3x cos 4x dx D e 3x cos 4x C
e sin 4x !
e 3x cos 4x dx
3
3 3
3
Z
1 3x
4 3x
16
D e cos 4x C e sin 4x !
e 3x cos 4x dx
3
9
9
Z
Solving this equation for e 3x cos 4x dx gives us
19.
Z
Z
e 3x cos 4x dx D
3 3x
4
1 3x
e cos 4x C e 3x sin 4x D
e .3 cos 4x C 4 sin 4x/ C C
25
25
25
x ln x dx
SOLUTION
Let u D ln x and v 0 D x. Then we have
u D ln x
v D 12 x 2
u0 D
v0 D x
1
x
Using Integration by Parts, we get
"
Z
Z ! "!
1
1
1 2
x ln x dx D x 2 ln x !
x
dx
2
x
2
Z
1
1
1
1
D x 2 ln x !
x dx D x 2 ln x !
2
2
2
2
Z
ln x
20.
dx
x2
SOLUTION
x2
2
!
CC D
1 2
x .2 ln x ! 1/ C C:
4
Let u D ln x and v 0 D x !2 . Then we have
u D ln x
u0 D
Using Integration by Parts, we get
Z
1
ln x
dx D ! ln x !
x
x2
D!
Z
1
x
1
x
!
v D !x !1
v 0 D x !2
!1
x
"
dx D !
1
ln x C
x
1
1
1
ln x ! C C D ! .ln x C 1/ C C:
x
x
x
Z
x !2 dx
S E C T I O N 7.1
21.
Z
Integration by Parts
x 2 ln x dx
SOLUTION
Let u D ln x and v 0 D x 2 . Then we have
u D ln x
v D 13 x 3
u0 D
v0 D x 2
1
x
Using Integration by Parts, we get
!
"
Z
Z
Z
1
1 1 3
1
1
x 2 ln x dx D x 3 ln x !
x
dx D x 3 ln x !
x 2 dx
3
x 3
3
3
!
!
"
1 3
1 x3
x3
1
D x ln x !
CC D
ln x !
C C:
3
3 3
3
3
Z
22.
x !5 ln x dx
SOLUTION
Let u D ln x and v 0 D x !5 . Then we have
u D ln x
u0 D
1
x
1
v D ! x !4
4
v D x !5
Using Integration by Parts, we get
Z
Z
Z
1
1 !4 1
1
1
x !5 ln x dx D ! x !4 ln x C
x
dx D ! x !4 ln x C
x !5 dx
4
4
x
4
4
!
"
1
1
1
1
D ! x !4 ln x ! x !4 C C D ! 4 ln x C
CC
4
16
4
4x
Z
23.
.ln x/2 dx
SOLUTION
Let u D .ln x/2 and v 0 D 1. Then we have
u D .ln x/2
vDx
u0 D
v0 D 1
2
ln x
x
Using Integration by Parts, we get
"
Z
Z !
Z
2
.ln x/2 dx D .ln x/2 .x/ !
ln x x dx D x.ln x/2 ! 2 ln x dx:
x
Z
We must apply Integration by Parts again to evaluate ln x dx. Using u D ln x and v 0 D 1, we have
Z
ln x dx D x ln x !
Z
1
" x dx D x ln x !
x
Z
dx D x ln x ! x C C:
Plugging this into the original equation, we get
Z
h
i
.ln x/2 dx D x.ln x/2 ! 2 .x ln x ! x/ C C D x .ln x/2 ! 2 ln x C 2 C C:
Z
24.
x.ln x/2 dx
SOLUTION
Let u D .ln x/2 , v 0 D x. Then we have
u D .ln x/2
u0 D
2 ln x
x
vD
1 2
x
2
v0 D x
Using Integration by Parts, we get
Z
Z
Z
1
ln x
1
x.ln x/2 dx D x 2 .ln x/2 ! x 2
dx D x 2 .ln x/2 ! x ln x dx
2
x
2
787
788
TECHNIQUES OF INTEGRATION
CHAPTER 7
Apply Integration by Parts again to this integral, with u D ln x, v 0 D x, to get
Z
Z
1
1
1
1
1
x ln x dx D x 2 ln x !
x 2 dx D x 2 ln x ! x 2
2
2
x
2
4
Plug this back into the first formula to get
!
"
!
"
Z
1
1 2
1
1
1
x.ln x/2 dx D x 2 .ln x/2 !
x ln x ! x 2 C C D x 2 .ln x/2 ! ln x C
CC
2
2
4
2
2
Z
25.
x sec2 x dx
SOLUTION
Let u D x and v 0 D sec2 x. Then we have
uDx
u0 D 1
v D tan x
v 0 D sec2 x
Using Integration by Parts, we get
Z
Z
x sec2 x dx D x tan x ! .1/ tan x dx D x tan x ! ln j sec xj C C:
26.
Z
x tan x sec x dx
SOLUTION
Let u D x and v 0 D tan x sec x. Then we have
uDx
0
u D1
v D sec x
v 0 D tan x sec x
Using Integration by Parts, we get
Z
Z
x tan x sec x dx D x sec x ! sec x dx D x sec x ! ln jsec x C tan xj C C
27.
Z
cos!1 x dx
SOLUTION
Let u D cos!1 x and v 0 D 1. Then we have
u D cos!1 x
!1
u0 D p
1 ! x2
vDx
v0 D 1
Using Integration by Parts, we get
We can evaluate
28.
Z
Z
sin!1 x dx
SOLUTION
Z
p
cos!1 x dx D x cos!1 x !
Z
p
!x
1 ! x2
dx:
!x
dx by making the substitution w D 1 ! x 2 . Then dw D !2x dx, and we have
1 ! x2
Z
Z
Z
1
!2x dx
1
cos!1 x dx D x cos!1 x !
p
D x cos!1 x !
w !1=2 dw
2
2
1 ! x2
p
1
D x cos!1 x ! .2w 1=2 / C C D x cos!1 x ! 1 ! x 2 C C:
2
Let u D sin!1 x and v 0 D 1. Then we have
u D sin!1 x
1
u0 D p
1 ! x2
vDx
v0 D 1
Using Integration by Parts, we get
Z
sin!1 x dx D x sin!1 x !
Z
p
x
1 ! x2
dx:
Integration by Parts
S E C T I O N 7.1
We can evaluate
29.
Z
Z
p
x
1 ! x2
Z
dx by making the substitution w D 1 ! x 2 . Then dw D !2x dx, and we have
Z
!2x dx
1
p
D x sin!1 x C
w !1=2 dw
2
1 ! x2
p
1
D x sin!1 x C .2w 1=2 / C C D x sin!1 x C 1 ! x 2 C C:
2
1
2
sin!1 x dx D x sin!1 x C
sec!1 x dx
SOLUTION
Z
We are forced to choose u D sec!1 x, v 0 D 1, so that u0 D
Z
sec!1 x dx D x sec!1 x !
Z
x
p1
x 2 !1
and v D x. Using Integration by parts, we get:
x dx
p
D x sec!1 x !
x x2 ! 1
Z
p
p
Via the substitution x 2 ! 1 D tan ! (so that x D sec ! and dx D sec ! tan !d!), we get:
Z
Z
Z
sec ! tan !d!
sec!1 x dx D x sec!1 x !
D x sec!1 x ! sec !d!
tan !
30.
Z
x5x dx
Z
x2 ! 1
p
:
x 2 ! 1j C C:
Let u D x and v 0 D 5x . Then we have
Using Integration by Parts, we get
Z
31.
dx
D x sec!1 x ! ln j sec ! C tan !j C C D x sec!1 x ! ln jx C
SOLUTION
!
5x
ln 5
uDx
vD
u0 D 1
v 0 D 5x
"
Z
5x
x 5x
1
dx D
!
5x dx
ln 5
ln 5
ln 5
! x "
!
"
x 5x
1
5
5x
1
D
!
CC D
x!
C C:
ln 5
ln 5 ln 5
ln 5
ln 5
x 5x dx D x
5x
ln 5
!
Z
.1/
3x cos x dx
SOLUTION
Let u D cos x and v 0 D 3x . Then we have
3x
ln 3
u D cos x
vD
u0 D ! sin x
v 0 D 3x
Using Integration by Parts, we get
Z
3x
1
3 cos x dx D
cos x C
ln 3
ln 3
x
Z
3x sin x dx
Apply Integration by Parts to the remaining integral, with u D sin x and v 0 D 3x ; then
Z
Z
3x
1
x
3 sin x dx D
sin x !
3x cos x dx
ln 3
ln 3
Plug this into the first equation to get
! x
"
Z
Z
3x
1
3
1
3x cos x dx D
cos x C
sin x !
3x cos x dx
ln 3
ln 3 ln 3
ln 3
Z
x
x
1
3
3
sin
x
!
3x cos x dx
D
cos x C
ln 3
.ln 3/2
.ln 3/2
Z
Solving for 3x cos x dx gives
Z
789
3x cos x dx D
3x sin x
3x
3x ln 3 cos x
C
CC D
.ln 3 cos x C sin x/ C C
2
2
1 C .ln 3/
1 C .ln 3/
1 C .ln 3/2
790
32.
CHAPTER 7
Z
TECHNIQUES OF INTEGRATION
x sinh x dx
SOLUTION
Let u D x, v 0 D sinh x. Then
uDx
u0 D 1
v D cosh x
v 0 D sinh x
Integration by Parts gives us
33.
Z
Z
x sinh x dx D x cosh x !
Z
cosh x dx D x cosh x ! sinh x C C
x 2 cosh x dx
SOLUTION
Let u D x 2 , v 0 D cosh x. Then
u D x2
v D sinh x
u0 D 2x
v 0 D cosh x
Integration by Parts gives us (along with Exercise 32)
Z
Z
2
2
x cosh x dx D x sinh x ! 2 x sinh x; dx D x 2 sinh x ! 2x cosh x C 2 sinh x C C
Z
34.
cos x cosh x dx
SOLUTION
Let u D cos x and v 0 D cosh x. Then
u D cos x
0
v D sinh x
u D ! sin x
v 0 D cosh x
Integration by Parts gives us
Z
Z
Z
cos x cosh x dx D cos x sinh x ! .! sin x/ sinh x dx D cos x sinh x C sin x sinh x dx:
We must apply Integration by Parts again to evaluate
Z
Z
sin x sinh x dx. Using u D sin x and v 0 D sinh x, we find
sin x sinh x dx D sin x cosh x !
Z
cos x cosh x dx:
Plugging this into the original equation, we have
Z
Z
cos x cosh x dx D cos x sinh x C sin x cosh x ! cos x cosh x dx:
Solving this equation for
35.
Z
Z
cos x cosh x dx yields
Z
cos x cosh x dx D
1
.cos x sinh x C sin x cosh x/ C C:
2
tanh!1 4x dx
SOLUTION
Using u D tanh!1 4x and v 0 D 1 gives us
u D tanh!1 4x
4
u0 D
1 ! 16x 2
Integration by Parts gives us
Z
vDx
v0 D 1
tanh!1 4x dx D x tanh!1 4x !
Z !
4
1 ! 16x 2
"
x dx:
For the integral on the right we’ll use the substitution w D 1 ! 16x 2 , dw D !32x dx. Then we have
Z
Z
1
1
dw
tanh!1 4x dx D x tanh!1 4x C
D x tanh!1 4x C ln jwj C C
8
w
8
D x tanh!1 4x C
1
ln j1 ! 16x 2 j C C:
8
S E C T I O N 7.1
36.
Z
Integration by Parts
sinh!1 x dx
Using u D sinh!1 x and v 0 D 1 gives us
SOLUTION
u D sinh!1 x
1
u0 D p
1 C x2
Integration by Parts gives us
Z
vDx
v0 D 1
sinh!1 x dx D x sinh!1 x !
Z !
1
p
1 C x2
"
x dx:
For the integral on the right we’ll use the substitution w D 1 C x 2 , dw D 2x dx. Then we have
Z
Z
p
1
dw
sinh!1 x dx D x sinh!1 x !
p D x sinh!1 x ! w C C
2
w
p
!1
D x sinh x ! 1 C x 2 C C:
In Exercises 37 and 38, evaluate using substitution and then Integration by Parts.
Z p
37.
e x dx Hint: Let u D x 1=2
Let w D x 1=2 . Then dw D 12 x !1=2 dx, or dx D 2 x 1=2 dw D 2w dw. Now,
Z p
Z
e x dx D 2 we w dw:
SOLUTION
Using Integration by Parts with u D w and v 0 D e w , we get
Z
2 we w dw D 2.we w ! e w / C C:
Substituting back, we find
38.
Z
Z
2
e
p
x
dx D 2e
p
x
p
. x ! 1/ C C:
x 3 e x dx
SOLUTION
Let w D x 2 . Then dw D 2x dx, and
Z
2
x 3 e x dx D
1
2
Z
we w dw:
Using Integration by Parts, we let u D w and v 0 D e w . Then we have
Z
Z
we w dw D we w ! .1/e w dw D we w ! e w C C:
Substituting back in terms of x, we get
Z
2
x 3 e x dx D
(
1 ' 2 x2
2
x e ! e x C C:
2
In Exercises 39–48, evaluate using Integration by Parts, substitution, or both if necessary.
Z
39.
x cos 4x dx
SOLUTION
Let u D x and v 0 D cos 4x. Then we have
uDx
u0 D 1
vD
1
4
sin 4x
v 0 D cos 4x
Using Integration by Parts, we get
"
!
Z
Z
1
1
1
1
1
x cos 4x dx D x sin 4x ! .1/ sin 4x dx D x sin 4x !
! cos 4x C C
4
4
4
4
4
D
1
1
x sin 4x C
cos 4x C C:
4
16
791
792
40.
TECHNIQUES OF INTEGRATION
CHAPTER 7
Z
ln.ln x/ dx
x
SOLUTION
Let w D ln x. Then dw D dx=x, and we have
Z
Z
ln.ln x/ dx
D ln w dw
x
Now we can use Integration by Parts, letting u D ln w and v 0 D 1. Then u0 D 1=w, v D w, and
Z
Z
1
ln w dw D w ln w !
.w/ dw D w ln w ! w C C:
w
Substituting back in terms of x, we get
41.
Z
x dx
p
xC1
SOLUTION
42.
Z
Z
ln.ln x/ dx
D .ln x/ ln.ln x/ ! ln x C C:
x
Let u D x C 1. Then du D dx, x D u ! 1, and
"
Z
Z
Z !
Z
x dx
.u ! 1/ du
u
1
p
D
p
D
p !p
du D .u1=2 ! u!1=2 / du
u
u
u
xC1
D
2 3=2
2
u
! 2u1=2 C C D .x C 1/3=2 ! 2.x C 1/1=2 C C:
3
3
x 2 .x 3 C 9/15 dx
SOLUTION
43.
Z
Note that .x 3 C 0/0 D 3x 2 , so use substitution with u D x 3 C 9, du D 3x 2 dx. Then
Z
Z
1
1 16
1 3
x 2 .x 3 C 9/15 dx D
u15 du D
u CC D
.x C 9/16 C C
3
48
48
cos x ln.sin x/ dx
SOLUTION
Let w D sin x. Then dw D cos x dx, and
Z
Z
cos x ln.sin x/ dx D ln w dw:
Now use Integration by Parts with u D ln w and v 0 D 1. Then u0 D 1=w and v D w, which gives us
Z
Z
cos x ln.sin x/ dx D ln w dw D w ln w ! w C C D sin x ln.sin x/ ! sin x C C:
Z
p
44.
sin x dx
SOLUTION
p
p
x and dw D dx=.2 x/. This gives us
p
p
Z
Z
Z
p
.2 x/ sin x dx
sin x dx D
p
D 2 w sin w dw:
.2 x/
First use substitution, with w D
Now use Integration by Parts, with u D w and v 0 D sin w. Then we have
!
"
Z
Z
Z
p
sin x dx D 2 w sin w dw D 2 !w cos w ! ! cos w dw
45.
Z
p
xe
SOLUTION
p
D 2.!w cos w C sin w/ C C D 2 sin
x
p
p
p
x ! 2 x cos x C C:
dx
Let w D
p
x. Then dw D
1
p
2 x
dx and
Z
p
xe
p
x
dx D 2
Z
w 2 e w dw:
Now, use Integration by Parts with u D w 2 and v 0 D e w . This gives
Z
Z
Z
p px
xe
dx D 2 w 2 e w dw D 2w 2 e w ! 4 we w dw:
S E C T I O N 7.1
Integration by Parts
We need to use Integration by Parts again, this time with u D w and v 0 D e w . We find
Z
Z
we w dw D we w ! e w dw D we w ! e w C C I
finally,
46.
Z
tan
p
x dx
p
x
SOLUTION
47.
Z
Z
Let u D
p
p
p
x
dx D 2w 2 e w ! 4we w C 4e w C C D 2xe
x and du D
Z
ln.ln x/ ln x dx
x
SOLUTION
xe
1 !1=2
.
2x
p
x
p
p p
! 4 xe x C 4e x C C:
Then
p
Z
p
tan x dx
p
D 2 tan u du D !2 ln j cos uj C C D !2 ln j cos xj C C
x
Let w D ln x. Then dw D dx=x, and
Z
Z
ln.ln x/ ln x dx
D w ln w dw:
x
Now use Integration by Parts, with u D ln w and v 0 D w. Then,
1 2
w
2
u D ln w
vD
u0 D w !1
v0 D w
and
Z
48.
Z
ln.ln x/ ln x dx
1
1
D w 2 ln w !
x
2
2
D
Z
1
1
w dw D w 2 ln w !
2
2
w2
2
!
CC
1
1
1
.ln x/2 ln.ln x/ ! .ln x/2 C C D .ln x/2 Œ2 ln.ln x/ ! 1" C C:
2
4
4
sin.ln x/ dx
SOLUTION
Let u D sin.ln x/ and v 0 D 1. Then we have
u D sin.ln x/
u0 D
cos.ln x/
x
vDx
v0 D 1
Using Integration by Parts, we get
Z
Z
Z
cos.ln x/
sin.ln x/ dx D x sin.ln x/ ! .x/
dx D x sin.ln x/ ! cos.ln x/ dx:
x
R
We must use Integration by Parts again to evaluate cos.ln x/ dx. Let u D cos.ln x/ and v 0 D 1. Then
%
&
Z
Z
sin.ln x/ dx D x sin.ln x/ ! x cos.ln x/ ! .! sin.ln x// dx
D x sin.ln x/ ! x cos.ln x/ !
Solving this equation for
R
Z
sin.ln x/ dx:
sin.ln x/ dx, we get
Z
x
sin.ln x/ dx D Œsin.ln x/ ! cos.ln x/" C C:
2
793
794
TECHNIQUES OF INTEGRATION
CHAPTER 7
In Exercises 49–54, compute the definite integral.
Z 3
49.
xe 4x dx
0
Let u D x, v 0 D e 4x . Then u0 D 1 and v D
SOLUTION
Z
3
0
50.
Z
xe 4x dx D
!=4
x sin 2x dx
!
1 4x
e . Using Integration by Parts,
4
"ˇ
Z
1 4x ˇˇ3 1 3 4x
3
1
1
11 12
1
xe
!
e dx D e 12 ! e 12 C
D
e C
ˇ
4
4 0
4
16
16
16
16
0
0
Let u D x and v 0 D sin 2x. Then u0 D 1 and v D ! 12 cos 2x. Using Integration by Parts,
SOLUTION
Z
51.
Z
!=4
0
2
ˇ!=4 Z !=4 !
"
!
! "
"ˇ
ˇ
1
1
1
1 sin 2x ˇˇ!=4
x sin.2x/ dx D ! x cos 2x ˇˇ
!
! cos 2x dx D ! x cos 2x C
ˇ
2
2
2
2
2
0
0
0
!
"
'
(
'
(
'
(
1 #
#
1
#
1
D !
cos
C sin
! .0 C 0/ D :
2 4
2
4
2
4
x ln x dx
1
Let u D ln x and v 0 D x. Then u0 D
SOLUTION
52.
Z
e
1
Z
Z
!
1
x ln x dx D
ln x dx
x2
!
and v D 12 x 2 . Using Integration by Parts gives
ˇ
" ˇ2
Z
ˇ
1 2
1 2
1 ˇ2
3
x ln x ˇˇ !
x dx D 2 ln 2 ! x 2 ˇˇ D 2 ln 2 !
2
2 1
4
4
1
1
Let u D ln x and v 0 D x !2 . Then u0 D x !1 and v D !x !1 . Using Integration by Parts gives
ˇ
ˇe
Z e
Z e
ˇ
ln x dx
ln x ˇˇe
2
!2
!1
!1 ˇ
D
!
C
x
dx
D
!e
!
x
ˇ
ˇ D1! e
2
x 1
x
1
1
1
SOLUTION
53.
2
1
x
e x sin x dx
0
Let u D sin x and v 0 D e x ; then u0 D cos x and v D e x . Integration by Parts gives
SOLUTION
Z
!
0
ˇ! Z
ˇ
e x sin x dx D e x sin x ˇˇ !
0
!
0
e x cos x dx D !
Z
!
e x cos x dx
0
Apply integration by parts again to this integral, with u D cos x and v 0 D e x ; then u0 D ! sin x and v D e x , so we get
ˇ
!
"
Z !
Z !
Z !
#
$ ˇ!
e x sin x dx D ! e x cos x ˇˇ C
e x sin x dx D e ! C 1 !
e x sin x dx
0
Solving for
54.
Z
0
Z
!
0
0
e x sin x dx gives
0
Z
!
0
1
0
e x sin x dx D
e! C 1
2
tan!1 x dx
SOLUTION
Let u D tan!1 x and v 0 D 1. Then we have
u D tan!1 x
1
u0 D 2
x C1
vDx
v0 D 1
Integration by Parts gives us
Z
tan!1 x dx D x tan!1 x !
Z !
"
1
x dx:
x2 C 1
Integration by Parts
S E C T I O N 7.1
For the integral on the right we’ll use the substitution w D x 2 C 1, dw D 2x dx. Then we have
Z
Z
1
dw
1
1
tan!1 x dx D x tan!1 x !
D x tan!1 x ! ln jwj C C D x tan!1 x ! ln jx 2 C 1j C C:
2
w
2
2
Now we can compute the definite integral:
Z
1
0
!
tan!1 x dx D
55. Use Eq. (5) to evaluate
Z
x tan!1 x !
x 4 e x dx.
"ˇ1 !
"
ˇ
1
1
#
1
ln jx 2 C 1j ˇˇ D .1/ tan!1 .1/ ! ln 2 ! .0/ D ! ln 2:
2
2
4
2
0
SOLUTION
Z
x 4 e x dx D x 4 e x ! 4
%
&
Z
x 3 e x dx D x 4 e x ! 4 x 3 e x ! 3 x 2 e x dx
Z
D x 4 e x ! 4x 3 e x C 12
Z
%
&
Z
x 2 e x dx D x 4 e x ! 4x 3 e x C 12 x 2 e x ! 2 xe x dx
D x 4 e x ! 4x 3 e x C 12x 2 e x ! 24
Z
%
&
Z
xe x dx D x 4 e x ! 4x 3 e x C 12x 2 e x ! 24 xe x ! e x dx
)
*
D x 4 e x ! 4x 3 e x C 12x 2 e x ! 24 xe x ! e x C C:
Thus,
Z
x 4 e x dx D e x .x 4 ! 4x 3 C 12x 2 ! 24x C 24/ C C:
Z
56. Use substitution and then Eq. (5) to evaluate x 4 e 7x dx.
SOLUTION
Let u D 7x. Then du D 7dx, and
Z
Z
Z
1
1
x 4 e 7x dx D 5 .7x/4 e 7x .7dx/ D 5
u4 e u du:
7
7
Now use the result from Exercise 55:
Z
1
x 4 e 7x dx D 5 e u Œu4 ! 4u3 C 12u2 ! 24u C 24" C C
7
1 7x
e Œ.7x/4 ! 4.7x/3 C 12.7x/2 ! 24.7x/ C 24" C C
75
1
D 5 e 7x Œ2401x 4 ! 1372x 3 C 588x 2 ! 168x C 24" C C:
7
D
57. Find a reduction formula for
SOLUTION
Z
x n e !x dx similar to Eq. (5).
Let u D x n and v 0 D e !x . Then
u D xn
u0 D nx n!1
v D !e !x
v 0 D e !x
Using Integration by Parts, we get
Z
Z
Z
x n e !x dx D !x n e !x ! nx n!1 .!e !x / dx D !x n e !x C n x n!1 e !x dx:
58. Evaluate
SOLUTION
R
x n ln x dx for n ¤ !1. Which method should be used to evaluate
Let u D ln x and
v0
D
x n.
Then we have
u D ln x
u0 D
1
x
vD
R
x !1 ln x dx?
x nC1
nC1
v0 D x n
and
Z
x n ln x dx D
x nC1
ln x !
nC1
Z
1 x nC1
x nC1
1
"
dx D
ln x !
x nC1
nC1
nC1
Z
x n dx
795
796
CHAPTER 7
For n D !1,
R
TECHNIQUES OF INTEGRATION
!
"
x nC1
1
x nC1
x nC1
1
D
ln x !
"
D
ln x !
C C:
nC1
nC1 nC1
nC1
nC1
x !1 ln x dx, use the substitution u D ln x, du D dx=x. Then
Z
Z
u2
1
x !1 ln x dx D u du D
C C D .ln x/2 C C:
2
2
In Exercises 59–66, indicate a good method for evaluating the integral (but do not evaluate). Your choices are algebraic manipulation, substitution (specify u and du), and Integration by Parts (specify u and v 0 ). If it appears that the techniques you have learned
thus far are not sufficient, state this.
Z
p
59.
x ln x dx
p
SOLUTION Use Integration by Parts, with u D ln x and v 0 D x.
Z 2 p
x ! x
60.
dx
2x
SOLUTION Use algebraic manipulation:
p
x2 ! x
x
1
D ! p :
2x
2 2 x
Z
x 3 dx
61.
p
4 ! x2
SOLUTION Use substitution, followed by algebraic manipulation: Let u D 4 ! x 2 . Then du D !2x dx, x 2 D 4 ! u, and
"
Z
Z
Z
Z !
x3
1
.x 2 /.!2x dx/
1
.4 ! u/.du/
1
4
u
p
dx D !
p
D!
p
D!
p !p
du:
2
2
2
u
u
u
u
4 ! x2
Z
dx
62.
p
4 ! x2
SOLUTION The techniques learned so far are insufficient. This problem requires the technique of trigonometric substitution.
Z
xC2
63.
dx
x 2 C 4x C 3
SOLUTION Use substitution. Let u D x 2 C 4x C 3; then du D 2x C 4 dx D 2.x C 2/ dx, and
Z
Z
xC2
1
1
dx
D
du
2
u
x 2 C 4x C 3
Z
dx
64.
.x C 2/.x 2 C 4x C 3/
SOLUTION
65.
x sin.3x C 4/ dx
Z
x cos.9x 2 / dx
SOLUTION
66.
The techniques learned so far are insufficient. This problem requires the technique of trigonometric substitution.
Z
Use Integration by Parts, with u D x and v 0 D sin.3x C 4/.
Use substitution, with u D 9x 2 and du D 18x dx.
Z
67. Evaluate .sin!1 x/2 dx. Hint: Use Integration by Parts first and then substitution.
SOLUTION
SOLUTION
First use integration by parts with v 0 D 1 to get
Z
Z
x sin!1 x dx
p
:
1 ! x2
p
Now use substitution on the integral on the right, with u D sin!1 x. Then du D dx= 1 ! x 2 and x D sin u, and we get (using
Integration by Parts again)
Z
Z
p
x sin!1 x dx
p
D u sin u du D !u cos u C sin u C C D ! 1 ! x 2 sin!1 x C x C C:
1 ! x2
p
p
where cos u D 1 ! sin2 u D 1 ! x 2 . So the final answer is
Z
p
.sin!1 x/2 dx D x.sin!1 x/2 C 2 1 ! x 2 sin!1 x ! 2x C C:
.sin!1 x/2 dx D x.sin!1 x/2 ! 2
Integration by Parts
S E C T I O N 7.1
68. Evaluate
SOLUTION
Z
797
.ln x/2 dx
. Hint: Use substitution first and then Integration by Parts.
x2
Let w D ln x. Then dw D dx=x, e w D x, and
Z
.ln x/2 dx
D
x2
Z
w 2 dw
:
ew
Now use Integration by Parts, with u D w 2 and v 0 D e !w :
Z
Z
w 2 dw
2 !w
D
!w
e
!
2w.!e !w / dw D !w 2 e !w C 2.!we !w ! e !w / C C
ew
D !e !w .w 2 C 2w C 2/ C C D !e ! ln x ..ln x/2 C 2 ln x C 2/ C C:
The final answer is
69. Evaluate
SOLUTION
Z
Z
.ln x/2 dx
!Œ.ln x/2 C 2 ln x C 2"
D
C C:
2
x
x
x 7 cos.x 4 / dx.
First, let w D x 4 . Then dw D 4x 3 dx and
Z
Z
1
x 7 cos.x 4 / dx D
w cos x dw:
4
Now, use Integration by Parts with u D w and v 0 D cos w. Then
!
"
Z
Z
1
1
1
1
1
7
4
x cos.x / dx D
w sin w ! sin w dw D w sin w C cos w C C D x 4 sin.x 4 / C cos.x 4 / C C:
4
4
4
4
4
70. Find f .x/, assuming that
Z
SOLUTION
f .x/e x dx D f .x/e x !
We see that Integration by Parts was applied to
Z
f 0 .x/ D u0 D x !1 . Thus f .x/ D ln x C C for any constant C .
Z
x !1 e x dx
f .x/ e x dx with u D f .x/ and v 0 D e x , and that therefore
71. Find the volume of the solid obtained by revolving the region under y D e x for 0 # x # 2 about the y-axis.
SOLUTION
By the Method of Cylindrical Shells, the volume V of the solid is
V D
Z
b
a
.2# r/h dx D 2#
Z
2
xe x dx:
0
Using Integration by Parts with u D x and v 0 D e x , we find
ˇ2
)
*
ˇ
V D 2# .xe x ! e x /ˇ D 2# .2e 2 ! e 2 / ! .0 ! 1/ D 2#.e 2 C 1/:
0
72. Find the area enclosed by y D ln x and y D .ln x/2 .
SOLUTION
The two graphs intersect at x D 1 and at x D e, and ln x is above .ln x/2 , so the area is
Z e
Z e
Z e
ln x ! .ln x/2 dx D
ln x dx !
.ln x/2 dx
1
1
Using integration by parts for the second integral, let u D
.ln x/2 ,
1
v0
D 1; then u0 D
Z e
Z
'
( ˇˇe
2
2 ˇ
.ln x/ dx D x.ln x/ ˇ ! 2
1
1
1
e
ln x D e ! 2
1
Substituting this back into the original equation gives
Z e
Z
ln x ! .ln x/2 dx D 3
2 ln x
x
Z
and v D x, so that
e
ln x
1
e
1
ln x dx ! e
We use integration by parts to evaluate the remaining integral, with u D ln x and v 0 D 1; then u0 D
ˇe Z e
Z e
ˇ
ln x dx D x ln x ˇˇ !
1 dx D e ! .e ! 1/ D 1
1
1
1
1
x
and v D x, so that
798
TECHNIQUES OF INTEGRATION
CHAPTER 7
and thus, substituting back in, the value of the original integral is
Z e
Z
ln x ! .ln x/2 dx D 3
1
e
1
ln x dx ! e D 3 ! e
73. Recall that the present value (PV) of an investment that pays out income continuously at a rate R.t/ for T years is
Z T
R.t/e !rt dt, where r is the interest rate. Find the PV if R.t/ D 5000 C 100t $/year, r D 0:05 and T D 10 years.
0
The present value is given by
SOLUTION
PV D
Z
T
0
R.t/e !rt dt D
Z
10
0
.5000 C 100t/e !rt dt D 5000
Z
0
10
e !rt dt C 100
Z
10
0
te !rt dt:
Using Integration by Parts for the integral on the right, with u D t and v 0 D e !rt , we find
"!
#
!
"ˇ10
"ˇ10 Z 10
ˇ
ˇ
1
t
!1 !rt
P V D 5000 ! e !rt ˇˇ C 100
! e !rt ˇˇ !
e
dt
r
r
r
0
0
0
ˇ
!
"ˇ10
ˇ
5000 !rt ˇˇ10 100
1
D !
e ˇ !
te !rt C e !rt ˇˇ
r
r
r
0
0
%!
" !
"&
5000 !10r
100
1
1
D!
.e
! 1/ !
10e !10r C e !10r ! 0 C
r
r
r
r
%
&
5000 1000 100
5000
100
D e !10r !
!
! 2 C
C 2
r
r
r
r
r
D
5000r C 100 ! e !10r .6000r C 100/
:
r2
74. Derive the reduction formula
Z
Z
.ln x/k dx D x.ln x/k ! k
.ln x/k!1 dx
6
Use Integration by Parts with u D .ln x/k and v 0 D 1. Then u0 D k.ln x/k!1 =x, v D x, and we get
SOLUTION
Z
.ln x/k dx D x.ln x/k ! k
75. Use Eq. (6) to calculate
Z
Z
.ln x/k!1 x dx
D x.ln x/k ! k
x
Z
.ln x/k!1 dx:
.ln x/k dx for k D 2; 3.
SOLUTION
Z
Z
.ln x/2 dx D x.ln x/2 ! 2
.ln x/3 dx D x.ln x/3 ! 3
Z
Z
ln x dx D x.ln x/2 ! 2.x ln x ! x/ C C D x.ln x/2 ! 2x ln x C 2x C C I
h
i
.ln x/2 dx D x.ln x/3 ! 3 x.ln x/2 ! 2x ln x C 2x C C
D x.ln x/3 ! 3x.ln x/2 C 6x ln x ! 6x C C:
76. Derive the reduction formulas
Z
Z
SOLUTION
For
R
x n cos x dx D x n sin x ! n
Z
x n sin x dx D !x n cos x C n
x n!1 sin x dx
Z
x n!1 cos x dx
x n cos x dx, let u D x n and v 0 D cos x. Then we have
u D xn
0
u D nx
n!1
v D sin x
v 0 D cos x
Using Integration by Parts, we get
Z
x n cos x dx D x n sin x ! n
Z
x n!1 sin x dx:
S E C T I O N 7.1
For
Z
Integration by Parts
799
x n sin x dx, let u D x n and v 0 D sin x. Then we have
u D xn
0
u D nx
n!1
v D ! cos x
v 0 D sin x
Using Integration by Parts, we get
77. Prove that
SOLUTION
Z
xb x dx D b x
!
Z
x n sin x dx D !x n cos x C n
x
1
!
ln b ln2 b
"
Z
x n!1 cos x dx:
C C.
Let u D x and v 0 D b x . Then u0 D 1 and v D b x = ln b. Using Integration by Parts, we get
!
"
Z
Z
xb x
1
xb x
1
bx
x
1
x b x dx D
!
b x dx D
!
"
C C D bx
!
C C:
ln b
ln b
ln b
ln b ln b
ln b .ln b/2
78. Define Pn .x/ by
Z
x n e x dx D Pn .x/ e x C C
Use Eq. (5) to prove that Pn .x/ D x n ! nPn!1 .x/. Use this recursion relation to find Pn .x/ for n D 1; 2; 3; 4. Note that P0 .x/ D 1.
SOLUTION
Use induction on n. Clearly for n D 0, we have
Z
Z
x 0 e x dx D e x dx D e x C C D .1/e x C C
so we may take P0 .x/ D 1 D x 0 ! 0. Now assume that
Z
x n e x dx D Pn .x/e x C C
where Pn .x/ D x n ! nPn!1 .x/. Then using Eq. (5) with n C 1 in place of n gives
Z
Z
x nC1 e x dx D x nC1 e x ! .n C 1/ x n e x dx D x nC1 e x ! .n C 1/.Pn .x/e x C C1 /
D .x nC1 ! .n C 1/Pn .x//e x C C
Thus we may define PnC1 .x/ D x nC1 ! .n C 1/Pn .x/ and we get
Z
x nC1 e x dx D PnC1 .x/e x C C
as required.
Further Insights and Challenges
79. The Integration by Parts formula can be written
Z
Z
u.x/v.x/ dx D u.x/V .x/ ! u0 .x/V .x/ dx
7
where V .x/ satisfies V 0 .x/ D v.x/.
(a) Show directly that the right-hand side of Eq. (7) does not change if V .x/ is replaced by V .x/ C C , where C is a constant.
Z
(b) Use u D tan!1 x and v D x in Eq. (7) to calculate x tan!1 x dx, but carry out the calculation twice: first with V .x/ D 12 x 2
and then with V .x/ D 12 x 2 C 12 . Which choice of V .x/ results in a simpler calculation?
SOLUTION
R
(a) Replacing V .x/ with V .x/ C C in the expression u.x/V .x/ ! V .x/u0 .x/ dx, we get
Z
Z
Z
u.x/.V .x/ C C / ! .V .x/ C C /u0 .x/ dx D u.x/V .x/ C u.x/C ! V .x/u0 .x/ dx ! C u0 .x/ dx
D u.x/V .x/ !
Z
%
&
Z
V .x/u0 .x/ dx C C u.x/ ! u0 .x/ dx
800
CHAPTER 7
TECHNIQUES OF INTEGRATION
D u.x/V .x/ !
D u.x/V .x/ !
(b) If we evaluate
Z
Z
Z
V .x/u0 .x/ dx C C Œu.x/ ! u.x/"
V .x/u0 .x/ dx:
x tan!1 x dx with u D tan!1 x and v 0 D x, and if we don’t add a constant to v, Integration by Parts gives us
Z
x tan!1 x dx D
x2
1
tan!1 x !
2
2
Z
x 2 dx
:
x2 C 1
The integral on the right requires algebraic manipulation in order to evaluate. But if we take V .x/ D 12 x 2 C 12 instead of V .x/ D
1 2
2 x , then
!
"
Z
Z 2
1 2 1
1
x C1
1
1
x tan!1 x dx D
x C
tan!1 x !
dx D .x 2 C 1/ tan!1 x ! x C C
2
2
2
2
2
x2 C 1
D
1 2 !1
.x tan x ! x C tan!1 x/ C C:
2
80. Prove in two ways that
Z
a
0
f .x/ dx D af .a/ !
Z
a
0
xf 0 .x/ dx
8
First use Integration by Parts. Then assume f .x/ is increasing. Use the substitution u D f .x/ to prove that
to the area of the shaded region in Figure 1 and derive Eq. (8) a second time.
y
Z
a
xf 0 .x/ dx is equal
0
y = f(x)
f (a)
f (0)
0
a
x
FIGURE 1
SOLUTION
Let u D f .x/ and v 0 D 1. Then Integration by Parts gives
ˇa Z a
Z a
Z
ˇ
f .x/ dx D xf .x/ˇˇ !
xf 0 .x/ dx D af .a/ !
0
0
0
a
0
xf 0 .x/ dx:
Alternately, let u D f .x/. Then du D f 0 .x/ dx, and if f .x/ is either increasing or decreasing, it has an inverse function, and
x D f !1 .u/. Thus,
Z xDa
Z f .a/
0
xf .x/ dx D
f !1 .u/ du
xD0
f .0/
which is precisely the area of the shaded region in Figure 1 (integrating
along the vertical axis). Since the area of the entire rectangle
R
is af .a/, the difference between the areas of the two regions is 0a f .x/ dx.
81. Assume that f .0/ D f .1/ D 0 and that f 00 exists. Prove
Z 1
Z
f 00 .x/f .x/ dx D !
0
0
1
f 0 .x/2 dx
9
Use this to prove that if f .0/ D f .1/ D 0 and f 00 .x/ D $f .x/ for some constant $, then $ < 0. Can you think of a function
satisfying these conditions for some $?
SOLUTION
Z
Let u D f .x/ and v 0 D f 00 .x/. Using Integration by Parts, we get
ˇ1 Z
ˇ
f 00 .x/f .x/ dx D f .x/f 0 .x/ˇ !
1
0
0
1
0
f 0 .x/2 dx D f .1/f 0 .1/ ! f .0/f 0 .0/ !
Now assume that f 00 .x/ D $f .x/ for some constant $. Then
Z 1
Z 1
Z
f 00 .x/f .x/ dx D $
Œf .x/"2 dx D !
0
Since
Z
1
0
0
1
0
Z
1
0
f 0 .x/2 dx D !
Z
0
1
f 0 .x/2 dx:
f 0 .x/2 dx < 0:
Œf .x/"2 dx > 0, we must have $ < 0. An example of a function satisfying these properties for some $ is f .x/ D sin #x.
S E C T I O N 7.1
82. Set I.a; b/ D
Z
0
1
Integration by Parts
x a .1 ! x/b dx, where a; b are whole numbers.
(a) Use substitution to show that I.a; b/ D I.b; a/.
1
(b) Show that I.a; 0/ D I.0; a/ D
.
aC1
(c) Prove that for a $ 1 and b $ 0,
I.a; b/ D
a
I.a ! 1; b C 1/
bC1
(d) Use (b) and (c) to calculate I.1; 1/ and I.3; 2/.
aŠ bŠ
(e) Show that I.a; b/ D
.
.a C b C 1/Š
SOLUTION
(a) Let u D 1 ! x. Then du D !dx and
Z
I.a; b/ D
uD0
uD1
a b
.1 ! u/ u .!du/ D
(b) I.a; 0/ D I.0; a/ by part (a). Further,
I.a; 0/ D
Z
1
0
Z
x a .1 ! x/0 dx D
1
ub .1 ! u/a du D I.b; a/:
0
Z
1
0
x a dx D
1
:
aC1
(c) Using Integration by Parts with u D x a and v 0 D .1 ! x/b gives
!ˇ1
Z 1
.1 ! x/bC1 ˇˇ
a
a
a
I.a; b/ D !x
x a!1 .1 ! x/bC1 D
I.a ! 1; b C 1/
ˇ C
ˇ
bC1
bC1 0
bC1
0
.
(d)
I.1; 1/ D
I.3; 2/ D
1
1
1 1
1
I.1 ! 1; 1 C 1/ D I.0; 2/ D " D
1C1
2
2 3
6
1
1 1
1 1
1
I.4; 2/ D " I.5; 0/ D
" D
:
2
2 5
10 6
60
(e) We proceed as follows:
I.a; b/ D
::
:
D
D
a
a
a!1
I.a ! 1; b C 1/ D
"
I.a ! 2; b C 2/
bC1
bC1 bC2
a
a!1
1
"
"""
I.0; b C a/
bC1 bC2
bCa
a.a ! 1/ " " " .1/
1
"
.b C 1/.b C 2/ " " " .b C a/ b C a C 1
bŠ aŠ
aŠ bŠ
D
:
bŠ .b C 1/.b C 2/ " " " .b C a/.b C a C 1/
.a C b C 1/Š
Z
Z
83. Let In D x n cos.x 2 / dx and Jn D x n sin.x 2 / dx.
D
(a) Find a reduction formula that expresses In in terms of Jn!2 . Hint: Write x n cos.x 2 / as x n!1 .x cos.x 2 //.
(b)
Use the result of (a) to show that In can be evaluated explicitly if n is odd.
(c) Evaluate I3 .
SOLUTION
(a) Integration by Parts with u D x n!1 and v 0 D x cos.x 2 / dx yields
Z
1
n!1
1
n!1
In D x n!1 sin.x 2 / !
x n!2 sin.x 2 / dx D x n!1 sin.x 2 / !
Jn!2 :
2
2
2
2
(b) If n is odd, the reduction process will eventually lead to either
Z
x cos.x 2 / dx or
both of which can be evaluated using the substitution u D x 2 .
Z
x sin.x 2 / dx;
801
802
CHAPTER 7
TECHNIQUES OF INTEGRATION
(c) Starting with the reduction formula from part (a), we find
Z
1
2
1
1
I3 D x 2 sin.x 2 / !
x sin.x 2 / dx D x 2 sin.x 2 / C cos.x 2 / C C:
2
2
2
2
7.2 Trigonometric Integrals
Preliminary Questions
1. Describe the technique used to evaluate
Z
sin5 x dx.
SOLUTION Because the sine function is raised to an odd power, rewrite sin5 x D sin x sin4 x D sin x.1 ! cos2 x/2 and then
substitute u D cos x.
Z
2. Describe a way of evaluating sin6 x dx.
Repeatedly use the reduction formula for powers of sin x.
Z
3. Are reduction formulas needed to evaluate sin7 x cos2 x dx? Why or why not?
SOLUTION
No, a reduction formula is not needed because the sine function is raised to an odd power.
Z
4. Describe a way of evaluating sin6 x cos2 x dx.
SOLUTION
SOLUTION Because both trigonometric functions are raised to even powers, write cos2 x D 1 ! sin2 x and then apply the
reduction formula for powers of the sine function.
5. Which integral requires more work to evaluate?
Z
sin798 x cos x dx
or
Z
sin4 x cos4 x dx
Explain your answer.
SOLUTION The first integral can be evaluated using the substitution u D sin x, whereas the second integral requires the use of
reduction formulas. The second integral therefore requires more work to evaluate.
Exercises
In Exercises 1–6, use the method for odd powers to evaluate the integral.
1.
Z
cos3 x dx
SOLUTION
Use the identity cos2 x D 1 ! sin2 x to rewrite the integrand:
Z
Z '
(
cos3 x dx D
1 ! sin2 x cos x dx:
Now use the substitution u D sin x, du D cos x dx:
Z
Z '
(
1
1
cos3 x dx D
1 ! u2 du D u ! u3 C C D sin x ! sin3 x C C:
3
3
Z
2.
sin5 x dx
SOLUTION
Use the identity sin2 x D 1 ! cos2 x to rewrite the integrand:
Z
Z '
Z '
(2
(2
sin5 x dx D
sin2 x sin x dx D
1 ! cos2 x sin x dx:
Now use the substitution u D cos x, du D ! sin x dx:
Z
Z '
Z '
(2
(
2
1
sin5 x dx D !
1 ! u2 du D !
1 ! 2u2 C u4 du D !u C u3 ! u5 C C
3
5
D ! cos x C
2
1
cos3 x ! cos5 x C C:
3
5