CHAPTER 68 INTEGRATION BY PARTS
EXERCISE 272 Page 739
1. Determine:
∫ xe
∫ xe
Let u = x then
2x
dx
2x
dx
du
=1
dx
and dv = e 2 x d x
 1 2x 
2x
Hence, =
∫ xe d x ( x)  2 e  − ∫
2. Determine:
4x
∫e
3x
and du = 1 dx
from which,
1 2x 1 2x
1
11

xe − ∫ e dx =xe 2 x −  e 2 x  + c
2
2
2
22

=
e2 x 
1
1 2x 1 2x
xe − e +c =
x− +c
2 
2
2
4
4x
∫e
−3 x
dx=
2x
3x
1 2x
e
2
 1 2x 
 e  (1) d x
2

dx
Let u = 4x then
and dv = e −3 x d x
∫ 4x e
∫e
=
d x = ∫ 4 x e −3 x d x
Hence,
v=
du
=4
dx
and du = 4 dx
from which,
v=
∫e
−3 x
1
d x = − e −3 x
3
 1

 1

d x= (4 x)  − e −3 x  − ∫  − e −3 x  4 d x
 3

 3

4
4
4
4 1

= − x e −3 x + ∫ e −3 x d x =
− x e −3 x +  − e −3 x  + c
3
3
3
3 3

4
1
4
4

= − x e −3 x − e −3 x + c = − e −3 x  x +  + c
3
3
3
9

3. Determine:
∫ x sin x d x
∫ x sin x d x
Let u = x then
du
=1
dx
and du = 1 dx
1086
© 2014, John Bird
and dv = sin x
Hence,
from which,
v=
∫ sin x d x =
− cos x
( x) ( − cos x ) − ∫ ( − cos x ) (1) d x
∫ x sin x d x =
= − x cos x + ∫ cos x d x
= –x cos x + sin x + c
4. Determine:
∫ 5θ cos 2θ d θ
du
=5
dθ
Let u = 5θ then
and du = 5 dθ
and dv = cos 2θ dθ from which,
v=
1
∫ cos 2θ d θ = 2 sin 2θ
1

1

5θ cos 2θ d θ ( 5θ )  sin 2θ  − ∫  sin 2θ  5d θ
∫=
2

2

5. Determine:
=
5
5
5
5 1

θ sin 2θ − ∫ sin 2=
θ dθ
θ sin 2θ −  − cos 2θ  + c
2
2
2
2 2

=
5
1
5
5

θ sin 2θ + cos 2θ + c =  θ sin 2θ + cos 2θ  + c
2
2
2
4

∫ 3t
2
e 2t d t
du
= 6t
dt
Let u = 3t 2 then
and dv = e 2t dt
2 2t
Hence, =
∫ 3t e d t
=
( 3t )  12 e
2
∫ te
2t
from which,
2t
v=
∫e
dt =
2t
1 2t
e
2

 1 2t 
 − ∫  e  (6t ) d t

2 
3 2 2t 
t e − 3 ∫ t e 2t d t 


2
For ∫ t e 2t d t , let u = t then
Hence,
and du = 6t dt
du
=1
dt
and du = dt
and dv = e 2t dt
from which,
1 
d t = (t)  e 2t  –
2 
dt
∫  2 e =

1
2t
(1)

v=
∫e
2t
dt =
1 2t
e
2
1 2t 1 2t
te − e
2
4
1087
© 2014, John Bird
Substituting in (1) gives:
∫ 3t
2
3
1 
1
e 2t d t = t 2 e 2t − 3  t e 2t − e 2t 
2
4 
2
3 2 2t 3 2t 3 2t
t e − te + e
2
2
4
=
=
3 2t  2
1
e t − t +  + c
2 
2
∫
6. Evaluate, correct to 4 significant figures:
Let u = 2x then
du
=2
dx
2
0
2x ex d x
and du = 2 dx
and dv = e x d x
from which,
∫
( )
∫e
v=
x
dx = e x
( )
2 x e x d x = (2 x) e x − ∫ e x 2 d x = 2 x e x − 2 e x + c
Hence,
∫
2
0
2
2 x e x d x=  2 x e x − 2 e x  =
0
( 4 e − 2 e ) − ( 0 − 2 e ) = 2 e + 2 = 16.78
2
7. Evaluate, correct to 4 significant figures:
∫ x sin 2 x d x
Hence,
2
∫
π /4
0
2
x sin 2 x d x
Let u = x then
du
=1
dx
and dv = sin 2x
from which,
∫ x sin 2 x d x =
0
and du = 1 dx
v=
∫ sin 2 x d x =
1
− cos 2 x
2
 1

 1

( x)  − cos 2 x  − ∫  − cos 2 x  (1) d x
 2

 2

1
1
= − x cos 2 x + ∫ cos 2 x d x
2
2
1
1
= − x cos 2 x + sin 2 x + c
2
4
Hence,
∫
π /4
0
π /4
1
 1

x sin 2 x d x =
 − 2 x cos 2 x + 4 sin 2 x 
0
 1  π 
1
π  1
 π  

=  −   cos 2   + sin 2    −  0 + sin 0  
4
4 4
 4  

 2  4 
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© 2014, John Bird
 π

π 1 π
=  − cos + sin  − ( 0 )  = 0.2500
2 4
2
 8

∫
8. Evaluate, correct to 4 significant figures:
Let u = t 2 then
du
= 2t
dt
and dv = cos t dt
from which,
π /2
0
t 2 cos t d t
and du = 2t dt
v=
∫ cos t d t = sin t
cos t d t ( t ) ( sin t ) − ∫ ( sin t ) 2t d t
∫ t=
2
Hence,
2
= t 2 sin t − 2  ∫ t sin t d t 


For ∫ t sin t d t , let u = t then
du
=1
dt
and dv = sin t dt
∫ t sin t d t
Hence,
∫t
2
and du = dt
from which,
= (t)(– cos t) –
Substituting in (1) gives:
(1)
v=
∫ (− cos t ) d t
∫ sin t d t =
− cos t
= –t cos t + sin t
cos t =
d t t 2 sin t − 2 [ −t cos t + sin t ] + c
= t 2 sin t + 2t cos t − 2sin t + c
Thus,
∫
π /2
0
π /2
t 2 cos t d t = t 2 sin t + 2t cos t − 2sin t 
0
 π  2
π 2π
π
π
=   sin +
cos − 2sin  − [ 0 + 0 − 0]
2 2
2
2 
 2 
π2
π 
=   + 0 − 2=
− 2 = 0.4674
4
2
2
9. Evaluate, correct to 4 significant figures:
Let u = 3x 2 then
x
2
and dv = e d x
∫
2
1
x
3x 2 e 2 d x
du
= 6x and du = 6x dx
dx
from which,
x
2
x
2
x
e
2
v = ∫ e d=
x = 2e
1
2
1089
© 2014, John Bird
x
2
Hence, ∫=
3x e d x
2
 2x 
 2x 
3x  2 e  − ∫  2 e  6 x d x




( )
2
x
x


= 6 x 2 e 2 − 12  ∫ x e 2 d x 


For
∫ xe
x
2
d x , let u = x then
x
2
and dv = e d x
∫
Hence,
du
=1
dx
(1)
and du = dx
x
2
v = ∫ e d x = 2e
from which,
x
2
x
x
x
x
 x
x e2 d x = ( x) 2 e2  − ∫ 2 e2 d x =
2x e 2 − 4 e 2


Substituting in (1) gives:
∫
x
x
x
x


3 x 2 e 2 d x = 6 x 2 e 2 − 12  2 x e 2 − 4 e 2  + c


x
x
x
= 6 x 2 e 2 − 24 x e 2 + 48e 2 + c
Thus,
∫
2
1
2
x
x
x


3 x e d x = 6 x 2 e 2 − 24 x e 2 + 48e 2 

1
2
x
2
1
1
 1

= ( 24 e1 − 48e1 + 48e1 ) −  6 e 2 − 24 e 2 + 48e 2 


1


= ( 24 e1 ) −  30 e 2  = 15.78


1090
© 2014, John Bird
EXERCISE 273 Page 741
1. Determine:
Let u = ln x
∫ 2x
2
ln x d x
du 1
=
dx x
then
and
and dv = 2 x 2 d x from which,
Hence,
v=
2
2 x ln x d x ( ln x )  x
∫=
3
2
=
3
1
dx
x
du =
∫ 2x
2
dx=
2 3
x
3

2 31
−∫  x  dx

3 x
2 3
2
2 3
2  x3 
2 3
2
2
x ln x − ∫ x=
dx
x ln x −  =
x ln x − x 3 + c
+c
3
3
3
3 3 
3
9
=
2. Determine:
∫ 2 ln 3x d x
∫ 2 ln 3x d x
Let u = ln 3x, from which,
2 3
1
x  ln x −  + c
3 
3
du
1
dx
= , i.e. du =
x
dx
x
and let dv = 2dx, from which, v = ∫ 2 d x = 2x
Substituting into
∫udv
= uv – ∫ v d u gives:
∫ 2 ln 3x d x = (ln 3x)(2x) – ∫ 2 x
dx
x
= 2x ln 3x – ∫ 2 d x = 2x ln 3x – 2x + c
Hence
∫ 2 ln 3x d x ln x dx = 2x(ln 3x – 1) + c
3. Determine:
∫x
Let u = x 2 then
2
sin 3x d x
du
= 2x and du = 2x dx
dx
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© 2014, John Bird
and dv = sin 3x dx
Hence,
∫x
2
from which,
v=
( x )  − 13 cos 3x  − ∫
sin 3 x d x=
1
− cos 3 x
3
∫ sin 3x d x =
2
 1

 − cos 3 x  2 x d x
 3

1
2
= − x 2 cos 3 x +  ∫ x cos 3 x d x 

3
3
For
∫ x cos 3x d x ,
let u = x then
du
=1
dx
and du = dx
and dv = cos 3x dx from which,
Hence,
1

∫ x cos 3x d x = ( x )  3 sin 3x  − ∫
Substituting in (1) gives:
∫x
2
(1)
v=
1
∫ cos 3x d x = 3 sin 3x
1
1
1

 sin 3x  d x = x sin 3 x + cos 3 x
3
9
3

1
2 1
1

− x 2 cos 3 x +  x sin 3 x + cos 3 x  + c
sin 3 x d x =
3
3 3
9

1
2
2
= − x 2 cos 3 x + x sin 3 x + cos 3 x + c
3
9
27
or
4. Determine:
∫ 2e
5x
du
= 5e5x i.e. du = 5e5x dx
dx
and let dv = cos 2x dx from which, v = ∫ cos 2 x d x =
∫udv
∫ 2e
5x
)
cos 2 x d x
Let u = e5x, from which
Substituting into
(
cos 3 x
2
2 − 9 x 2 + x sin 3 x + c
27
9
1
sin 2x
2
= uv – ∫ v d u gives:


1
 1

cos 2 x d x = 2 ( e5 x )  sin 2 x  − ∫  sin 2 x  ( 5e5 x ) d x 
2
 2



= e5x sin 2x – 5  ∫ e5 x sin 2 x d x 


∫e
5x
(1)
sin 2 x d x is now determined separately using integration by parts again:
Let u = e5x then du = 5e5x dx
1
and let dv = sin 2x dx from which, v = ∫ sin 2 x d x = – cos 2x
2
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© 2014, John Bird
Substituting into the integration by parts formula gives:
∫e
5x
 1

 1

sin 2 x d x = (e5x)  − cos 2 x  – ∫  − cos 2 x  ( 5e5 x ) d x
 2

 2

=–
1 5x
5
e cos 2x + ∫ e5 x cos 2 x d x
2
2
Substituting this result into equation (1) gives:
∫ 2e
5x
5
 1

cos 2 x d x = e5x sin 2x – 5  − e5 x cos 2 x + ∫ e5 x cos 2 x d x 
2
 2

5
25 5 x
= e5x sin 2x + e5x cos 2x –
e cos 2 x d x
2 ∫
2
The integral on the far right of this equation is the same as the integral on the left hand side and thus
they may be combined.
∫ 2e
5x
cos 2 x d x +
5
25 5 x
e cos 2 x d x = e5x sin 2x + e5x cos 2x
∫
2
2
25  5 x
5 5x

5x
 2 +  ∫ e cos 2 x d x = e sin 2x + e cos 2x
2 
2

i.e.
i.e.
5
 29  5 x
5x
  ∫ e cos 2 x d x = e (sin 2x + cos 2x)
2
 2 
Hence
∫e
and
∫ 2e
5x
5x
5
 2 
cos 2 x d x =   e5x (sin 2x + cos 2x)
2
 29 
5
 4 
cos 2 x d x =   e5x (sin 2x + cos 2x)
2
 29 
=
5. Determine:
Let u = 2θ
∫ 2θ sec
then
and dv = sec 2 θ d θ
2
2 5x
e (2 sin 2x + 5 cos 2x) + c
29
θ dθ
du
=2
dθ
and
from which,
du = 2 dθ
v = ∫ sec 2 θ d θ = tan θ
1093
© 2014, John Bird
Hence,
sec θ d θ ( 2θ )( tan θ ) − ∫ ( tan θ ) 2 d θ
∫ 2θ=
2
= 2θ tan θ − 2 ln(sec θ ) + c from Problem 9, page 713
= 2 [θ tan θ − ln(sec θ ) ] + c
6. Evaluate, correct to 4 significant figures:
∫ x ln x d x
Let u = ln x, from which,
∫
2
1
x ln x d x
du
1
=
dx
x
i.e. du =
and let dv = x dx, from which, v = ∫ 2 x d x =
Substituting into
∫udv
1
=
∫
2
1
1 2
x
2
= uv – ∫ v d u gives:
∫ x ln x d x = (ln x)  2 x
Hence
dx
x
2
1 2dx

 – ∫ x 
2  x

1
1
1 2
1
x ln x – ∫ x d x = x 2 ln x – x 2 + c
2
2
4
2
2
1 2
1
( 2 ln 2 − 1) − ( 0.5ln1 − 0.25 ) 
x ln x=
d x  x 2 ln x − x=
4  1 
2
= 0.6363
7. Evaluate, correct to 4 significant figures:
Let u = 2 e3 x
then
and dv = sin 2x dx
Hence,
∫ 2e
3x
du
= 6 e3 x
dx
and
from which,
v=
sin 2 x d=
x
∫
1
0
2 e3 x sin 2 x d x
du = 6 e3 x d x
∫ sin 2 x d x =
( 2 e )  − 12 cos 2 x  − ∫
3x
1
− cos 2 x
2
 1
 3x
 − cos 2 x  6 e d x
 2

= − e3 x cos 2 x + 3  ∫ e3 x cos 2 x d x 


For
∫e
3x
cos 2 x d x , let u = e3 x
then
du
= 3e3 x
dx
1094
and
(1)
du = 3e3 x d x
© 2014, John Bird
and dv = cos 2x dx
from which,
( e )  12 sin 2 x  − ∫
Hence, ∫=
e3 x cos 2 x d x
3x
=
1
∫ cos 2 x d x = 2 sin 2 x
1
 3x
 sin 2 x  3e d x
2

1 3x
3
e sin 2 x −  ∫ e3 x sin 2 x d x 

2
2
∫
Substituting in (1) gives:
v=
3
1

2 e3 x sin 2 x d x =
− e3 x cos 2 x + 3  e3 x sin 2 x − ∫ e3 x sin 2 x d x 
2
2

3
9
= − e3 x cos 2 x + e3 x sin 2 x − ∫ e3 x sin 2 x d x
2
2
Hence,
9  3x
3

− e3 x cos 2 x + e3 x sin 2 x
 2 +  ∫ e sin 2 x d x =
2
2

by combining the far left- and far
right-hand integrals
i.e.
3
 13  3 x
− e3 x cos 2 x + e3 x sin 2 x
  ∫ e sin 2 x d x =
2
2
∫e
and
3x
∫ 2e
Thus,
2  3x
3 3x

sin 2 x d x =
 − e cos 2 x + e sin 2 x 
13 
2

3x
4  3x
3 3x

sin 2 x d x =
 − e cos 2 x + e sin 2 x 
13 
2

1
and
∫
1
0
 4  3x
3 3x

2 e sin 2 x d x =
13  − e cos 2 x + 2 e sin 2 x  
 0
 
3x
4
3
3
  4 

=   − e3 cos 2 + e3 sin 2   −   − cos 0 + sin 0  
2
2
  13 

13 
4
 4

=  ( 8.3585 + 27.3956 )  −  (−1) 
13
 13

= (11.0013) + (0.30769) = 11.31
8. Evaluate, correct to 4 significant figures:
∫
π /2 t
0
e cos 3t d t
From Problem 9, page 740–741 of the textbook,
ax
∫ e cos bx d x =
Hence,
e ax
(b sin bx + a cos bx) + c
a 2 + b2
π /2
 et

=
e
cos
3
t
d
t
 2 2 ( 3sin 3t + 1cos 3t ) 
∫0
1 + 3
0
π /2 t
1095
© 2014, John Bird
 eπ /2 
3π
3π
= 
+ cos
 3sin
2
2
 10 
0

  e
−
   ( 3sin 0 + cos 0 )  
   10

= –1.543
9. Evaluate, correct to 4 significant figures:
Let u = ln x
3
and dv =
x dx
∫
4
1
and
from which,
v=
du =
∫
1
x3 ln x d x
1
dx
x
3
2
2 52
x dx= x
5
2
5

2 51
x2  − ∫  x2  d x
5 
5 x
( ln x ) 
x3 ln x d x
∫=
Hence,
Thus,
du 1
=
dx x
then
4
∫
=
3
2 5
2
x ln x − ∫ x 2 d x
5
5
=
2 5
22 5 
x ln x −  x 2  + c
5
55 
4
4
2 5

=
x ln x d x 
x ln x −
x5 
25
5
1
3
4
4 5
2 5
 2
=
4 ln 4 −
45  −  15 ln1 −
1 
25
25
5
 5

4
4 
2
 
=  ( 32 ) ln 4 − ( 32 )  −  0 − 
25
25 
5
 
= 12.78
10. In determining a Fourier series to represent f(x) = x in the range –π to π, Fourier coefficients are
given by:
an =
1
π
π
∫π
−
x cos nx d x and b n =
1
π
π
∫ π x sin nx d x
−
where n is a positive integer. Show by using integration by parts that a n = 0 and b n = –
For an =
1
π
π∫π
−
x cos nx d x , let u = x then
du
=1
dx
1096
and
2
cos nπ.
n
du = dx
© 2014, John Bird
and dv = cos nx dx
Hence,
1

( x )  sin nx  − ∫
∫ x cos nx d x =
n

π
1
Then an
=
∫π
π
−
from which,
v=
1
∫ cos nx d x = n sin nx
x
1
1

sin nx + 2 cos nx
 sin nx  d x =
n
n
n

π
1 x
1

x=
cos nx d x
sin nx + 2 cos nx 

n
π n
 −π
=
1  π
1
1
  π

 sin nπ + 2 cos nπ  −  − sin(− nπ ) + 2 cos(−nπ )  

π  n
n
n
  n

sin nπ = sin(–nπ) = 0 for all values of n
1 
1
1
 

 0 + 2 cos nπ  −  0 + 2 cos(−nπ )  

π 
n
n
 

Hence, a=
n
1
[cos nπ − cos(−nπ )] = 0
π n2
=
For bn =
π
1
π∫π
−
x sin nx d x ,
since cos nπ = cos(–nπ) for all values of n.
let u = x then
du
=1
dx
and dv = sin nx dx
Hence,
 1

( x )  − cos nx  − ∫
∫ x sin nx d x =
 n

and
from which,
du = dx
v=
∫ sin nx dx =
1
− cos nx
n
x
1
 1

− cos nx + 2 sin nx
 − cos nx  d x =
n
n
 n

π
1 π
1 x
1

Then bn =
− cos nx + 2 sin nx 
x sin nx d x =
∫

n
π −π
π n
 −π
=
1  π
1
1
 π

 − cos nπ + 2 sin nπ  −  cos(−nπ ) + 2 sin(−nπ )  

π  n
n
n
 n

sin nπ = sin(–nπ) = 0 for all values of n
1  π
 π

 − cos nπ + 0  −  cos(−nπ ) + 0  

π  n
 n

Hence, bn =
=
−π
[cos nπ + cos(−nπ )]
πn
cos nπ = cos(–nπ) hence cos nπ + cos(–nπ) ≡ 2 cos nπ
Thus,
bn = −
2
1
( 2 cos nπ ) = − cos nπ
n
n
1097
© 2014, John Bird
11. The equations C =
∫
1
0
e − 0.4θ cos1.2θ d θ and S =
∫
1
0
e − 0.4θ sin1.2θ d θ
are involved in the study of damped oscillations. Determine the values of C and S.
From Problem 9, page 740–741
∫
e ax
e cos bx
=
dx
( b sin bx + a cos bx ) + c
a 2 + b2
ax
−0.4 θ
Thus,
=
∫ e cos1.2θ d θ
e −0.4 θ
(1.2sin1.2θ − 0.4 cos1.2θ )
(−0.4) 2 + (1.2) 2
1
 e − 0.4θ

and C = ∫ =
e
cos1.2θ d θ 
(1.2sin1.2θ − 0.4 cos1.2θ )
0
 1.6
0
1
− 0.4 θ
 e −0.4

1
= 
(1.2sin1.2 − 0.4 cos1.2 ) −  (1.2sin 0 − 0.4 cos 0

 1.6
 1.6
= (0.40785) – (–0.25)
i.e.
C = 0.66
From equation (2) on page 740–741 (obtained in a similar way to that for Problem 7 above),
∫
e ax
( a sin bx − b cos bx ) + c
a 2 + b2
e − 0.4 θ
e − 0.4θ sin1.2θ d θ = (−0.4sin1.2θ − 1.2 cos1.2θ )
1.6
∫
Thus,
e ax sin =
bx d x
1
and
S=
∫
1
0
e
−0.4θ
 e − 0.4 θ

sin1.2θ d θ =
( −0.4sin1.2θ − 1.2 cos1.2θ )

 1.6
0
 e −0.4

1
=
( −0.4sin1.2 − 1.2 cos1.2 ) −  (−0.4sin 0 − 1.2 cos 0

 1.6
 1.6
= (–0.33836) – ( –0.75)
i.e.
S = 0.41
1098
© 2014, John Bird