CHAPTER 64 INTEGRATION USING ALGEBRAIC
SUBSTITUTIONS
EXERCISE 258 Page 712
1. Integrate with respect to x: 2 sin(4x + 9)
Let u = 4x + 9 then
Hence,
du
du
= 4 and dx =
dx
4
∫ 2sin(4 x + 9) d x =2∫ sin u
1
du 2
1
= ( − cos u ) + c =− cos u = − cos(4 x + 9) + c
2
4 4
2
2. Integrate with respect to θ: 3 cos(2θ – 5)
Let u = 2θ – 5 then
Hence,
du
du
= 2 and dθ =
dθ
2
∫ 3cos(2θ − 5) d θ= 3∫ cos u
du 3
3
=
( sin u ) + c = sin(2θ − 5) + c
2
2 2
3. Integrate with respect to t: 4 sec2(3t + 1)
Let u = 3t – 1 then
Hence,
du
du
= 3 and dt =
3
dt
∫ 4sec (3t + 1) d=t 4∫ sec
2
4. Integrate with respect to x:
Let u = 5x – 3 then
Hence,
∫
2
du 4
4
u =
( tan u ) + c = tan(3t + 1) + c
3 3
3
1
(5x – 3)6
2
du
du
= 5 and dx =
5
dx
1 6 du 1
1  u7
6 du
=
u
=
u

2
5 10 ∫
10  7
1

(5 x − 3)7 + c
+c =
70

1043
© 2014, John Bird
5. Integrate with respect to x:
Let u = 2x – 1 then
du
du
= 2 and dx =
dx
2
−3
−3 d u
∫ (2 x − 1) d x =
∫ u 2
Hence,
−3
(2 x − 1)
3 1
3
3
=
− ∫ du =
− ln u + c = − ln(2 x − 1) + c
2 u
2
2
6. Integrate with respect to θ: 3e3θ+5
du
du
= 3 and dθ =
3
dθ
Let u = 3θ + 5 then
∫ 3e θ
3 +5
Hence,
d θ= 3∫ eu
du 3 u
=
e d u= eu + c = e3θ +5 + c
3 3∫
7. Evaluate correct to 4 significant figures:
Let u = 3x + 1 then
∫
Hence,
Thus,
∫ 0 ( 3x + 1)
Let u = 2 x 2 + 1
Thus,
∫x
∫
0
5
dx
du 1 5
1  u6 
1
u5 =
u
=
d
u
+c
(3 x + 1)6 + c
 =
∫
3 3
3 6 
18
1
2
0
du
du
= 3 and dx =
dx
3
5
dx=
1 
1 6 6
6 1
3 x + 1)  =
4 − 1 ] = 227.5
(
 0 18 [
18 
8. Evaluate correct to 4 significant figures:
Hence,
∫ ( 3x + 1)
1
then
du
= 4x
dx
(2 x 2 + 1)=
dx
∫
0
dx=
i.e.
x
( 2 x 2 + 1) d x
du
4x
 3
3
1
du 1
1  u2 
1
1
2 +1 2 =
2=
x u=
u
d
u
c
2
x
=
+
+c
(
)


∫
4x 4
4 3 
6
6
 2 
1
x ( 2 x + 1) d x = 
6 
2
∫
2
( 2 x 2 + 1)
3
+c
2
1

( 2 x + 1)  = [ 27 − 1] = 4.333
0 6
2
3
1044
© 2014, John Bird
∫
9. Evaluate correct to 4 significant figures:
π /3
0
π

2sin  3t +  d t
4

π
∫
π /3
0
π
2 
π  3
2   3π π 
π

− cos  3t +   =
− cos 
+  − cos  (note angles are in
2sin  3t +  d t =
4
3 
4  0
3  3 4
4

radians)
=−
2
[ −0.70711 − 0.70711]
3
= 0.9428
10. Evaluate correct to 4 significant figures:
∫
1
0
3cos ( 4 x − 3) d x
1
1
3

( sin(4 − 3) ) − ( sin(0 − 3) ) 
=
) d x 3  sin(4 x − 3)
∫0 3cos ( 4 x − 3=

4
4
0
1
=
3
3
sin(−3) )
( sin1 −=
( 0.84147 − −0.14112 )
4
4
= 0.7369
11. The mean time to failure, M years, for a set of components is given by: M =
∫ (1 − 0.25t )
4
1.5
0
dt
Determine the mean time to failure.
Let u = 1 – 0.25t then
Hence,
∫
du
du
= −0.25 and dt =
dt
−0.25
(1 − 0.25t )1.5 d t =∫ u1.5
5
du
1  u 2.5 
1
1
2 +c =
=
+
c
=
u
−
u5 + c


−0.25 −0.25  2.5 
(−0.25)(2.5)
0.625
=−
Hence, mean time to failure, M =
1
(1 − 0.25t )5 + c
0.625
4
1
1.5


5
∫0 (1 − 0.25t ) d t =
 − 0.625 (1 − 0.25t ) 
0
4
1045
© 2014, John Bird
=−
1046
1 
0.625 
(
) (
(1 − 1)5 −
)
(1)5  = 1.6 years

© 2014, John Bird
EXERCISE 259 Page 714
1. Integrate with respect to x: 2x (2x2 – 3)5
Let u = 2 x 2 − 3
Hence,
∫
then
du
= 4x
dx
2 x ( 2 x 2 − 3) d=
x
5
∫
5 cos5 t sin t
∫
du
= − sin t
dt
5cos5 t sin t d t =
∫ 5u 5 sin t
du
4x
dx=
du 1
1  u6 
1 6
1
6
5 d=
2 xu 5 =
u
u
c
u +c =
( 2 x 2 − 3) + c
  +=
∫
4x 2
2 6 
12
12
2. Integrate with respect to t:
Let u = cos t then
i.e.
dt =
and
du
− sin t
du
5
 u6 
=
−5∫ u 5 d u =
−5   + c = – cos 6 t + c
− sin t
6
 6 
3. Integrate with respect to x: 3 sec2 3x tan 3x
Let u = tan 3x then
Hence,
∫ 3sec
2
du
= 3sec 2 3 x
dx
3 x tan 3 x d=
x
∫ 3sec
Alternatively, let u = sec 3x then
Hence,
∫
3sec 2 3 x tan=
3x d x
∫
2
du
3sec 2 3 x
i.e.
dx=
3 x (u )
du
=
3sec 2 3 x
∫ u d u=
du
= 3sec 3 x tan 3 x
dx
3u 2 tan 3 x
du
=
3sec 3 x tan 3 x
i.e.
∫
1
u2
+ c = tan 2 3 x + c
2
2
dx=
du
3sec 3 x tan 3 x
u2
=
du
sec 3 x
u2
=
∫ u du
=
4. Integrate with respect to t:
Let u = 3t 2 − 1
then
du
= 6t
dt
2t
∫ udu
1
u2
+ c = sec 2 3 x + c
2
2
( 3t 2 − 1)
and
dt =
du
6t
1047
© 2014, John Bird
∫ 2t ( 3t
2
dt
− 1)=
∫
 3
2
du 1
1 1
1u2 
2 3
2t u =
du
du
u=
u2 =
u +c =
+c
 3 =
∫
∫
9
6t 3
3
3 
9
 2 
5. Integrate with respect to θ:
Let u = ln θ
∫
Hence,
du 1
=
dθ θ
then
ln θ
θ
3
+c
ln θ
θ
and
dθ = θdu
u
u2
1
2
+ c = ( ln θ ) + c
2
2
u
∫ θ (θ d u=) ∫ u d=
θ
d=
( 3t 2 − 1)
6. Integrate with respect to t: 3 tan 2t
sin 2t
∫ 3 tan 2t d t = 3∫ cos 2t d t
Hence, 3∫
Let u = cos 2t then
du
= −2sin 2t
dt
i.e.
dt =
−du
2sin 2t
sin 2t
3 1
3
 sin 2t  − d u 
dt =
3∫ 
− ∫ du =
− ln u + c

=
cos 2t
2 u
2
 u  2sin 2t 
3
3
−1
= − ln=
cos 2t + c
ln ( cos 2t ) + c
2
2
3
ln(sec 2t ) + c
2
=
7. Integrate with respect to t:
Let u = et + 4
2e
t
∫ ( e + 4 ) d=t ∫
t
du
= et
dt
then
2e du
=
u et
t
∫
2 et
(et + 4)
and
du
et
dt =
2
1
d=
u 2∫ 1 d =
u 2∫ u
u
u2
1
−
2
 1
u2
d=
u 2
1

 2


c 4 u +c
 +=


=4
8. Evaluate correct to 4 significant figures:
∫
1
0
( e4 + 4 ) + c
3 x e( 2 x2 −1) d x
1048
© 2014, John Bird
Let u = 2 x 2 − 1
then
∫ 3 x e(
d x=
Hence,
∫
1
0
( 2 x 2 −1)
3x e
2 x 2 −1)
=
dx
du
= 4x
dx
∫ 3x e
u
dx=
i.e.
du 3 u
3 u
3 ( 2 x2 −1)
=
e d u=
e + c=
e
+c
∫
4x 4
4
4
3 ( 2 x2 −1) 1 3 1 −1
=
e
 0 4 [ e − e ] = 1.763
4
∫
9. Evaluate correct to 4 significant figures:
Let u = sin θ
∫ 3sin
4
Hence,
then
∫
0
du
= cos θ
dθ
∫ 3u
θ cos=
θ dθ
π /2
4
π /2
0
3sin 4 θ cos θ d θ
dθ =
and
du
cos θ
du
3
 u5 
4 du
cos θ = 3∫ u=
3   + c = sin 5 θ + c
cos θ
5
 5
π /2
5
3  π 
3
3
5
5
5
3 5 
3sin 4 θ cos θ d θ =
sin
θ
=
sin

 − ( sin 0 )  = (1) − ( 0 )  = = 0.6000
 5

5
5 
2
0
 5
10. Evaluate correct to 4 significant figures:
Let u = 4 x 2 − 1
∫ ( 4x
Hence,
du
4x
then
3x
2
− 1)
du
= 8x
dx
d x =∫
5
∫
3x
dx
(4 x 2 − 1)5
dx=
i.e.
du
8x
3 x  du  3 −5
3  u −4 
3
3
d
u
u
=
=
+ c =−
+c
 

 + c =−
4
∫
5
4
8  −4 
32u
u  8x  8
32 ( 4 x 2 − 1)
1
Thus,
∫
1
0

3 
1
3 1
1 
= 0.09259
dx=
− 
−  −
 =
5
4
32  ( 4 x 2 − 1) 
32  34 (−1) 4 
( 4 x 2 − 1)

0
3x
11. The electrostatic potential on all parts of a conducting circular disc of radius r is given by the
equation:
V = 2πσ ∫
9
0
R
dR
R2 + r 2
Solve the equation by determining the integral.
1049
© 2014, John Bird
Let u = R 2 + r 2
then
R
d R=
R2 + r 2
R
u
du
= 2R
dR
i.e.
dR =
du
2R
1
∫
∫
Hence, V = 2πσ ∫
9
0
1
1 u2
 du  1
−
2 d u=
=
u
+ c=


∫
2 1
 2R  2
2
R
=
d R 2πσ 

R2 + r 2
9
2 
( R 2 + r=
)0
u + c=
( R2 + r 2 )
+c
2πσ 

( 92 + r 2 ) −
r2 

= 2πσ
( 9 2 + r 2 ) − r}
{
12. In the study of a rigid rotor the following integration occurs:
Zr
=
∞
∫ ( 2 J + 1) e
− J ( J +1) h 2
8π 2 I k T
0
dJ
Determine Z r for constant temperature T assuming h, I and k are constants.
Let u = J(J + 1) = J 2 + J
∫ ( 2 J + 1) e
− J ( J +1) h 2
8π 2 I kT
du
= 2J +1
dJ
then
dJ =
∫ (2 J + 1) e
− u h2
8π 2 I k T
dJ =
i.e.
du
2J +1
2
2
uh
uh
−
−
du
1
8π 2 I k T d u =
8π 2 I k T + c
e
e
=
h2
(2 J + 1) ∫
−
8π 2 I k T
8π 2 I k T −8Jπ( J2 +I 1)k Th
+c
e
h2
2
=−
∞
Thus, Z r =
∫ ( 2 J + 1) e
− J ( J +1) h 2
8π 2 I kT
0
∞
8π 2 I k T  −8Jπ( J2 I+1)k Th 
8π 2 I k T −∞ 0
−
=
−
dJ =
e
[e − e ]


h2
h2

0
2
=−
π

13. In electrostatics, E = ∫ 
0
 2 ε

dθ 
( a 2 − x 2 − 2ax cos θ ) 
a 2σ sin θ
greater than a, and x is independent of θ. Show that E =
Let u = a 2 + x 2 − 2ax cos θ
then
du
= 2ax sin θ
dθ
1050
and
8π 2 I k T
8π 2 I k T
0 − 1] =
[
h2
h2
where a, σ and ε are constants, x is
a 2σ
εx
dθ =
du
2ax sin θ
© 2014, John Bird
a 2σ
Hence, E
=
2ε
∫
 1
sin θ  d u  a 2σ
du
a 2σ  1  − 1
a 2σ  1   u 2
=

 =
 =
 u 2 du


2ε  2ax  ∫
2ε  2ax   1
u  2ax sin θ  2ε ∫ 2ax u
 2
=
Hence, E =
aσ
2ε x
( a 2 + x 2 − 2ax cos θ ) + c
π
aσ  2
aσ  2
2 − 2ax cos
2

θ
a
x
+
=
 0 2 ε x  a + x − 2ax cos π −
2ε x 
a 2 + x 2 − 2ax cos 0 
=
aσ 
(a 2 + x 2 + 2ax) − (a 2 + x 2 − 2ax) 
2ε x 
=
aσ 
2 ε x 
=
aσ
aσ
[ ( x + a ) − ( x − a ) ] = [ 2a ]
2ε x
2ε x
( x + a)
2


+c


−
( x − a)
E=
i.e.
2


a 2σ
εx
14. The time taken, t hours, for a vehicle to reach a velocity of 130 km/h with an initial speed of
60 km/h is given by:
t=∫
130
60
dv
where v is the velocity in km/h. Determine t, correct to
650 − 3v
the nearest second.
Let u = 650 – 3v
Hence,
dv
du
= −3
dv
1  du 
dv =
and
1
du
−3
1
1
1
∫ 650 − 3v =∫ u  −3  =− 3 ∫ u d u =− 3 ln u + c =− 3 ln(650 − 3v) + c
130
dv
1
 1

Thus, t =
− ln ( 650 − 390 ) − ln ( 650 − 180 ) 
∫ 60 650 − 3v =
 − 3 ln(650 − 3v)  =
3
60
130
1
=
− ln ( 260 ) − ln ( 470 )  = 0.19735 hours
3
0.19735h = 0.19735 × 60 minutes = 11.841 min
0.841 min = 0.841 × 60 seconds = 50.46 s
1051
© 2014, John Bird
Hence, t = ∫
130
60
dv
= 11 min 50 s, correct to the nearest second
650 − 3v
1052
© 2014, John Bird