CHAPTER 65 INTEGRATION USING TRIGONOMETRIC AND
HYPERBOLIC SUBSTITUTIONS
EXERCISE 260 Page 717
1. Integrate with respect to x: sin2 2x
cos 2 x = 1 − 2sin 2 x
Hence,
∫ sin
2
2 x=
dx
and
cos 4 x = 1 − 2sin 2 2 x
1
∫ 2 (1 − cos 4 x ) d x
=
from which,
sin 2 =
2x
1
(1 − cos 4 x)
2
1
sin 4 x 
x−
+c
2
4 
2. Integrate with respect to t: 3 cos2 t
=
cos 2t 2 cos 2 t − 1
from which,
2t
cos=
1
t d t 3∫ (1 + cos 2t ) d t
∫ 3cos=
2
Hence,
2
=
1
(1 + cos 2t )
2
3  sin 2t 
t +
+c
2
2 
3. Integrate with respect to θ: 5 tan2 3θ
Since 1 + tan2 x = sec2 x, then 1 + tan23θ = sec23θ and tan23θ = sec23θ – 1
1

2 3θ d θ
Hence 3 ∫ 5 tan
=
5∫ ( sec 2 3θ − 1) d θ = 5  tan 3θ − θ  + c
3

4. Integrate with respect to t: 2 cot2 2t
∫ 2 cot
2
 1

2t d t = 2 ∫ ( cosec 2 2t − 1) d t = 2  − cot 2t − t  + c = –(cot 2t + 2t) + c
 2

5. Evaluate, correct to 4 significant figures:
cos 2 x = 1 − 2sin 2 x
and
∫
π /3
0
3sin 2 3 x d x
cos 6 x = 1 − 2sin 2 3 x
1053
from which,
sin 2 =
3x
1
(1 − cos 6 x)
2
© 2014, John Bird
Hence,
∫
π /3
0
π /3
3
3  sin 6 x 
3sin 3 x d x =
x−
∫0 2 (1 − cos 6 x) d x =
2 
6  0
π /3
2

6π
sin


3 π
3
=  −
2  3
6

6. Evaluate, correct to 4 significant figures:


 3 π  π

or 1.571
 − (0 − sin0)  =  =
2
2
3






∫
π /4
0
cos 2 4 x d x
=
cos 2 x 2 cos 2 x − 1 and =
cos8 x 2 cos 2 4 x − 1
Hence,
∫
π /4
0
from which,
cos 2 =
4x
1
(1 + cos8 x )
2
π /4
π /4 1
1  sin 8 x 
cos 2 4 x d x =
x+
∫0 2 (1 + cos8 x ) d x =
2 
8  0


 π 
sin  8 ×  


1 π
sin 0   π
4  

=  +
or 0.3927
−0+
 =
2  4
8
8  8
 






7. Evaluate, correct to 4 significant figures:
∫
1
0
2 tan 2t d=
t
2
∫
1
0
∫
1
0
2 tan 2 2t d t
1
 tan 2t 
2 ( sec 2t − 1) d=
t 2
− t
 2
0
2
 tan 2   tan 0

= 2 
− 1 − 
− 0   = tan 2 – 2 = –4.185
  2

 2
(note that ‘tan 2’ means ‘tan 2 radians’)
8. Evaluate, correct to 4 significant figures:
π /3
∫π
/6
cot 2 θ d θ =
π /3
∫π ( cosec
/6
2
π /3
∫π
/6
cot 2 θ d θ

π

3
θ − 1) d θ =
[ − cot θ − θ ] π 6 =
 − cot
π /3
1054
−
π 
π π 
 −  − cot −  
3 
6 6 
© 2014, John Bird

 


1
π 
1
π 
=  −
− −−
− 
 tan π 3   tan π 6  
3
6
 
 

= (–1.624547) – (–2.2556496) = 0.6311
1055
© 2014, John Bird
EXERCISE 261 Page 718
1. Integrate with respect to θ: sin3 θ
Since cos2 θ + sin2 θ = 1 then
∫ sin
Hence
3
sin2 θ = (1 – cos2 θ)
θ dθ =
∫ sin θ (sin 2 θ ) d θ =
∫ sin θ (1 − cos2 θ ) d θ =
∫ (sin θ − sin θ cos2 θ ) d θ
= – cos θ +
cos3 θ
+c
3
2. Integrate with respect to x: 2 cos3 2x
∫ 2 cos
3
2 x d x =2 ∫ cos 2 x cos 2 2 x d x =2 ∫ cos 2 x (1 − sin 2 2 x ) d x =2 ∫ ( cos 2 x − cos 2 x sin 2 2 x ) d x
 sin 2 x sin 3 2 x 
= 2
−
+c
6 
 2
using the algebraic substitution u = sin 2x
= sin 2 x −
sin 3 2 x
+c
3
3. Integrate with respect to t: 2 sin3 t cos2 t
∫ 2sin
3
t=
cos 2 t d t
∫ 2sin t sin
2
∫ 2sin t (1 − cos t ) cos
t=
cos 2 t d t
2
2
t dt
 cos3 t cos5 t 
2
+
+c
= 2 ∫ ( sin t cos 2 t − sin t cos 4 t ) d t =−
3
5 

using the algebraic substitution u = cos t
2
2
= − cos3 t + cos5 t + c
3
5
4. Integrate with respect to x: sin3 x cos4 x
∫ sin
3
4 xd x
x cos
=
∫ sin x sin
2
∫ sin x (1 − cos x ) cos
4 xd x
x cos
=
2
=
∫ ( sin x cos
=−
4
4
xd x
x − sin x cos 6 x ) d x
cos5 x cos 7 x
+
+c
5
7
1056
© 2014, John Bird
5. Integrate with respect to θ: 2 sin4 2θ
∫
1
2
 1 − cos 4θ 
2
2sin 4 2θ d θ =
2 ∫ ( sin 2 2θ ) d θ =
2∫ 
 d θ = ∫ (1 − 2 cos 4θ + cos 4θ ) d θ
2
2


2
=
1 
 1 + cos8θ  
1 − 2 cos 4θ + 
  dθ

∫
2 
2


=
1
sin 4θ θ sin 8θ 
3θ 1
1
+ +
+c =
θ−
− sin 4θ + sin 8θ + c


2
2
2
16 
4 4
32
6. Integrate with respect to t: sin2 t cos2 t
∫ sin
2
1
1   1 + cos 4t  
 1 − cos 2t   1 + cos 2t 
t cos 2 t d t =
(1 − cos 2 2t ) d t =
1− 
 d t
∫  2   2  d t =
∫
4
4 ∫  
2

1  1 cos 4t 
1  t sin 4t 
= ∫  −
+c
dt =  −
4 2
2 
4 2
8 
=
t 1
− sin 4t + c
8 32
1057
© 2014, John Bird
EXERCISE 262 Page 719
1. Integrate with respect to t:
sin 5t cos 2t
1
∫ sin 5t cos 2t d t = ∫ 2 [sin(5t + 2t) + sin(5t – 2t)] dt, from 6 of Table 65.1,
which follows from section 44.4, page 494,
=
1  − cos 7t cos 3t 
1  cos 7t cos 3t 
1
−
+
(sin 7t + sin 3t ) d t = 
+c = − 
+c
∫
2 7
3 
2 7
3 
2
2. Integrate with respect to x:
2 sin 3x sin x
1
2 ∫ − ( cos 4 x − cos 2 x ) d x
∫ 2sin 3x sin x d x =
2
from 9 of Table 65.1
sin 2 x sin 4 x
 sin 4 x sin 2 x 
= −
−
+c =
−
+c

2 
2
4
 4
3. Integrate with respect to x:
3 cos 6x cos x
1
6 x cos x d x 3∫ ( cos 7 x + cos 5 x ) d x
∫ 3cos=
2
=
3  sin 7 x sin 5 x 
+
+c
2  7
5 
4. Integrate with respect to θ:
1
θ sin 2θ d θ
∫ 2 cos 4=
=
5. Evaluate:
∫
π /2
0
from 8 of Table 65.1
1
cos 4θ sin 2θ
2
1 1
(sin 6θ − sin 2θ ) d θ
2∫ 2
from 7 of Table 65.1
1  cos 6θ cos 2θ 
1  cos 2θ cos 6θ
−
+
+c = 
−


4
6
2 
4 2
6

+c

cos 4 x cos 3 x d x
1058
© 2014, John Bird
∫
π /2
0
∫
cos 4 x cos 3 x d x =
π /2
0
π /2
1
1  sin 7 x

(cos 7 x + cos x) d x = 
+ sin x 
2
2 7
0
from 7 of Table 65.1



π 
sin 7  




1
 2  + sin π  −  sin 0 + sin 0  
= 


2 
7
2  7







=
6. Evaluate:
∫
1
0
∫
1
0
 3
1  1 
 − + 1 − ( )  = 7 or 0.4286

2  7 

2sin 7t cos 3t d t
1
1
 cos10t cos 4t 
2sin 7t cos 3t d t =
2 ∫ (sin10t + sin 4t ) d t =
 − 10 − 4 
2
0
from 6 of Table 65.1
 cos10 cos 4   cos 0 cos 0 
−
−
= −
−−

10
4   10
4 

= (0.24732) – (–0.35)
= 0.5973
7. Evaluate: − 4 ∫
−4 ∫
π /3
0
π /3
0
sin 5θ sin 2θ d θ
π /3
−4 ∫
sin 5θ sin 2θ d θ =
0
−
1
[cos 7θ − cos 3θ ] d θ
2
from 9 of Table 65.1

7π
3π
 sin 3 sin 3
 sin 7θ sin 3θ 
= 2
− =
−
2 
3  0
3
 7
 7

π /3
8. Evaluate:
∫
2
1
∫
2
1




 − (0 − 0)  = 0.2474




3cos8t sin 3t d t
3cos8
=
t sin 3t d t 3∫
2
1
1
[sin11t − sin 5t ] d t
2
from 7 of Table 65.1
1059
© 2014, John Bird
3  cos11t cos 5t 
3  cos 22 cos10   cos11 cos 5  
= −
+
=
+
+
−
−−


2
11
5  1 2 
11
5  
11
5  
2
=
3
(−0.07690877 − 0.0563301) = –0.1999
2
1060
© 2014, John Bird
EXERCISE 263 Page 721
1. Determine:
∫
5
=
dt
(4 − t 2 )
2. Determine:
∫
3
=
dx
(9 − x 2 )
3. Determine:
∫
5
dt
(4 − t 2 )
t
t

5  sin −1  + c = 5sin −1 + c
2
2

5
dt
∫ ( 4=
−t )
2
∫
3
dx
(9 − x 2 )
3
dt
∫ ( 3=
−t )
2
2
x
x

3  sin −1  + c = 3sin −1 + c
3
3

∫ (4 − x ) d x
2
∫ ( 4 − x 2 ) d =x ∫ ( 22 − x 2=)
22
x x
sin −1 +
2
2 2
= 2sin −1
4. Determine:
x x
+
2 2
( 22 − x 2 )
from 11 of Table 65.1
( 4 − x2 ) + c
∫ (16 − 9t ) d t
2
∫ (16 − 9t ) d=t ∫
2
from 10 on page 716
  16 2  
t
9  9 − t   d=

 
∫
  4 2
 
t
t
3
= 3∫    sin −1
+
 2
4 2
 

3

 4  2

9   − t 2  d t
 3 


 4 

2 +c
−
t
 

 3 
 


2
from 11 of Table 65.1
2

8 −1 3t 3t  4 
= sin
+
  − t 2  + c
3
4 2  3 

1061
© 2014, John Bird
 4  2

8 −1 3t t
8 −1 3t t
2
2 +c
= sin
3   − t=
sin
(42 − 9t 2 ) + c
+
+

3
4 2
3
3
4
2




=
5. Evaluate:
∫
4
0
∫
0
(16 − 9t 2 ) + c
1
dx
(16 − x 2 )
4
0
1
=
dx
(16 − x 2 )
6. Evaluate:
1
∫
8 −1 3t t
sin
+
3
4 2
∫
4
0
4
x

1
=
d x sin −=
4  0

( 42 − x 2 )
1
[sin −1 1 − sin −1 0] =
π
2
or 1.571
∫ (9 − 4x ) d x
1
2
0
( 9 − 4 x 2 ) d x=
∫
1
0
1
 9

2
 4  4 − x   d x= 2 ∫ 0

 
  3 2
 
x
x
2
= 2    sin −1
+
 2
3 2
 

2

 3  2

  − x 2  d x
 2 

1

2
 3 

2 
−
x
 

 2 
 

0
from 11 of Table 65.1
 9

2 1

= 2  sin −1 +
1.25  − (0 + 0)  = 2.760
3 2

 8

1062
© 2014, John Bird
EXERCISE 264 Page 722
3
∫ 4+t
1. Determine:
2
dt
3
1
t
3
t
1
=
∫ 4 + t 2 d t 3=
∫ 22 + t 2 d t 3  2 tan −1 2  + c = 2 tan −1 2 + c
5
∫ 16 + 9θ
2. Determine:
5
=
∫ 16 + 9θ 2 d θ
2
dθ
5
dθ
∫=
 16

9 +θ 2 
9

5
1
dθ
2
∫
9 4
2
  +θ
3
 



5 1
θ 
−
1

+c
=
tan
94
4
 
3
3
 
=
3. Evaluate:
∫
1
0
∫
from 12 of Table 65.1
from 12 of Table 65.1
5
3θ
tan −1
+c
12
4
3
dt
0 1+ t2
1
1
1
3
1
1 −1 x 
=
d t 3∫ =
d t 3  tan=
3 [ tan −1 1 − tan −1 0]

0 12 + t 2
1+ t2
1
1

0
=
4. Evaluate:
∫
3
0
from 12 of Table 65.1
3π
or 2.356
4
5
dx
4 + x2
3
3
3
5
1
x
1
=
=
d
x
5
∫ 0 4 + x2
∫ 0 22 + x 2 d x 5  2 tan 2  0
=
from 12 of Table 65.1
5  −1 3

− tan −1 0  = 2.457
tan

2
2

1063
© 2014, John Bird
EXERCISE 265 Page 723
∫
1. Find:
∫
2
dx
( x + 16)
2
2
=
d x 2∫
( x 2 + 16)
2. Find:
∫
1
x
x

=
d x 2  sinh −1  + c = 2sinh −1 + c
4
4

( x 2 + 42 )
3
dx
(9 + 5 x 2 )
3
=
∫ (9 + 5x2 ) d x
3
3
5  + x  

 5
 3  2

2
5 
 +x 
 5 

dx ∫
∫=
 9

2
=
=
3. Find:
∫
∫
2
=
4. Find:
∫
∫
3
x
sinh −1
+c
 3 
5


 5
dx
from 13 of Table 65.1
3
5
sinh −1
x+c
3
5
( x 2 + 9) d x
 ( 3)2
x x
( x + 3 ) d t =  sinh −1 +
3 2
 2
2
from 13 of Table 65.1
9
x x
sinh −1 +
2
3 2

[ x 2 + 32 ]  + c

from 14 of Table 65.1
[ x 2 + 9] + c
(4t 2 + 25) d t
( 4t 2 + 25) d=t
∫
  2 25  
t
 4  t + 4   d=

 
∫
2

5 
4 t 2 +    d t
 2  

1064
© 2014, John Bird
  5 2
 
t
t
2
= 2    sinh −1
+
 2
5 2
 

2




5 
t 2 +     + c
 2   



2
from 14, page 716 of textbook
2

25
2t t
5 
−
1
2
=
sinh
4 t +    + c
+
4
5 2
 2  

5. Evaluate:
∫
3
0
∫
0
25
2t t
sinh −1 +
4
5 2
[ 4t 2 + 52 ] + c
=
25
2t t
sinh −1 +
4
5 2
[ 4t 2 + 25] + c
4
dt
(t + 9)
3
0
2
3
t
1

−1
=
=
d t 4 sinh
4 [sinh −1 1 − sinh −1 0] = 3.525

2 + 32
3

t
)
(
0
3
4
=
d t 4∫
0
(t 2 + 9)
6. Evaluate:
1
∫
=
∫
1
0
(16 + 9θ 2 ) d θ
(16 + 9θ 2 ) d θ=
∫
1
0
  16

2
9  9 + θ   d θ=

 
  4 2
 
θ
θ
3
= 3    sinh −1
+
 2
4 2
 

3

∫
1
0
 4  2

9   + θ 2  d θ
 3 

1

2
 4 

2 
+
θ
 

 3 
 

0
from 14 of Table 65.1
 16
3 1
  16

= 3  sinh −1 +
2.777777  −  sinh 0 + 0  
4 2
  18

 18
= 4.348
1065
© 2014, John Bird
EXERCISE 266 Page 725
1. Find
∫
∫
1
dt
(t − 16)
2
1
dt = ∫
(t − 16)
2
2. Find
∫
=
=
∫
∫
( t 2 − 42 )
d t = cosh −1
3
dx ∫
∫=
 
9 
2
4  x − 4 

 
∫ ( 4θ
2
3
2

3 
2
2 x −   
 2  

3
x
cosh −1
+c
2
3
 
2
dx
from 15 of Table 65.1
3
2x
cosh −1
+c
2
3
∫ (θ 2 − 33 ) d θ
=
=
∫
from 15 of Table 65.1
(θ 2 − 9) d θ
(θ 2 − 9) d θ =
4. Find
t
+c
4
3
dx
(4 x 2 − 9)
3
=
∫ ( 4 x2 − 9) d x
3. Find
1
θ
2
θ
2
(θ 2 − 32 ) −
(θ 2 − 9 ) −
32
θ
cosh −1 + c
2
3
9
θ
cosh −1 + c
2
3
(4θ 2 − 25) d θ
− 25 ) d θ =
∫
  2 25  
4 θ − 4  d θ =

 
∫
2

5 
2
2 θ −    d θ
 2  

1066
© 2014, John Bird


θ
= 2
2


2

5
2




θ 
5  2
cosh −1
+c
θ 2 −    −
2
 5 
 2  

 
 2 
from 16 of Table 65.1
25  25
2θ

= θ  θ 2 −  − cosh −1
+c
4  4
5

5. Evaluate
∫
2
1
∫
2
2
2
1
( x 2 − 1)
2
2
=
d x 2∫
1
( x 2 − 1)
6. Evaluate
3
∫
(t
2
∫
3
2
− 4 ) d=
t
dx
2
x
1

−1
=
=
d x 2 cosh
2 [ cosh −1 2 − cosh −1 1] = 2.634

2
2
1

1
( x −1 )
(t 2 − 4) d t
∫
3
2
t
( t − 2 ) d=t 
2
2
2
3
22
t
( t − 2 ) − cosh −1 
2
22
2
2
3
3
5 − 2 cosh −1  − ( 0 − 2 cosh −1 1)
=
2
2
= 1.429
1067
© 2014, John Bird