LESSON 8 TRIGONOMETRIC INTEGRALS
Recall the half-angle formulas from trigonometry.
sin 2 x 
1  cos 2 x 1
1  cos 2 x 1
 (1  cos 2 x ) and cos 2 x 
 (1  cos 2 x )
2
2
2
2
Also, recall that these formulas come from the double angle formula for cosine.
cos 2 x  cos 2 x  sin 2 x
2
2
2
Replacing cos x by 1  sin x , we obtain that cos 2 x  1  2 sin x 
sin 2 x 
1  cos 2 x
2
2
. Replacing sin x by 1  cos x , we obtain that
2
2
cos 2 x  2 cos 2 x  1  cos x 
1  cos 2 x
.
2
From these half-angle formulas, we obtain the following two integral formulas.
2
 sin x dx 
x sin 2 x

 c
2
4
2
 cos x dx 
x sin 2 x

 c
2
4
These formulas are obtained by the following calculations.
2
 sin x dx 
1
1
1
x sin 2 x

(
1

cos
2
x
)
dx

x

sin
2
x

c


 c


2
2
2
2
4

2
 cos x dx 
1
1
1
x sin 2 x

(
1

cos
2
x
)
dx

x

sin
2
x

c


 c


2
2
2
2
4

It is helpful to know the following integral formulas, where k is a constant:
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
1.
 sin
2.
 cos
2
2
k x dx 
1  k x sin 2 k x 
x sin 2 k x

c

c 
k 2
4 
2
4k
k x dx 
1  k x sin 2 k x 
x sin 2 k x

c

c 
k 2
4 
2
4k
Examples Evaluate the following integrals.
1.
 cos
 cos
3
8 x dx
3
8 x dx =
 cos
2
8 x cos 8 x dx =
 (1  sin
2
8 x ) cos 8 x dx
2
2
2
2
NOTE: cos 8 x  sin 8 x  1  cos 8 x  1  sin 8 x
Let u  sin 8 x . Then du  8 cos 8 x dx
 cos
3
8 x dx =
 cos
2
8 x cos 8 x dx =
 (1  sin
2
8 x ) cos 8 x dx =
1
u3 
1
1
2
2
c =
( 1  sin 8 x ) 8 cos 8 x dx =  ( 1  u ) du =  u 
8
3 
8
8
1
1
u (3  u 2 )  c =
sin 8 x ( 3  sin 2 8 x )  c
24
24
Answer:
2.

4 / 3
  /6
1
sin 8 x ( 3  sin 2 8 x )  c
24
sin 5  d
5
Since the function f ( )  sin  is continuous on its domain of definition,
which is the set of all real numbers, then it is continuous on the closed
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
  4 
interval   ,
. Thus, the Fundamental Theorem of Calculus can be
 6 3 
applied.
4 / 3
5
   / 6 sin  d =

4 / 3
  /6
4 / 3
4
   / 6 sin  sin  d =

4 / 3
( sin 2  ) 2 sin  d =
  /6
( 1  cos 2  ) 2 sin  d
2
2
2
2
NOTE: cos   sin   1  sin   1  cos 
Let u  cos  . Then du   sin  d
4 / 3
4 / 3
2
2
2
2
   / 6 (1  cos  ) sin  d =     / 6 (1  cos  ) (  1) sin  d =

 1/ 2

(1  u 2 ) 2 du =
3 /2
3 /2
 1/ 2
(1  u 2 ) 2 du
3


  
 u  cos     cos 
and
6
2
6
 6
4
4

1
 
 u  cos
  cos  
3
3
3
2
NOTE: u  cos  and   
3 /2

3 /2
 1/ 2
(1  u 2 ) 2 du =

1
u (15  10 u 2  3 u 4 )
15
1
15

3 /2
 1/ 2
2u 3 u 5 
2
4

(1  2 u  u ) du = u 
3
5   1 / 2 =
3 /2
 1/ 2
=
 3 
1
5
3  
3
 9 
15

10

3

15











2
2 16   =
 2 
4
 16  
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
1 
30 
15 27 
5
3 

3  15 

  15  
=
2
16 
2 16 

1 
30 
1  147 3
203 
240 40
3 
 240 120 27 

3




 
= 30  16
16  =
16
16 
16
16 16 
 16

147 3  203
480
Answer:
3.
 sin
4
147 3  203
480
10 t dt
2
1

(
1

cos
20
t
)
sin
10
t
dt
(
sin
10
t
)
dt
=
=
  2


 dt
4
2
2
2
NOTE: By the half-angle formula, sin 10 t 
1
( 1  cos 20 t )
2
2
1
1

(
1

cos
20
t
)
dt
( 1  cos 20 t ) 2 dt =
=
  2


4
1
(1  2 cos 20 t  cos 2 20 t ) dt =

4
1
4
1
1  20 t sin 40 t  

t

2

sin
20
t



 c =

20
20  2
4  

NOTE: We applied the integral formula
 cos
2
k t dt =
1  kt
sin 2 k t 


  c with k  20 .
k  2
4 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
1
sin 20 t
t sin 40 t 
13
sin 20 t sin 40 t 
 

t 
 c =  t 
 c =
4
10
2
80 
42
10
80 
1
( 120 t  8 sin 20 t  sin 40 t )  c
320
Answer:
4.
 sin
2
1
( 120 t  8 sin 20 t  sin 40 t )  c
320
3 x cos 2 3 x dx
2
Since sin 3 x 
1
1
( 1  cos 6 x ) and cos 2 3 x  (1  cos 6 x ) , we obtain that
2
2
2
2
 sin 3 x cos 3 x dx =
1
1

(
1

cos
6
x
)
  2
  2 (1  cos 6 x )  dx =
1
1
1
2
(
1

cos
6
x
)
(
1

cos
6
x
)
dx
(
1

cos
6
x
)
dx
sin 2 6 x dx =
=
=



4
4
4
2
2
2
2
NOTE: cos 6 x  sin 6 x  1  1  cos 6 x  sin 6 x
1
4
 1  6 x sin 12 x  
1  x sin 12 x 
1


c


c
( 12 x  sin 12 x )  c




=
=
6 2
4  
4 2
24 
96
 
2
NOTE: We applied the integral formula  sin k x dx =
1  k x sin 2 k x 


  c with k  6 .
k 2
4 
Answer:
1
( 12 x  sin 12 x )  c
96
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Integrals of the Form
a.
 sin
m
x cos n x dx
n is an odd positive integer.
If n  1 , then
 sin
m
x cos n x dx =
 sin
m
x cos x dx . Let u  sin x . Then
um1
sin m  1 x
c =
 c,
du  cos x dx .  sin x cos x dx =  u du =
m 1
m 1
m
m
where m   1 . If m   1 , then
 sin
m
x cos x dx =
 cot x dx
=
ln sin x  c .
If n is greater than one, then write the integral
 sin
m
 sin
m
x cos n x dx as
x cos n  1 x cos x dx . Since n is an odd integer greater than one, then
n  1 is an even positive integer. Thus, n  1  2 k for some positive integer
n 1
x  cos 2 k x  ( cos 2 x ) k . Since
k. Then we may write cos
cos 2 x  sin 2 x  1 , then cos 2 x  1  sin 2 x . Thus, cos n  1 x  ( cos 2 x ) k =
(1  sin 2 x ) k . Then
 sin
m
x cos n  1 x cos x dx =
 sin
m
x (1  sin 2 x ) k cos x dx .
m
2
k
Let u  sin x . Then du  cos x dx . Thus,  sin x (1  sin x ) cos x dx =
u
m
(1  u 2 ) k du . Since the expression (1  u 2 ) k can be expanded by the
Binomial Expansion Theorem, which says that ( a  b ) 
k
k

i  0
 k k  i i
b ,
  a
i
 
k
 k k  i
 k
k!
2 k
(
1

u
)

  1 (  u 2 ) i =

where   
, then
i  0 i 
 i  i ! ( k  i )!
k
k
 k
i 2i
m
i  k  2i
m
2
k
(

1
)
u
u
(

1
)


  u

 
. Thus, u (1  u ) =
=
i  0
i  0 i 
i 
 k
(  1) i   u 2 i u m =

i  0
i 
k
 sin
m
k
 (  1)
i  0
x cos n  1 x cos x dx =
 sin
m
i
 k  2i
  u
i 
 m
m
n
. Thus,  sin x cos x dx =
x (1  sin 2 x ) k cos x dx =
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
u
m
(1  u 2 ) k du =
k
 k

i  k  2i  m
i  k
(

1
)
u
du


(

1
)
   u 2 i  m du =




=
 i  0
i 
i  0
 
i 

2i  m  1
2i  m  1
k
k
x
i  k u
i  k  sin
(  1)  
 c =  (  1)  
 c.

i
i
2
i

m

1
2
i

m

1
i  0
i  0
 
 
NOTE: In this case, m can be any rational number.
b.
m is an odd positive integer
If m  1 , then
du   sin x dx .
 sin x cos x dx =  sin x cos x dx .
 sin x cos x dx =   (  1) sin x cos
m
n
n
n
n
Let u  cos x . Then
x dx =   u n du =
un 1
cos n  1 x

c = 
 c , where n   1 . If n   1 , then
n 1
n 1
 sin x cos
n
x dx =
 tan x dx
= ln sec x  c .
If m is greater than one, then write the integral
 sin
m 1
 sin
m
x cos n x dx as
x cos n x sin x dx . Since m is an odd integer greater than one, then
m  1 is an even positive integer. Thus, m  1  2 k for some positive
m 1
x  sin 2 k x  ( sin 2 x ) k . Since
integer k. Then we may write sin
cos 2 x  sin 2 x  1 , then sin 2 x  1  cos 2 x . Thus, sin m  1 x  ( sin 2 x ) k =
 sin
(1  cos 2 x ) k . Then
m 1
x cos n x sin x dx =
Let u  cos x . Then du   sin x dx . Thus,
 (1  cos
 (1  cos
2
2
x ) k cos n x sin x dx .
x ) k cos n x sin x dx =
  (1  cos 2 x ) k cos n x (  1) sin x dx =   (1  u 2 ) k u n du . From Part a above,
 k  2i
  u . Thus, (1  u 2 ) k u n =
i  0
i 
k
k
i  k  2i n
i  k  2i  n
(

1
)
u
u
(

1
)


  u


=
=
. Thus,
i 
i  0
i  0
 
i 
k
we have that (1  u )
2
i  k  2i
(

1
)
  u

i  0
i 
k
u
n
 sin
m
x cos n x dx =
k
 sin
=
 (  1)
m 1
i
x cos n x sin x dx =
 (1  cos
2
x ) k cos n x sin x dx
2
k
n
2 k n
=   (1  cos x ) cos x (  1) sin x dx =   (1  u ) u du =
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
k
 k

 k
 k
    (  1) i   u 2 i  n  du =  (  1) i  1    u 2 i  n du =
i  0
i 
i 
i  0

2i  n  1
2i  n  1
k
k
x
i 1  k u
i  1  k  cos
(  1)  
 c =  (  1)  
 c.

i
i
2
i

n

1
2
i

n

1
i  0
i  0
 
 
NOTE: In this case, n can be any rational number.
c.
both m and n are even positive integers
2
Use the half-angle formulas sin x 
cos 2 x 
1  cos 2 x 1
 (1  cos 2 x ) and
2
2
1  cos 2 x 1
 (1  cos 2 x ) to reduce the exponents by one-half.
2
2
Examples Evaluate the following integrals.
1.
 sin
5
4 x cos 7 4 x dx
The exponent on both the sine and cosine function is odd. So, you can apply
Part a or Part b above. It will be the easiest to work with the smallest odd
number. So, apply Part b.
 sin
5
 ( sin
4 x cos 7 4 x dx =
2
 sin
4
4 x cos 7 4 x sin 4 x dx =
4 x ) 2 cos 7 4 x sin 4 x dx =
 (1  cos
2
4 x ) 2 cos 7 4 x sin 4 x dx
Let u  cos 4 x . Then du   4 sin 4 x dx
 (1  cos

2
4 x ) 2 cos 7 4 x sin 4 x dx = 
1
(1  cos 2 4 x ) 2 cos 7 4 x (  4 ) sin 4 x dx =

4
1
1
1
(1  u 2 ) 2 u 7 du =   ( 1  2 u 2  u 4 ) u 7 du =   ( u 7  2 u 9  u 11 ) du =

4
4
4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
 u8
u 10
u 12 
11
1
1


  c =   cos 8 4 x  cos 10 4 x 


cos 12 4 x   c
48
5
12
5
12 

 8
1

4
Answer: 
11
1
1

8
10
cos 12 4 x   c
 cos 4 x  cos 4 x 
48
5
12

 sin
NOTE: Here is the work to evaluate
 sin
 sin
 sin
5
4 x cos 7 4 x dx =
5
5
4 x ( cos 2 4 x ) 3 cos 4 x dx =
5
4 x cos 7 4 x dx using Part a above.
4 x cos 6 4 x cos 4 x dx =
 sin
5
4 x (1  sin 2 4 x ) 3 cos 4 x dx
Let u  sin 4 x . Then du  4 cos 4 x dx
 sin
5
4 x (1  sin 2 4 x ) 3 cos 4 x dx =
1
sin 5 4 x (1  sin 2 4 x ) 3 4 cos 4 x dx =

4
1 5
1 5
2 3
u
(
1

u
)
du
u ( 1  3 u 2  3 u 4  u 6 ) du =
=


4
4
1  u 6 3u 8 3u 10 u 12 
1
5
7
9
11
c =



( u  3 u  3 u  u ) du = 
4 6
8
10
12 
4
11 6
3 8
3
1

sin 10 4 x 
sin 12 4 x   c
 sin 4 x  sin 4 x 
46
8
10
12

2.

3
sin y cos 3 y dy
The exponent on the cosine function is odd. So, you can apply Part a above.

3
sin y cos 3 y dy =

3
sin y cos 2 y cos y dy =

3
Copyrighted by James D. Anderson, The University of Toledo
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sin y (1  sin 2 y ) cos y dy
Let u  sin y . Then du  cos y dy

sin y (1  sin 2 y ) cos y dy =
3
1/ 3
7/3
 ( u  u ) du =

3
u (1  u 2 ) du =
u
1/ 3
(1  u 2 ) du =
3 4/3
3 10 / 3
3 4/3
u 
u
 c =
u (5  2u 2 )  c =
4
10
20
3
( sin y ) 4 / 3 ( 5  2 sin 2 y )  c
20
Answer:
3.
t
2
3
( sin y ) 4 / 3 ( 5  2 sin 2 y )  c
20
sin 7 ( t 3 ) cos 4 ( t 3 ) dt
The exponent on the sine function is odd. So, you can apply Part b above.
t
t
t
2
sin 7 ( t 3 ) cos 4 ( t 3 ) dt =
2
sin 6 ( t 3 ) cos 4 ( t 3 ) sin ( t 3 ) dt =
2
[ sin 2 ( t 3 ) ] 3 cos 4 ( t 3 ) sin ( t 3 ) dt =
t
2
[ 1  cos 2 ( t 3 ) ] 3 cos 4 ( t 3 ) sin ( t 3 ) dt
3
2
3
Let u  cos t . Then du   3t sin t dt
t
2
[ 1  cos 2 ( t 3 ) ] 3 cos 4 ( t 3 ) sin ( t 3 ) dt =

1
[ 1  cos 2 ( t 3 ) ] 3 cos 4 ( t 3 ) (  3 ) t 2 sin ( t 3 ) dt =

3

1
1
2 3 4
(
1

u
)
u
du

(1  3 u 2  3 u 4  u 6 ) u 4 du =
=


3
3
1  u 5 3u 7 u 9 u 11 
1
4
6
8
10
c =



  ( u  3 u  3 u  u ) du =  
3 5
7
3
11 
3
Copyrighted by James D. Anderson, The University of Toledo
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
1 1
3
1
1
5
3
7
3
9
3
11
3 
cos
(
t
)

cos
(
t
)

cos
(
t
)

cos
(
t
)  c
3  5
7
3
11

Answer: 
4.
 sin
2
1 1
3
1
1

5
3
7
3
9
3
cos
(
t
)

cos
(
t
)

cos
(
t
)

cos 11 ( t 3 )   c

3 5
7
3
11

x cos 4 x dx
The exponent on both the sine and cosine function is even. So, you will need
to apply Part c above.
2
Since sin x 
 sin
2
1
1
( 1  cos 2 x ) and cos 2 x  (1  cos 2 x ) , we obtain that
2
2
x cos 4 x dx =
 sin
2
x ( cos 2 x ) 2 dx =
2
1
1

(
1

cos
2
x
)
  2
  2 (1  cos 2 x )  dx =
1
1
(1  cos 2 x ) ( 1  cos 2 x ) 2 dx =  ( 1  cos 2 x ) (1  cos 2 x ) (1  cos 2 x ) dx =

8
8
1
1
(1  cos 2 2 x ) (1  cos 2 x ) dx =  (1  cos 2 x  cos 2 2 x  cos 3 2 x ) dx

8
8
NOTE:
 cos 2 x dx 
1
sin 2 x  c ;
2
2
 cos 2 x dx 
1  2 x sin 4 x 


 c =
2 2
4 
x sin 4 x

 c.
2
8
We will have to do the following work in order to evaluate
 cos
3
2 x dx =
 cos
2
2 x cos 2 x dx =
 (1  sin
2
2 x ) cos 2 x dx
Copyrighted by James D. Anderson, The University of Toledo
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 cos
3
2 x dx .
Let u  sin 2 x . Then du  2 cos 2 x dx
2
 (1  sin 2 x ) cos 2 x dx =
1
1
2
(
1

sin
2
x
)
2
cos
2
x
dx
( 1  u 2 ) du =
=


2
2
1
u3 
1
sin 3 2 x 
1
1
 u 
  c =  sin 2 x 
  c = sin 2 x  sin 3 2 x  c .
2
3 
2
3 
2
6
Thus,
1
1
sin 2 x  sin 3 2 x  c . Then
2
6
3
 cos 2 x dx =
2
4
 sin x cos x dx =
1
(1  cos 2 x  cos 2 2 x  cos 3 2 x ) dx =

8
1
1
1 3 
 x sin 4 x   1
x  sin 2 x   
   sin 2 x  sin 2 x    c =

8
2
8  2
6
2

1
1
1
1
1
1 3 
 x  sin 2 x  x  sin 4 x  sin 2 x  sin 2 x   c =
8
2
2
8
2
6

11
1
1 3 
 x  sin 4 x  sin 2 x   c
82
8
6

Answer:
11
1
1 3 
 x  sin 4 x  sin 2 x   c
82
8
6

Examples Evaluate the following integrals.
1.
 sec
 sec
4
5 d
4
5 d =
 sec
2
5 sec 2 5 d =
 (1  tan
2
5 ) sec 2 5 d
2
2
2
2
NOTE: sec 5  tan 5  1  sec 5  1  tan 5
Copyrighted by James D. Anderson, The University of Toledo
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2
Let u  tan 5 . Then du  5 sec 5 d
2
2
 (1  tan 5 ) sec 5 d =
1
1
(1  tan 2 5 ) 5 sec 2 5 d =  ( 1  u 2 ) du =

5
5
1
u3 
1
1
 u 
  c =
u (3  u 2 )  c =
tan 5 ( 3  tan 2 5 )  c
5
3 
15
15
Answer:
2.
1
tan 5 ( 3  tan 2 5 )  c
15
2
 cot
4x
dx
7
2
 cot
4x
dx =
7
1
4x
( 7 cot
 4x)  c
4
7

2
NOTE: csc
Answer: 
4 / 3
3.
7
4x
 2 4x

  csc 7  1  dx =  4 cot 7  x  c =

/2
csc 6
4x
4x
4x
4x
 cot 2
 1  cot 2
 csc 2
1
7
7
7
7
1
4x
( 7 cot
 4x)  c
4
7
t
dt
2
t
t
is defined and is continuous if 0    since
2
2
sin 0  0 and sin   0 would imply that the cosecant function is undefined if
the angle argument is 0 or  . For this definite integral, we have that
6
The function f ( t )  csc
Copyrighted by James D. Anderson, The University of Toledo
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
4

4

t
2
t 
t 

 
. Thus,
. Since the closed
2
3
2
3
4 2
3
  2 
interval  ,
is a subset of the set ( 0 ,  ) , then the function f is
 4 3 
  4 
continuous on the closed interval  ,
. Thus, the Fundamental
 2 3 
Theorem of Calculus can be applied.
4 / 3

t
csc dt =
2
6
/2

/2
t
t
csc csc 2 dt =
2
2
4
4 / 3

/2
2
 2t 
2 t
 csc  csc dt =
2
2

2

2 t 
2 t
 1  cot  csc dt
2
2

4 / 3

4 / 3
/2
2
NOTE: csc
t
t
t
t
 cot 2  1  csc 2  1  cot 2
2
2
2
2
t
1
2 t
u

cot
Let
. Then du   csc dt
2
2
2
2

/2
2
2
4 / 3 
1 2t

2 t 
2 t
2 t  
 1  cot  csc dt =  2   / 2  1  cot     csc dt =
2
2
2  2
2


4 / 3
 1/ 3
1
(1  u 2 ) 2 du
t


4
 u  cot  1 and t 

and t 
2
2
4
3
2
1
2
2


u  cot
  tan   3 , then cot
. Since tan
.
3
3
3
3
3
NOTE: u  cot
2
 1/ 3
1
1
(1  u 2 ) 2 du = 2 
 1/
1

2 u 3 u5  
 
2  u 

3
5

   1/
1
3
(1  u 2 ) 2 du = 2 
 1/
3
( 1  2 u 2  u 4 ) du =
1
3
2
2
4 
= 15 u (15  10 u  3u ) 
  1/
Copyrighted by James D. Anderson, The University of Toledo
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=
3
3
2 
1 
10 1  
2 
  =
 15 
 15  10  3 
 28 
15 
3
3  
15 
3
3 
 56  
  =
 3  
2  252 56 3 
504  112 3

=
15  9
9 
135
Answer:
4.
 tan
 tan
504  112 3
135
7
y sec 3 y dy
7
y sec 3 y dy =
 ( tan
2
 tan
6
y sec 2 y sec y tan y dy =
y ) 3 sec 2 y sec y tan y dy =
 ( sec
2
y  1) 3 sec 2 y sec y tan y dy
2
2
2
2
NOTE: sec y  tan y  1  tan y  sec y  1
Let u  sec y . Then du  sec y tan y dy
 ( sec
 (u
6
2
y  1) 3 sec 2 y sec y tan y dy =
 3u 4  3u 2  1) u 2 du =
 (u
 (u
8
2
 1) 3 u 2 du =
 3u 6  3u 4  u 2 ) du =
1
3
3
1
u 9 3u 7 3u 5 u 3



 c = sec 9 y  sec 7 y  sec 5 y  sec 3 y  c
9
7
5
3
9
7
5
3
Answer:
1
3
3
1
sec 9 y  sec 7 y  sec 5 y  sec 3 y  c
9
7
5
3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Integrals of the Form
a.
 tan
m
x sec n x dx
n is an even positive integer.
If n  2 , then
 tan
m
 tan
x sec n x dx =
m
x sec 2 x dx . Let u  tan x . Then
um1
tan m  1 x
c =
 c,
du  sec x dx .  tan x sec x dx =  u du =
m 1
m 1
m
2
2
m
 tan
where m   1 . If m   1 , then

m
 cot x sec
x sec 2 x dx =
sec 2 x
dx . Let u  tan x . Then du  sec 2 x dx .
tan x

sec 2 x
dx =
tan x
2
x dx =

du
=
u
ln u  c = ln tan x  c .
If n is greater than two, then write the integral
 tan
m
 tan
m
x sec n x dx as
x sec n  2 x sec 2 x dx . Since n is an even integer greater than two, then
n  2 is an even positive integer. Thus, n  2  2 k for some positive integer
n2
x  sec 2 k x  ( sec 2 x ) k . Since sec 2 x  tan 2 x  1 ,
k. Then we may write sec
2
2
n2
x  ( sec 2 x ) k = (1  tan 2 x ) k . Then
then sec x  1  tan x . Thus, sec
 tan
m
x sec n  2 x sec 2 x dx =
2
Then du  sec x dx . Thus,
u
m
 tan
 tan
m
m
x (1  tan 2 x ) k sec 2 x dx . Let u  tan x .
x (1  tan 2 x ) k sec 2 x dx =
(1  u 2 ) k du . Since the expression (1  u 2 ) k can be expanded by the
 k k  i i
   a b ,
i  0 i 
k
 k k  i 2 i
 k
  1 ( u ) =    u 2 i .
i  0 i 
i 
Binomial Expansion Theorem, which says that ( a  b ) 
k
k
 k
k!
2 k
where   
, then (1  u ) 
 i  i ! ( k  i )!
k
 k  2i
m
m
2 k
u
  u
u
(
1

u
)

Thus,
=
=
i  0 i 
Thus,
 tan
 tan
x (1  tan 2 x ) k sec 2 x dx =
m
m
x sec n x dx =
 tan
m
k

i  0
k

i  0
 k  2i m
  u u =
i 
x sec n  2 x sec 2 x dx =
u
m
(1  u 2 ) k du =
Copyrighted by James D. Anderson, The University of Toledo
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k

i  0
 k  2i
  u
i 
 m
.
k
 k  k  2i  m 
u
du



= 
 i  0  i 
i  0

k
 k  tan 2 i  m  1 x
 c.
 

i
2
i

m

1
i  0  
 k  2i
   u
i 
 m
k
du =

i  0
 k  u 2i  m  1
 c =
 
i
2
i

m

1
 
NOTE: In this case, m can be any rational number.
b.
m is an odd positive integer.
If m  1 , then
 tan
m
Let u  sec x . Then
 tan x sec
du  sec x tan x dx .  sec
x sec n x dx =
n
x dx =
n 1
 sec
n 1
x sec x tan x dx .
x sec x tan x dx =
u
n 1
du =
un
sec n x
 c =
 c , where n  0 . If n  0 , then  tan x sec n x dx =
n
n
 tan x dx = ln sec x  c .
If m is greater than one, then write the integral
 tan
m 1
 tan
m
x sec n x dx as
x sec n  1 x sec x tan x dx . Since m is an odd integer greater than one,
then m  1 is an odd positive integer. Thus, m  1  2 k for some positive
m 1
x  tan 2 k x  ( tan 2 x ) k . Since
integer k. Then we may write tan
sec 2 x  tan 2 x  1 , then tan 2 x  sec 2 x  1. Thus, tan m  1 x  ( tan 2 x ) k =
( sec 2 x  1) k . Then
 ( sec
2
 tan
2
x sec n  1 x sec x tan x dx =
x  1) k sec n  1 x sec x tan x dx . Let u  sec x . Then
du  sec x tan x dx . Thus,
 (u
m 1
 ( sec
2
x  1) k sec n  1 x sec x tan x dx =
 1) k u n  1 du . Since the expression ( u 2  1) k can be expanded by the
Binomial Expansion Theorem, which says that ( a  b ) 
k
 k
k!
2
k
where   
, then ( u  1) 
 i  i ! ( k  i )!
k

i  0
k

i  0
 k k  i i
b ,
  a
i 
 k 2 k  i
  ( u ) (  1) i =
i 
Copyrighted by James D. Anderson, The University of Toledo
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k
 k  2(k  i )
n 1
2
k
n

1
(  1)   u
u
(  1) i
( u  1) u


.
Thus,
=
i  0
i  0
i 
k
k
 k
i  k  2(k  i ) n  1
(  1)   u
u
(  1) i   u 2 ( k  i )  n  1 . Thus,


=
i  0
i  0
i 
i 
k
i
 tan x sec x dx =  ( sec x  1) sec x sec x tan x dx =
 ( sec x  1) sec x sec x tan x dx =  ( u  1) u du
m
2
n
2
k
n 1
k
n 1
2
 k

i  k  2(k  i )  n  1
(

1
)
u


 du =
 i  0
i 
 

k
u 2(k  i )  n
i  k
 (  1)  i  2 ( k  i )  n  c =
i  0
 
 k  2(k  i )
  u
=
i 
k
n 1
=
i  k
(

1
)
   u 2 ( k  i )  n  1 du =

i  0
i 
2(k  i )  n
k
x
i  k  sec
(  1)  
 c.

i
2
(
k

i
)

n
i  0
 
k
NOTE: In this case, n can be any rational number.
c.
n is an odd integer and m is an even positive integer.
There is not a standard process for evaluating these integrals.
Integrals of the Form
a.
 cot
m
x csc n x dx
n is an even positive integer.
 cot x csc x dx =  cot x csc x dx . Let u  cot x . Then
x dx .  cot x csc x dx =   cot x (  1) csc x dx =   u du =
If n  2 , then
du   csc 2
m
n
m
m
2
2
m
2
m
um 1
cot m  1 x

 c = 
 c , where m   1 . If m   1 , then
m 1
m 1
 cot
m
2
x csc x dx =
du   csc x dx .
2

 tan x csc
2
x dx =

csc 2 x
dx . Let u  cot x . Then
cot x
csc 2 x
 csc 2 x
du
dx =  
dx =  
=  ln u  c =
cot x
cot x
u
 ln cot x  c = ln tan x  c .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
If n is greater than two, then write the integral
 cot
m
 cot
m
x csc n x dx as
x csc n  2 x csc 2 x dx . Since n is an even integer greater than two, then
n  2 is an even positive integer. Thus, n  2  2 k for some positive integer
n2
x  csc 2 k x  ( csc 2 x ) k . Since
k. Then we may write csc
csc 2 x  cot 2 x  1 , then csc 2 x  1  cot 2 x . Thus, csc n  2 x  ( csc 2 x ) k =
 cot
(1  cot 2 x ) k . Then
 cot
 cot
u
m
m
m
m
x csc n  2 x csc 2 x dx =
x (1  cot 2 x ) k csc 2 x dx . Let u  cot x . Then du   csc 2 x dx . Thus,
x (1  cot 2 x ) k csc 2 x dx =   cot m x (1  cot 2 x ) k (  1) csc 2 x dx =
(1  u 2 ) k du . Since the expression (1  u 2 ) k can be expanded by the
 k k  i i
b ,
  a
i
i  0  
k
 k k  i 2 i
 k
  1 ( u ) =    u 2 i .
i  0 i 
i 
k

k
Binomial Expansion Theorem, which says that ( a  b ) 
 k
k!
2 k
where   
, then (1  u ) 
 i  i ! ( k  i )!
k
 k  2i
m
m
2 k
u
  u

Thus, u (1  u ) =
=
i  0 i 

i  0
k

i  0
 k  2i m
  u u =
i 
Thus,
 cot
 cot
x (1  cot 2 x ) k csc 2 x dx =   u m (1  u 2 ) k du =
m
m
x csc n x dx =
 cot
k
m
k

i  0
 k  2i
  u
i 
 m
.
x csc n  2 x csc 2 x dx =
k
 k  k  2i  m 
 k  2i  m
      u
du

du =
   u


=
i
i
i  0  
i  0  

k
k
 k  u 2i  m  1
 k  cot 2 i  m  1 x
   
 c =    
 c.
i  0  i  2i  m  1
i  0  i  2i  m  1
NOTE: In this case, m can be any rational number.
b.
m is an odd positive integer.
If m  1 , then
 cot
m
Let u  csc x . Then
 cot x csc x dx =  csc x csc x cot x dx .
x csc x cot x dx =
du   csc x cot x dx .  csc
x csc n x dx =
n 1
n
n 1
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
un
csc n x
c = 
 c,
  csc
x (  1) csc x cot x dx =   u du = 
n
n
n
where n  0 . If n  0 , then  cot x csc x dx =  cot x dx = ln sin x  c .
n 1
n 1
If m is greater than one, then write the integral
 cot
m 1
 cot
m
x csc n x dx as
x csc n  1 x csc x cot x dx . Since m is an odd integer greater than one,
then m  1 is an odd positive integer. Thus, m  1  2 k for some positive
m 1
x  cot 2 k x  ( cot 2 x ) k . Since
integer k. Then we may write cot
csc 2 x  cot 2 x  1 , then cot 2 x  csc 2 x  1 . Thus, cot m  1 x  ( cot 2 x ) k =
 cot
( csc 2 x  1) k . Then
 ( csc
2
m 1
x csc n  1 x csc x cot x dx =
x  1) k csc n  1 x csc x cot x dx . Let u  csc x . Then
du   csc x cot x dx . Thus,
 ( csc
2
x  1) k csc n  1 x csc x cot x dx =
  ( csc 2 x  1) k csc n  1 x (  1) csc x cot x dx =   ( u 2  1) k u n  1 du . Since
2
k
the expression ( u  1) can be expanded by the Binomial Expansion
k
 k k  i i
k
(
a

b
)

   a b , where
Theorem, which says that
i  0 i 
k
 k 2 k  i
 k
k!
2
k
(
u

1
)

  ( u ) (  1) i =
  

, then
i  0 i 
 i  i ! ( k  i )!
k
k
i  k  2(k  i )
n 1
2
k
n

1
(  1)   u
u
(  1) i
( u  1) u


.
Thus,
=
i  0
i  0
i 
 k  2(k  i ) n  1
(

1
)
u
  u

=
i
i  0
 
k
i
i  k  2(k  i )  n  1
(

1
)
  u

. Thus,
i  0
i 
k
 cot x csc x dx =  ( csc x  1) csc x csc x cot x dx
 ( csc x  1) csc x csc x cot x dx =   ( u  1) u
m
2
2
n
k
 k  2(k  i )
  u
=
i 
k
n 1
n 1
 k

 k
    (  1) i   u 2 ( k  i )  n  1  du =
i 
i  0

k
u 2(k  i )  n
i 1 k
 (  1)  i  2 ( k  i )  n  c =
i  0
 
2
k
=
n 1
du =
i 1  k
(

1
)
   u 2 ( k  i )  n  1 du =

i  0
i 
2(k  i )  n
k
x
i  1  k  csc
(  1)  
 c.

i  0
 i  2( k  i )  n
k
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
NOTE: In this case, n can be any rational number.
c.
n is an odd integer and m is an even positive integer.
There is not a standard process for evaluating these integrals.
Examples Evaluate the following integrals.
1.

 /6
  / 12
tan 5 2 x sec 4 2 x dx
5
4
The function f ( x )  tan 2 x sec 2 x is defined and is continuous if



  
  2x 
since cos     0 and cos  0 would imply that both the
2
2
2
 2

tangent and secant functions are undefined if the angle argument is  . For
2




 x  . Thus, 
 x 

this definite integral, we have that 
12
6
12
6
   


  2 x  . Since the closed interval   ,  is a subset of the set
6
3
 6 3
   
  ,  , then the function f is continuous on the closed interval
 2 2
   
  6 , 3  . Thus, the Fundamental Theorem of Calculus can be applied.
The exponent on the secant function is even. So, Part a above can be applied.
The exponent on the tangent function is odd. So, Part b above can be
applied. It will be the easiest to work with the smallest exponent. However,
we will work this problem both ways. This will allow us to check the
correctness of our answer.
Using Part a above, we have that
 /6
5
4
   / 12 tan 2 x sec 2 x dx =

 /6
tan 5 2 x sec 2 2 x sec 2 2 x dx =
  / 12
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

 /6
tan 5 2 x ( 1  tan 2 2 x ) sec 2 2 x dx
  / 12
2
Let u  tan 2 x . Then du  2 sec 2 x dx

1
2
 /6
tan 5 2 x ( 1  tan 2 2 x ) sec 2 2 x dx =
  / 12
 /6
5
2
2
   / 12 tan 2 x (1  tan 2 x ) 2 sec 2 x dx =
NOTE: u  tan 2 x and x  


 u  tan 
6
3
x 
1
2

3
 1/ 3
u 5 ( 1  u 2 ) du

1

  
 u  tan      tan  
and
6
12
3
 6
3
3
1
2

3
 1/ 3
1
u 5 ( 1  u 2 ) du =
2
3
1 6

u ( 4  3 u 2 )
48
  1/
=
3
1
48

3
 1/ 3
1 u6 u8 
5
7


( u  u ) du = 
2 6
8    1 /
1
1 
5 


27
(
4

9
)

(
4

1
)
351


 =
=


27
48 
27 
1  9472 
1  592 
592

 = 
 =
48  27 
3  27 
81
Using Part b above, we have that
 /6
5
4
   / 12 tan 2 x sec 2 x dx =

 /6
tan 4 2 x sec 3 2 x sec 2 x tan 2 x dx =
  / 12
 /6

( tan 2 2 x ) 2 sec 3 2 x sec 2 x tan 2 x dx =
  / 12

 /6
  / 12
=
3
( sec 2 2 x  1) 2 sec 3 2 x sec 2 x tan 2 x dx
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Let u  sec 2 x . Then du  2 sec 2 x tan 2 x dx

1
2
 /6
  / 12
( sec 2 2 x  1) 2 sec 3 2 x sec 2 x tan 2 x dx =
 /6
2
2
3
   / 12 ( sec 2 x  1) sec 2 x 2 sec 2 x tan 2 x dx =
NOTE: u  sec 2 x and x  


2
2/ 3
( u 2  1) 2 u 3 du

2

  
 u  sec     sec 
and
6
12
3
 6


 u  sec  2
6
3
x 
1
2
1
2
2
2/ 3
( u 2  1) 2 u 3 du =
1
2

2
2/ 3
( u 4  2 u 2  1) u 3 du =
2
1
2

2
2/ 3
1 u8 u6 u4 
7
5
3



( u  2 u  u ) du = 
2 8
3
4   2 /
2
1 4

u ( 3 u 4  8 u 2  6 )
48
 2/
1
48
16 

16
(
22
)



9 

=
3
1 
16  16 32

16
(
48

32

6
)



6

 =
48 
9  3
3

2
16 
1
22

 =

3 
48 
9
2
1
2 
2
1 
  =  22 

 =  11 
3 
3
27 
3
27 
2  296 
592

 =
3  27 
81
Answer:
2.

=
3
592
81
csc 8 t
dt
cot 4 t
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

csc 8 t
dt =
cot 4 t
 ( cot t )
4
csc 8 t dt
The exponent on the cosecant function is even. So, Part a above can be
applied.
 ( cot t )
4
csc 8 t dt =
 ( cot t )
 ( cot t )
4
(1  cot 2 t ) 3 csc 2 t dt
4
csc 6 t csc 2 t dt =
 ( cot t )
4
( csc 2 t ) 3 csc 2 t dt =
2
Let u  cot t . Then du   csc t dt
 ( cot t )
4
(1  cot 2 t ) 3 csc 2 t dt =   ( cot t )  4 (1  cot 2 t ) 3 (  1) csc 2 t dt =
  u  4 (1  u 2 ) 3 du =   u  4 (1  3u 2  3u 4  u 6 ) du =
  (u
4
 3u
2
 u  3 3u  1
u3

 3u 
 3  u ) du =  
1
3
 3
2

1 3
u (  1  9u 2  9u 4  u 6 )  c =
3

1
( cot t )  3 ( cot 6 t  9 cot 4 t  9 cot 2 t  1)  c =
3
cot 6 t  9 cot 4 t  9 cot 2 t  1

c
3 cot 3 t
cot 6 t  9 cot 4 t  9 cot 2 t  1
c
Answer: 
3 cot 3 t
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

  c =

3.
 cot
5
6 w csc 5 6 w dw
The exponent on the cotangent function is odd. So, Part b above can be
applied.
 cot
5
 ( cot
 ( csc
6 w csc 5 6 w dw =
 cot
4
6 w csc 4 6 w csc 6 w cot 6 w dw =
2
6 w ) 2 csc 4 6 w csc 6 w cot 6 w dw =
2
6 w  1) 2 csc 4 6 w csc 6 w cot 6 w dw
Let u  csc 6 w . Then du   6 csc 6 w cot 6 w dw
 ( csc
2
6 w  1) 2 csc 4 6 w csc 6 w cot 6 w dw =

1
6
 ( csc

1
6
 (u
1

6

2
2
6 w  1) 2 csc 4 6 w (  6 ) csc 6 w cot 6 w dw =
 1) 2 u 4 du = 
1
6
 (u
4
 2 u 2  1) u 4 du =
1  u 9 2u 7 u 5 
c =


( u  2 u  u ) du =  
6 9
7
5 
8
6
4
1  35 u 9 90 u 7 63u 5 
1
  c = 
 


u 5 ( 35 u 4  90 u 2  63 )  c =
6  315
315
315 
1890

1
csc 5 6 w ( 35 csc 4 6 w  90 csc 2 6 w  63 )  c
1890
Answer: 
1
csc 5 6 w ( 35 csc 4 6 w  90 csc 2 6 w  63 )  c
1890
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
4.

 /3
  /4
tan 3  d
3
The function f (  )  tan  is defined and is continuous if 


 
2
2

  
since cos     0 and cos  0 would imply that the tangent function is
2
 2

undefined if the angle argument is  . For this definite integral, we have
2
   






that
. Since the closed interval   ,  is a subset of the
4
3
 4 3
   
set   ,  , then the function f is continuous on the closed interval
 2 2
   
  4 , 3  . Thus, the Fundamental Theorem of Calculus can be applied.
The exponent on the tangent function is odd. So, Part b above can be
applied.
 /3
3
   / 4 tan  d =
 /3
2
   / 4 tan  tan  d =
 /3
2
   / 4 ( sec  tan   tan  ) d =
To find the value of the integral
 /3
   / 4 tan  d = ln sec 
ln
2
2
= ln

 /3
  /4


 /3
  /4
 /3
  /4

 /3
  /4
( sec 2   1) tan  d =
sec 2  tan  d 

  /4
tan  d
tan  d :
= ln sec

  
 ln sec    = ln 2  ln 2
3
 4
2
To evaluate the integral

 /3
 /3
sec 2  tan  d :
  /4
Copyrighted by James D. Anderson, The University of Toledo
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=
2
Let u  tan  . Then du  sec  d .

3
 /3
  /4
sec 2  tan  d =

 /3
3
Thus,    / 4 tan  d =
Answer: 1  ln
5.
 cot
5
3
1

u2 
1
( 3  1) = 1
u du =
=

2  1
2
 /3
( sec 2  tan   tan  ) d = 1  ln
  /4
2
2
x dx
The exponent on the cotangent function is odd. So, Part b above can be
applied.
 cot
5
 ( csc
x dx =
4
 cot
4
x cot x dx =
 ( cot
4
x dx  2  cot x csc 2 x dx 
We know that
 cot x dx
To evaluate the integral
x ) 2 cot x dx =
 ( cot x csc
x  2 csc 2 x  1) cot x dx =
 cot x csc
2
4
 ( csc
2
x  1) 2 cot x dx =
x  2 cot x csc 2 x  cot x ) dx =
 cot x dx
 ln sin x  c .
 cot x csc
2
x dx :
2
Since D x cot x   csc x , then this integral is evaluated by simple
2
substitution. Let u  cot x . Then du   csc x dx .
1
u2
 c =  cot 2 x  c
  cot x (  1) csc x dx =   u du = 
2
2
2
Copyrighted by James D. Anderson, The University of Toledo
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To evaluate the integral
 csc
4
x cot x dx :
Since the exponent on the cotangent function is odd, then Part b above can be
applied. Since the exponent on the cosecant function is even, then you could
also apply Part a above.
 cot x csc
4
x dx =
 csc
3
x csc x cot x dx
Let u  csc x . Then du   csc x cot x dx
u4
 csc x csc x cot x dx =   csc x (  1) csc x cot x dx =   u du =  4  c =
3

5
x dx =
 cot x csc
4
x dx  2  cot x csc 2 x dx 
 cot x dx
=
1 4
1
 1

csc x  2   cot 2 x   ln sin x  c =  csc 4 x  cot 2 x  ln sin x  c
4
4
 2

Answer: 
6.
3
1 4
csc x  c
4
 cot

3
 tan
4
1 4
csc x  cot 2 x  ln sin x  c
4
t dt
The exponent on the tangent function is even. So, Part b above does not
apply.
 tan
4
t dt =
 ( tan
2
t ) 2 dt =
 ( sec
2
t  1) 2 dt =
 ( sec
2
We know that  sec t dt  tan t  c .
Copyrighted by James D. Anderson, The University of Toledo
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4
t  2 sec 2 t  1) dt
To evaluate the integral
 sec
4
t dt :
Since the exponent on the secant function is even, then Part a above can be
applied.
 sec
4
t dt =
 sec
2
t sec 2 t dt =
 (1  tan
2
t ) sec 2 t dt
2
Let u  tan t . Then du  sec t dt
1
u3
3
tan
t

tan
t  c
u


c
(
1

tan
t
)
sec
t
dt
(
1

u
)
du
=
=
=


3
3
2
2
2
4
2
 ( sec t  2 sec t  1) dt = tan t 
1
tan 3 t  2 tan t  t  c =
3
1
tan 3 t  tan t  t  c
3
Answer:
7.
 csc
3
1
tan 3 t  tan t  t  c
3
6 d
The exponent on the cosecant function is odd. So, Part a above does not
apply. This problem is the last example in Lesson 7. It was evaluated using
the Integration by Parts technique.
8.
 sec
5
y dy
u  sec 3 y
Let
3
Then du  3 sec y tan y dy
and
dv  sec 2 y dy
v  tan y
3
3
2
3
NOTE: D y sec y  D y ( sec y )  3 ( sec y ) sec y tan y  3sec y tan y
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
 sec
5
y dy =
 u dv
= uv 
 v du
=
sec 3 y tan y  3  sec 3 y tan 2 y dy = sec 3 y tan y  3  sec 3 y ( sec 2 y  1) dy =
sec 3 y tan y  3  ( sec 5 y  sec 3 y ) dy = sec 3 y tan y  3  sec 5 y dy  3  sec 3 y dy
Thus, we have that
 sec
5
y dy = sec 3 y tan y  3  sec 5 y dy  3  sec 3 y dy
5
Adding 3  sec y dy to both sides of the equation, we obtain that
4  sec 5 y dy = sec 3 y tan y  3  sec 3 y dy . Dividing both sides of the equation
by 4, we obtain that
1
3
sec 3 y tan y   sec 3 y dy .
4
4
5
 sec y dy =
Now, to evaluate the integral
 sec
3
y dy , we apply the Integration by Parts
technique.
u  sec y
Let
Then du  sec y tan y dy
and
 sec
 v du
3
y dy =
 u dv
= uv 
sec y tan y 
 sec y tan
sec y tan y 
 ( sec
sec y tan y 
 sec
3
3
2
dv  sec 2 y dy
v  tan y
=
y dy = sec y tan y  3  sec y ( sec 2 y  1) dy =
y  sec y ) dy = sec y tan y 
 sec
y dy  ln sec y  tan y
Thus, we have that
Copyrighted by James D. Anderson, The University of Toledo
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3
y dy 
 sec y dy
=
 sec
3
y dy = sec y tan y 
 sec
3
y dy  ln sec y  tan y
3
Adding  sec y dy to both sides of the equation, we obtain that
2  sec 3 y dy = sec y tan y  ln sec y  tan y 
 0 dy .
Since
 0 dy  C , then
3
we have that 2  sec y dy = sec y tan y  ln sec y  tan y  C . Dividing
both sides of the equation by 2, we obtain that
3
 sec y dy =
C
1
1
sec y tan y  ln sec y  tan y  c , where c  . Thus,
2
2
2
 sec
1
3
sec 3 y tan y   sec 3 y dy =
4
4
5
y dy =
1
31
1

sec 3 y tan y   sec y tan y  ln sec y  tan y   c =
4
42
2

1
3
3
sec 3 y tan y  sec y tan y  ln sec y  tan y  c
4
8
8
1
( 2 sec 3 y tan y  3 sec y tan y  3 ln sec y  tan y )  c
8
Answer:
1
( 2 sec 3 y tan y  3 sec y tan y  3 ln sec y  tan y )  c
8
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860