Download Trigonometric Integrals

Trigonometric Integrals
I. Integrating Powers of the Sine and Cosine Functions
A. Useful trigonometric identities
1. sin 2 x  cos 2 x  1
2. sin 2x  2 sin x cos x
3. cos 2 x  cos 2 x  sin 2 x  2 cos 2 x  1  1  2 sin 2 x
1  cos 2 x
2
1  cos 2 x
cos 2 x 
2
1
sin x cos y  [sin( x  y )  sin( x  y )]
2
1
sin x sin y  [cos( x  y )  cos( x  y )]
2
1
cos x cos y  [cos( x  y )  cos( x  y )]
2
4. sin 2 x 
5.
6.
7.
8.
B. Reduction formulas
1.

1
n 1
sin n x dx   sin n1 x cos x 
sin n2 x dx
n
n

2.

cos n x dx 

1
n 1
cos n1 x sin x 
cos n2 x dx
n
n
C. Examples
1. Find
 sin
2
x dx .
 sin x (sin x dx) . Let
u  sin x and dv  sin x dx  du  cos x dx and v  sin x dx 

Method 1(Integration by parts):

1
sin 2 x dx 
 cos x . Thus,
 sin
2

x dx  (sin x)( cos x)  cos 2 x dx   sin x cos x 
 (1  sin x) dx   sin x cos x  1dx   sin x dx   sin x cos x  x 
 sin x dx  2 sin x dx   sin x cos x  x   sin x dx 
2
2
2
2
2
1
1
 sin x cos x  x  C .
2
2
Method 2(Trig identity):

sin 2 x dx 
Method 3(Reduction formula):


1
1
1
(1  cos 2 x) dx  x  sin 2 x  C .
2
2
4

1
1
sin 2 x dx   sin x cos x 
1dx 
2
2
1
1
 sin x cos x  x  C .
2
2
2. Find
 cos x dx .
3
Use the reduction formula:


1
2
cos 3 x dx  cos 2 x sin x 
cos x dx 
3
3
1
2
1
2
cos 2 x sin x  sin x  C  sin x(1  sin 2 x)  sin x  C 
3
3
3
3
1
sin x  sin 3 x  C .
3

3. Find sin 3 x cos 2 x dx .
 sin x cos x dx   sin x sin x cos x dx   (1  cos x) cos x sin xdx 
 (cos x  cos x)(sin x dx) . Let u  cos x  du   sin x dx . Thus,
3
2
2

2
2
2
2
4
1

1
1
(cos 2 x  cos 4 x)(sin x dx)   (u 2  u 4 ) du   u 3  u 5  C 
3
5
1
1
 cos 3 x  cos 5 x  C .
3
5
2
4. Find

 sin x cos x dx .
2
2



1
 1  cos 2 x  1  cos 2 x 
(1  cos 2 2 x) dx 


 dx 
2
2
4



1
1  1  cos 4 x 
1
1
sin 2 2 x dx 
1 dx 
cos 4 x dx 

 dx 
4
4 
2
8
8

1
1
x  sin 4 x  C .
8
32
sin 2 x cos 2 x dx 

5. Find


 sin 4x cos 3x dx .
Method 1(Integration by parts): Let u  sin 4x and dv  cos 3x dx  du =
1
4 cos 4 x dx and v  sin 3x . Thus, sin 4 x cos 3x dx 
3
1
1
 4
(sin 4 x) sin 3x  
cos 4 x sin 3x dx  sin 4 x sin 3x 
3
3
 3
4
cos 4 x sin 3x dx . Find cos 4 x sin 3x dx . Let u  cos 4x and dv =
3
1
sin 3x dx  du  4 sin 4 x dx and v   cos 3 x . Thus,
3
1
4
cos 4 x sin 3x dx   cos 4 x cos 3x 
sin 4 x cos 3x dx . Returning to
3
3
1
the original integral, sin 4 x cos 3x dx = sin 4 x sin 3 x 
3
4 1
4
 1
sin 4 x cos 3x dx   sin 4 x sin 3x 
 cos 4 x cos 3x 
3 3
3
 3
4
16
7
cos 4 x cos 3x 
sin 4 x cos 3x dx  
sin 4 x cos 3x dx 
9
9
9
1
4
sin 4 x sin 3x  cos 4 x cos 3x  sin 4 x cos 3x dx =
3
9
3
4
 sin 4 x sin 3x  cos 4 x cos 3x  C .
7
7






Method 2(Trig identity):






sin 4 x cos 3x dx 
1
1
 cos x  cos 7 x  C .
2
14
3
1
2
 sin x  sin 7x dx 
II. Integrating Powers of the Tangent and Secant Functions
A. Useful trigonometric identity: tan 2 x  1  sec 2 x
B. Useful integrals
1.
 sec x tan x dx  sec x  C
2.
 sec x dx  tan x  C
3.
 tan x dx  ln sec x  C   ln cos x  C
4.
 sec x dx  ln sec x  tan x  C
2
C. Reduction formulas
1.
2.

sec n  2 x tan x n  2
sec x dx 

sec n  2 x dx
n 1
n 1

tan n 1 x
tan x dx 
 tan n  2 x dx
n 1

n

n
D. Examples
1. Find

2
x dx .
tan 2 x dx 
2. Find

 tan

(sec 2 x  1) dx 


sec 2 x dx  1dx  tan x  x  C .
 tan xdx .
3
tan 3 xdx 
tan 2 x

2

tan x dx 
1
tan 2 x  ln sec x  C .
2
4
3. Find

 sec xdx .
3
sec 3 x dx 
4. Find

sec x tan x 1
1
1

sec x dx  sec x tan x  ln sec x  tan x  C .
2
2
2
2
 tan x sec x dx .
2
Let u  tan x  du  sec 2 xdx 

tan x sec 2 x dx 

udu 
1 2
u C 
2
1
tan 2 x  C .
2
5. Find

tan x sec 4 x dx .
 tan x sec x dx   tan x sec x sec x dx   tan x(1  tan x) sec x dx 
 tan x sec x dx   tan x sec dx . Let u  tan x  du  sec x dx . Thus,
1
1
1
1
tan x sec x dx  udu  u du  u  u  C  tan x  tan x  C .

  2 4
2
4
4
2
2
3

2
2
2
2
4
6. Find
2
3
2
4
2
4
tan x sec 3 x dx .
 tan x sec x dx   sec x (sec x tan x dx) . Let u  sec x  du  sec x tan xdx .
1
1
Thus, tan x sec x dx  u du  u  C  sec x  C .

 3
3
3
2
3
7. Find

2
3
3
tan 2 x sec 3 x dx .
 tan x sec x dx   (sec x 1) sec x dx   sec x dx   sec x dx . Using
1
3
the reduction formula, sec x dx  sec tan x 
sec x dx . Thus,

4
4
2
3
2
3
5
5
3
5
3
3





1
3
sec 5 x dx  sec 3 x dx  sec 3 x tan x 
sec 3 xdx 
4
4
1
1
1
1
sec 3 x dx  sec 3 x tan x 
sec 3 x dx  sec 3 x tan x  sec x tan x 
4
4
4
8
tan 2 x sec 3 x dx 

1
ln sec x  tan x  C .
8
8. Find


tan x sec 4 x dx .
tan x sec 4 x dx 

tan x sec 2 x sec 2 x dx 
Let u  tan x  du  sec 2 x dx 

tan x tan 2 x sec 2 x dx 



tan x (1  tan 2 x) sec 2 xdx .
tan x sec 4 x dx 


tan x sec 2 x dx 
5
1
2 3
2 7
u 2 du  u 2 du  u 2  u 2  C 
3
7
3
7
2
2
(tan x) 2  (tan x) 2  C .
3
7
9. Find

sec x tan x dx .
Let u  sec x  u 2  sec x  2udu  sec x tan xdx  u 2 tan xdx 
2udu 2
2 
tan x dx  2  du . Thus,
sec x tan x dx  u du   2 1du 
u
u
u 

2u  C  2 sec x  C .
6


Practice Sheet forTrigonometric Integrals
(1) Prove the reduction formula:

1
n 1
sin n x dx   sin n1 x cos x 
sin n2 x dx
n
n
(2) Prove the reduction formula:

cos n x dx 
(3) Prove the reduction formula:

sec n  2 x tan x n  2
sec x dx 

sec n  2 x dx
n 1
n 1
(4) Prove the reduction formula:

tan n x dx 

(5)


n
4

tan 3 (3 x ) dx =
0

(6)
4

cos 2 (2 x) dx =
0

8

sin( 5 x) cos(3x) dx =
(8)

tan 3 x sec 3 x dx =
(9)

sin x cos 3 x dx =
(7)

1
n 1
cos n1 x sin x 
cos n2 x dx
n
n
0
7
tan n 1 x
 tan n  2 x dx
n 1

(10)


(11)
cos 3 x sin 2 x dx =
2

sin 3 x
cos x
dx =
0
(12)

sin 2 x cos 2 xdx 
(13)

tan 5 x sec xdx 
Solution Key for Trigonometric Integrals
(1)

sin n x dx 

sin n1 x sin x dx . Use integration by parts with u  sin n 1 x and
dv  sin x dx  du  (n  1) sin n2 x cos x dx and v 
 sin x dx   sin
n
n 1
 sin x dx   cos x 

x sin x dx =  sin n1 x cos x  (n  1) sin n2 x cos 2 x dx 



 sin n1 x cos x  (n  1) sin n2 x 1  sin 2 x dx   sin n1 x cos x 



(n  1) sin n2 x dx  (n  1) sin n x dx  n sin n x dx   sin n1 x cos x 



1
n 1
(n  1) sin n2 x dx  sin n x dx   sin n1 x cos x 
sin n2 x dx .
n
n
(2)
 cos x dx   cos
n
n 1
x cos x dx . Use integration by parts with u  cos n 1 x and
8
dv  cos x dx  du  (n  1) cos n2 x ( sin x) dx and v 
 cos x dx  sin x 


cos n x dx 

cos n1 x cos x dx = cos n1 x sin x  (n  1) cos n2 x sin 2 x dx 



cos n1 x sin x  (n  1) cos n2 x 1  cos 2 x dx  cos n1 x sin x 



(n  1) cos n2 x dx  (n  1) cos n x dx  n cos n x dx  cos n1 x sin x 


(n  1) cos n2 x dx  cos n x dx 
(3)


1
n 1
cos n1 x sin x 
cos n2 x dx .
n
n

sec n x dx  sec n2 x sec 2 x dx . Use integration by parts with u  sec n  2 x and
dv  sec 2 x dx  du  (n  2) sec n3 x (sec x tan x dx) and v 
 sec x dx   sec
n
n2

sec 2 x dx  tan x 

x sec 2 x dx = sec n2 x tan x  (n  2) sec n2 x tan 2 x dx 




sec n2 x tan x  (n  2) sec n2 x sec 2 x  1 dx  sec n2 x tan x  (n  2) sec n xdx 



(n  2) sec n2 x dx  (n  1) sec n x dx  sec n2 x tan x  (n  2) sec n2 x dx 

(4)
sec n x dx 
sec n  2 x tan x n  2

sec n  2 x dx .
n 1
n 1

 tan x dx   tan
n

tan n  2 x dx 
n2
x tan 2 x dx 
tan n 1 x

n 1
 tan
n2
 tan
n2
x dx .
9


x sec 2 x  1 dx 
 tan
n2
x sec 2 xdx 
(5)
Let u  3x  du  3 dx 

tan 3 (3x) dx 
reduction formula #4 above to get

1
1
tan 2 u  ln sec u 
6
3
4



1
1
tan 3 (3x) 3dx 
tan 3 u du . Use
3
3
1
1  tan 2 u  1

tan 3u du  
tan u du 
3
3  2  3



1
1
 4
tan (3 x ) dx =  tan 2 (3x)  ln sec(3x)  
3
6
0
3
0
1
1
 3  
2  3 
 tan    ln sec   
 4  3
 4 
6
1 2 1
1 1
(0)  ln 1   ln
6
3
6 3
1
1
1
 1
2
2
 tan 0  ln sec0   (1)  ln  2 
3
3
6
 6
 2 .
(6) Use the trigonometric identity cos 2  


1  cos 2
to get
2


cos 2 (2 x) dx 
1  cos(4 x)
1
1
1
1
dx 
1dx 
cos(4 x) dx  x  sin 4 x  
2
2
2
2
8

4

cos 2 (2 x) dx =
0
1    1
 1
1

    sin     (0)  sin( 0)   .
8
8
2  4  8
 2
1
(7) Use the trigonometric identity sin x cos y  [sin( x  y )  sin( x  y )] to get
2


8

0
sin( 5x) cos(3x) dx 


1
1
1
1
sin( 2 x) dx 
sin( 8x) dx   cos( 2 x)  cos(8x) 
2
2
4
16
 1
  1
1
  1

sin( 5 x) cos(3x) dx   cos   cos    cos 0  cos 0 
16
 4  16

 4
  4
1 2  1 1 1 3 2
  
 

4  2  16 4 16
8
10
(8)
(9)

tan 3 x sec 3 x dx =

(sec 2 x  1) sec 2 x (sec x tan x dx)  sec 4 x (sec x tan x dx) 

1
1
sec 2 x (sec x tan x dx)  sec 5 x  sec 3 x  C .
5
3
(10)


1

5
2
cos x dx 
 cos x sin
cos 3 x sin 2 x dx =
2
2
1
2
1  sin xcos xdx 
2
3
7
2
2
(sin x) 2  (sin x) 2  C .
3
7
x cos x dx  
 1  sin
2


x sin 2 x cos x dx 

sin 3 x
cos x
(cos x)

dx 
1
2

(cos x)
1

2
sin 2 x (sin x dx 

(cos x)
1
3
(sin x dx)  (cos x) 2 (sin x dx)  2(cos x)
2
1
1  cos xsin x dx 
2
2
5
2
 (cos x) 2 
5
5


5 
8
2

   2     2  
dx =  2 cos    cos      2 cos 0  cos 0  2   .
5
cos x
 2  5   2   
 5



sin 3 x
(12) Use the trigonometric identities cos 2  


(sin x)
1
1
sin 2 x (cos x dx)  sin 4 x (cos x dx)  sin 3 x  sin 5 x  C .
3
5
2
0

sin x cos 2 x ( cos x dx) 
(sin x) 2 cos x dx  (sin x)



sin x cos 3 x dx =


(11)




tan 2 x sec 2 x (sec x tan x dx) 
sin 2 x cos 2 xdx 

1  cos 2
1  cos 2
and sin 2  
.
2
2
1
 1  cos 2 x  1  cos 2 x 


 dx 
2
2
4



11
 1  cos 2xdx 
2




1
1
1
1  1  cos 4 x 
1
1
1 dx 
cos 2 2 x dx  x 
1 dx 

 dx  x 
4
4
4
4 
2
4
8


1
1
1
1
1
1
cos 4 x dx  x  x  sin 4 x  C  x  sin 4 x  C .
8
4
8
32
8
32
(13)

tan 5 x sec xdx 
 sec

2

tan 4 x tan x sec x dx 

2
x  1 sec x tan x dx 
 tan x
2
2
tan x sec x dx 
 sec x  2 sec x  1sec x tan x dx 
4
2


sec 4 x sec x tan x dx   2 sec 2 xsec x tan x dx   sec x tan x dx 
1
2
sec 5 x  sec 3 x  sec x  C .
5
3
12