Download Lesson 9 Integration by Trigonometric Substitution

LESSON 9 INTEGRATION BY TRIGONOMETRIC SUBSTITUTION
2
2 n/2
If the integrand of an integral contains an expression of the form ( a  x ) ,
where a is a positive real number and n is an odd integer, then the domain of the
integrand is either the closed interval [  a , a ] or is a subset of this interval. Then


try the trigonometric substitution: Let x  a sin  , where     . Then
2
2
dx  a cos  d . Note that the range of the function x  a sin  is the closed
2
2
2
2
2
2
2
interval [  a , a ] . Also, we have that a  x = a  ( a sin  ) = a  a sin  =
a 2 (1  sin 2  ) = a 2 cos 2  . Thus, ( a 2  x 2 ) n / 2 = ( a 2 cos 2  ) n / 2 = ( a 2 cos 2  ) n =
( a cos  ) n = a cos  . Since a  0 , then a  a . Since cos   0 whenever


    , then cos   cos  . Thus, ( a 2  x 2 ) n / 2 = a n cos n  .
2
2
n
n
If x  a sin  , then sin  
x
. Using right triangle trigonometry, we have that
a
a
x

a2  x2
cos  
a2  x2
, sec  
a
a
a2  x2
, tan  
x
a2  x2
, cot  
a2  x2
x
2
2 n/2
If the integrand of an integral contains an expression of the form ( a  x ) ,
where a is a positive real number and n is an integer, then the domain of the
integrand is the set of all real numbers. Then try the trigonometric substitution:


2
Let x  a tan  , where     . Then dx  a sec  d . Note that the range
2
2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
of the function x  a tan  is the set of all real numbers. Also, we have that
a 2  x 2 = a 2  ( a tan  ) 2 = a 2  a 2 tan 2  = a 2 (1  tan 2  ) = a 2 sec 2  . Thus,
( a 2  x 2 ) n / 2 = ( a 2 sec 2  ) n / 2 = ( a 2 sec 2  ) n = ( a sec  ) n = a sec  .


Since a  0 , then a  a . Since sec   1  0 whenever     , then
2
2
2
2 n/2
n
n
sec   sec  . Thus, ( a  x )
= a sec  .
n
If x  a tan  , then tan  
n
x
. Using right triangle trigonometry, we have that
a
a2  x2
x

a
cos  
a
a2  x2
, sec  
a2  x2
, sin  
a
x
a2  x2
, csc  
a2  x2
x
2
2 n/2
If the integrand of an integral contains an expression of the form ( x  a ) ,
where a is a positive real number and n is an odd integer, then the domain of the
integrand is the set (   ,  a ]  [ a ,  ) or is a subset of this set. Then try the

3
trigonometric substitution: Let x  a sec  , where 0   
or    
.
2
2
Then dx  a sec  tan  d . Note that the range of the function x  a sec  is the set
(   ,  a ]  [ a ,  ) . Also, we have that x 2  a 2 = ( a sec  ) 2  a 2 =
a 2 sec 2   a 2 = a 2 ( sec 2   1) = a 2 tan 2  . Thus, ( x 2  a 2 ) n / 2 = ( a 2 tan 2  ) n / 2
tan  . Since a  0 , then a  a . Since

3




tan   0 whenever 0   
or
, then tan   tan  . Thus,
2
2
( x 2  a 2 ) n / 2 = a n tan n  .
2
2
n
n
= ( a tan  ) = ( a tan  ) = a
n
n
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
If x  a sec  , then sec  
x
. Using right triangle trigonometry, we have that
a
x
x2  a2
a
sin  
x2  a2
, csc  
x
x
x2  a2
, tan  
x2  a2
, cot  
a
a
x2  a2
Examples Evaluate the following integrals.
1.

dx
x2  9
Note that we do not have the x in the integral to do the simple substitution to
2
let u  x  9 . We will use the technique of Trigonometric Substitution.
Let x  3 tan  , where 


   . Then dx  3sec 2  d .
2
2
2
2
2
2
Also, x  9  9 tan   9 = 9 ( tan   1 ) = 9 sec  .
x 2  9  9 sec 2   3 sec   3sec  since sec   1  0


   . Thus, we have that
when 
2
2
Thus,

dx
x2  9
=

3 sec 2  d
=
3 sec 
 sec  d
= ln sec  tan   c1
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Since x  3 tan  , then tan  
have that sec  
x
. Using right triangle trigonometry, we
3
x2  9
.
3
x2  9
x

3
Thus,
dx

x2  9
x2  9
ln
 sec  d
= ln sec   tan   c1 =
x 
x

 c1 = ln
3
3
ln x 
=
x2  9
x2  9
3
 ln 3  c1 = ln x 
x2  9
c  c 1  ln 3
Answer: ln x 
2.

3
2x3
2
25  x
2
x2  9
 c1 =
 c
dx
Copyrighted by James D. Anderson, The University of Toledo
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 c , where
Since the function f ( x ) 
2x3
25  x 2
is continuous on its domain of
definition, which is the open interval (  5 , 5 ) , then it is continuous on the
closed interval [  2 , 3 ] . Thus, the Fundamental Theorem of Calculus can be
applied.

2x3
dx using the
25  x 2
technique of Trigonometric Substitution. Then we will apply the
Fundamental Theorem of Calculus with one of these antiderivatives.
We will first evaluate the indefinite integral
Let x  5 sin  , where 


   . Then dx  5 cos  d .
2
2
2
2
2
2
Also, 25  x  25  25 sin  = 25 ( 1  sin  ) = 25 cos  .
25  x 2  25 cos 2   5 cos  5 cos since cos  0






when
. Thus, we have that
2
2
Thus,

2x3
25  x 2
2 ( 125 sin 3  ) ( 5 cos  d )
3
dx = 
= 250  sin  d =
5 cos 
250  sin 2  sin  d = 250  ( 1  cos 2  ) sin  d
Let u  cos  . Then du   sin  d
250  ( 1  cos 2  ) sin  d =  250  ( 1  cos 2  ) (  1 ) sin  d =
 u3


u3 
  c = 250 
 u   c =
 250  ( 1  u ) du =  250  u 
3 

 3

2
250
250
u ( u 2  3)  c =
cos  ( cos 2   3 )  c
3
3
Copyrighted by James D. Anderson, The University of Toledo
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Since x  5 sin  , then sin  
have that cos  
x
. Using right triangle trigonometry, we
5
25  x 2
.
5
5
x

25  x 2
Thus,
250

3
2

3

2x3
25  x 2
25  x 2
5
25  x 2
1
dx = 250 sin 3  d =

250
cos  ( cos 2   3 )  c =
3
2
 25  x 2
75 
50 25  x   x 2  50 

  c =

  c =


3
1
25
25
25




  x 2  50 
2

  c =  ( x 2  50 ) 25  x 2  c
1
3


Now, that we have all the antiderivatives of the integrand f ( x ) 
2x3
25  x 2
we can apply the Fundamental Theorem of Calculus to evaluate the definite
3
2x3
dx .
integral   2
2
25  x


3
2
2x3
25  x 2
,
3
2
dx =  ( x 2  50 ) 25  x 2 
=
3
 2
2
2
4
( 59 16  54 21 ) =  ( 236  54 21 ) =  ( 118  27 21 ) =
3
3
3
Copyrighted by James D. Anderson, The University of Toledo
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4
( 27 21  118 )
3
Answer:
3.

0
7
4
( 27 21  118 )
3
49  t 2 dt
Note that we do not have the t in the integral to do the simple substitution to
2
let u  49  t .
2
Since the function f ( t )  49  t is continuous on its domain of
definition, which is the closed interval [  7 , 7 ] , then it is continuous on the
closed interval [  7 , 0 ] . Thus, the Fundamental Theorem of Calculus can be
applied.
However, the value of this definite integral can be obtained using the area of
2
the region bounded by the graph of the integrand y  49  t and the
horizontal t-axis on the closed interval [  7 , 0 ] . Since the graph of the
2
function y  49  t is the top-half of the circle given by the equation
t 2  y 2  49 . The center of the circle is the origin and the radius is 7. Thus,
2
the region bounded by the graph of y  49  t and the t-axis on the
interval [  7 , 0 ] is the upper left quarter of the circle. Since the area of the
2
circle is given by  r and r  7 , then the area of the circle is 49 . Thus,
0
49 
49 
2
the area of the quarter circle is
. Thus,   7 49  t dt =
.
4
4
To apply the Fundamental Theorem of Calculus, we will first evaluate the

49  t 2 dt using the technique of Trigonometric
Substitution in order to obtain all the antiderivatives of the integrand
indefinite integral
f (t ) 
49  t 2 .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Let t  7 sin  , where 


   . Then dt  7 cos  d .
2
2
2
2
2
2
Also, 49  t  49  49 sin  = 49 ( 1  sin  ) = 49 cos  .
49  t 2  49 cos 2   7 cos  7 cos since cos  0


   . Thus, we have that
when 
2
2
Thus,

49  t 2 dt =
 ( 7 cos  ) ( 7 cos  d )
2
= 49  cos  d =
sin 2  
 
49 

  c
4
 2

Since t  7 sin  , then sin  
t
1 t


sin
and
. Using right triangle
7
7
trigonometry, we have that cos 
49  t 2
.
7
7
t

49  t 2
By the double angle formula for sine, we have that sin 2   2 sin  cos  .
 t  
Thus, sin 2  2  7  
 
49  t 2 
2t
 =
7

49  t 2
49
Copyrighted by James D. Anderson, The University of Toledo
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.
Thus,
sin 2  
 

49  t 2 dt = 49  cos 2  d = 49 
  c =
4
 2


 1
1  2 t
1 t
49  sin

2
7
4



 1
1  t
1 t
49  sin

2
7
2




 sin  1 t  t

7

49
2
49  t 2  
  c =

49

49  t 2  
  c =

49

49  t 2 
 c

49

2
Now, that we have all the antiderivatives of the integrand f ( t )  49  t ,
we can apply the Fundamental Theorem of Calculus to evaluate the definite
integral

0
7
49  t 2 dt .
0

0
7
49  t 2
t 49  t 2  
49   1 t

sin

dt =
=

2 
7
49

 7
49
49
[ sin  1 0  0  ( sin  1 (  1)  0 ) ] =
[ sin  1 0  sin  1 (  1) ] =
2
2
49
2

 
0



 2

49   

49 
 =
  =
2 2
4

49 
Answer:
4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
4.

6x3
8
x  16
2
5
dx
Since the function f ( x ) 
6x3
is continuous on its domain of
x 2  16
definition, which is the set (   ,  4 )  ( 4 ,  ) , then it is continuous on the
closed interval [ 5 , 8 ] . Thus, the Fundamental Theorem of Calculus can be
applied.
We will first evaluate the indefinite integral

6x3
dx using the
x  16
technique of Trigonometric Substitution. Then we will apply the
Fundamental Theorem of Calculus with one of these antiderivatives.
Let x  4 sec  , where 0   
dx  4 sec  tan  d .
2

3
or    
. Then
2
2
2
2
2
2
Also, x  16  16 sec   16 = 16 ( sec   1) = 16 tan  .
x 2  16  16 tan 2   4 tan   4 tan  since tan   0

3
when 0   
or    
. Thus, we have that
2
2
Thus,

6 ( 64 sec 3  ) ( 4 sec  tan  d )
4
dx = 
384
sec
 d =
=

4 tan 
x 2  16
6x3
384  sec 2  sec 2  d = 384  ( 1  tan 2  ) sec 2  d
2
Let u  tan  . Then du  sec  d

u3 
  c =
384  ( 1  tan  ) sec  d = 384  ( 1  u ) du = 384  u 
3


2
2
2
Copyrighted by James D. Anderson, The University of Toledo
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384
u ( u 2  3 )  c = 128 tan  ( tan 2   3 )  c
3
Since x  4 sec  , then sec  
have that tan  
x
. Using right triangle trigonometry, we
4
x 2  16
.
4
x
x 2  16

4
6x3
Thus,

128 
x 2  16  x 2  16 48 

  c = 32 

4
16
16


2
x  16
2
dx = 384 sec 4  d = 128 tan  ( tan 2   3 )  c =

x 2  16  x 2  32 

  c =
1
16


x 2  16  x 2  32 

  c = 2 ( x 2  32 )
1
1


x 2  16  c
Now, that we have all the antiderivatives of the integrand f ( x ) 
6x3
x 2  16
we can apply the Fundamental Theorem of Calculus to evaluate the definite
8
6x3
dx .
integral  5
2
x  16

8
5
6x3
x 2  16
dx = 2 ( x 2  32 )
x  16
2

8
5
=
Copyrighted by James D. Anderson, The University of Toledo
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,
2 ( 96 48  57 9 ) = 2 ( 96 16  3  171 ) = 2 ( 384 3  171 ) =
768 3  342
Answer: 768 3  342
5.

y 5 ( 9  7 y 2 ) 3 / 2 dy
We want to rewrite the integral before we use the technique of Trigonometric
Substitution.
y
y
5
5
( 9  7 y 2 ) 3 / 2 dy =
7
3/ 2
Let y 
9
2 
  y 
7

3
7

 9

y 7   y2 

 7
3/ 2
5
3/ 2
dy = 7
3/ 2

dy =
9

y   y2 
7

3/ 2
5
dy


   . Then dy 
2
2
tan  , where 
3
7
sec 2  d .
9
9
9 2
9
9
2
2
2

y


tan

(
1

tan

)
Also,
=
= sec  .
7
7
7
7
7
9
2 
Thus,   y 
7

3/ 2
9

  sec 2  
7

since sec   1  0 when 
y
5
( 9  7 y 2 ) 3 / 2 dy = 7
3/ 2

27
sec 
7 3/ 2
3

27
sec 3 
3/ 2
7


   . Thus, we have that
2
2
3/ 2

9

y   y2 
7

3/ 2
5
Copyrighted by James D. Anderson, The University of Toledo
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dy =
7
3/ 2

35
tan 5 
5/ 2
7
7 3/ 2 3 9
7 9/2
39
= 3
7


39
73
 ( sec
 3
 1 / 2 sec 2  d =
7
39
tan  sec  d = 3
7
5
 tan
39
73
 33
 3 / 2 sec 3 
7
4
5
 tan
5
 sec 5  d =
 sec 4  sec  tan  d =
( tan 2  ) 2 sec 4  sec  tan  d =
2
  1 ) 2 sec 4  sec  tan  d
Let u  sec  . Then du  sec  tan  d
39
73
39
( sec   1 ) sec  sec  tan  d = 3
7

39
73
 (u
39
73
 u9
2u 7
u5 

  c =


9
7
5


2
4
2
 2u
2
4
39
 1 ) u du = 3
7
4
 (u
8
 (u
2
 1 ) 2 u 4 du =
 2 u 6  u 4 ) du =
39  1
2
1

sec 9   sec 7   sec 5    c
3 
7 9
7
5

Since y 
3
7
tan  , then tan  
trigonometry, we have that sec  
7 y
. Using right triangle
3
9  7y2
3
.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
9  7y2
7y

3
Thus,
y
5
( 9  7 y 2 ) 3 / 2 dy = 7

9

y   y2 
7

5
3/ 2
dy =
39  1
2
1

tan  sec  d = 3  sec 9   sec 7   sec 5    c =
7 9
7
5

39
73

39
73
 1 ( 9  7 y 2 ) 9/2
2 ( 9  7 y 2 ) 7/2
1 ( 9  7 y 2 ) 5/ 2
 
 
 
9
7
9
3
7
3
5
35


  c =

39 1  1 ( 9  7 y 2 ) 9/2
2 ( 9  7 y 2 ) 7/2
1 ( 9  7 y 2 ) 5/ 2
 

 
 
7 3 3 5  9
34
7
32
5
1

  c =

34
73
5
5
 1 ( 9  7 y 2 ) 9/2
2 ( 9  7 y 2 ) 7/2
( 9  7 y 2 ) 5/ 2
 
 

9
81
7
9
5

81
343
 ( 9  7 y 2 ) 9/2
2( 9  7 y 2 ) 7/2
( 9  7 y 2 ) 5/ 2



729
63
5

81
Answer:
343
6.
3/ 2
 5x

  c =


  c

 ( 9  7 y 2 ) 9/2
2( 9  7 y 2 ) 7/2
( 9  7 y 2 ) 5/ 2



729
63
5

4 x 2  11 dx
Copyrighted by James D. Anderson, The University of Toledo
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
  c

This integral can be evaluated by simple substitution which is FASTER than
trigonometric substitution.
2
Let u  4 x  11
Then du  8 x dx

5x
5
8

4 x 2  11 dx = 5  x
u du =
5
8

u 1 / 2 du =
4 x 2  11 dx =
5
8
 8x
4 x 2  11 dx =
5 2 3/ 2
5
 u
 c =
( 4 x 2  11 ) 3 / 2  c
8 3
12
If you missed the simple substitution and proceeded with the technique of
Trigonometric Substitution, then your work would have been the following
(and the following, and the following.)

5x
4 x 2  11 dx =
11
Let x 
dx 
2
11
2
2
Also, x 
 5x
11 

4 x 2 
 dx = 10  x
4 

sec  , where 0   
x2 
11
dx
4

3
or    
. Then
2
2
sec  tan  d .
11
11
11
11
11

sec 2  
( sec 2   1 ) =
tan 2  .
=
4
4
4
4
4
11
11
11
2
tan


tan


tan  since
Thus,
4
2
2

3




tan   0 when 0   
or
. Thus, we have that
2
2
x2 

5x
11

4
4 x 2  11 dx = 10  x
x2 
11
dx =
4
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10 
11
2
55 11
4

sec  


 tan
2
11
2

tan  


11
2
sec  tan  d =
 sec 2  d
2
Let u  tan  . Then du  sec  d
55 11
4
55 11
12
 tan
2
 sec  d =
2
55 11
4
u
2
du =
55 11
tan 3   c
Since x 
11
2
sec  , then sec  
trigonometry, we have that tan  
2x
11
. Using right triangle
4 x 2  11
11
.
2x
4 x 2  11

11

5x
55 11

Thus,
4
4
u3

 c =
3
4 x 2  11 dx = 10  x
tan 2  sec 2  d =
55 11
12
x2 
11
dx =
4
tan 3   c =
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55
11
12

( 4 x 2  11 ) 3 / 2
11 11
 c =
5
( 4 x 2  11 ) 3 / 2  c
12
5
( 4 x 2  11 ) 3 / 2  c
Answer:
12
7.
t
3
( 6  t 2 ) 5 / 2 dt
6 sin  , where 
Let t 


   . Then dt 
2
2
6 cos  d .
2
2
2
2
Also, 6  t  6  6 sin  = 6 ( 1  sin  ) = 6 cos  .
2 5/ 2
 ( 6 cos 2  ) 5 / 2  6 5 / 2 cos 
Thus, ( 6  t )
cos  0 when 
t
3
 sin
6 9/2

3
 6 5 / 2 cos 5  since


   . Thus, we have that
2
2
( 6  t 2 ) 5 / 2 dt =
6 9/2
5
6
3/ 2
sin 3  ( 6 5 / 2 cos 5  ) 6 1 / 2 cos  d =
 cos 6  d = 6 9 / 2
 sin
2
 cos 6  sin  d =
( 1  cos 2  ) cos 6  sin  d
Let u  cos  . Then du   sin  d
6 9/2
 (1 
cos 2  ) cos 6  sin  d =
 6 9/2
 (1 
cos 2  ) cos 6  (  1) sin  d =  6 9 / 2
 6 9/2
 (u
 u 8 ) du = 6 9 / 2
6
 (u
8
 u 6 ) du =
Copyrighted by James D. Anderson, The University of Toledo
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 (1 
u 2 ) u 6 du =
6
9/2
 u9
u7 
1
1


  c = 6 9 / 2  cos 9   cos 7    c

7 
7
9

 9
Since t 
t
6 sin  , then sin  
we have that cos  
. Using right triangle trigonometry,
6
6  t2
.
6
6
t

6  t2
Thus,
t
3
( 6  t 2 ) 5 / 2 dt = 6 9 / 2
 sin
3
 cos 6  d =
1
1

6 9 / 2  cos 9   cos 7    c =
7
9

6
9/2
 1 ( 6  t 2 ) 9/2
1 ( 6  t 2 ) 7/2
 
 
9/2
9
6
7
6 7/2


  c =

1
6
( 6  t 2 ) 9/2  ( 6  t 2 ) 7/2  c
9
7
Answer:
1
6
( 6  t 2 ) 9/2  ( 6  t 2 ) 7/2  c
9
7
Copyrighted by James D. Anderson, The University of Toledo
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8.
 (9 y
2
dy
 25 ) 3 / 2
Note that we do not have the y in the integral to do the simple substitution to
2
let u  9 y  25 . We will use the technique of Trigonometric
Substitution. First, we will need to rewrite the integral.
 (9 y
1
27
2
dy

25  
 2
9
y




9  
 
3/ 2
=
dy

9
3/ 2
25 
 2
y 

9 

3/ 2
dy

y

Let y 
dy 
dy
=
 25 ) 3 / 2
25 


9 
2
3/ 2

3
5
sec  , where 0   
or    
. Then
3
2
2
5
sec  tan  d .
3
2
Also, y 
25
25
25
25
25

sec 2  
( sec 2   1 ) =
tan 2  .
=
9
9
9
9
9
25 
 2

Thus,  y 
9 

3/ 2
3/ 2
25 3 / 2
125
3
 25

2
 
tan  
 3 / 2 tan  
tan 3 
9
27
 9


3
since tan   0 when 0   
or    
. Thus, we have that
2
2
 (9 y
2
1
dy
= 27
 25 ) 3 / 2

y

dy
2
25 


9 
3/ 2
=
Copyrighted by James D. Anderson, The University of Toledo
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=
1
27

5
sec  tan  d
1 5 1
3
= 27  3  125
125
3
tan 
27
27
1 5 27
 
27 3 125
1
75



sec 
d =
tan 2 
1 1 1
sec  cot  d =  
1 3 25
2

1
cos 2 

d =
cos  sin 2 
cos 
d
sin 2 
Let u  sin  . Then du  cos  d
1
75


cos 
1
d =
2
sin 
75

du
1
2 =
u
75
u
2
1 u 1

 c =
du =
75  1
1
1
1 1
1

 c = 
  c = 
csc   c
75 sin 
75 u
75
5
3y
sec  , then sec  
. Using right triangle trigonometry,
3
5
3y
csc


we have that
.
9 y 2  25
Since y 
3y
9 y 2  25

5
Thus,
 (9 y
2
dy
1
=
 25 ) 3 / 2
75

cos 
1
d


csc   c =
=
sin 2 
75
Copyrighted by James D. Anderson, The University of Toledo
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
1

75
3y
9y
Answer: 
9.
2
 c = 
 25
y
25
9y
 25
2

( x 2  4) 2
dx
3x

( x 2  4) 2
1
dx =
3x
3
1
3


y
25
9y
2
 25
 c
 c
x 4  8 x 2  16
dx =
x

1  x4
16 
 3

 4 x 2  16 ln x   c =
 x  8x 
 dx =
3 4
x 


1 4
4
16
x  x2 
ln x  c
12
3
3
If you evaluate the integral using the technique of Trigonometric
Substitution, then your work would have been the following.
Let x  2 tan  , where 


   . Then dx  2 sec 2  d .
2
2
2
2
2
2
Also, x  4  4 tan   4 = 4 ( tan   1 ) = 4 sec  .
2
2
2
2
4
Thus, ( x  4 )  ( 4 sec  )  16 sec  . Thus, we have that

( x 2  4) 2
dx =
3x

16 sec 4  ( 2 sec 2  d )
16
=
3 ( 2 tan  )
3
Copyrighted by James D. Anderson, The University of Toledo
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 ( tan  )
1
sec 6  d =
16
3

16
3
 ( tan  )
16
3
( tan  )  1 sec 4  sec 2  d =
1
 ( tan  )
( sec 2  ) 2 sec 2  d
1
( 1  tan 2  ) 2 sec 2  d
2
Let u  tan  . Then du  sec  d
16
3

( tan  )  1 ( 1  tan 2  ) 2 sec 2  d =
16
3

u  1 ( 1  2 u 2  u 4 ) du =
16
3
 u4


 u 2  ln u   c =
 4

16
3
 (u
3
16
3
u
1
( 1  u 2 ) 2 du =
 2 u  u  1 ) du =
16  1

4
2
 tan   tan   ln tan    c
3 4

Since x  2 tan  , then tan  
Thus,

( x 2  4) 2
16
dx =
3x
3
x
.
2
 ( tan  )
1
sec 6  d =
16  1

4
2
 tan   tan   ln tan    c 1 =
3 4

16
3
 1 x4
x2
x
 

 ln
4
2
 4 16

  c 1 =

16
3
 x4

x2


 ln x  ln 2   c 1 =
4
 64

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1 4
4 2
16
16
x 
x 
ln x 
ln 2  c 1 =
12
3
3
3
1 4
4 2
16
16
x 
x 
ln x  c , where c  c 1 
ln 2
12
3
3
3
1 4
4
16
x  x2 
ln x  c
12
3
3
Answer:
Example If a is a positive real number, b is any real number, m is a rational
number, and n is an odd integer, then determine the trigonometric integral for the
integral
 bx
m
( a 2  x 2 ) n / 2 dx if you do the trigonometric substitution to let
x  a sin  , where 


  .
2
2
Since x  a sin  , then dx  a cos  d .
a 2  x 2 = a 2  ( a sin  ) 2 = a 2  a 2 sin 2  = a 2 (1  sin 2  ) = a 2 cos 2  .
2
2
n
n
2
2 n/2
2
2
n/2
Thus, ( a  x )
= ( a cos  )
= ( a cos  ) = ( a cos  ) =


n
n
a cos  . Since a  0 , then a  a . Since cos   0 whenever     ,
2
2
2
2 n/2
n
n
then cos   cos  . Thus, ( a  x )
= a cos  .
Thus,
 bx (a  x
b  sin  cos
m
am  n 1
Answer:
2
m
 bx
m
2
) n / 2 dx = b  a m sin m  ( a n cos n  ) a cos  d =
n 1
 d .
( a 2  x 2 ) n / 2 dx = a m  n  1 b
 sin
m
 cos n  1  d
For example, in our Example 2 above, we had the integral
Copyrighted by James D. Anderson, The University of Toledo
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
3
2x3
2
25  x
2
dx .
For this integral, a  5 , b  2 , m  3 , and n   1 . Thus,
5 3 2  sin 3  d = 250
 sin
3
2x3

25  x
2
dx =
 d .
Example If a is a positive real number, b is any real number, m is a rational
number, and n is an integer, then determine the trigonometric integral for the
integral
 bx
m
( a 2  x 2 ) n / 2 dx if you do the trigonometric substitution to let
x  a tan  , where 


  .
2
2
2
Since x  a tan  , then dx  a sec  d .
a 2  x 2 = a 2  ( a tan  ) 2 = a 2  a 2 tan 2  = a 2 (1  tan 2  ) = a 2 sec 2  .
2
2
n
n
2
2 n/2
2
2
n/2
Thus, ( a  x )
= ( a sec  )
= ( a sec  ) = ( a sec  ) =
sec  . Since a  0 , then a  a . Since sec   1  0 whenever


    , then sec   sec  . Thus, ( a 2  x 2 ) n / 2 = a n sec n  .
2
2
a
n
Thus,
n
 b x (a  x
b  tan  sec
m
a m  n 1
Answer:
2
m
 bx
m
2
) n / 2 dx = b  a m tan m  ( a n sec n  ) a sec 2  d =
n2
 d .
( a 2  x 2 ) n / 2 dx = a m  n  1 b  tan m  sec n  2  d
For example, in our Example 1 above, we had the integral
integral, a  3 , b  1 , m  0 , and n   1 . Thus,
3 0 (1)
 tan
0
 sec  d =

 sec  d .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

dx
x2  9
dx
=
x2  9
. For this
Example If a is a positive real number, b is any real number, m is a rational
number, and n is an odd integer, then determine the trigonometric integral for the
integral
 bx
m
( x 2  a 2 ) n / 2 dx if you do the trigonometric substitution to let
x  a sin  , where 0   

3
or    
.
2
2
Since x  a sec  , then dx  a sec  tan  d .
x 2  a 2 = ( a sec  ) 2  a 2 = a 2 sec 2   a 2 = a 2 ( sec 2   1) = a 2 tan 2  .
2
2
n
n
2
2 n/2
2
2
n/2
Thus, ( x  a )
= ( a tan  )
= ( a tan  ) = ( a tan  ) =
a
n
n
tan  . Since a  0 , then a  a . Since tan   0 whenever 0   
or    
Thus,

3
2
2 n/2
n
n
, then tan   tan  . Thus, ( x  a )
= a tan  .
2

2
b x m ( x 2  a 2 ) n / 2 dx = b  a m sec m  ( a n tan n  ) a sec  tan  d =
a m  n  1 b  tan n  1  sec m  1  d .
Answer:
 bx
m
( x 2  a 2 ) n / 2 dx = a m  n  1 b  tan n  1  sec m  1  d
For example, in our Example 8 above, we had the integral

dy
25  
 2
9  y  9 

 
the integral

3/ 2
=
dy

9
3/ 2
dy
25 

27  y 2 

9 

3/ 2
25 
 2
y 

9 

,a 
3/ 2
=

 (9 y
2
dy
=
 25 ) 3 / 2
dy
25 

27  y 2 

9 

3/ 2
. For
5
1
,b 
, m  0 , and n   3 . Thus,
3
27
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

dy
3/ 2
5
=  
3
2
25 

27  y 2 

9 

1
9 1
2

cot

sec

d

=
75
25 27 
1
27

 ( tan  )
2
sec  d =
1
cos 2 
1

d

=
75
sin 2  cos 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

cos 
d
sin 2 