Inverse Trigonometric Functions:
Integration
Lesson 5.8
Review
• Recall derivatives of inverse trig functions
d
1 du
1
sin u 
, u 1
dx
1  u 2 dx
d
1 du
1
tan u 
dx
1  u 2 dx
d
1
du
1
sec u 
, u 1
2
dx
u u  1 dx
2
Integrals Using Same
Relationships
du
u
 a 2  u 2  arcsin a  C
du
1
u
 a 2  u 2  a arctan a  C
du
1
u
 u u 2  a 2  a arcsec a  C
When given
integral problems,
look for these
patterns
3
Identifying Patterns
• For each of the integrals below, which
inverse trig function is involved?
4dx
 13  16 x 2

x
dx
25 x 2  4
dx
9 x
2
4
Warning
• Many integrals look like the inverse trig
forms
• Which of the following are of the inverse
trig forms?

x dx
 1  x2
dx
 1  x2

x dx
1  x2
dx
1 x
2
If they are not,
how are they
integrated?
5
Simplify.
1.
2.

x
dx
4  x2
dx
1
4 x  64
2
dx
The hardest part is getting the integral into the proper form.

dx
4 x

2
a2

du
a u
2
2
 sin 1  ua   C
ux
du  dx
du
a2  u2
 sin
  C
1 u
a
 x
 sin    C
2
1
The hardest part is getting the integral into the proper form.

dx
9 x

2
a 3

du
a u
2
2
 sin 1  ua   C
ux
du  dx
du
a2  u2
 sin
  C
1 u
a
 x
 sin    C
3
1
The hardest part is getting the integral into the proper form.

xdx
1 x
4

du
a u
a 1
1
du

2
a2  u2
1 1 u
 sin  a   C
2
2
2
 sin 1  ua   C
ux
du  2xdx
2
 
1 1 2
 sin x  C
2
The hardest part is getting the integral into the proper form.

dx
3  4x
2

du
a u
a 3
1
du

2
a2  u2
1 1 u
 sin  a   C
2
2
2
 sin 1  ua   C
u  2x
du  2dx
1 1  2 x 
 sin 
C
2
 3
The hardest part is getting the integral into the proper form.
1
dx
0 1  x 2
du
1
1 u
 a 2  u 2  a tan  a   C
a 1
ux
du  dx
1
 tan x
1
0
 tan 1 1  tan 1 0


4
0 

4
Simplify.
1.
2.

x
dx
4  x2
dx
1
4 x  64
2
dx
Simplify.
1.
2.
x
 4  x 2 dx  arcsin  2   C
a2 ux
du  dx
dx
x
1
4 x  64
2
dx  

1
 2x    8 
2
2
u  2x
du  2dx
a8
x
dx
1
2x
 2x    8 
2
2
2dx
 x 
 2x 
1
1
 arc sec 
  C  arc sec    C
8
8
4
 8 
Simplify.
3.
4.

t
16  t
4
dt
1
 9   2x  3 
2
dx
Simplify.
3.

t
16  t
4
dt  
t
 4  t 
2
a4
4.
1
 9   2x  3 
2
dx  
2
2
1
dt  
2
u  t2
du  2tdt
3
a3
  2x  3 
 4
2
 
 t
2
2
dt
 t2 
1
 arcsin    C
2
4
1
2
2t
2
dx
u  2x  3
du  2dx
1
1
 
2dx
2
2
2  3    2x  3 
1
 2x  3 
 arctan 
C

6
 3 
Try These
• Look for the pattern or how the expression
can be manipulated into one of the
patterns

8dx
 1  16 x 2

x dx
1  25 x 2
dx
4 x 2  4 x  15

x 5
x  10 x  16
2
16
dx
Completing the Square
• Often a good strategy when quadratic
functions are involved in the integration
dx
 x 2  2 x  10
• Remember … we seek
(x – b)2 + c
– Which might give us an integral resulting in
the arctan function
Example
dx
 x 2  2 x  10
COMPLETE
THE
SQUARE!!!!
dx
 x  12  9
u  x 1
du
 u2  a2
a 3
du  dx
1
 x 1 
 tan 1 
C
3
 3 
x

x  2x  1 10 1
2
 2 x  __  10  __
2
x  12  9
1.
dx
 x 2  6 x  15
2.
dx
 2x 2  8 x  10
1.
dx
1
dx

 x 2  6 x  15   x 2  6 x  9  15  9   x  3 2  6 dx
u  x 3 a  6
du  dx
 x 3

arctan 
 C
6
 6 
1
2.
dx
1
1
1
1

 2x 2  8 x  10  2  x 2  4 x  5 dx 2  x 2  4 x  4  5  4 dx
1
1
 
dx
2
2  x  2  1
u  x 2 a 1
du  dx
1
 arctan  x  2   C
2
3.

dx
3x  x 2
3.

dx
3x  x
2

dx
x  3x
2




dx

 x 2  3x

dx


 x 2  3 x  94  94
dx
9
4

 x 2  3 x  94

dx
   x  
3 2
2
3 2
2
 x  32 
 arcsin  3   C
 2 
 2x  3 
 arcsin 
C

 3 
a  32
u  x  32
du  dx
Rewriting as Sum of Two Quotients
• The integral may not appear to fit basic
integration formulas
– May be possible to split the integrand into two
portions, each more easily handled

4x  3
1 x
2
dx  
4x
1 x
2

3
1 x
2
4.

2x  3
4x  x
2
dx
4.

2x  3
4x  x 2
dx  
u  4x  x 2
du   4  2x  dx
2x  3  1  1
dx
4x  x 2
2x  4
1

dx  
dx
4x  x 2
4x  x 2
  2x  4 
1
 
dx  
dx
2
2
4x  x
x  4x
1
 12
   u du  
dx
 x 2  4x  4  4
1
 12
  u du  
dx a  2
2
u  x 2
4   x  2
du  dx
1
 x 2
2
 2u  arcsin 
C

 2 
1
 x 2
2 2
 2  4 x  x   arcsin 
C

 2 


Basic Integration Rules
• Note table of basic rules
• Most of these should be committed to
memory
• Note that to apply these, you must create
the proper du to correspond to the u in the
formula
cos
u
du

sin
u

C

1.

x 2
dx
x 1
u
x 2
1. 
dx   2
 2udu 
x 1
u  2 1
u  x 2
u2
 2 2
du
u2  2  x
u 3
2udu  dx
u2  3  3
 2
du
2
u 3
u2  3
1
 2 2
du  6 2
du
u 3
u 3
1
 2 du  6  2
du
u 3
1
 u 
 2u  6 
arctan 
 C
3
 3
 x 2 
6
 2 x 2
arctan 
C



3
3




2.
Find the area of the region bounded by
1
y 2
, y  0, x  1, and x  3.
x  2x  5
2.
Find the area of the region bounded by
1
y 2
, y  0, x  1, and x  3.
x  2x  5
3
3
1
1

1 x 2  2x  5 dx 1 x 2  2x  1  5  1dx
1
0.5
3

2
1

fnInt 1/  x  2x  5  , x,1,3
2

1
 x  1
2
4
dx
3
1
 x  1
  arctan 

 2  1
2
1
1
arctan 1  arctan  0 
2
2
1 
  0
2 4 



8
 .251?
3.
dy
xy

dx 1  x 2