1
Section 4.8: Inverse Trigonometric Funtions: Integration
Practice HW from Larson Textbook (not to hand in)
p. 291 # 1-29 odd
Recall that the inverse sine function determines the angle that one must take the sine of to
obtain a given quantity.
Notation: y  f ( x)  sin 1 x  arcsin x , 

2
 y

2
1
1
Example 1: Compute sin 1    arcsin   .
2
2
Solution:
█
There are other inverse trigonometric functions with similar interpretations
Integral Formulas Involving the Inverse Trigonometric Functions
1
u
C
a
1.

2.
 a 2  u 2 du  a arctan a  C
3.

a2  u2
du  arcsin
1
1
1
u u2  a2
du 
u
1
|u|
arc sec
C
a
a
We illustrate these formula in the following examples
2
Example 2: Integrate

1
16  x 2
dx
Solution:
█
Example 3: Integrate
1
x
 9  4 x 2 dx and compare with  9  4 x 2 dx
Solution:
█
3
Example 4: Integrate
t

16  t
4
dt
Solution: This formula that best fits this integral is

1
du  arcsin
u
C
a
a2  u2
(however, this function has the variable t in the numerator). We can integrate using a udu substitution as follows:

t
16  t 4
t
dt  
(4) 2  (t 2 ) 2
dt
Let u  t 2

1
du
2
(4) 2  (u ) 2
du  2t dt
1
du  t dt
2
1
(Bring outside integral)
2

1
1
du

2
(4) 2  (u ) 2

1
u
arcsin  C
2
4
(Apply integratio n formula with a  4)

1
t2
arcsin
C
2
4
(Substitut e and write answer in terms of t )
█
4
Example 5: Integrate
ex
 4  e2x
dx
Solution: This formula that best fits this integral is
1
1
u
 a 2  u 2 du  a arctan a  C
(however, this function has the e x term in the numerator). We can integrate using a u-du
substitution as follows:
ex
 4  e2x
dx  
ex
(2)  (e )
2
x 2
dx
(Note e 2 x  (e x ) 2
)
Let u  e x



1
(2)  (u )
2
2
1
u
arctan  C
2
2
1
ex
arctan
C
2
2
du
du  e x dx
(Apply integratio n formula with a  2)
(Substitut e and write answer in terms of x)
█
5
Example 6: Integrate
x3  2
 x 2  1 dx
Solution:
█
6
Integrating By Completing the Square
Involves completing the square to rewrite function a form where it can be integrated
using an integration formula.
Steps for Completing the Square
1. Make sure the coefficient of the x 2 term is 1 and the quadratic term is of the form
x 2  bx  c .
1
2. Take
of the coefficient of the x term, square it, and add and subtract the term to the
2
1
polynomial expression. That is, for x 2  bx  c , take of the coefficient of the x
2
term,
b
, square to get
2
2
b
  , and compute
2
2
2
2
2
b
b
x 2  bx     c   
2
 2
b

b
3. Write as perfect square of the form  x    c    and try to apply an integration
2

2
formula.
7
1
Example 7: Integrate

3
1
x  6 x  13
2
dx
Solution:
█
8

Example 8: Integrate
1
 x  2x
2
dx
Solution: Note that in this problem, the coefficient of x 2 is -1 instead of 1. We perform
the following steps to integrate the function.

1
 x  2x
2
dx  

1
 ( x  2 x)
2
(Factor understood  1 )
dx
1
 ( x 2  2 x  1)  1
dx
Take
1
(2)  1 and square to get (-1) 2  1
2
Note how the 1 is added and subtracted




1
 ( x  1)  1
2
1
1  ( x  1) 2
1
1  ( x  1) 2
1
(1) 2  (u ) 2
 arcsin
u
C
1
 arcsin( x  1)  C
dx
Write as a perfect square
dx
Rearrange terms under radical
dx
Let u  x  1, du  dx
du
Let u  x  1, du  dx
(Apply arcsin integratio n formula with a  1)
(Substitut e and write answer in terms of x)