5
Logarithmic, Exponential, and
Other Transcendental Functions
Copyright © Cengage Learning. All rights reserved.
Calculus Warm-Up
2

d
x 
 arcsin

dx 
5 
Think of
x
2
5
as
d
arcsin u 
dx
u
a
, then
d
dx
u
1 u
arcsin
u
a
2

2x

25  x
u
a u
2
2
4
5.7
Inverse Trigonometric Functions:
Integration
Copyright © Cengage Learning. All rights reserved.
Objectives
■ Integrate functions whose antiderivatives involve
inverse trigonometric functions.
■ Use the method of completing the square to
integrate a function.
■ Review the basic integration rules involving
elementary functions.
Integrals Involving Inverse Trigonometric Functions
The derivatives of the six inverse trigonometric functions fall into
three pairs. In each pair, the derivative of one function is the
negative of the other.
For example,
and
Integrals Involving Inverse Trigonometric Functions
When listing the antiderivative that corresponds to each
of the inverse trigonometric functions, you need to use
only one member from each pair.
It is conventional to use arcsin x as the antiderivative
of
rather than –arccos x.
Inverse Trigonometric
Function Integrals:

a
u
du
a u
2
du
2
u
2
 arcsin
2
1

u a
C
a
arctan

2
u
C
a
a
du
2
u
1
a
arc sec
u
a
C
Integrals Involving Inverse Trigonometric Functions
Example

dx
4x
2
 arcsin
x
2
C
Example
dx
 2  9x
2


dx
 2   3 x 
2
Let u = 3x
du = 3 dx
2
du
3
1
 
3
du
 
2
2
 u 
2

1
3 2
arctan
3x
2
C
 dx
Ex. Integrate by substitution.
dx

e
2x
1


u 



1
2


e 
x
2
1
du
Let u = ex
x
e
du = ex dx du
du
u 
2
1
 dx
 dx
u
1 u
du
u
dx
 arc sec
u
1
 C  arc sec
e
x
1
C
Ex. Rewriting the integrand as the sum of two quotients.
x2

4x
dx
2
Let u = 4 – x2
du = -2x dx
du
2
2 

 x dx

x

4x
1

2

1
2
dx
2 x
2
2
1
2
du  
2
u
 2u   
 2 arcsin
1 2
x
2
u

4x
1
u
a u
2
a
2

dx 
du

1 2
 arcsin
du
2
u
2

2
1
arctan
a
du
u u a

2
1
a
2
du
2
 4  x  2 arcsin
x
2
u
C
arc sec
Final Answer
2
C
a
dx
  4x
u
a
2
C
u
a
C
Ex. Integrating an improper rational function.
3x  2
3

x 4
2
dx
Do long division and then rewrite the
integrand as the sum of two quotients.
12 x  2 

   3x  2
dx
x 4 

12 x
  3 xdx   2
dx 
x 4

 3 xdx  6  x

3x
2
2
2x
2
4

a u
2
a
x

2 dx
2
dx  2 
 6 ln  x  4   arctan
2
du
4
dx
x 4
2
x
2
C
du
2
u
2
 arcsin

1
arctan
a
du
2
C
a
2
u u a
u

2
u
C
a
1
a
arc sec
u
a
C
Ex:
get into the form of
a
du
2

u
a

x

2


 4x  7
dx

x
dx
x  2 
2
a u
2
2
by com pleting the square
dx
du
2
du
2
u
2
 arcsin

2
 4 x  4  7  4



du
u 
2

 
3
2

1
3
arctan
x2
3
arctan

Let u = x – 2
du = dx
3
1
a
du
2
C
a
2
u u a
u
C
u
C
a
1
a
arc sec
u
a
C
Ex. Completing the square when the
leading coefficient is not 1.
 2x


2

2 x
dx
2
 4x  5
1

2 x
 4x  4
Let u = x – 2
du = dx
2
u
2
u
C
a
2
1

arctan
a
du
u u a
2

2
u
C
a
1
arc sec
a
u
C
a
Now complete the square
in the denominator.
1

2 x

dx
2
du
 arcsin
First, factor out a 1/2
 8 x  10
1
a u
2
a

dx
du


1
 54
du

2 (u )
2
1
dx
2
 4x 


1
5
dx

2 ( x  2)
1
2
arctan
2
1
( x  2)
1
C
Adding and Subtracting Common Denominators

Ex: Find the integral:
2x  5
2
x  2x  2
dx
The derivative of x2 + 2x + 2 is 2x + 2, so to get it, add
7
and subtract
x 2 x  2
2

2x  5
2
x  2x  2

dx 

7
2
x  2x  2
dx 

7
2
x  2x  2
dx
Adding and Subtracting Common Denominators
Now, combine the first two integrals.



2x  2
2
x  2x  2
2x  2
2
x  2x  2
dx 
2x  2
2
dx 
x  2x  2
dx 



7
2
x  2x  2
dx
7
x
2
 2 x  1  2  1
7
 x  1
1
2
1
2
dx
 ln x  2 x  2  7  arctan
1
2
dx
x 1
1
C
Review of Basic Integration Rules
You have now completed the introduction of the basic
integration rules. To be efficient at applying these rules, you
should have practiced enough so that each rule is committed to
memory.
Example 6 – Comparing Integration Problems
Find the following integrals using the formulas and
techniques you have studied so far.
Example 6 – Solution
a. You can find this integral (it fits the Arcsecant Rule).
Example 6 – Solution
b. You can find this integral (it fits the Power Rule).
Example 6 – Solution
c. You cannot find this integral using the techniques
you have studied so far.
Review of Basic Integration Rules
Homework Day 1
Pg. 385 1-37 odds + 53, 63
Homework 5.7 Day 2
MMM pgs. 207, 208
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athematics/blackboard/conte
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CT=a&TYPE=se
HWQ 1/27
1
Evaluate the integral:
2

0
arcsin x
1 x
2
dx