5
Logarithmic, Exponential, and
Other Transcendental Functions
Copyright © Cengage Learning. All rights reserved.
Calculus HWQ 11/4
• Find the equation of the tangent line to the
graph of the function at the given point:
x
 
y  arctan   at  2, 
2
 4

1
y    x  2
4 4
Calculus Warm-Up
d 
x2 
 arcsin 
dx 
5
d
u
arcsin u 
dx
1 u2

2x
25  x
x2
u
d
u
u
Think of
as , then
arcsin 
5
a
dx
a
a2  u 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:
du
u
 a 2  u 2  arcsin a  C
du
1
u
 a 2  u 2  a arctan a  C
u
du
1

arc
sec

C
 u u2  a2 a
a
Integrals Involving Inverse Trigonometric Functions
Example

dx
x
 arcsin  C
2
2
4 x
Example
dx
 2  9x2 
1
 
3

dx
du
 
2
Let u = 3x
du = 3 dx
 2   3x
2
2  u
2

2
1
3 2
arctan
du
 dx
3
3x
2
C
Ex. Integrate by substitution.
dx

e 1
2x





dx
e  1
x 2
1
du
2
u   1 u
du
u u   1
2
u
du
 dx
x
e
du = ex dx du
 dx
u
Let u = ex
x
e
 arc sec  C  arc sec  C
1
1


4 x
2
dx  
x
4 x
2
dx  
2
4 x
 arcsin
u
C
a
a u
du
1
u
 a 2  u 2  a arctan a  C
u
du
1
 u u 2  a 2  a arc sec a  C
Ex. Rewriting the integrand as the sum of two quotients.
x2
du
2
2
2
dx
1 1
1 1 2
 
du    u du
2
2
u
du = -2x dx
du
1
12
2
 x dx


2
u


u


4

x


2
2
Let u = 4 – x2
2
dx
22  x 2
x
 2 arcsin
2
Final Answer
x
 4  x  2 arcsin  C
2
2
Ex. Integrating an improper rational function.
3x  2
Do long division and then rewrite the
dx
2
 x 4
integrand as the sum of two quotients.
du
u

arcsin
C
12 x  2 
 a u

a
   3x  2
dx
du
1
u

arctan
C

x 4 

a
a
a u
u
du
1

arc
sec
C
u u a a
12 x
2dx
a
  3 xdx   2
dx   2
x 4
x 4
2x
dx
  3 xdx  6 2
dx  2 2
x 4
x 4
3
2
2
2
2
2
3x 2
x
2

 6 ln  x  4   arctan  C
2
2
2
Ex:
du
get into the form of  2
a  u2
by completing the square
dx
 x2  4x  7 
dx
 
2
x  2  3

du

2
u  3


du
 arcsin
a2  u 2
du
1
u
 a 2  u 2  a arctan a  C
u
du
1

arc
sec
C
 u u2  a2 a
a
dx
2
x  4 x  4 7  4
Let u = x – 2
du = dx
 
2

u
C
a
1
3
arctan
x2
3
C
Ex. Completing the square when the
leading coefficient is not 1.
dx
 2 x 2  8 x  10

du
 arcsin
u
C
a
a2  u 2
du
1
u

arctan
C
 a2  u2 a
a
u
du
1
 u u 2  a 2  a arc sec a  C
First, factor out a 1/2
Now complete the square
in the denominator.
1
dx
1
dx
  2
  2
2 x  4x  5 2 x  4x 
5
1
dx
1
dx
  2
 
2
2 x  4 x  4  5  4 2 ( x  2)  1
1
du
1
( x  2)
Let u = x – 2
  2
 arctan
C
du = dx
2
1
2 (u )  1
Adding and Subtracting Common Denominators
Ex: Find the integral:

2x  5
dx
2
x  2x  2
The derivative of x2 + 2x + 2 is 2x + 2, so to get it, add
and subtract 2 7
x 2 x  2

2x  5
7
dx   2
dx 
2
x  2x  2
x  2x  2


7
dx
2
x  2x  2
Adding and Subtracting Common Denominators
Now, combine the first two integrals.

2x  2
dx 
2
x  2x  2

2x  2
dx 
2
x  2x  2

2x  2
dx 
2
x  2x  2

7
dx
2
x  2x  2

7
dx
2
x  2x 1 2 1

7
dx
2
2
x  1  1
1
x 1
 ln x  2 x  2  7  arctan
C
1
1
2
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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HWQ 1/27
1
Evaluate the integral:

0
2
arcsin x
1  x2
dx