Trigonometric Substitution
We can use right triangles and the pythagorean
theorem to simplify some problems.
1

dx
4 x
2
These are
in the same
form.
a4 x x
ln sec  tan  C
4 x
x
 C
2
2
2
ln
x

2sec2  d
 2sec
 sec d
22
2
a2
4  x2
sec 
2
2sec  4  x
2
x
tan  
2
2 tan   x
2sec2  d  dx

We can use right triangles and the pythagorean
theorem to simplify some problems.
1

dx
4  x2
2sec2  d
 2sec
ln
4  x2  x
C
2
ln 4  x2  x  ln 2  C
 sec d
ln sec  tan  C
ln
4  x2 x
 C
2
2
This is a constant.
ln 4  x2  x  C

This method is called Trigonometric Substitution.
a x
2
a x ,
2
If the integral contains
2
2
x

we use the triangle at right.
a
If we need
a2  x2 , we
move a to the hypotenuse.
If we need
move x to the hypotenuse.
x
a
x

a x
2
x2  a2 , we
2
x2  a2

a

Trigonometric Substitution
Use trigonometric substitution to evaluate integrals involving
the radicals
The objective with trigonometric substitution is to eliminate the
radical in the integrand. You do this by using the Pythagorean
identities
2

2
x dx
9  x2
3

9  x2
9sin 2   3cos d

3cos
1  cos 2
9
d
2
9
1  cos 2 d

2
9
9 1
   sin 2  C
2
2 2
x
x sin   3
3sin   x
9  x2
cos 
3
3cos  9  x2
3cos d  dx
x
sin  
3
1 x
  sin
3
double angle
formula
9 1 x 9
sin
  2sin  cos   C
2
3 4
9 1 x 9 x 9  x 2
sin
  
C
2
3 2 3
3

2

2
x dx
9  x2
3

9  x2
x
x sin   3
3sin   x
9  x2
cos 
3
3cos  9  x2
3cos d  dx
x
sin  
3
1 x
  sin
3
9 1  x  x
sin   
9  x2  C
2
3 2
double angle
formula
9 1 x 9
sin
  2sin  cos   C
2
3 4
9 1 x 9 x 9  x 2
sin
  
C
2
3 2 3
3

3


dx
2 x  x2
2x  x 2
dx
1   x  1

2
Let u  x  1


 x2  2x

du  dx

 x2  2 x  1  1
du
1 u2
cos  d
 cos 
 d
We can get 2x  x 2 into the necessary
form by completing the square.
  x  1  1
2
1   x  1
2
1

sin
u C
 C
 sin1  x 1  C
1
u

1  u2
1 u2
 1  u2
cos 
1
sin   u cos  d  du

4
dx
 4x2  4x  2
4 x2  4 x  2
4 x2  4 x  1  1
dx
  2x  1
Complete the square:
2
1
1 du
2  u2 1
1 sec 2  d
2  sec2 
 2 x  1
2
1
Let u  2 x  1
du  2 dx
1
du  dx
2
1
1
1
1

tan
u C



C
d

2
2
2
1

tan 1  2 x  1  C
2
u2 1
u

1
tan   u
sec2  d  du
sec  u2 1
sec2   u 2  1

Here are a couple of shortcuts that are result from
Trigonometric Substitution:
du
1
1 u
 u 2  a 2  a tan a  C

du
u
 sin
C
2
2
a
a u
1
These are on your list of
formulas. They are not
really new.
p
8.5 day 1 Partial Fractions
The Empire Builder, 1957
Greg Kelly, Hanford High School, Richland, Washington
Partial Fractions
Motivation
The method of partial fractions is a procedure for
decomposing a rational function into simpler rational
functions to which you can apply the basic integration
formulas.
To see the benefit of the method of partial fractions,
consider the integral
Partial Fractions
To evaluate this integral without
partial fractions, you can complete
the square and use trigonometric
substitution (see Figure 8.13) to
obtain
Figure 8.13
Partial Fractions
Partial Fractions
Now, suppose you had observed that
Then you could evaluate the integral easily, as follows.
This method is clearly preferable to trigonometric substitution.
However, its use depends on the ability to factor the denominator,
x2 – 5x + 6, and to find the partial fractions
Partial Fractions
Linear Factors
1
Write the partial fraction decomposition for
Solution:
Because x2 – 5x + 6 = (x – 3)(x – 2), you should include one
partial fraction for each factor and write
where A and B are to be determined.
Multiplying this equation by the least common denominator
(x – 3)(x – 2) yields the basic equation
1 = A(x – 2) + B(x – 3).
Basic equation.
Because this equation is to be true for all x, you can substitute
any convenient values for x to obtain equations in A and B.
The most convenient values are the ones that make particular
factors equal to 0.
To solve for A, let x = 3 and obtain
1 = A(3 – 2) + B(3 – 3)
1 = A(1) + B(0)
A=1
Let x = 3 in basic equation.
To solve for B, let x = 2 and obtain
1 = A(2 – 2) + B(2 – 3)
1 = A(0) + B(–1)
B = –1
cont
’d
Let x = 2 in basic equation
So, the decomposition is
as shown at the beginning of this section.
2

5x  3
dx
2
x  2x  3
This would be a lot easier if we could
re-write it as two separate terms.
5x  3
A
B


 x  3 x  1 x  3 x  1
These are called nonrepeating linear factors.
You may already know a
short-cut for this type of
problem. We will get to
that in a few minutes.

1

5x  3
dx
2
x  2x  3
This would be a lot easier if we could
re-write it as two separate terms.
5x  3
A
B


 x  3 x  1 x  3 x  1
Multiply by the common
denominator.
5x  3  A x 1  B  x  3
5 x  3  Ax  A  Bx  B  3
5x  Ax  Bx
5  A B
Set like-terms equal to
each other.
3  A  B  3
3  A  3B
Solve two equations with
two unknowns.

1

5  A B
5x  3
dx
2
x  2x  3
3  A  3B
3   A  3B
8  4B
5x  3
A
B


 x  3 x  1 x  3 x  1
5x  3  A x 1  B  x  3
5 x  3  Ax  A  Bx  B  3
5x  Ax  Bx
5  A B
3  A  B  3
3  A  3B
2B
5  A2
3 A
3
2
 x  3  x  1 dx
3ln x  3  2ln x 1  C
This
is called
Solvetechnique
two equations
with
Fractions
twoPartial
unknowns.

1

5x  3
dx
2
x  2x  3
The short-cut for this type of problem is
called the Heaviside Method, after
English engineer Oliver Heaviside.
5x  3
A
B


 x  3 x  1 x  3 x  1
5x  3  A x 1  B  x  3
Multiply by the common
denominator.
Let x = - 1
8  A  0  B  4
2B
12  A   4  B  0
3 A
Let x = 3

1

5x  3
dx
2
x  2x  3
The short-cut for this type of problem is
called the Heaviside Method, after
English engineer Oliver Heaviside.
5x  3
A
B


 x  3 x  1 x  3 x  1
5x  3  A x 1  B  x  3
8  A  0  B  4
2B
12  A   4  B  0
3 A
3
2
 x  3  x  1 dx
3ln x  3  2ln x 1  C

Good News!
The AP Exam only requires non-repeating linear factors!
The more complicated methods of partial fractions are
good to know, and you might see them in college, but they
will not be on the AP exam or on my exam.

2
6x  7
 x  2
2
A
B


x  2  x  2 2
Repeated roots: we must
use two terms for partial
fractions.
6 x  7  A  x  2  B
6 x  7  Ax  2 A  B
6x  Ax
7  2A  B
6 A
7  26  B
7  12  B
6
5

x  2  x  2 2
5  B

3
2 x3  4 x 2  x  3
x2  2 x  3
If the degree of the numerator is
higher than the degree of the
denominator, use long division first.
2x
x 2  2 x  3 2 x3  4 x 2  x  3
2 x3  4 x 2  6 x
5x  3
5x  3
2x  2
x  2x  3
(from example one)
5x  3
3
2
2x 
 2x 

 x  3 x  1
 x  3  x  1

You Try:

5x  20 x  6
dx
3
2
x  2x  x
2
5 x  20 x  6
2

x  x  1
2
A
B
C
 

2
x x  1 x  1
5x  20x  6  Ax  1  Bxx 1  Cx
2
2
A  6, B  1, C  9

6
1
9
5x  20 x  6

dx
dx   
2
3
2
x x  1 ( x  1)
x  2x  x
2
x  1

 6ln x  ln x  1  9
1
1

x6
9
 ln

C
x 1 x 1
C
Quadratic Factors
4
Distinct Linear and Quadratic Factors
Find
Solution:
Because (x2 – x)(x2 + 4) = x(x – 1)(x2 + 4) you should include one
partial fraction for each factor and write
Multiplying by the least common denominator x(x – 1)(x2 + 4)
yields the basic equation:
2x3 – 4x – 8 = A(x – 1)(x2 + 4) + Bx(x2 + 4) + (Cx + D)(x)(x – 1)
4
To solve for A, let x = 0 and obtain
–8 = A(–1)(4) + 0 + 0
2=A
To solve for B, let x = 1 and obtain
–10 = 0 + B(5) + 0
–2 = B
At this point, C and D are yet to be determined.
You can find these remaining constants by choosing two other
values for x and solving the resulting system of linear equations.
4
If x = –1, then, using A = 2 and B = –2, you can write
–6 = (2)(–2)(5) + (–2)(–1)(5) + (–C + D)(–1)(–2)
2 = –C + D
If x = 2, you have
0 = (2)(1)(8) + (–2)(2)(8) + (2C + D)(2)(1)
8 = 2C + D
Solving the linear system by subtracting the first equation from
the second
–C + D = 2
2C + D = 8
yields C = 2
4
Consequently, D = 4, and it follows that
Homework
• Section 8.5 Day 1: pg. 559, 7-31 odd
• Day 2: MMM BC pgs. 115-116