Calculus II
MAT 146
Additional Methods of Integration
The methods of substitution and integration by parts are widely used
methods of integration. Each of these methods is associated with a derivative
rule. Substitution relies on undoing the chain rule and integration by parts
results from undoing the product rule. Additional methods of integration are
associated with particular types of functions. Here, we explore how to
integrate certain trigonometric functions as well as rational functions.
Trigonometric Integrals: Using Trig Identities
Shown below are four examples to illustrate integration of certain families of
trigonometric integrals.
Within these examples,
Example 1: Evaluate  sin 3 xdx .
3
notice the use of one or
 sin xdx
more of the following
2
  sin x  sin x dx
trig identities:
 
2
  sin x 1  cos x dx
  sin x  sin x  cos2 x dx
  sin xdx   sin x cos 2 x dx
The first integral in the last line can be solved by
inspection and the second by using the
substitution u = cosx. With this in mind, the final
result is
1
3
3
 sin xdx   cosx  cos x  C
3
Pythagorean Identities
cos2 x  sin 2 x  1
1  tan 2 x  sec 2 x
Half - Angle Identities
1
1  cos2x 
2
1
sin 2 x = 1  cos2x 
2
2
cos x =
Example 2: Evaluate  cos2 xdx .
2
 cos xdx
1
  1  cos2x dx
2
1
  1  cos2x dx
2
1
1
  dx   cos2x dx
2
2

1
1 1
 x   sin 2x   C

2
2 2
1
1
 x  sin2x   C
2
4
Example 3: Evaluate  sin 5 xdx .
5
 sin xdx
 2  2 
  sin x 1  cos 2 x 1  cos2 x dx
  sin x 1  2 cos2 x  cos 4 x dx
  sin x  2sin x cos2 x  sin x cos 4 x dx
  sin x  sin x sin x dx
  sin xdx  2  sin x cos 2 xdx   sin x cos4 xdx
The left integral in the last line can be solved by inspection, while each of
the other two require a u-substitution of u = cosx. This gives us
5
2
4
 sin xdx   sin xdx  2 sin x cos xdx   sin x cos xdx
2
1
  cos x  cos3 x  cos5 x  C
3
5
Example 4: Evaluate  sin 4 xdx .
4
 sin xdx
  sin 2 x  sin 2 xdx
  12 1 cos2x  dx
2


1
2
1  2 cos2x   cos 2x  dx
4
1
1
1
  dx   cos2x dx   cos2 2x dx
4
2
4
1
1
1 1
  dx   cos2x dx   1  cos4x dx
4
2
4 2
1
1
1
  dx   cos2x dx   1  cos4x dx
4
2
8
1
1
1
1
  dx   cos2x dx   dx   cos4x dx
4
2
8
8
Each of the four integrals in the last line can be solved by inspection or by a
straightforward substitution. This gives us
1
1
1
1
4
 sin xdx   dx   cos2x dx   dx   cos4x dx
4
2
8
8
1
1
1
1
 x  sin 2x   x  sin4 x   C
4
4
8
32
3
1
1
 x  sin 2x   sin4 x   C
8
4
32

As we look back on these four examples of integrals involving powers of
trigonometric functions, we can make some useful observations about
strategies for evaluating such integrals. With integrals of the form  sin n xdx
or  cosn xdx , where n is a positive integer greater than 1, we follow one of
two strategies depending on whether the exponent is odd or even.
Situation A: If n = 2k + 1 for k some positive integer (i.e., n is an odd
exponent) [see Examples 1 and 3 above]:
1. Factor the integrand from  sin 2k 1 xdx to  sin xsin 2k xdx .
2. Using the trig identity sin 2 x  cos2 x  1, rewrite  sin xsin 2k xdx as
k
2

sin
x

1

cos
x
dx .

3.


k
Expand 1 cos2 x  to get
2
k 1
2k2
k
2k
1  k cos x   1 k cos
x  1 cos x .
4. Multiply each term in the above expansion by sinx to get the integral
2
k 1
2k 2
k
2k
x  1 sin x cos xdx
 sin x  ksin x cos x   1 ksin xcos
5. Integrate this expression term by term:
 sin xdx   ksin xcos xdx 
2
k 1
 1
2k2
 ksin x cos
x  1  sin xcos
k
2k
xdx
6. Evaluate the first term by inspection and make a substitution in all
other terms, using u = cosx.
7. Evaluate the remaining integral expressions.
Situation B: If n = 2k for k some positive integer (i.e., n is an even exponent)
[see Examples 2 and 4 above]:
1
1. Using the trig identity cos2 x  1  cos2x , rewrite  cos2k xdx as
2
k
1

1  cos2x  dx ,
 
2

1
k
2. Expand  1  cos2x  to get
2

k1
k
1  kcos2x   k cos 2 x  cos 2x ,
3. Integrate the previous expression term by term to get
1 k
1 k
   dx     k  cos2x dx 
2 
2 
1 k
1k
k 1
   k  cos 2x dx     cosk 2x dx
2 
2
4. For each term in the previous sum, either (a) evaluate by inspection,
(b) return to step (1) here and use the same trig identity (even
exponents), or (c) return to Situation (A) previously described and
follow those steps (odd exponents). Continue this process until each
integral created can be solved by inspection.
Rational-Function Integrals: Using Partial Fraction Decomposition
Which of the following integrals is easier to evaluate?
1
1
1
(A)  2
dx or (B) 17 
dx  17 
dx
x 6
x1
x  5x  6
For most of us, it is (B). Each of the two integrals in (B) is of the form
du
C  , where C is some constant and u is a function of x. The integral in (A)
u
does not have a straightforward strategy for solution unless the denominator
matches a special case.
It turns out, however, that (A) and (B) are equivalent, because
1
1
1


2
x  5x  6 7x  6 7 x 1
You can verify this by adding the two fractions in the right-side expression.
After simplifying the sum, it will be the left-side expression.
In order to evaluate integrals such as those illustrated in (A) above, it is
useful to be able to transform a rational expression into a sum of difference
of simpler rational expressions. This transformation process, essentially
reversing the process of determining a common denominator and then
adding or subtracting, is called partial fraction decomposition. Here we
describe how to carry out partial fraction decomposition and then use that
process to transform integrals of rational expressions into integrals that are
equivalent yet more easily evaluated.
The format for a partial fraction decomposition depends on the
characteristics of the denominator in a rational expression. In the expression
1
, we can factor the denominator into two linear terms:
x 2  5x  6
1
1

2
x  5x  6 x  6  x  1
If a rational expression contains two linear terms in its denominator—linear
terms that are not multiples of one another—the expression can be
transformed into an equivalent expression as follows:
1
A
B


x  6 x  1 x  6  x 1
If we multiply each side of this equation by (x-6)(x+1), we get:
1  A x 1  B x  6
0x  1  Ax  A  Bx  6B
0x  1   A  Bx   A  6B
Note that we have rewritten the left side expression as 0x+1. We do this in
order to equate the linear and constant coefficients on the two sides of the
equal sign:
0x  1   A  Bx   A  6B
 A B 0
 A  6B  1
The last two equations for a system that we can solve using techniques
learned in algebra:
A  6B  1 A  1  6B
1
A  B  0  (1  6B)  B  0  7B  1 B  
7
 1 
1
 A     0  A 
 7 
7
This shows us that
1
1
 17
1
1
7




x  6 x  1 x  6  x 1 7 x  6 7x  1
Example 1: Use partial fraction decomposition to transform the integral
5x  4
dx into a form more easily integrated and then carry out the
 2
2x  x 1
integration.
5x  4
5x  4
A
B



2
2x  x  1  x  12x 1  x  1 2x  1
 5x  4  A2x  1  B x  1
 5x  4  2 Ax  A  Bx  B
 5x  4  2A  Bx   B  A
 2A  B  5 and B  A  4
The final two equations form a linear system whose solution is A = 3 and
 3
5x  4
1 
B = -1. Therefore,  2
dx   

dx .
x  1 2x 1
2x  x 1
We can evaluate the new integral by inspection or by using a separate usubstitution for each fraction. In so doing, the final result is
 3
1 

dx  3ln x  1  12 ln 2x 1  C .
 
x 1 2x  1
Example 2: Evaluate 
10x  3
dx .
x2  x
10x  3 10x  3 A
B



x 2  x x x  1 x x 1
 10x  3  A x 1  B x 
 10x  3  Ax  A  Bx
 10x  3   A  Bx   A
 A  B  10 and B  3
A7
The linear system has solution A = 7 and B = 3. Therefore,
7
10x  3
3 
dx    
dx . We can evaluate the new integral by
 2
x x 1
x x
7
3 
inspection. The final result is   
dx  7 ln x  3 ln x  1  C .
x x  1
1
Example 3: Evaluate 
x
3
dx .
2
x 1
Because the numerator is a higher-power polynomial than the denominator,
we first carry out long division on the rational expression. This gives us
3
x
x
 x 2
. We now explore how to solve the new integrals:
2
x 1
x 1
1 
1
1
x 
x

x

dx

xdx

dx .


 2
2
x 1
0 
0
0 x 1
We can evaluate the first integral by inspection. In the second, letting
u = x2+1 with du = 2xdx or 1/2du = xdx leads to an integral we can solve by
inspection.
0
We get
1
x
x2 1
xdx

dx

 ln x 2  1

 2
2 2
0
0 x 1
0
1
1
1 1
  1 
   ln 2  0  ln 1
2 2
  2 
1
 1  ln 2
2
Note 1: On pages 406-407 in our textbook, the author describes various formats and
requirements for use of partial fraction decomposition that depend on the characteristics
of the denominator in the initial rational expression.
Note 2: The expand function on your TI-89 or TI-92 calculator will perform a partial
fraction decomposition before your eyes! Type expand( and then type in the rational
expression and close the parentheses. When you hit return, the decomposition will be
shown on your screen. Among other locations, the expand( command can be found under
the algebra menu. Touch the F2 key and then select choice 3.