Chapter 8
8.2 Integration By Parts
Let G(x) be any antiderivative of g(x). In this case G’(x)=g(x), then by the product rule,
d
[ f ( x)G ( x)]  f ( x) g ( x)  f '( x)G ( x)
dx
Which implies
 [ f ( x) g ( x)  f '( x)G( x)]dx  f ( x)G( x)
Or, equivalently, as
 f ( x) g ( x)dx  f ( x)G( x)   f '( x)G ( x)dx
(1)
The application of this formula is called integration by parts.
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
In practice, we usually rewrite (1) by letting
u  f ( x),
v  G ( x),
du  f '( x)dx
dv  G '( x)dx  g ( x)dx
This yields the following alternative form for (1)
 udv  uv   vdu
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example: Use integration by parts to evaluate
Solution: Let
 x cos xdx
u  x, dv  cos xdx
du  dx, v   dv   cos xdx  sin x
 x cos xdx  x sin x   sin xdx
 x sin x  ( cos x)  C
 x sin x  cos x  C
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Guidelines for Integration by Parts
The main goal in integration by parts is to choose u and dv to obtain a new integral
That is easier to evaluate than the original.
A strategy that often works is to choose u and dv so that u becomes “simpler” when
Differentiated, while leaving a dv that can be readily integrated to obtain v.
There is another useful strategy for choosing u and dv that can be applied when the
Integrand is a product of two functions from different categories in the list.
Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential
In this case, you will often be successful if you take u to be the function whose
Category occurs earlier in the list and take dv to the rest of the integrand (LIATE).
This method does not work all the time, but it works often enough to b e useful.
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example. Evaluate
x
xe
 dx
Solution: According to LIATE, we should let
u  x, dv  e x dx
du  dx, v   e x dx  e x
x
x
x
xe
dx

udv

uv

vdu

xe

e



 dx
 xe x  e x  C
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example: Evaluate
 ln xdx
Solution: Let u  ln x, dv  dx
du 
1
dx, v   dx  x
x
 ln xdx   udv  uv   vdu  x ln x   dx
 x ln x  x  C
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Repeated Integration by Parts
Example: Evaluate
Solution: Let
2 x
x
 e dx
u  x 2 , dv  e  x dx
du  2 xdx, v   e  x dx  e  x
2 x
2
x
x
x
e
dx

udv

uv

vdu

x
(

e
)


e



 (2 x)dx
  x 2 e  x  2 xe  x dx
Apply integration by parts to
x
xe
 dx
u  x, dv  e  x dx
du  dx, v   e  x dx  e  x
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
x
x
x
xe
dx

udv

uv

vdu

x
(

e
)


e



 dx
  xe x   e  x dx
  xe x  e x  C
Finally,
2 x
2 x
x
2 x
x
x
x
e
dx


x
e

2
xe
dx


x
e

2(

xe

e
)C


 ( x 2  2 x  2)e x  C
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

x
Example: Evaluate e cos xdx
x
Solution: Let u  cos x, dv  e dx,
du   sin xdx, v   e x dx  e x
x
x
x
e
cos
xdx

udv

uv

vdu

e
cos
x

e



 sin xdx
Let u  sin x,
(a)
dv  e x dx
du  cos dx, v   e x dx  e x
x
x
x
e
sin
xdx

udv

uv

vdu

e
sin
x

e



 cos xdx
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Together with (a), we have
x
x
x
x
e
cos
xdx

e
cos
x

e
sin
x

e

 cos xdx
Solve for the unknown integral, we have
2 e x cos xdx  e x cos x  e x sin x
1 x
1 x
x
e
cos
xdx

e
cos
x

e sin x  C

2
2
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Integration by Parts for Definite Integrals
For definite integrals, the formula corresponding to

b
a
 udv  uv   vdu
is
udv  uv a   vdu
b
b
a
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example: Evaluate

2
0
Solution: Let
xe2 x dx
u  x, dv  e2 x ,
1
du  dx, v   e2 x dx  e2 x
2

2
0
xe dx   udv  uv 0   vdu
2x
2
2
0
2
0
1 2x 2 1 2 2x
 xe ]0   e dx
2
2 0
1
1
 xe 2 x ]02  e 2 x ]02
2
4
1 4
4
 (e  0)  (e  1)
4
3 4 1
 e 
4
4
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
8.3 Trigonometric Integrals
Integrating powers of sine and cosine
By applying the integration by parts , we have two reduction formulas
1 n 1
n 1
n
n2
sin
dx


sin
x
cos
x

sin
xdx


n
n
1
n 1
n
n 1
n2
cos
dx

cos
x
sin
x

cos
xdx


n
n
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
In particular,
1
1
 sin xdx  2 x  4 sin 2 x  C
1
1
2
cos
xdx

x

sin 2 x  C

2
4
2
1
3
3
sin
xdx

cos
x  cos x  C

3
1 3
3
cos
xdx

sin
x

sin x  C

3
……
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Integrating Products of Sines and Cosines
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example: Evaluate
4
5
sin
x
cos
xdx

Solution: since n=5 is odd,
4
5
4
4
sin
x
cos
xdx

sin
x
cos
x cos xdx


  sin 4 x(1  sin 2 x) 2 cos xdx
  u 4 (1  u 2 ) 2 du
1 5 2 7 1 9
 u  u  u C
5
7
9
1
2
1
 sin 5 x  sin 7 x  sin 9 x  C
5
7
9
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Integrating Powers of Tangent and Secant
There are similar reduction formulas to integrate powers of tangent and secant.
tan n 1 x
n2
tan
xdx


tan
xdx


n 1
secn  2 x tan x n  2
n
n2
sec
xdx


sec
xdx


n 1
n 1
n
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
In particular,
 tan xdx  ln | sec x | C
 sec xdx  ln | sec x  tan x | C
2
tan
 xdx  tan x  x  C
2
sec
 xdx  tan x  C
……
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Integrating Products of Tangents and Secents (optional)
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Section 8.4 Trigonometric Substitutions
We are concerned with integrals that contain expressions of the form
a2  x2 , x2  a2 , x2  a2
In which a is a positive constant. The idea is to make a substitution for x that
will eliminate the radical.
For example, to eliminate the radical in a 2  x2, we can make the substitution
x = a sinx, - /2    /2
Which yields
a 2  x 2  a 2  a 2 sin 2   a 2 (1  sin 2  )
 a 2 cos 2   a | cos  | a cos  .
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example
x
Example 1. Evaluate
dx
2
4  x2
Solution. To eliminate the radical we make the substitution
x= 2sin, dx= 2cosd
This yields
x
dx

2 cos  d
4  x2
(2sin  ) 2 4  4sin  2
2 cos  d
1 d


(2sin  ) 2 (2 cos  ) 4  sin 2 
1
1
  csc 2  d   cot   C
4
4
2
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example Cont.
2
4

x
From the left figure, we obtain cot  
x
Substituting this in the previous answer yields
x
1 4  x2

C
2
4
x
4 x
dx
2
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example
There are two methods to evaluate the definite integral.
• Make the substitution in the indefinite integral and then evaluate the definite
integral using the x-limits of integration.
• Make the substitution in the definite integral and convert the x-limits to the
corresponding -limits.
Example. Evaluate

1
2
dx
x2 4  x2
Solution: Since the substitution can be expressed as =sin-1(x/2), the -limits of
the integration are
x=1: = sin-1(1/2)=/6
x= 2 : = sin-1( 2 /2)=/4
Thus
2
dx
1
1

1
x2
  [cot  ] /4
  [1  3]
/6
4
4
4  x2
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Method of Trigonometric Substitution
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example
x 2  25
dx, assuming that x5.
Example. Evaluate 
x
Solution. Let x=5 sec, 0  </2
dx= 5sec tan d.
Thus,
x 2  25
25sec 2   25
 x dx   5sec 5sec tan  d
5 | tan  |

5sec  tan  d
5sec 
 5 tan 2  d
 5 (sec 2   1)d  5 tan   5  C
Note that
tan  

x 2  25
x
and   sec 1 ( ),
5
5
x 2  25
x 2  25
x
dx 
 sec1 ( )  C
5
5
5
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
8.5 Integrating Rational Functions by Partial Fractionas
Suppose that P(x) / Q(x) is a proper rational function, by which we mean that the
Degreee of the numerator is less than the degree of the denominator. There is a
Theorem in advanced algebra which states that every proper rational function can
Be expressed as a sum
P( x)
 F1 ( x)  F2 ( x) 
Q( x)
Fn ( x)
Where F1(x), F2(x), …, Fn(x) are rational functions of the form
A
Ax  B
or
(ax  b) k
(ax 2  bx  c) k
In which the denominators are factors of Q(x). The sum is called the partial fraction
Decomposition of P(x)/Q(x), and the terms are called partial fraction.
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Finding the Form of a Partial Fraction Decomposition
The first step in finding the form of the partial fraction decomposition of a proper
Rational function P(x)/Q(x) is to factor Q(x) completely into linear and irreducible
Quadratic factors, and then collect all repeated factors so that Q(x) is expressed as
A product of distinct factors of the form
(ax  b)m and (ax 2  bx  c)m
From these factors we can determine the form of the partial fraction decomposition
Using two rules that we will now discuss.
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Linear Factors
If all of the factors of Q(x) are linear, then the partial fraction decomposition of
P(x)/Q(x) can be determined by using the following rule:
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example: Evaluate
dx
 x2  x  2
1
1

Solution: x 2  x  x ( x  1)( x  2)
By the linear factor rule, the decomposition has the form
1
A
B


x 2  x  x ( x  1) ( x  2)
Where A and B are constants to be determined. Multiplying this expression
Through by (x-1)(x+2) yields
1=A(x+2)+B(x–1)
Let x=1, we have 1= 3A which implies A = 1/3; and let x = -2, we have 1 = -3B,
Which means B = -1/3. Thus,
1
1

1
3 
3

2
x  x  x ( x  1) ( x  2)

Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
The integration can now be completed as follows:
dx
1 dx 1 dx
 ( x  1)( x  2)  3  x  1  3  x  2
1
1
 ln | x  ln | x  2 | C
3
3
1
x 1
 ln |
| C
3
x2
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
2x  4
dx
Example: Evaluate  3
x  2 x2
Solution:
2x  4
2x  4

x3  2 x 2 x 2 ( x  2)
By the linear factor rule,
2x  4
A B
C



x 2 ( x  2) x x 2 x  2
Multiplying by x 2 ( x  2) yields
2 x  4  Ax( x  2)  B( x  2)  Cx 2 , which is
2 x  4  ( A  C ) x 2  (2 A  B) x  2 B
(1)
(2)
Setting x=0 in (1) yields B = -2, and setting x = 2 in (1) yields C = 2. Then
By equating the coefficient of x2 on the two sides to obtain
A + C = 0 or A = -C = -2
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Substituting A = -2, B = -2, and C = 2 yields the partial fraction decomposition
2x  4
2 2
2



x 2 ( x  2) x x 2 x  2
Thus
2x  4
dx
dx
dx
dx


2

2

2
 x2 ( x  2)
 x  x2  x  2
2
 2 ln | x |   2 ln | x  2 | C
x
x2 2
 2 ln |
| C
x
x
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Quadratic Factors
If some of the factors of Q (x) are irreducible quadratics, then the contribution of
Those facctors to the partial fraction decomposition of P(x)/Q(x) can be determined
From the following rule:
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Examples
x2  x  2
dx
Example. Evaluate  3 2
3x  x  3x  1
Solution. 3x3-x2+3x-=x2(3x-1)+(3x-1)=(3x-1)(x2+1)
By the linear factor rule and the quadratic factor rule, the partial fraction
decomposition is
x2  x  2
A
Bx  C

 2
3
2
3x  x  3x  1 3x  1 x  1
Multiplying by (3x-1)(x2+1) yields
X2+x-2=A(x2+1)+(Bx+C)(3x-1).
Thus X2+x-2=(A+3B)x2+(-B+3C)x+(A-C)
Equating corresponding coefficients gives
A+3B
=1
-B+3C=1
A
-C=-2
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example Cont.
By solving this, we get A=-7/5, B=4/5 and C=3/5.
Thus
7
4
3
x

x2  x  2
5 5
5

3x3  x 2  3x  1 3x  1 x 2  1

x2  x  2
7 dx
4
x
3 dx
dx



dx

 3x3  x 2  3x  1
5  3x  1 5  x 2  1
5  x2  1
7
2
3
  ln | 3 x  1|  ln( x 2  1)  tan 1 x  C
15
5
5
Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.