4.5
Integration by Substitution
Antidifferentiation of a Composite Function
Let f and g be functions such that f og and g’ are
continuous on an interval I. If F is an antiderivative
of f on I, then
Inside

f ( g ( x)) g ' ( x)dx  F ( g ( x))  C
Outside
Derivative
of Inside
 x
Evaluate
2
 1 2x dx
2
du
 u 2x 2x
3

2
3
x  1
u

 C 
C
3
3
2
Evaluate
 5 cos 5 x dx
du
 5 cos u
5
 sin u  C  sin 5x  C
Let u = x2 + 1
du = 2x dx
du
 dx
2x
Let u = 5x
du = 5 dx
du
 dx
5
Multiplying and dividing by a constant
 xx
2
 1 dx
2
x  1
du u

 C 
  xu 
C
2x
6
6
2

3
2
3
du
1 2 du
 u
u
2
2
2 x 1dx  

32
32
1 2u
2x 1


C 
C
2 3
3
Let u = x2 + 1
du = 2x dx
du
 dx
2x
Let u = 2x - 1
du = 2dx
du
 dx
2
Substitution and the General Power Rule
What would you let u = in the following examples?
 3(3x  1) dx
 2 x  1x  x  dx
 3x x  2 dx
4
2
2
3
 4x
dx
2
1 2x


 cos x sin x dx
2
u = 3x - 1
u = x2 + x
u = x3 - 2
u = 1 – 2x2
u = cos x
A differential equation, a point, and slope field are given. Sketch
the solution of the equation that passes through the given point.
Use integration to find the particular solution of the differential
equation. dy
2
3
2
dx

x 2 x 3 1 dx
2
du
x u 3x 2
2

2
 x (x 1)
u  x 3 1
du  3x 2 dx
du
 dx
2
3x
x 1

u du u
 C
C
3
9
9
2
3
1,0
3
3
13 1
3
0
C
3
x 1

9
y
C0
9
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Day 1 stop (1-41 odd)
x
2
3
u = x3
sin x dx
du = 3x2 dx
du
  x sin u du
 dx
2
2
3x
3x
1
1
  sin u du  ( cos u )  C
3
3
2
1
3
  cos x  C
3
 sin 3x cos 3x dx
Let u = sin 3x
du = 3cos 3x dx
 sin 3x cos3x dx
du
 dx
3 cos 3 x
2
rewritten as
2
  u cos 3 x
2
du
3 cos 3x
1 2
  u du
3
sin 3 3x
u3
C
 C 
9
9
Day 2 stop (43-56 all, 57-61 odd)
 x 2 x 1dx
Let u = 2x - 1
du = 2dx
 u  1  u du
 

2
 2 
du
 dx
2
1

4
 u
32
u
12
u + 1 = 2x
u 1
x
2
52
32

1 2u
2u 
  C
du  

4 5
3 


2 x  1

52
10

2 x  1

32
6
C
Evaluate
u = x2 + 1
du = 2x dx
1
 x( x
2
 1) dx
3
0
du
  x(u )
2x
3
2
u 
 
8 1
4
16 1 15
  
8 8 8
du
 dx
2x
Note that there are no upper and lower
limits of integration.
We must determine new upper and
lower limits by substituting the old
ones in for x in u = x2 + 1
Or, we could use the old limits if we
substitute x2 + 1 back in.

x  1
16 1 15


  

8 8 8
8 
0
2
4 1
5
Area  
1
x
dx
2x 1
u 2  1 udu 

2u
u  2x 1
u  2x 1
2
2u du = 2 dx
u du = dx
3
u 1
x
2
2
What limits are
we going to use?

1u
1
2
  u  1 du    u 
2 3
2
 1
1  27   1  16
   3     1 
3
2  3
  3 


3
See area comparisons when using different upper and lower
limits. Page 302
Integration of Even and Odd Functions
If f is an even function, then
2
a

a
a
f ( x)dx  2 f ( x)dx
0
Ex.
 x dx  0
3
2
Odd or Even?
If f is an odd function, then
a
 f ( x)dx  0
a