AP CALCULUS AB
CHAPTER 6:
DIFFERENTIAL EQUATIONS AND
MATHEMATICAL MODELING
SECTION 6.2:
ANTIDIFFERENTIATION BY
SUBSTITUTION
What you’ll learn about
 Indefinite Integrals
 Leibniz Notation and Antiderivatives
 Substitution in Indefinite Integrals
 Substitution in Definite Integrals
… and why
Antidifferentiation techniques were historically
crucial for applying the results of calculus.
Section 6.2 – Antidifferentiation by
Substitution
 Definition: The set of all antiderivatives of a function
f(x) is the indefinite integral of f with respect to x and is
denoted by

f x dx.
Section 6.2 – Antidifferentiation by
Substitution
integrand
 f x dx  F x   C
integral sign
constant of
integration
variable of
integration
is read “The indefinite integral of f with respect to x is
F(x) + C.”
 Example:
 2 xdx  x
2
C
Section 6.2 – Antidifferentiation by
Substitution

Integral Formulas:
Indefinite Integral
Corresponding
Derivative Formula
n 1
x
n
1.
x
 dx  n  1  C for n  1
1
2.
 x dx  ln x  C
3.
 e dx  e
x
x
C
kx
e
kx
e
 dx  k  C
d n
x  nx n 1
dx
d
1
ln x 
dx
x
d x
e  ex
dx
d kx
e  kekx
dx
Section 6.2 – Antidifferentiation by
Substitution

More Integral Formulas:
Indefinite Integral
Corresponding
Derivative Formula
4. sin  x dx   cos x  C

cos kx
 sin kxdx   k  C
5.
 cosx dx  sin x  C
sin kx
 coskxdx  k  C
d
cos x   sin x
dx
d
cos kx   k sin kx
dx
d
sin x  cos x
dx
d
sin kx  k cos kx
dx
Section 6.2 – Antidifferentiation by
Substitution

More Integral Formulas:
Indefinite Integral
Corresponding
Derivative Formula
d
2


sec
x
dx

tan
x

C
tan
x

sec
x

dx
d
2
7.
csc  x dx   cot x  C
cot x   csc 2 x
dx
d
sec x  sec x tan x
8. sec x tan x dx  sec x  C
dx
d
csc x   csc x cot x
9.  csc x cot x dx   csc x  C
dx
6.
2


Trigonometric Formulas
 cos udu  sin u  C
 sin udu   cos u  C
 sec udu  tan u  C
 csc udu   cot u  C
 sec u tan udu  sec u  C
 csc u cot udu   csc u  C
2
2
Section 6.2 – Antidifferentiation by
Substitution

1.
Properties of Indefinite Integrals:
Let k be a real number.
Constant multiple rule:




kf
x
dx

k
f
x
dx


2.
Sum and Difference Rule:
  f x   g x dx   f x dx   g x dx
Section 6.2 – Antidifferentiation by
Substitution
 Example:
3
 2 3
2
  2 x  x  dx   2 x dx   x dx
1
2
 2  x dx  3 dx
x
 x3 
 2    3ln x  C
 3
C’s can be combined into
one big C at the end.
Example Evaluating an Indefinite Integral
Evaluate  2x  cos xdx.
2
 2x  cos xdx  x  sin x  C
Section 6.2 – Antidifferentiation by
Substitution
 Remember the Chain Rule for Derivatives:
d  u n1 
n du

u
dx  n  1 
dx
 By reversing this derivative formula, we obtain the
integral formula
n 1
u
 n du 
  u dx  dx  n  1  C
Section 6.2 Antidifferentiation by
Substitution
 Power Rule for Integration
If u is any differentiable function of x, then
n 1
u
n
 u du  n  1  C,
n  1.
Exponential and Logarithmic Formulas
 e du  e  C
u
u
u
a
a
du

C

ln a
u
 ln udu  u ln u  u  C
 log udu  
a
ln u
u ln u  u
du 
C
ln a
ln a
Section 6.2 – Antidifferentiation by Substitution
 A change of variable can often turn an unfamiliar
integral into one that we can evaluate. The method
for doing this is called the substitution method of
integration.
Section 6.2 – Antidifferentiation by Substitution
 Example:
2
x
cos
2
x

 dx

Let u  2 x 2
 1

  x cos  u   du 
 4x 
1
  cos  u  du
4
1
 sin u  C
4
1
 sin  2x 2   C
4
du
 4x
dx
du  4 xdx
1
du  dx
4x
 Rules For Substitution-- A mnemonic help:





L - Logarithmic functions: ln x, logb x, etc.
I - Inverse trigonometric functions: arctan x, arcsec x, etc.
A - Algebraic functions: x2, 3x50, etc.
T - Trigonometric functions: sin x, tan x, etc.
E - Exponential functions: ex, 19x, etc.
 The function which is to be dv is whichever comes last in the
list: functions lower on the list have easier antiderivatives
than the functions above them.
 The rule is sometimes written as "DETAIL" where D stands
for dv.
To demonstrate the LIATE rule,
consider the integral:
Following the LIATE rule, u = x
and dv = cos x dx, hence du = dx
and v = sin x, which makes the
integral become
which equals
Section 6.2 – Antidifferentiation by Substitution
 Example:

9r 2 dr
Let u  1  r 3
1 r3
 
  9r u
2
 3 u
1
1
2
2
 du 

2 
 3r 
du
1
u 2
 3
+C
1
2
 2  12
 3   u  C
1
 6 1  r 3  C
du
 3r 2
dr
du  3r 2 dr
du
 dr
2
3r
Example Paying Attention to the
Differential
Let f ( x)  x2 1 and let u  x3. Find each
of the following antiderivatives in terms of x.
a.  f ( x)dx
b.  f (u)du
c.  f (u)dx
a.  f ( x)dx    x2 1dx  1 x  x  C
3
b.  f (u)du    u 1du  1 u  u  C
3
 1 x    x  C  1 x  x C
3
3
c.  f (u)dx    u 1dx     x  1 dx
   x 1 dx  1 x  x  C
7
3
2
3 3
3
9
2
6
3
7
3
3
2
Example Using Substitution
Evaluate  x e dx.
2
Let u  x . Then
3
x3
du
 3 x , from which we conclude that
dx
2
1
du  x dx. We rewrite the integral and proceed as follows
3
1
 x e dx   e du
3
1
 e C
3
1
 e C
3
2
2
x3
u
u
x3
Example Using Substitution
Evaluate  6 x 1  x dx.
2
Let u  1  x . Then du  2 xdx. Rewrite the integral in terms of u :
2
 6 x 1  x dx  3 udu
2
2 
 3 u   C
3 
3
2
 2 1  x
2
 C
3
2
Example Setting Up a Substitution with a
Trigonometric Identity
Evaluate  sin xdx.
3
Hint: let u = cos x and –du = sinxdx
 sin xdx    sin x  sin xdx
3
2
  1  cos x  sin xdx
2
   1  u du
2
let u  cos x and - du  sin xdx
3
u
 u   C
3
cos x
  cos x 
C
3
3
Section 6.2 – Antidifferentiation by Substitution
 Substitution in Definite Integrals
 
Substitute
u  g x , du  g 
and integrate with respect to u from
 x  dx,
u  g  a  to u  g  b  .

b
a
f  g  x    g   x  dx  
g b 
ga
f  u  du
Section 6.2 – Antidifferentiation by Substitution
 Ex:

1
0
x 1  x 2 dx

u 1
u  0
 
x u
1
2
Let u  1  x 2
 du 


 2 x 
1 0 12
   u du
2 1
when x  0, u  1  0  1
du  2 xdx
when x  1, u  1  1  0
0
1u 2


2 3 
2 1
1  2  32 0
   u
1
23
3
1  32
   0   1 2 

3 
1
1
   1 
3
3
3
du
 2 x
dx
du
 dx
2 x
Example Evaluating a Definite Integral by
Substitution
x
Evaluate 
dx.
x 9
2
0
2
Let u  x  9 and du  2 xdx. Then u (0)  0  9  9 and
2
2
u (2)  2  9  5. So,
2
1 du
x

dx


2 u
x 9
2
5
0
9
2
1
= ln u
2
5
9
1
 ln 5  ln 9 
2
1 5
 ln  
2 9

Section 6.2 – Antidifferentiation by Substitution
 You try:
6 cos t
  2  sin t 
2
dt
Section 6.2 – Antidifferentiation by Substitution
 You try:



4
4
tan 2 x sec2 xdx
Section 6.2 – Antidifferentiation by Substitution
 You try:

7
0
dx
x2
Section 6.2 – Antidifferentiation by Substitution
 You try:



cos x
dx
4  3sin x