CHAPTER 6:DIFFERENTIAL EQUATIONS AND
MATHEMATICAL MODELINGSECTION 6.2:
ANTIDIFFERENTIATION BY SUBSTITUTION
AP CALCULUS AB
What you’ll learn about
Indefinite IntegralsLeibniz Notation and AntiderivativesSubstitution in Indefinite IntegralsSubstitution in Definite Integrals
… and whyAntidifferentiation 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
.dxxf
Section 6.2 – Antidifferentiation by Substitution
is read “The indefinite integral of f with respect to x is F(x) + C.”
Example:
CxFdxxf
Cxxdx 22
constant of integration
integral sign
integrand
variable ofintegration
Section 6.2 – Antidifferentiation by Substitution
Integral Formulas:Indefinite Integral Corresponding
Derivative Formula
1.
2.
3.
1for 1
1
nCn
xdxx
nn 1 nn nxx
dx
d
Cxdxx
ln1
xx
dx
d 1ln
Ck
edxe
Cedxe
kxkx
xx
kxkx
xx
keedx
d
eedx
d
Section 6.2 – Antidifferentiation by Substitution
More Integral Formulas:Indefinite Integral Corresponding
Derivative Formula
4.
5.
Ck
kxdxkx
Cxdxx
cos
sin
cossin
kxkkxdx
d
xxdx
d
sincos
sincos
Ck
kxdxkx
Cxdxx
sincos
sincos
kxkkxdx
d
xxdx
d
cossin
cossin
Section 6.2 – Antidifferentiation by Substitution
More Integral Formulas:Indefinite Integral Corresponding
Derivative Formula
6.
7.
8.
9.
Cxdxx tansec2
Cxdxx cotcsc2
Cxdxxx sectansec
Cxdxxx csccotcsc
xxdx
d 2sectan
xxdx
d 2csccot
xxxdx
dtansecsec
xxxdx
dcotcsccsc
Trigonometric Formulas
2 2
cos sin sin cos
sec tan csc cot
sec tan sec csc cot csc
udu u C udu u C
udu u C udu u C
u udu u C u udu u C
Section 6.2 – Antidifferentiation by Substitution
Properties of Indefinite Integrals:Let k be a real number.
1. Constant multiple rule:
2. Sum and Difference Rule:
dxxfkdxxkf
dxxgdxxfdxxgxf
Section 6.2 – Antidifferentiation by Substitution
Example:
2 2
2
3
3 32 2
12 3
2 3ln3
x dx x dx dxx x
x dx dxx
xx C
C’s can be combined intoone big C at the end.
Example Evaluating an Indefinite Integral
Evaluate 2 cos .x xdx
22 cos sinx xdx x x C
Section 6.2 – Antidifferentiation by Substitution
Remember the Chain Rule for Derivatives:
By reversing this derivative formula, we obtain the integral formula
1
1
nn du uu dx Cdx n
1
1
nnd u duu
dx n dx
Section 6.2 Antidifferentiation by Substitution
Power Rule for IntegrationIf u is any differentiable function of x, then
1
, 1.1
nn uu du C n
n
Exponential and Logarithmic Formulas
ln
ln ln
ln lnlog
ln ln
u u
u
u
a
e du e C
aa du C
a
udu u u u C
u u u uudu du C
a 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
2
cos 2
Let 2
cos 4
1cos 4
41
sin
1
4
1
4
41
si4
x x dx
u x
dux u x
dx
u du du xdx
u C d
d
xx
ux
du
2n 2x C
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:
23
3
12 22
1
2
22
12
9 Let 1
1
9 3
3 3
3 +C 1 3
2
3
du
r
du
r dru r
r
dur u r
dr
u du du r dr
u
12
3
2
23
1
6 1
drr
u C
r C
Example Paying Attention to the Differential
2 3Let ( ) 1 and let . Find eachof the following antiderivatives in terms of .a. ( )b. ( )c. ( )
f x x u xx
f x dxf u duf u dx
3
2 3
33 3 9 3
22 3
6 7
2 1a. ( ) 131b. ( ) 13
1 13 3
c. ( ) 1 1
117
f x dx x dx x x C
f u du u du u u C
x x C x x C
f u dx u dx x dx
x dx x x C
Example Using Substitution
32Evaluate .xx e dx
3
3
3 2
2
2
Let . Then 3 , from which we conclude that
1. We rewrite the integral and proceed as follows
31
x31
31
3
x u
u
x
duu x x
dx
du x dx
e dx e du
e C
e C
Example Using Substitution
2Evaluate 6 1 .x x dx
2
2
3
2
32 2
Let 1 . Then 2 . Rewrite the integral in terms of :
6 1 3
2 3
3
2 1
u x du xdx u
x x dx udu
u C
x C
Example Setting Up a Substitution with a Trigonometric Identity
3Evaluate sin .xdx
3 2
2
2
3
3
let cos and - sin
sin sin sin
1 cos sin
1
3
cos cos
3
u x du xdx
xdx x xdx
x xdx
u du
uu C
xx C
Hint: let u = cos x and –du = sinxdx
Section 6.2 – Antidifferentiation by Substitution
Substitution in Definite IntegralsSubstituteand integrate with respect to u from
, ,u g x du g x dx
to .u g a u g b
b g b
a g af g x g x dx f u du
Section 6.2 – Antidifferentiation by Substitution
Ex:
1 2 2
0
1 12
0
12
3
1
2
1
0
0
1 Let 1
2 when 0, 1 0
1 2 when 1, 1 1
2
1
32
1
0
2
2
u
u
x x dx u x
dux u x x u
dx
u du du xd
u
du
u
x
x x
032
1
3 32 2
1 2
2 3
10 1
31
1
2
1
3 3
dud
u
xx
Example Evaluating a Definite Integral by Substitution
2
0 2Evaluate .
9
xdx
x
2 2
2
2 5
0 92
5
9
1ln
2
Let 9 and 2 . Then (0) 0 9 9 and
(2) 2 9 5. So,
1
9 2
=
1 ln5 ln9
21 5
ln2 9
u
u x du xdx u
u
x dudx
x u
Section 6.2 – Antidifferentiation by Substitution
You try:
2
6cos
2 sin
tdt
t
Section 6.2 – Antidifferentiation by Substitution
You try:
4 2 2
4
tan secx xdx
Section 6.2 – Antidifferentiation by Substitution
You try:
7
0 2
dx
x
Section 6.2 – Antidifferentiation by Substitution
You try:
cos
4 3sin
xdxx