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